3 years ago

Last question plz make sure it’s right and NO WEBSITES/LINKS will mark brainiest if right :)

Mathematics

1 answer:

Luba_88 [7]3 years ago

6 0

Answer:

4/5

Step-by-step explanation:

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Please help!!! answer the following questions, thanks:)

sp2606 [1]

Answer:

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2 years ago

6 times a number is 27 less than the square of that number. Find the positive solution.

Verdich [7]

Let x = your number.

6x = x^2 - 27. Subtract 6x from each side.

0 = x^2 - 6x - 27. Factor

0 = (x-9) (x+3). Set each term equal to zero

(x+3) = 0. Subtract 3 from each side.

x = -3. This is the negative solution.

(x-9) = 0. Add 9 to each side.

x = 9. This is the positive solution.

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3 years ago

If a=2b^3 and b= -1/2c ^-2,express a in terms of c.

ElenaW [278]

Answer:

Part 1) $a=-\frac{1}{4c^6}$

Part 2) $a=-\frac{1}{4c^{-6}}$

Step-by-step explanation:

Part 1) we have

$a=2b^{3}$ ----> equation A

$b=-\frac{1}{2}c^{-2}$ ----> equation B

substitute equation B in equation A

$a=2(-\frac{1}{2}c^{-2})^{3}$

Applying property of exponents

$(x^{m})^{n}=x^{m*n}$

$x^{-m} =\frac{1}{x^{m}}$

$a=2(-\frac{1}{2}c^{-2})^{3}=2(-\frac{1}{2})^3(c^{-2})^{3}=2(-\frac{1}{8})(c^{-6})=-\frac{1}{4c^6}$

therefore

$a=-\frac{1}{4c^6}$

Part 2) we have

$a=2b^{3}$ ----> equation A

$b=-\frac{1}{2c^{-2}}=-\frac{c^{2}}{2}$ ----> equation B

substitute equation B in equation A

$a=2(-\frac{c^{2}}{2})^{3}$

Applying property of exponents

$(x^{m})^{n}=x^{m*n}$

$x^{-m} =\frac{1}{x^{m}}$

$a=2(-\frac{c^{2}}{2})^{3}=2(-\frac{c^{6}}{8})$

simplify

$a=-\frac{c^{6}}{4}$

therefore

$a=-\frac{1}{4c^{-6}}$

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3 years ago

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