Whats a^2=4/25 in square root form, №18010792, 15.04.2022 17:28

Mathematics

Whats a^2=4/25 in square root form

Step-by-step answer

P Answered by Specialist

a = ± $\sqrt{\frac{4}{25} }$

Step-by-step explanation:

$a^{2} =\frac{4}{25}$

$\sqrt{a^{2} } =\sqrt{\frac{4}{25} } \\|a|=\sqrt{\frac{4}{25} \\$

a = ± $\sqrt{\frac{4}{25} }$

Mathematics

8.07a, part 1 1. find the vertex, focus, directrix, and focal width of the parabola. (1 point) negative 1 divided by 40 x squared equals y a) vertex: (0, 0); focus: (0, -10); directrix: y = 10; focal width: 160 b) vertex: (0, 0); focus: (-20, 0); directrix: x = 10; focal width: 160 c) vertex: (0, 0); focus: (0, -10); directrix: y = 10; focal width: 40 d) vertex: (0, 0); focus: (0, 10); directrix: y = -10; focal width: 10 2. find the standard form of the equation of the parabola with a focus at (0, -2) and a directrix at y = 2. a)y2 = -2x b) y2

Step-by-step answer

P Answered by PhD

41

Note: Please refer to the attached file for the explanation.

Answers:

1. None from the choices, it should be

Vertex: (0, 0); Focus: (0, -10); Directrix: y = 10; Focal width: 10

2. C) y equals negative 1 divided by 8 x squared

3. D) x equals negative 1 divided by 32 y squared

4. C) x2 = -5.3y

5. C) Center: (0, 0); Vertices: (0, -15), (0, 15); Foci: (0, -12), (0, 12)

6. None from the choices, it should be

Center: (0, 0); Vertices: the point negative two square root two comma zero and the point 2 square root two comma zero; Foci: Ordered pair negative square root 6 comma zero and ordered pair square root 6 comma zero

7. B) A vertical ellipse is shown on the coordinate plane centered at the origin with vertices at the point zero comma seven and zero comma negative seven. The minor axis has endpoints at negative six comma zero and six comma zero.

8. D) x squared divided by 16 plus y squared divided by 25 equals 1

9. D) Vertices: (1, -1), (-11, -1); Foci: (-15, -1), (5, -1)

Mathematics

8.07a, part 2 10. graph the hyperbola with equation quantity x minus 1 squared divided by 49 minus the quantity of y plus 3 squared divided by 9 equals 1 . a) a horizontal hyperbola is shown on the coordinate plane centered at the origin with vertices at the point negative seven comma zero and seven comma zero. b) a vertical hyperbola is shown on the coordinate plane centered at the origin with vertices at the point zero comma seven and zero comma negative seven. c) a horizontal hyperbola is shown on the coordinate plane centered at the point one

