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2 years ago

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A polygon has coordinates A(-7, 8), B(-4, 6), C(-4, 3), D(-8, 3), and E(-9, 6). What are the coordinates of its image, polygon A

′B′C′D′E′, after a 270° counterclockwise rotation about the origin and a translation 2 units to the left and 3 units up?

1 answer:

Alla [95]2 years ago

4 0

Okay so the answer is E6

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B) Evaluate the algebraic expression 2x2 + 5x +6 if x is equal to (-1).

Korvikt [17]

Step-by-step explanation:

if x is equal to - 1

Then

2x² + 5x + 6

= 2* ( -1)² + 5 * (- 1) + 6

= 2 - 5 + 6

= 8 - 5

= 3

Hope it will help :)❤

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2 years ago

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How do you turn -3 1/3 into a negative fraction decimal

lina2011 [118]

Dear Joanna243, what do you mean by a "negative fraction decimal". You can't have a fraction and a decimal at the same time. I'll solve it anyway. Take 1/3 and do 1÷3 to get approximately 0.33, then you plug the -3 with 0.33 to get -3.33

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2 years ago

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Find the equation for intercept x 4 and y -7

nadezda [96]

Answer:

y = $\frac{-7x}{4} - 14$

Step-by-step explanation:

The equation used to find the equation for a line with two intercepts is called point slope form. $y - y_{1} = m(x - x_{1})$ , where $y_{1}$ and $x_{1}$ are the intercept points and m is the slope.

However, we need the slope to use this formula.

Use $m = \frac{y_{1} - y_{2} }{x_{1} - x_{2}}$

m = \frac{y_{1} - y_{2} }{x_{1} - x_{2}}

$m = \frac{((0) - (7))}{((4) - (0))}$

$m = \frac{-7}{4}$

$y - y_{1} = m(x - x_{1})$

y - ( -7 ) = m ( x - ( 4 ) )

y - ( -7 ) = -7/4 ( x - ( 4 ) )

y + 7 = -7/4 ( x + 4 )

y + 7 = $\frac{-7(x+4)}{4}$

y + 7 = $\frac{-7x-28}{4}$

y + 7 = $\frac{-7x}{4} - 7$

y + 7 = $\frac{-7x}{4} - 7$ - 7

y = $\frac{-7x}{4} - 14$

8 0

3 years ago

What is this problem

kobusy [5.1K]

Answer:

28 basketballs

Step-by-step explanation:

basketballs : soccer balls

7 : 6

6 units rep 24

1 unit rep 4

7 units rep 28

7 0

1 year ago

Please solve this, will rate 5 stars and mark as STAR!

Nina [5.8K]

Answer:

$\boxed{5 \cdot \sqrt{2} \cdot \sqrt[6]{5} }$

Step-by-step explanation:

$\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}$

$\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}$

$\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$250=2 \cdot 5^3$

$\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5 \sqrt[3]{2}$

Once

$\sqrt[6]{2} \cdot \sqrt[6]{5} = \sqrt[6]{10}$

We have

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5}$

We can proceed considering the common base of exponentials

$\sqrt[3]{2} \cdot \sqrt[6]{2} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}$

Therefore,

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2} \cdot \sqrt[6]{5}$

7 0

2 years ago