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2 years ago

12.

Mathematics

1 answer:

g100num [7]2 years ago

7 0

Answer:

Step-by-step explanation:

$tan^{-1} (\frac{\sqrt{3} }{3} )=x\\tan x=\frac{\sqrt{3} }{3} =tan (\frac{\pi}{6} )\\x=\frac{\pi}{6}$

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3 years ago

What’s the distributive property of 17 x 8

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Answer:

Step-by-step explanation:

Remember simple multiplication? Write the problem with 17 at the bottom:

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2 years ago

Read 2 more answers

Select the correct answer from the drop-down menu.

ELEN [110]

Missing term = –2xy

Solution:

Let us first find the quotient of $-8x^2y^3 \div xy$ .

$-8x^2y^3 \div xy=\frac{-8x^2y^3 }{xy}$

$=\frac{-8\times x\times x\times y\times y\times y}{xy}$

Taking common term xy outside in the numerator.

$=\frac{xy(-8\times x\times y\times y)}{xy}$

Both xy in the numerator and denominator are cancelled.

$=-8xy^2$

Thus, the quotient of $-8x^2y^3 \div xy$ is $-8xy^2$ .

Given the quotient of $-8x^2y^3 \div xy$ is same as the product of 4xy and ____.

$-8xy^2=4xy$ × missing term

Divide both sides by 4xy, we get

⇒ missing term = $\frac{-8xy^2}{4xy}$

Cancel the common terms in both numerator and denominator.

⇒ missing term = –2xy

Hence the missing term of the product is –2xy.

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