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3 years ago

Please help. I need help with # 36. Thanks

Mathematics

1 answer:

Xelga [282]3 years ago

8 0

I would say that the answer is B. First because the line AB is descending from left to right which shows that it is negative. Secondly, the gradient 3 would be steep and the gradient 1/3 would not be as steep which is why iI would say it is 1/3. Overall, the answer would be negative and 1/3 which would be B.

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ElenaW [278]

Answer:

12−3xStep-by-step explanation:

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2 years ago

Write the expression using rational exponents. Then simplify and convert back to radical notation.

ioda

Answer:

The radical notation is $3x\sqrt[3]{y^2z}$

Step-by-step explanation:

Given

$\sqrt[3]{27 x^{3} y^{2} z}$

Step 1 of 1

Write the expression using rational exponents.

$\sqrt[n]{a^{m}}=\left(a^{m}\right)^{\frac{1}{n}}$

$=a^{\frac{m}{n}}:\left({27 x^{3} y^{2} z})^{\frac{1}{3}}$

$(a \cdot b)^{r}=a^{r} \cdot b^{r}:(27)^{\frac{1}{3}}\left(x^{3}\right)^{\frac{1}{3}} \cdot\left(y^{2}\right)^{\frac{1}{3}} \cdot(z)^{\frac{1}{3}}$

$=$(3^3)^{\frac{1}{3}}\left(x^{3}\right)^{\frac{1}{3}} \cdot\left(y^{2}\right)^{\frac{1}{3}} \cdot(z)^{\frac{1}{3}}$$

$=\left(3\right)\left(x}\right)} \cdot\left(y}\right)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

$=3x \cdot(y)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

Simplify $3 x \cdot(y)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

$=3 x \sqrt[3]{y^{2} z}$

Learn more about radical notation, refer :

brainly.com/question/15678734

4 0

3 years ago

Given the following triangle, if b = 9 and B = 15°, find a.<br><br> 0.27<br> 2.41<br> 33.59

SVEN [57.7K]

SOH,CAH, TOA so it would be TANGENT since it would be Opposite over Adjacent. Tan(15) = 9/a put Tan(15) onto your calculator it would be .267949 round that to .27 = 9/a cross multiply
a = 33.588. C is the correct answer.

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3 years ago

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Answer:

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Step-by-step explanation:

i am small child from American

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I am born in Nepal

but develop in America

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3 years ago

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Tpy6a [65]

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Step-by-step explanation:

Rise is 10 run is 5 Rate of Change is rise over run.

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2 years ago