Step-by-step answer

P Answered by Specialist

Question 1
We are given horizontal hyperbola in its conic form:
$\frac{(x-1)^2}{49}-\frac{(y+3)^2}{9}=1$
The general formula for hyperbola in conic form is:
$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$
This is a horizontal hyperbola if we want to get vertical hyperbola we simply switch places of y and x:
$\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1$
If we look at our parabola we can see that a=7, b=3, h=1 and k=-3.
We know that center of a parabola is at the point (h,k) and that vertices are at (h+a,k) and (h-a,k).
We that mind we find:
$Center:(1,-3)\\ Vertices:(8,-3),(-6,-3)$
The answer is C. Please click this link to see that graph(https://www.desmos.com/calculator/g2pbnxs9ad)
Question 2
Standard form is the same as the conic form that we mention in part 1.
We are given:
$Vertices:(0,\pm 2)\\ Foci:(0,\pm 11)$
What we can notice right away is that this is a vertical hyperbola.
We know that vertices are at:
$(h,k\pm a)$
And that foci are at:
$(h,k\pm c)$
We know that h=0. We can also conclude that k=0. We know that vertices are a point far away from the center. So our vertices are:
$k+a=2\\ k-a=-2$
We solve this system and we get that a=2 and k=0. That means that our center is at (0,0).
Our foci are given with these two equations:
$k+c=11\\ k-c=-11\\$
Since we know that k=0, c=11. We that in mind we can find b:
$c^2=a^2+b^2\\ b^2=c^2-a^2\\ b^2=117\\ b=\sqrt{117}$
Now we have all the information to write the equation for the parabola:
$\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1\\ \frac{y^2}{4}-\frac{x^2}{117}=1$
The answer is A. Link to the graph: https://www.desmos.com/calculator/gw6fnf8pg0 .
Question 3
We are given:
$Vertices:(0,\pm 8)\\ Asymptotes:y=\pm \frac{x}{2}$
We notice that this is a vertical hyperbola (look at the coordinates of vertices, they are changing along the y-axis).
We know that our vertices are given by the following formula:
$k+a=8\\ k-a=8$
From this, we can conclude that k=8, and a=8. We know that our center is at (0,0).
Asymptote for the vertical hyperbola is given by this formula:
$y=\pm \frac{a}{b}(x-h)+k$
For the horizontal hyperbola, you just swap a and b.
We knot that our h and k are zeros, and if we look at our asymptotes we notice that:
$\frac{a}{b}=\frac{1}{2}$
We know that a=8 so we can find b from the above equation:
$\frac{8}{b}=\frac{1}{2}\\ b=16$ .
This is all we need to write down the equation of this hyperbola:
$\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1\\ \frac{y^2}{64}-\frac{x^2}{256}=1$
The answers is B. Link to the graph(https://www.desmos.com/calculator/tbecgjct6g).
Question 4
We are given these two equations:
$x=t-3\\ y=\frac{2}{t+5}$
To eliminate t, we will express t, from the first equation, in terms of x and then plug it back into the second equation:
$x=t-3\\ y=\frac{2}{t+5}\\ t=x+3\\ y=\frac{2}{x+3+5}\\ y=\frac{2}{5+8}$
The answer is D.
Question 5
We are given polar coordinates:
$(7,\frac{2\pi}{3})$
To conver polar cordinates in rectangular one with use the following formula:
$x=r\cos(\theta)\\ y=r\sin(\theta)\\$
In our case r=7 and $\theta =\frac{2\pi}{3}$ .
Let us calculate the rectangular coordinates:
$x=7\cos(\frac{2\pi}{3})\\ y=7\sin(\frac{2\pi}{3})\\ x=\frac{-7}{2}\\ y=\frac{7\sqrt{3}}{2}}$
Coordinates are:
$(\frac{-7}{2},\frac{7\sqrt{3}}{2}})$
The answer is A.
Question 6
We are given a point P with the following coordinates:
$P(1,\frac{\pi}{3})$
What you should keep in mind is that for any point there are two pairs of polar coordinates, because you can reach the same point rotating your point vector clock-wise or counter-clockwise. If you are given a pair of polar coordinates this is the formula you can use to get the second pair:
$(r,\theta),(-r,\theta+\pi)$
Reason for this is because you rotate your pointing vector by 180 degrees. This means you are doing a reflection on the y=x line and therefore you end up with the same point if you use -r instead of r.With that in mind the second pair for our points is:
$(-1,\frac{\pi}{3}+\pi)\\ (-1,\frac{4\pi}{3})$
If we rotate our pointing vector by 360 degrees(2pi radians) we will always end up at the same point. In fact, if you any number of full circles you end with the same point.So, the final answer is:
$(-1,\frac{4\pi}{3}+2n\pi),(1,\frac{\pi}{3}+2n\pi)$
None of the answers you provided are correct. You can change r to -r and end up at the same point without changing the angle.

Mathematics

8.07a, part 1 1. find the vertex, focus, directrix, and focal width of the parabola. (1 point) negative 1 divided by 40 x squared equals y a) vertex: (0, 0); focus: (0, -10); directrix: y = 10; focal width: 160 b) vertex: (0, 0); focus: (-20, 0); directrix: x = 10; focal width: 160 c) vertex: (0, 0); focus: (0, -10); directrix: y = 10; focal width: 40 d) vertex: (0, 0); focus: (0, 10); directrix: y = -10; focal width: 10 2. find the standard form of the equation of the parabola with a focus at (0, -2) and a directrix at y = 2. a)y2 = -2x b) y2

Step-by-step answer

P Answered by PhD

41

Note: Please refer to the attached file for the explanation.

