Mathematics: The vertices of a triangle are (1, - 2); (- 2, - 2) , and (- 2, - 1) . Graph the image of the triangle after ot is rotated 270° clockwise about the orgin

The vertices of a triangle are (1, - 2); (- 2,, №13626936, 21.09.2021 22:33

Mathematics

Find the coordinates of the vertices of each figure after the given transformation. 1) translation: (x, y) = (1 + 1, y - 5) D(2.0), E(1.2), F(2.3). G(4,2) A) D(-1, 1), E(-2, 3), F(-1,4), G(1,3) B) D(-4,-1), E(-5, 1), F(-4,2), G(-2, 1) C) D(3,-5), E(2, -3), F(3,-2), G(5,-3) D) D(-4,1), E(-5,3), F(-4,4), G(-2,3) ОА О. B С O O D

Step-by-step answer

P Answered by Master

3

The answer is C
(X+1,y-5) you would add 1 to all x values and subtract 5 from all h values

Mathematics

Find the coordinates of the vertices of each figure after the given transformation. 1) translation: (x, y) = (1 + 1, y - 5) D(2.0), E(1.2), F(2.3). G(4,2) A) D(-1, 1), E(-2, 3), F(-1,4), G(1,3) B) D(-4,-1), E(-5, 1), F(-4,2), G(-2, 1) C) D(3,-5), E(2, -3), F(3,-2), G(5,-3) D) D(-4,1), E(-5,3), F(-4,4), G(-2,3) ОА О. B С O O D

Step-by-step answer

P Answered by Master

3

The answer is C
(X+1,y-5) you would add 1 to all x values and subtract 5 from all h values

Mathematics

Need . just need my answers checked. in advance. my answers are in brackets. 1. the coordinates of the vertices of δjkl are j(-5, -1) , k(0, 1) , and l(2, -5). which statement correctly describes whether δjkl is a right triangle? a. δjkl is a right triangle because jk is perpendicular to jl. b. δjkl is a right triangle because jl is perpendicular to kl. c. δjkl is a right triangle because jk is perpendicular to kl. [ d. δjkl is not a right triangle because no two of its sides are perpendicular. ] 2. the coordinates of the vertices of δjkl are j(0,

Step-by-step answer

P Answered by Specialist

6

#1) d. ΔJKL is not a right triangle because no two of its sides are perpendicular; #2) -1/3, 3, -7, is, two of these slopes have a product of -1; #3) a. Quadrilateral DEFG is a rhombus because opposite sides are parallel and all four sides have the same length; #4) 1, -1/6, 1, -2/5, is not, only one pair of opposite sides is parallel; #5) c. Quadrilateral PQRS is not a rectangle because it has only one right angle.

Step-by-step explanation:

#1) The slope of any line segment is found using the formula

$m=\frac{y_2-y_1}{x_2-x_1}$

For JK, this gives us (1-1)/(-5-0) = 0/-5 = 0. For KL this gives us (1--5)/(0-2) = 6/-2 = -3. For LJ this gives us (-5-1)/(2--5) = -6/7. None of these slopes are negative reciprocals, so none of the angles are right angles and this is not a right triangle.

#2) The slope of JK is (2-1)/(0-3) = 1/-3 = -1/3. The slope of KL is (1--5)/(3-1) = 6/2 = 3. The slope of LJ is (2--5)/(0-1) = 7/-1 = -7. Two of these slopes have a product of -1, 3 and -1/3. This means they are negative reciprocals so this has a right angle; this means JKL is a right triangle.

#3) The slope of DE is (5-4)/(-2-2) = 1/-4 = -1/4. The slope of EF is (4-0)/(2-0) = 4/2 = 2. The slope of FG is (0-1)/(0--4) = -1/4. The slope of GD is (1-5)/(-4--2) = -4/-2 = 2. Opposite sides have the same slope so they are parallel.

Next we use the distance formula to find the length of each side:

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Using our points, the length of DE is

$\sqrt{(5-4)^2+(-2-2)^2}=\sqrt{1^2+(-4)^2}=\sqrt{1+16}=\sqrt{17}$

The length of EF is

$d=\sqrt{(4-0)^2+(2-0)^2}=\sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20}$

The length of FG is

$d=\sqrt{(0-1)^2+(0--4)^2}=\sqrt{(-1)^2+(4)^2}=\sqrt{1+16}=\sqrt{17}$

The length of GD is

$d=\sqrt{(1-5)^2+(-4--2)^2}=\sqrt{(-4)^2+(-2)^2}=\sqrt{16+4}=\sqrt{20}$

Opposite sides have the same length and are parallel, so this is a parallelogram.

