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3 years ago

11

Given AE ≈ EC, for what value of x is quadrilateral ABCD a parallelogram?

1 answer:

DerKrebs [107]3 years ago

7 0

Answer:

X = 1/3 (to make it short) plz brainliest

Step-by-step explanation:

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If the point (x, y) is in Quadrant IV, which of the following must be true?

gayaneshka [121]

Answer:

Any coordinate point with (positive x, negative y)

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2 years ago

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

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Answer:

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Step-by-step explanation:

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Answer:

x + 50 × 10 = 150

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3 years ago

Read 2 more answers

6) Determine the dimensions of a rectangle if:<br><br> A = x2 – 19x – 90

Sunny_sXe [5.5K]

Answer:

Step-by-step explanation:

Determine the dimensions of a rectangle if:

A = x2 – 19x – 90

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3 years ago