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2 years ago

7

a carnival ride is a drawn on a cordinate plane so that the first car is located at the point 30,0 what are the coordinates of t

he first car after a rotation of 270 about the orgin

1 answer:

dexar [7]2 years ago

5 0

your answer would be (0,30)

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$(8x\:-\:4x\:+\:1) ÷ (x\:+\:6)$
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$\frac{4x\:+\:1}{x\:+\:6}$

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3 years ago

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2 years ago

Read 2 more answers

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Katena32 [7]

Answer:

(a)

$5^{-3}=\frac{1}{125}$

(b)

$-5^{-3}=-\frac{1}{125}$

(c)

$(-5^{-3})^{-1}=-125$

(d)

$(-5^{-3})^{0}=1$

Step-by-step explanation:

(a)

$5^{-3}$

we can use property of exponent

$a^{-n}=\frac{1}{a^n}$

we get

$5^{-3}=\frac{1}{5^3}$

$5^{-3}=\frac{1}{5\times 5\times 5}$

$5^{-3}=\frac{1}{125}$ ........Answer

(b)

$-5^{-3}$

we can use property of exponent

$a^{-n}=\frac{1}{a^n}$

we get

$-5^{-3}=-\frac{1}{5^3}$

$-5^{-3}=-\frac{1}{5\times 5\times 5}$

$-5^{-3}=-\frac{1}{125}$ ........Answer

(c)

$(-5^{-3})^{-1}$

we can use property of exponent

$(a^{n})^m=a^{m\times n}$

we get

$(-5^{-3})^{-1}=(-5)^{-3\times -1}$

$(-5^{-3})^{-1}=(-5)^3$

$(-5^{-3})^{-1}=(-5)\times (-5)\times (-5)$

$(-5^{-3})^{-1}=-125$ ........Answer

(d)

$(-5^{-3})^{0}$

we can use property of exponent

$(a^{n})^m=a^{m\times n}$

we get

$(-5^{-3})^{0}=(-5)^{-3\times 0}$

$(-5^{-3})^{-1}=(-5)^0$

we can use property

$a^0=1$

$(-5^{-3})^{0}=1$ ........Answer

4 0

2 years ago