#4) The slope of AB is (-1-2)/(-4--1) = -3/-3 = 1. The slope of BC is (2-1)/(-1-5) = 1/-6 = -1/6. The slope of CD is (1--3)/(5-1) = 4/4 = 1. The slope of DA is (-3--1)/(1--4) = -2/5. Only one pair of opposite sides is parallel, so this is not a parallelogram.

#5) The slope of PQ is (2-4)/(-4-3) = -2/-7 = 2/7. The slope of QR is (4-0)/(3-5) = 4/-2 = -2. The slope of RS is (0--2)/(5--3) = 2/8 = 1/4. The slope of SP is (-2-2)/(-3--4) = -4/1 = -4. Only one pair of sides has slopes that are negative reciprocals; this means this figure only has 1 right angle, so it is not a rectangle.

Mathematics

Need . just need my answers checked. in advance. my answers are in brackets. 1. the coordinates of the vertices of δjkl are j(-5, -1) , k(0, 1) , and l(2, -5). which statement correctly describes whether δjkl is a right triangle? a. δjkl is a right triangle because jk is perpendicular to jl. b. δjkl is a right triangle because jl is perpendicular to kl. c. δjkl is a right triangle because jk is perpendicular to kl. [ d. δjkl is not a right triangle because no two of its sides are perpendicular. ] 2. the coordinates of the vertices of δjkl are j(0,

Step-by-step answer

P Answered by Specialist

6

#1) d. ΔJKL is not a right triangle because no two of its sides are perpendicular; #2) -1/3, 3, -7, is, two of these slopes have a product of -1; #3) a. Quadrilateral DEFG is a rhombus because opposite sides are parallel and all four sides have the same length; #4) 1, -1/6, 1, -2/5, is not, only one pair of opposite sides is parallel; #5) c. Quadrilateral PQRS is not a rectangle because it has only one right angle.

Step-by-step explanation:

#1) The slope of any line segment is found using the formula

$m=\frac{y_2-y_1}{x_2-x_1}$

For JK, this gives us (1-1)/(-5-0) = 0/-5 = 0. For KL this gives us (1--5)/(0-2) = 6/-2 = -3. For LJ this gives us (-5-1)/(2--5) = -6/7. None of these slopes are negative reciprocals, so none of the angles are right angles and this is not a right triangle.

#2) The slope of JK is (2-1)/(0-3) = 1/-3 = -1/3. The slope of KL is (1--5)/(3-1) = 6/2 = 3. The slope of LJ is (2--5)/(0-1) = 7/-1 = -7. Two of these slopes have a product of -1, 3 and -1/3. This means they are negative reciprocals so this has a right angle; this means JKL is a right triangle.

#3) The slope of DE is (5-4)/(-2-2) = 1/-4 = -1/4. The slope of EF is (4-0)/(2-0) = 4/2 = 2. The slope of FG is (0-1)/(0--4) = -1/4. The slope of GD is (1-5)/(-4--2) = -4/-2 = 2. Opposite sides have the same slope so they are parallel.

Next we use the distance formula to find the length of each side:

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Using our points, the length of DE is

$\sqrt{(5-4)^2+(-2-2)^2}=\sqrt{1^2+(-4)^2}=\sqrt{1+16}=\sqrt{17}$

The length of EF is

$d=\sqrt{(4-0)^2+(2-0)^2}=\sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20}$

The length of FG is

$d=\sqrt{(0-1)^2+(0--4)^2}=\sqrt{(-1)^2+(4)^2}=\sqrt{1+16}=\sqrt{17}$

The length of GD is

$d=\sqrt{(1-5)^2+(-4--2)^2}=\sqrt{(-4)^2+(-2)^2}=\sqrt{16+4}=\sqrt{20}$

Opposite sides have the same length and are parallel, so this is a parallelogram.

#4) The slope of AB is (-1-2)/(-4--1) = -3/-3 = 1. The slope of BC is (2-1)/(-1-5) = 1/-6 = -1/6. The slope of CD is (1--3)/(5-1) = 4/4 = 1. The slope of DA is (-3--1)/(1--4) = -2/5. Only one pair of opposite sides is parallel, so this is not a parallelogram.

#5) The slope of PQ is (2-4)/(-4-3) = -2/-7 = 2/7. The slope of QR is (4-0)/(3-5) = 4/-2 = -2. The slope of RS is (0--2)/(5--3) = 2/8 = 1/4. The slope of SP is (-2-2)/(-3--4) = -4/1 = -4. Only one pair of sides has slopes that are negative reciprocals; this means this figure only has 1 right angle, so it is not a rectangle.