Answers:

1. None from the choices, it should be

Vertex: (0, 0); Focus: (0, -10); Directrix: y = 10; Focal width: 10

2. C) y equals negative 1 divided by 8 x squared

3. D) x equals negative 1 divided by 32 y squared

4. C) x2 = -5.3y

5. C) Center: (0, 0); Vertices: (0, -15), (0, 15); Foci: (0, -12), (0, 12)

6. None from the choices, it should be

Center: (0, 0); Vertices: the point negative two square root two comma zero and the point 2 square root two comma zero; Foci: Ordered pair negative square root 6 comma zero and ordered pair square root 6 comma zero

7. B) A vertical ellipse is shown on the coordinate plane centered at the origin with vertices at the point zero comma seven and zero comma negative seven. The minor axis has endpoints at negative six comma zero and six comma zero.

8. D) x squared divided by 16 plus y squared divided by 25 equals 1

9. D) Vertices: (1, -1), (-11, -1); Foci: (-15, -1), (5, -1)

Mathematics

8.08, part 2 11. find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9). a) y squared over 45 minus x squared over 36 = 1 b) y squared over 81 minus x squared over 36 = 1 c) y squared over 36 minus x squared over 81 = 1 d) y squared over 36 minus x squared over 45 = 1 12. find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ± 1 divided by 4. x. a) y squared over 16 minus x squared over 64 = 1 b) y squared over 16 minus x squared over 256 = 1 c) y squared over

Step-by-step answer

P Answered by PhD

2

11. Equation of hyperbola having vertices at (0, ±6) and foci at (0, ±9).

is given by $\frac{y^2}{a^2}- \frac{x^2}{b^2}=1$

also b²= c²-a²

b²=81-36

b²=45

So, Equation becomes , $\frac{y^2}{36}- \frac{x^2}{45}=1$

Option (D) is correct.

12.Vertices (0, ± 4), Asymptotes = ±1/4.x

equation of asymptote is given by, $x = \pm y \frac{b}{a}$

$\frac{4}{b}=\frac{1}{4}$

So a=4 and b=16,

So , equation becomes $\frac{y^2}{16}- \frac{x^2}{256}=1$

Option (B) is correct.

13. x= t-3, and y = t²+ 5

Replace t by x+3, we get

y= (x+3)² +5

y=x²+ 6x +14

Option (A) is correct.

14. Polar coordinates is given by (r,∅)

Polar coordinates of point is (3, 2π/3)

So, r =3, ∅ =2π/3

x= r cos∅ and y = r sin∅

x=3 Cos (2π/3) and y= 3 Sin (2π/3)

x= -3/2 and y=3√3/2

Option (A) is correct.

14. Polar coordinate is given by (r,∅)

Here , r=1, and ∅ = -π/6

x= r Cos ∅ and y =r Sin∅

x= 1 Cos (-π/6) and y= 1 Sin (-π/6)

x =√3/2 and y =1/2

Option (B) which is B) (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + 2nπ) is correct.

16. Two pairs of polar coordinates for the point (4, 4) with 0° ≤ θ < 360°.

is option (B) which is B) (4 square root 2 , 45°), (-4 square root 2 , 225°).

17. A circular graph with inner loop on the left of a limacon curve is given by

r = a + b Cos∅.

In this case a> b.

So , Option (D) r = 4 + cos θ as well as (A) r = 3 + 2 cos θ looks correct.

18. Equation of limacon curve is given by r = -2 + 3 cos θ, here , a<b

So it is symmetric about y-axis only.Option (B) is correct.

