Ask question

Ask question

All categories

Citrus2011 [14]

2 years ago

15

The triangle JKL shown on the coordinate grid below is reflected once to map onto triangle J'K'L': Triangle JKL is drawn on a 4

quadrant coordinate grid with vertices J is at 8, 1. K is at 5, 7. L is at 3, 4. If vertex L' is at (−3, 4), what are the coordinates of vertex J'?

2 answers:

Reil [10]2 years ago

5 0

Answer:

The vertex J' is at (-8,1).

Step-by-step explanation:

If is given that the triangle JKL reflected once to map onto triangle J'K'L'.

The vertices of the triangle JKL are J(8,1), K(5,7) and L(3,4). The vertex L'(-3,4).

$L(3,4)\rightarrow L'(-3,4)$

It is in the form of

$(x,y)\rightarrow (-x,y)$

Since the y-coordinate remains same and the sign of x coordinate is changed. It shows the reflection across the y axis.

The vertices of image are,

$K(5,7)\rightarrow K'(-5,7)$

$J(8,1)\rightarrow J'(-8,1)$

Therefore the vertex J' is at (-8,1).

vfiekz [6]2 years ago

3 0

If vertex L' is at (−3, 4), what are the coordinates of vertex J'?
(1, −8)
(−8, 1)
(−1, 8)
(8, −1)

You might be interested in

ΔABC underwent a sequence of rigid transformations to give ΔA′B′C′. Which transformations might have taken place?

Papessa [141]

Answer: The correct option is second, a rotation $90^{\circ}$ clockwise about the origin followed by a reflection across the x-axis.

Explanation:

From the given figure it is noticed that the vertices of ΔABC are A(-6,4), B(-4,6), C(-2,2) and vertices of ΔA'B'C' are A'(4,-6), B(6,-4), C(2,-2).

It means if the point is P(x,y) then after transformation it will be P'(y,x).

If a point P(x,y) reflection across the y-axis followed by a reflection across the x-axis, then the image of point after transformation will be P'(-x,-y), therefore it is not the correct option.

If a shape is rotated $90^{\circ}$ clockwise about the origin then the point P(x,y) will be P'(y,-x) and after that reflect across the x-axis, so the point after transformation will be P'(y,x), therefore it is the correct option.

If a shape is rotated $270^{\circ}$ clockwise about the origin then the point P(x,y) will be P'(-y,x) and after that reflect across the x-axis, so the point after transformation will be P'(-y,-x), therefore it is not the correct option.

If a point P(x,y) reflection across the x-axis followed by a reflection across the y-axis, then the image of point after transformation will be P'(-x,-y), therefore it is not the correct option.

Hence, the correct option is second, a rotation $90^{\circ}$ clockwise about the origin followed by a reflection across the x-axis.

4 0

2 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

2 years ago

A hole is drilled in a sheet-metal component, and then a shaft is inserted through the hole. The shaft clearance is equal to dif

VashaNatasha [74]

Answer:

(A)

$P(X \geq 0.8) = \int\limits_{0.8}^{\infty} f(x) \, dx = \int\limits_{0.8}^{1} 1.25(1-x^4) \, dx = 0.08192$

(B)

Then the cumulative function would be

$CF(x) = 1.25x - 0.25x^5$ if 0<x<1

0 otherwise.

Step-by-step explanation:

(A)

We are looking for the probability that the random variable X is greater than 0.8.

$P(X \geq 0.8) = \int\limits_{0.8}^{\infty} f(x) \, dx = \int\limits_{0.8}^{1} 1.25(1-x^4) \, dx = 0.08192$

(B)

For any $x$ you are looking for the probability $P(X \geq x)$ which is

$P(X \geq x) = \int\limits_{-\infty}^{x} 1.25(1-t^4) dt = \int\limits_{0}^{x} 1.25(1-t^4) dt = 1.25x - 0.25x^2$

Then the cumulative function would be

$CF(x) = 1.25x - 0.25x^5$ if 0<x<1

0 otherwise.

5 0

2 years ago

Lola says these two expressions have the same value.

Delvig [45]

Answer:

It should be A.

Step-by-step explanation:

When you're using exponents, anything that has an exponent of 0 will be 1.

You can use a calculator to confirm on your own.

8 0

2 years ago

Read 2 more answers

2,438,783 divided 893

Naya [18.7K]

Answer:

quotient=2,731

Remainder=0

Step-by-step explanation:

893) 2,438,783 (2,731

- 1 786

---------

6527

- 6251

---------

2768

- 2679

-----------

893

- 893

---------

0

3 0

2 years ago