What is the horizontal distance from the origin, №10302104, 14.11.2020 10:32

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 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 1 1. find the vertex, focus, directrix, and focal width of the parabola. negative 1 divided by 16 times x squared = y a) vertex: (0, 0); focus: (0, -4); directrix: y = 4; focal width: 16 b) vertex: (0, 0); focus: (-8, 0); directrix: x = 4; focal width: 64 c) vertex: (0, 0); focus: (0, 4); directrix: y = -4; focal width: 4 d) vertex: (0, 0); focus: (0, -4); directrix: y = 4; focal width: 64 2. find the standard form of the equation of the parabola with a focus at (0, -8) and a directrix at y = 8. a) y = negative 1 divided by 32 x2 b)

Step-by-step answer

P Answered by Master

Hello,
Please, see the detailed solutions in the attached files.
Thanks.

8.08, part 1 1. find the vertex, focus, directrix, and focal width of the parabola. negative 1 divi

Mathematics

8.08, part 1 1. find the vertex, focus, directrix, and focal width of the parabola. negative 1 divided by 16 times x squared = y a) vertex: (0, 0); focus: (0, -4); directrix: y = 4; focal width: 16 b) vertex: (0, 0); focus: (-8, 0); directrix: x = 4; focal width: 64 c) vertex: (0, 0); focus: (0, 4); directrix: y = -4; focal width: 4 d) vertex: (0, 0); focus: (0, -4); directrix: y = 4; focal width: 64 2. find the standard form of the equation of the parabola with a focus at (0, -8) and a directrix at y = 8. a) y = negative 1 divided by 32 x2 b)

Step-by-step answer

P Answered by Specialist

Hello,
Please, see the detailed solutions in the attached files.
Thanks.

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 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

Check if my answers are correct. will mark brainliest to whoever comments first! you, and god bless. 2. how is the graph of y equals -8x^2 - 2. different from the graph of y equals negative 8x squared.? (1 point) it is shifted 2 units to the left it is shifted 2 units to the right. *** it is shifted 2 units up. it is shifted 2 units down. a model rocket is launched from a roof into a large field. the path of the rocket can be modeled by the equation y equals negative 0.06 x squared plus 9.6 x plus 5.4 where x is the horizontal distance, in meters,

Step-by-step answer

P Answered by PhD

29

1. It is shifted 2 units down.

The graph of y=-8x^2 -2 is shifted 2 units down with respect to the graph of y=-8x^2 . We can prove this by taking, for instance, x=0, and calculating the value of y in the two cases. In the first function:

y=-8*0^2 -2=-2

In the second function:

y=-8*0^2 =0

So, the first graph is shifted 2 units down.

2. 160.56 m

The path of the rocket is given by:

y=-0.06 x^2 +9.6 x +5.4

The problem asks us to find how far horizontally the rocket lands - this corresponds to find the value of x at which the height is zero: y=0. This means we have to solve the following equation

-0.06 x^2 +9.6 x+5.4 =0

Using the formula,

$x=\frac{-9.6 \pm \sqrt{(9.6)^2-4(-0.06)(5.4)}}{2(-0.06)}$

which has two solutions: x_1 = 160.56 m and x_2 = -0.56 m . The second solution is negative, so it has no physical meaning, therefore the correct answer is 160.56 m.

3. 27.43 m

The path of the rock is given by:

y=-0.02 x^2 +0.8 x +37

The problem asks us to find how far horizontally the rock lands - this corresponds to find the value of x at which the height is zero: y=0. This means we have to solve the following equation

-0.02 x^2 +0.8 x+37 =0

Using the formula,

$x=\frac{-0.8 \pm \sqrt{(0.8)^2-4(-0.02)(37)}}{2(-0.02)}$

which has two solutions: x_1 = 67.43 m and x_2 = -27.43 m . In this case, we have to choose the second solution (27.43 m), since the rock was thrown backward from the initial height of 37 m, so the negative solution corresponds to the backward direction.