Mathematics

8.07a, part 2 10. graph the hyperbola with equation quantity x minus 1 squared divided by 49 minus the quantity of y plus 3 squared divided by 9 equals 1 . a) a horizontal hyperbola is shown on the coordinate plane centered at the origin with vertices at the point negative seven comma zero and seven comma zero. b) a vertical hyperbola is shown on the coordinate plane centered at the origin with vertices at the point zero comma seven and zero comma negative seven. c) a horizontal hyperbola is shown on the coordinate plane centered at the point one

Step-by-step answer

P Answered by Specialist

Question 1
We are given horizontal hyperbola in its conic form:
$\frac{(x-1)^2}{49}-\frac{(y+3)^2}{9}=1$
The general formula for hyperbola in conic form is:
$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$
This is a horizontal hyperbola if we want to get vertical hyperbola we simply switch places of y and x:
$\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1$
If we look at our parabola we can see that a=7, b=3, h=1 and k=-3.
We know that center of a parabola is at the point (h,k) and that vertices are at (h+a,k) and (h-a,k).
We that mind we find:
$Center:(1,-3)\\ Vertices:(8,-3),(-6,-3)$
The answer is C. Please click this link to see that graph(https://www.desmos.com/calculator/g2pbnxs9ad)
Question 2
Standard form is the same as the conic form that we mention in part 1.
We are given:
$Vertices:(0,\pm 2)\\ Foci:(0,\pm 11)$
What we can notice right away is that this is a vertical hyperbola.
We know that vertices are at:
$(h,k\pm a)$
And that foci are at:
$(h,k\pm c)$
We know that h=0. We can also conclude that k=0. We know that vertices are a point far away from the center. So our vertices are:
$k+a=2\\ k-a=-2$
We solve this system and we get that a=2 and k=0. That means that our center is at (0,0).
Our foci are given with these two equations:
$k+c=11\\ k-c=-11\\$
Since we know that k=0, c=11. We that in mind we can find b:
$c^2=a^2+b^2\\ b^2=c^2-a^2\\ b^2=117\\ b=\sqrt{117}$
Now we have all the information to write the equation for the parabola:
$\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1\\ \frac{y^2}{4}-\frac{x^2}{117}=1$
The answer is A. Link to the graph: https://www.desmos.com/calculator/gw6fnf8pg0 .
Question 3
We are given:
$Vertices:(0,\pm 8)\\ Asymptotes:y=\pm \frac{x}{2}$
We notice that this is a vertical hyperbola (look at the coordinates of vertices, they are changing along the y-axis).
We know that our vertices are given by the following formula:
$k+a=8\\ k-a=8$
From this, we can conclude that k=8, and a=8. We know that our center is at (0,0).
Asymptote for the vertical hyperbola is given by this formula:
$y=\pm \frac{a}{b}(x-h)+k$
For the horizontal hyperbola, you just swap a and b.
We knot that our h and k are zeros, and if we look at our asymptotes we notice that:
$\frac{a}{b}=\frac{1}{2}$
We know that a=8 so we can find b from the above equation:
$\frac{8}{b}=\frac{1}{2}\\ b=16$ .
This is all we need to write down the equation of this hyperbola:
$\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1\\ \frac{y^2}{64}-\frac{x^2}{256}=1$
The answers is B. Link to the graph(https://www.desmos.com/calculator/tbecgjct6g).
Question 4
We are given these two equations:
$x=t-3\\ y=\frac{2}{t+5}$
To eliminate t, we will express t, from the first equation, in terms of x and then plug it back into the second equation:
$x=t-3\\ y=\frac{2}{t+5}\\ t=x+3\\ y=\frac{2}{x+3+5}\\ y=\frac{2}{5+8}$
The answer is D.
Question 5
We are given polar coordinates:
$(7,\frac{2\pi}{3})$
To conver polar cordinates in rectangular one with use the following formula:
$x=r\cos(\theta)\\ y=r\sin(\theta)\\$
In our case r=7 and $\theta =\frac{2\pi}{3}$ .
Let us calculate the rectangular coordinates:
$x=7\cos(\frac{2\pi}{3})\\ y=7\sin(\frac{2\pi}{3})\\ x=\frac{-7}{2}\\ y=\frac{7\sqrt{3}}{2}}$
Coordinates are:
$(\frac{-7}{2},\frac{7\sqrt{3}}{2}})$
The answer is A.
Question 6
We are given a point P with the following coordinates:
$P(1,\frac{\pi}{3})$
What you should keep in mind is that for any point there are two pairs of polar coordinates, because you can reach the same point rotating your point vector clock-wise or counter-clockwise. If you are given a pair of polar coordinates this is the formula you can use to get the second pair:
$(r,\theta),(-r,\theta+\pi)$
Reason for this is because you rotate your pointing vector by 180 degrees. This means you are doing a reflection on the y=x line and therefore you end up with the same point if you use -r instead of r.With that in mind the second pair for our points is:
$(-1,\frac{\pi}{3}+\pi)\\ (-1,\frac{4\pi}{3})$
If we rotate our pointing vector by 360 degrees(2pi radians) we will always end up at the same point. In fact, if you any number of full circles you end with the same point.So, the final answer is:
$(-1,\frac{4\pi}{3}+2n\pi),(1,\frac{\pi}{3}+2n\pi)$
None of the answers you provided are correct. You can change r to -r and end up at the same point without changing the angle.