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

4. the vertices of a triangle are a(-3, 1), b(-3, 4), and c(1, 2). after translation, a is the point (-1, -4). find the translation rule and the coordinates of b . 1 point (x, y) -- (x - 2, y - 5) ; b (-5, -1) (x, y) -- (x + 4, y + 5); b (1, 9) (x, y) -- (x - 4, y - 3); b (-7, 1) (x, y) -- (x + 2, y - 5); b (-1, -1)

Step-by-step answer

P Answered by PhD

2

A(-3, 1) to A' is the point (-1, -4)
rule
(x, y) --> (x + 2, y - 5); B' (-1, -1)

Mathematics

4. the vertices of a triangle are a(-3, 1), b(-3, 4), and c(1, 2). after translation, a is the point (-1, -4). find the translation rule and the coordinates of b . 1 point (x, y) -- (x - 2, y - 5) ; b (-5, -1) (x, y) -- (x + 4, y + 5); b (1, 9) (x, y) -- (x - 4, y - 3); b (-7, 1) (x, y) -- (x + 2, y - 5); b (-1, -1)

Step-by-step answer

P Answered by PhD

2

A(-3, 1) to A' is the point (-1, -4)
rule
(x, y) --> (x + 2, y - 5); B' (-1, -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

Find the coordinates of the vertices of the figure after the given transformation, translation: (x, y)→(x - 4, y + 4) a) j (1, 0), n (5, 0), m (5, 3), p (2, 5) b) j (-2, 5), n (2, 5), m (2, 2), p (-1, 0) c) j (-3, 4), n (1, 4), m (1, 1), p (-2, -1) d) j (-1, 0), n (-5, 0), m (-5, -3), p (-2, -5)

Step-by-step answer

P Answered by PhD

5

Option C J'(-3, 4), N'(1, 4), M'(1, 1), P'(-2, -1)

Step-by-step explanation:

we know that

The rule of the translation is

(x,y) > (x-4,y+4)

That means > The translation is 4 units at left and 4 units up

Find out the coordinates of the pre-image

Observing the figure

J(1,0),N(5,0),M(5,-3),P(2,-5)

Find out the coordinates of the image

Applying the rule of the translation at the coordinates of pre-image

J(1,0) > J'(1-4,0+4)

J(1,0) > J'(-3,4)

N(5,0) ---> N'(5-4,0+4)

N(5,0) ---> N'(1,4)

M(5,-3)> M'(5-4,-3+4)

M(5,-3)> M'(1,1)

P(2,-5) > P'(2-4,-5+4)

P(2,-5) > P'(-2,-1)

therefore

The coordinates of the image (after the transformation) are

J'(-3, 4), N'(1, 4), M'(1, 1), P'(-2, -1)

Mathematics

Find the coordinates of the vertices of the figure after the given transformation, translation: (x, y)→(x - 4, y + 4) a) j (1, 0), n (5, 0), m (5, 3), p (2, 5) b) j (-2, 5), n (2, 5), m (2, 2), p (-1, 0) c) j (-3, 4), n (1, 4), m (1, 1), p (-2, -1) d) j (-1, 0), n (-5, 0), m (-5, -3), p (-2, -5)

Step-by-step answer

P Answered by PhD

5

Option C J'(-3, 4), N'(1, 4), M'(1, 1), P'(-2, -1)

Step-by-step explanation:

we know that

The rule of the translation is

(x,y) > (x-4,y+4)

That means > The translation is 4 units at left and 4 units up

Find out the coordinates of the pre-image

Observing the figure

J(1,0),N(5,0),M(5,-3),P(2,-5)

Find out the coordinates of the image

Applying the rule of the translation at the coordinates of pre-image

J(1,0) > J'(1-4,0+4)

J(1,0) > J'(-3,4)

N(5,0) ---> N'(5-4,0+4)

N(5,0) ---> N'(1,4)

M(5,-3)> M'(5-4,-3+4)

M(5,-3)> M'(1,1)

P(2,-5) > P'(2-4,-5+4)

P(2,-5) > P'(-2,-1)

therefore

The coordinates of the image (after the transformation) are

J'(-3, 4), N'(1, 4), M'(1, 1), P'(-2, -1)

The vertices of a triangle are (1, - 2); (- 2, - 2) , and (- 2, - 1) . Graph the image of the triangle after ot is rotated 270° clockwise about the orgin

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