4. (-2, 16) and (1, -2)

The system is:

y=x^2 -5x +2 (1)

y=-6x+4 (2)

We can equalize the two equations:

x^2 -5x+2 = -6x +4

which becomes:

x^2 + x -2 =0

Solving it with the formula, we find two solutions: x=-2 and x=1. Substituting both into eq.(2):

x=-2 --> y=-6 (-2) +4 = 12+4 = 16

x=1 --> y=-6 (1) +4 = -6+4 =-2

So, the solutions are (-2, 16) and (1, -2).

5. (-1, 1) and (7, 33)

The system is:

y=x^2 -2x -2 (1)

y=4x+5 (2)

We can equalize the two equations:

x^2 -2x-2 = 4x +5

which becomes:

x^2 -6x -7 =0

Solving it with the formula, we find two solutions: x=7 and x=-1. Substituting both into eq.(2):

x=7 --> y=4 (7) +5 = 28+5 = 33

x=-1 --> y=4 (-1) +5 = -4+5 =1

So, the solutions are (-1, 1) and (7, 33).

6. 2.30 seconds

The height of the object is given by:

h(t)=-16 t^2 +85

The time at which the object hits the ground is the time t at which the height becomes zero: h(t)=0, therefore

-16t^2 +85 =0

By solving it,

16t^2 = 85

$t^2 = \frac{85}{16}$

$t=\sqrt{\frac{85}{16}}=2.30 s$

7. Reaches a maximum height of 19.25 feet after 0.88 seconds.

The height of the ball is given by

h(t)=-16t^2 + 28t + 7

The vertical velocity of the ball is equal to the derivative of the height:

v(t)=h'(t)=-32t+28

The maximum height is reached when the vertical velocity becomes zero: v=0, therefore when

-32t + 28 =0

from which we find t=0.88 s

And by substituting these value into h(t), we find the maximum height:

h(t)=-16(0.88)^2 + 28(0.88) + 7 = 19.25 m

8. Reaches a maximum height of 372.25 feet after 4.63 seconds.

The height of the boulder is given by

h(t)=-16t^2 + 148t + 30

The vertical velocity of the boulder is equal to the derivative of the height:

v(t)=h'(t)=-32t+148

The maximum height is reached when the vertical velocity becomes zero: v=0, therefore when

-32t + 148 =0

from which we find t=4.63 s

And by substituting these value into h(t), we find the maximum height:

h(t)=-16(4.63)^2 + 148(4.63) + 30 = 372.25 m

9. 12 m

Let's call x the length of the side of the original garden. The side of the new garden has length (x+3), so its area is

(x+3)^2 = 225

Solvign this equation, we find

$x+3 = \sqrt{225}=15$

x=15-3=12 m

10. 225/4

In fact, if we write $x^2 +15 x + \frac{225}{4}$ , we see this is equivalent to the perfect square:

$(x+\frac{15}{2})^2 = x^2 +15 x +\frac{225}{4}$

11. -11.56, 1.56

The equation is:

x^2 +10 x -18 =0

By using the formula:

$x=\frac{-10 \pm \sqrt{(10)^2-4(1)(-18)}}{2*1}$

which has two solutions: x=-11.56 and 1.56.

12. -10.35, 1.35

The equation is:

x^2 +9 x -14 =0

By using the formula:

$x=\frac{-9 \pm \sqrt{(9)^2-4(1)(-14)}}{2*1}$

which has two solutions: x=-10.35 and 1.35.

Mathematics

Check if my answers are correct. will mark brainliest to whoever comments first! you, and god bless. 2. how is the graph of y equals -8x^2 - 2. different from the graph of y equals negative 8x squared.? (1 point) it is shifted 2 units to the left it is shifted 2 units to the right. *** it is shifted 2 units up. it is shifted 2 units down. a model rocket is launched from a roof into a large field. the path of the rocket can be modeled by the equation y equals negative 0.06 x squared plus 9.6 x plus 5.4 where x is the horizontal distance, in meters,

Step-by-step answer

P Answered by PhD

29

1. It is shifted 2 units down.

The graph of y=-8x^2 -2 is shifted 2 units down with respect to the graph of y=-8x^2 . We can prove this by taking, for instance, x=0, and calculating the value of y in the two cases. In the first function:

y=-8*0^2 -2=-2

In the second function:

y=-8*0^2 =0

So, the first graph is shifted 2 units down.