Mathematics

8.08, part 2 11. find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9). a) y squared over 45 minus x squared over 36 = 1 b) y squared over 81 minus x squared over 36 = 1 c) y squared over 36 minus x squared over 81 = 1 d) y squared over 36 minus x squared over 45 = 1 12. find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ± 1 divided by 4. x. a) y squared over 16 minus x squared over 64 = 1 b) y squared over 16 minus x squared over 256 = 1 c) y squared over

Step-by-step answer

P Answered by PhD

3

11. The technique to answering this problem quickly is by examining the choices. We know that the denominator for y and x should be equal to the square of the foci (from the fact that $c^{2}=a^{2}+b^{2}$ ). This would mean that the denominators should add up to 81. From the choices, only two satisfy this and those are options A and D. We know it should be D since the smaller denominator always has to be in the first term.

ANSWER: D. y squared over 36 minus x squared over 45 = 1.

12. You can also quickly identify the equation of the hyperbola given the vertices and the asymptotes. The square root of the denominator of the FIRST TERM in the equation is the numerator of the asymptote while the square root of the denominator of the SECOND TERM is the denominator of the asymptote. HOWEVER, we have to consider that the vertices are at (0,4) and (0,-4) so the asymptote must be in the lowest term. Considering the vertices, we can arrive at the asymptote $y=+- \frac{4}{16}x$ . This means that the first term will have a denominator of 16 while the second term will have a denominator of 256. This equation is option B.

ANSWER: B. y squared over 16 minus x squared over 256 = 1.

13. To eliminate the parameter, t, we just need to equate both equations such that t is equal.

t=x+3

$t=\sqrt{y-5}$

$x+3=\sqrt{y-5}$
$x^{2}+6x+9=y-5$
$y=x^{2}+6x+14$

ANSWER: A. $y=x^{2}+6x+14$

14. To find the x coordinate we just need to multiply r and cos θ while for the y coordinate we would need to multiply r and sin θ.

$x=(3)[cos(\frac{2\pi}{3})]=-\frac{3}{2}$
$y=(3)[sin(\frac{2\pi}{3})]=\frac{3\sqrt{3}}{2}$

ANSWER: A. ordered pair negative 3 divided by 2 comma 3 square root 3 divided by 2.

15. The same polar coordinate as point P will be arrived if: (1) a full rotation of 2nπ is performed or (2) a rotation of (2n+1)π is performed and r is negated. Among the choices, we can see exactly one option with coordinates that follow these two rules, and it is choice D.

ANSWER: D. (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + (2n+1)π)

16. To convert the coordinate (4, 4) to polar, we just let x be rcosθ and y be rsinθ. We can get r by solving $r^{2}=x^{2}+y^{2}$ while we can get θ by solving for $\theta=arctan(\frac{y}{x})$ .

$r^{2}=(4)^{2}+(4)^{2}$
r=4sqrt(2)

$\theta=arctan(\frac{4}{4})=45degrees$

Therefore, one pair of polar coordinates would be (4 square root of 2, 45 degrees) and another one would be (-4 square root of 2, 225 degrees) [note the rule we stated in number 15].

ANSWER: B. (4 square root of 2, 45 degrees), (-4 square root of 2, 225 degrees)

17. Based on your definition that the graph is "circular" with an inner loop on the left, we can only deduce that the limacon with the form r = a + bcosθ has a value of b that is greater than a. Looking at the choices, we only have one option following this criteria, thus we can be sure that it is the correct answer.

ANSWER: B. r = 2 + 3cos θ

18. To test for symmetry about the x-axis, we replace the variables r and θ with r and -θ respectively or -r and π-θ. The equation will be symmetric if it will be unchanged (i.e. the same points will still satisfy the new equation).

$r=-2+3cos\theta$
$r=-2+3cos(-\theta)$
$-r=-2+3cos(\pi-\theta)$

If you examine closely, the same set of points will satisfy the equation above. Therefore, the equation is symmetric about the x axis.

For symmetry about the y-axis, we replace r and θ with -r and -θ.

$r=-2+3cos\theta$
$-r=-2+3cos(-\theta)$

Unfortunately, the equation was changed upon substitution therefore we know that it is not symmetric about the y-axis.

For symmetry about the origin, we just replace r and θ with -r and θ respectively.

$r=-2+3cos\theta$
$-r=-2+3cos\theta$

These two equations are not also similar so no symmetry about the origin is exhibited.

ANSWER: C. x-axis only

19. Let's consider the general form of the ellipse: $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ .

From the problem we know that a is equal to 54 since we are given the fact that the height of the tunnel is 54ft. To find b, we use the fact that the point (8,18) is a point on the ellipse as stated in the problem.

$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{54^{2}}=1$
$\frac{8^{2}}{b^{2}}+\frac{18^{2}}{54^{2}}=1$
$\frac{8^{2}}{b^{2}}+\frac{1}{9}=1$
$64=(\frac{8}{9})b^{2}$
$b^{2}=72$

ANSWER: $\frac{x^{2}}{72}+\frac{y^{2}}{2916}=1$

20. For this item, we follow similar rules stated in item #18. To test for symmetry we need to examine if the equation will remain unchanged after performing substitutions.

$r=2cos3\theta$

x-axis:
$r=2cos(-3\theta)$
$-r=2cos[3(\pi-\theta)]$

y-axis:
$-r=2cos(-3\theta)$

origin:
$-r=2cos3\theta$

If you examine the equations, you'll see that it is only symmetric at the x-axis.

ANSWER: The equation is symmetric about the x-axis.

Mathematics

8.08, part 2 11. find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9). a) y squared over 45 minus x squared over 36 = 1 b) y squared over 81 minus x squared over 36 = 1 c) y squared over 36 minus x squared over 81 = 1 d) y squared over 36 minus x squared over 45 = 1 12. find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ± 1 divided by 4. x. a) y squared over 16 minus x squared over 64 = 1 b) y squared over 16 minus x squared over 256 = 1 c) y squared over

Step-by-step answer

P Answered by PhD

8

11. Ans: (D)

Since all the vertices and the foci lie along the y axis, therefore, we would need the following equation for vertical hyperbola:

$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$

Since (h,k) = (0,0)
Therefore, the above equation becomes,
$\frac{(y)^2}{a^2} - \frac{(x)^2}{b^2} = 1$

Now the distance between the vertices is:
2a = 12
=> a = 6

And the distance between the foci is:
2c = 18
=> c = 9

Since,
c^2 = a^2 + b^2

=>

Hence, the equation becomes,
$\frac{(y)^2}{36} - \frac{(x)^2}{45} = 1$ (Option D:y squared over 36 minus x squared over 45 = 1)

12. Ans: (B)
The hyperbola's standard form is(as it is a vertical):
$\frac{y^2}{16} - \frac{x^2}{b^2} = 1$ -- (X)

=> $y^2 = ( \frac{16}{b^2})*(b^2 + x^2)$

=> y = ± $( \frac{4}{b} ).x$ --- (A)
Since asymptotes at y = ± $( \frac{1}{4} ).x$ . --- (B)
Compare (A) and (B), you would get,
$\frac{4}{b} = \frac{1}{4}$

=> b=16

The equation (X) would become:
$\frac{y^2}{16} - \frac{x^2}{256} = 1$ (Option-B)