2. 160.56 m

The path of the rocket is given by:

y=-0.06 x^2 +9.6 x +5.4

The problem asks us to find how far horizontally the rocket lands - this corresponds to find the value of x at which the height is zero: y=0. This means we have to solve the following equation

-0.06 x^2 +9.6 x+5.4 =0

Using the formula,

$x=\frac{-9.6 \pm \sqrt{(9.6)^2-4(-0.06)(5.4)}}{2(-0.06)}$

which has two solutions: x_1 = 160.56 m and x_2 = -0.56 m . The second solution is negative, so it has no physical meaning, therefore the correct answer is 160.56 m.

3. 27.43 m

The path of the rock is given by:

y=-0.02 x^2 +0.8 x +37

The problem asks us to find how far horizontally the rock lands - this corresponds to find the value of x at which the height is zero: y=0. This means we have to solve the following equation

-0.02 x^2 +0.8 x+37 =0

Using the formula,

$x=\frac{-0.8 \pm \sqrt{(0.8)^2-4(-0.02)(37)}}{2(-0.02)}$

which has two solutions: x_1 = 67.43 m and x_2 = -27.43 m . In this case, we have to choose the second solution (27.43 m), since the rock was thrown backward from the initial height of 37 m, so the negative solution corresponds to the backward direction.

4. (-2, 16) and (1, -2)

The system is:

y=x^2 -5x +2 (1)

y=-6x+4 (2)

We can equalize the two equations:

x^2 -5x+2 = -6x +4

which becomes:

x^2 + x -2 =0

Solving it with the formula, we find two solutions: x=-2 and x=1. Substituting both into eq.(2):

x=-2 --> y=-6 (-2) +4 = 12+4 = 16

x=1 --> y=-6 (1) +4 = -6+4 =-2

So, the solutions are (-2, 16) and (1, -2).

5. (-1, 1) and (7, 33)

The system is:

y=x^2 -2x -2 (1)

y=4x+5 (2)

We can equalize the two equations:

x^2 -2x-2 = 4x +5

which becomes:

x^2 -6x -7 =0

Solving it with the formula, we find two solutions: x=7 and x=-1. Substituting both into eq.(2):

x=7 --> y=4 (7) +5 = 28+5 = 33

x=-1 --> y=4 (-1) +5 = -4+5 =1

So, the solutions are (-1, 1) and (7, 33).

6. 2.30 seconds

The height of the object is given by:

h(t)=-16 t^2 +85

The time at which the object hits the ground is the time t at which the height becomes zero: h(t)=0, therefore

-16t^2 +85 =0

By solving it,

16t^2 = 85

$t^2 = \frac{85}{16}$

$t=\sqrt{\frac{85}{16}}=2.30 s$

7. Reaches a maximum height of 19.25 feet after 0.88 seconds.

The height of the ball is given by

h(t)=-16t^2 + 28t + 7

The vertical velocity of the ball is equal to the derivative of the height:

v(t)=h'(t)=-32t+28

The maximum height is reached when the vertical velocity becomes zero: v=0, therefore when

-32t + 28 =0

from which we find t=0.88 s

And by substituting these value into h(t), we find the maximum height:

h(t)=-16(0.88)^2 + 28(0.88) + 7 = 19.25 m

8. Reaches a maximum height of 372.25 feet after 4.63 seconds.

The height of the boulder is given by

h(t)=-16t^2 + 148t + 30

The vertical velocity of the boulder is equal to the derivative of the height:

v(t)=h'(t)=-32t+148

The maximum height is reached when the vertical velocity becomes zero: v=0, therefore when