13. Ans: (A) $y = x^{2} + 6x + 14$
Equations given:
x = t - 3 --- (equation-1)
y = $t^{2}$ + 5 --- (equation-2)

From equation-1,
t = x + 3

Put the value of t in (equation-2),
$y = (x+3)^{2} + 5$
y = x^2 + 9 + 6x + 5

Hence, the correct option is (A)

14. Ans: (A)

The polar coordinates given: $(3, \frac{2 \pi }{3} )$ = (r, θ)
Since,
x = r*cosθ,
y = r*sinθ

Plug-in the values of r, and θ in the above equations:
x = (3) * cos(120°); since $\frac{2 \pi }{3}$ = 120°
=> x = $- \frac{3}{2}$

y = (3) * sin(120°);
=> y = $\frac{3 \sqrt{3} }{2}$

Ans: (x,y) = $(- \frac{3}{2} ,\frac{3 \sqrt{3} }{2})$ (Option A)

15. Ans: (D)
The general forms of finding all the polar coordinates are:
1) When r >= 0(meaning positive): (r, θ + 2n $\pi$ ) where, n = integer
2) When r < 0(meaning negative): (-r, θ + (2n+1) $\pi$ ) where, n = integer

Since r is not mentioned in the question, but in options every r slot has the value r=1, therefore, I would take r = +1, -1(plus minus 1)

θ(given) = $\frac{- \pi }{6}$

When r = +1(r>0):
(1, $\frac{- \pi }{6}$ + 2n $\pi$ )

When r = -1(r<0):
(-1, $\frac{- \pi }{6}$ + (2n+1) $\pi$ )

Therefore, the correct option is (D): (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + (2n + 1)π)

16. Ans: (B)
In polar coordinates,
$r = \sqrt{x^{2} + y^{2}}$

Since x = 4, y=4; therefore,
$r = \sqrt{16 + 16} = 4 \sqrt{2}$

To find the angle,
tanθ = y/x = 4/4 = 1

=> θ = 45° (when $r =4 \sqrt{2}$ )
If r = - $r =4 \sqrt{2}$ , then,

θ = 45° + 180° = 225°
Therefore, the correct option is (B) (4 square root 2 , 45°), (-4 square root 2 , 225°)

17. Ans: (B)

(Question-17 missing Image is attached below) The general form of the limacon curve is:
r = b + a cosθ

If b < a, the curve would have inner loop. As you can see in the image attached(labeled Question-17), the limacon curve graph has the inner loop. Therefore, the correct option is (B) r = 2 + 3 cosθ, since b = 2, and a = 3; and the condition b < a (2 < 3) is met.

18. Ans: (C)
Let's find out!
1. If we replace θ with -θ, we would get:
r = -2 + 3*cos(-θ )
Since, cos(-θ) = +cosθ, therefore,
r = -2 + 3*cos(θ)

Same as the original, therefore, graph is symmetric to x-axis.

2. If we replace r with -r, we would get:
-r = -2 + 3*cos(θ )
r = 2 - 3*cos(θ)

NOT same as original, therefore, graph is NOT symmetric to its origin.

3. If we replace θ with -θ and r with -r, we would get:
-r = -2 + 3*cos(-θ )
Since, cos(-θ) = +cosθ, therefore,
r = 2 - 3*cos(3θ)

NOT same as original, therefore, graph is NOT symmetric to y-axis.

Ans: The graph is symmetric to: x-axis only!

19. (Image is attached below) As the question suggests that it is a horizontal ellipse, therefore, the equation for the horizontal ellipse is:

$\frac{x^{2}}{a^{2}} + \frac{y_{2}}{b_{2}} = 1$ -- (A)

Since, x = 8f,
y = 18ft,
b = 54ft,
$a^{2}$ = ?

Plug-in the values in equation (A),
(A)=> $\frac{64}{a^{2}} + \frac{324}{2916} = 1$

=> $a^{2}$ = 72

Therefore, the equation becomes,
Ans: $\frac{x^{2}}{72} + \frac{y_{2}}{2916} = 1$

20. Ans: x-axis only
Let's find out!

1. If we replace θ with -θ, we would get:
r = 2*cos(-3θ )
Since, cos(-θ) = +cosθ, therefore,
r = +2*cos(3θ) = Same as original

Therefore, graph is symmetric to x-axis.

2. If we replace r with -r, we would get:
-r = 2*cos(3θ )
r = -2*cos(3θ) = Not same

3. If we replace θ with -θ and r with -r, we would get:
-r = 2*cos(-3θ )
Since, cos(-θ) = +cosθ, therefore,
r = -2*cos(3θ) = Not Same

Ans: The graph is symmetric to: x-axis only!
8.08, part 2 11. find an equation in standard form for the hyperbola with vertices at (0, ±6) and fo

8.08, part 2 11. find an equation in standard form for the hyperbola with vertices at (0, ±6) and fo

Mathematics

8.08, part 2 11. find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9). a) y squared over 45 minus x squared over 36 = 1 b) y squared over 81 minus x squared over 36 = 1 c) y squared over 36 minus x squared over 81 = 1 d) y squared over 36 minus x squared over 45 = 1 12. find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ± 1 divided by 4. x. a) y squared over 16 minus x squared over 64 = 1 b) y squared over 16 minus x squared over 256 = 1 c) y squared over

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P Answered by PhD

2

11. Equation of hyperbola having vertices at (0, ±6) and foci at (0, ±9).

is given by $\frac{y^2}{a^2}- \frac{x^2}{b^2}=1$

also b²= c²-a²

b²=81-36

b²=45

So, Equation becomes , $\frac{y^2}{36}- \frac{x^2}{45}=1$

Option (D) is correct.

12.Vertices (0, ± 4), Asymptotes = ±1/4.x

equation of asymptote is given by, $x = \pm y \frac{b}{a}$

$\frac{4}{b}=\frac{1}{4}$

So a=4 and b=16,

So , equation becomes $\frac{y^2}{16}- \frac{x^2}{256}=1$

Option (B) is correct.

13. x= t-3, and y = t²+ 5

Replace t by x+3, we get

y= (x+3)² +5

y=x²+ 6x +14

Option (A) is correct.

14. Polar coordinates is given by (r,∅)

Polar coordinates of point is (3, 2π/3)

So, r =3, ∅ =2π/3

x= r cos∅ and y = r sin∅

x=3 Cos (2π/3) and y= 3 Sin (2π/3)

x= -3/2 and y=3√3/2

Option (A) is correct.

14. Polar coordinate is given by (r,∅)

Here , r=1, and ∅ = -π/6

x= r Cos ∅ and y =r Sin∅

x= 1 Cos (-π/6) and y= 1 Sin (-π/6)

x =√3/2 and y =1/2

Option (B) which is B) (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + 2nπ) is correct.

16. Two pairs of polar coordinates for the point (4, 4) with 0° ≤ θ < 360°.

is option (B) which is B) (4 square root 2 , 45°), (-4 square root 2 , 225°).

17. A circular graph with inner loop on the left of a limacon curve is given by

r = a + b Cos∅.

In this case a> b.

So , Option (D) r = 4 + cos θ as well as (A) r = 3 + 2 cos θ looks correct.

18. Equation of limacon curve is given by r = -2 + 3 cos θ, here , a<b

So it is symmetric about y-axis only.Option (B) is correct.

Mathematics

Look at the cups shown below (images are not drawn to scale): a cone is shown with width 2 inches and height 3 inches, and a cylinder is shown with width 2 inches and height 7 inches. how many more cubic inches of juice will cup b hold than cup a when both are completely full? round your answer to the nearest tenth. 18.8 cubic inches 21.9 cubic inches 25.1 cubic inches 32.6 cubic inches question 9(multiple choice worth 1 points) (03.02 lc) abcd is a rectangle. what is the value of x? rectangle with length 45 feet, width x feet, and diagonal 53

Step-by-step answer

P Answered by PhD

8) 18.8 cubic inches;
9) 28 feet;
10) 1.7 centimeters;
11) 13;
12) 10 centimeters;
13) it is incorrect because the length of the unknown side is the square root of 7,744;
14) square root of 18;
15) 12 miles;
16) 364 centimeters.

Whats a^2=4/25 in square root form

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