-32t + 148 =0

from which we find t=4.63 s

And by substituting these value into h(t), we find the maximum height:

h(t)=-16(4.63)^2 + 148(4.63) + 30 = 372.25 m

9. 12 m

Let's call x the length of the side of the original garden. The side of the new garden has length (x+3), so its area is

(x+3)^2 = 225

Solvign this equation, we find

$x+3 = \sqrt{225}=15$

x=15-3=12 m

10. 225/4

In fact, if we write $x^2 +15 x + \frac{225}{4}$ , we see this is equivalent to the perfect square:

$(x+\frac{15}{2})^2 = x^2 +15 x +\frac{225}{4}$

11. -11.56, 1.56

The equation is:

x^2 +10 x -18 =0

By using the formula:

$x=\frac{-10 \pm \sqrt{(10)^2-4(1)(-18)}}{2*1}$

which has two solutions: x=-11.56 and 1.56.

12. -10.35, 1.35

The equation is:

x^2 +9 x -14 =0

By using the formula:

$x=\frac{-9 \pm \sqrt{(9)^2-4(1)(-14)}}{2*1}$

which has two solutions: x=-10.35 and 1.35.

Mathematics

This is my final project for math and I need help Graphing Radical Functions, Radical Equations and Extraneous Roots, Solving Equations Containing Two Radicals You are on a team of architects. You are charged with building a scale-model replica of one section of a new roller coaster before construction gets underway. Certain reinforcement cables and struts are required to make the roller coaster sturdier. The goal for this project is for your team to determine where to place these cables or struts. The mathematical models for these reinforcements

Step-by-step answer

P Answered by Master

16

You have the first 2 here's most of the rest. Write the equation that models the height of the roller coaster. Start by writing the equation of the circle . (Recall that the general form of a circle with the center at the origin isx2+ y2 = r2. x^2+y^2=30^2, The radius is 30 ftx^2+y^2=900 Now solve this equation for y. Remember the roller coaster is above ground, so you are only interested in the positive root. y = √(900 -x^2) Graph the model of the roller coaster using the graphing calculator. Take a screenshot of your graph and paste the image below, or sketch a graph by hand. (Put this in you should be good ) y= root (900)-x^2 x^2+y^2=900 all your points should equal 30 and look like a semi circle, but on the top part of the top not the bottom. Write the equation you would need to solve to find the horizontal distance each beam is from the origin.(x=√30^2-25^2) Algebraically solve the equation you found in step 3. Round your answer to the nearest hundredth. 16.584.Explain where to place the two beams x= -16.58x=16.58 Graph the cable and the strut on the model of the roller coaster using the graphing calculator. Take a screenshot of your graph and paste the image below, or sketch a graph by hand 900= x^2 +y^2 y=root(900-x^2) y= ^2x +8 y= x-8 the intersection should be -10 and positive 10 enter those equation into your calculator. That's all I know so far

Step-by-step explanation:

Mathematics

The graph shows a journey in a car. Which of the statements most likely describes the journey at the portion of the graph labeled J? (1 point) A line graph is drawn on the first quadrant of a coordinate plane. The x-axis is labeled Time in seconds, and the y-axis is labeled Distance in miles. The line graph is divided into 7 segments labeled I, J, K, L, M, N, and O. I starts at the origin and is a straight line slanting up. J is a line segment that starts at the end of I and is horizontal. K is a curve that starts at the end of J and curves up.

Step-by-step answer

P Answered by Specialist

8

The car stops at a distance from the starting point because the portion shows a constant function away from the starting point.

Step-by-step explanation:

For this case, the first thing you should keep in mind is the following definition:

d = v * t

Where,

d: distance

v: speed

t: time

For the portion of the chart labeled J we have:

The time increases

The distance is the same

We have then that the speed is zero.

Therefore, the car is stopped at a distance far from the starting point.

The car stops at a distance from the starting point because the portion shows a constant function away from the starting point.

What is the horizontal distance from the origin to the point (1, 4)?

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