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3 years ago

13

6a+5a=-11 whats the breakdown to get the answers

1 answer:

kenny6666 [7]3 years ago

8 0

add 6a and 5a and get 11a

then:

11a=-11

a=-1

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Answer:

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Step-by-step explanation:

$\frac{mass \: of \: earth}{mass \: of \: human} = \frac{5.97 \times {10}^{24} }{6.35 \times {10}^{1} } \\ \\ = 0.94015748 \times {10}^{24 - 1} \\ \\ = 0.94015748 \times {10}^{23} \\ \\ = 9.4\times {10}^{22} \\ \\ \therefore \: mass \: of \: earth \\ =( 9.4\times {10}^{22} ) \times mass \: of \: human$

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3 years ago

27 POINTS » Answer this simple question!

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Write 6(x - 5)^4+ 4(x - 5)^2 + 6 = 0 in the form of a quadratic by using substitution

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Answer:

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Step-by-step explanation:

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3m^2+2m+3=0

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4 years ago

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3 years ago

Please help me with Algebra 2 questions (2).

user100 [1]

8. The simplified form of the expression is $- 5 x^{\frac{4}{5}}y^{3}$

11. The value of x is 4

<h3>Simplifying an expression</h3>

From the question, we are to simplify the give expression

The given expression is

$y^{3}\sqrt[5]{32x^{4} } - 7\sqrt[5]{x^{4}y^{15} }$

$y^{3} (32x^{4} )^{\frac{1}{5} } - 7(x^{4}y^{15} )^{\frac{1}{5}$

$y^{3} \times 32^{\frac{1}{5} } x^{\frac{4}{5} } - 7 \times x^{\frac{4}{5}}y^{\frac{15}{5}}$

$y^{3} \times 2^{5 \times} ^{\frac{1}{5} } \times x^{\frac{4}{5} } - 7 \times x^{\frac{4}{5}}y^{3}$

$y^{3} \times 2 \times x^{\frac{4}{5} } - 7 \times x^{\frac{4}{5}}y^{3}$

$2 x^{\frac{4}{5} } y^{3} - 7 x^{\frac{4}{5}}y^{3}$

$= - 5 x^{\frac{4}{5}}y^{3}$

11. $\frac{1}{2}(5x+7 )^{\frac{2}{3} } =\frac{9}{2}$

Multiply both sides by 2

$2 \times \frac{1}{2}(5x+7 )^{\frac{2}{3} } =2 \times \frac{9}{2}$

$(5x+7 )^{\frac{2}{3} } =9$

Multiply both sides by a power of 3/2

$(5x+7 )^{\frac{2}{3} \times {\frac{3}{2} } } =9^{\frac{3}{2} }$

$(5x+7 )=3^{2 \times} ^{\frac{3}{2} }$

$5x+7 =3^{3}$

$5x+7 =27$

$5x =27-7$

$5x = 20$

$x = \frac{20}{5}$

x = 4

Hence,

8. The simplified form of the expression is $- 5 x^{\frac{4}{5}}y^{3}$

11. The value of x is 4

Learn more on Simplifying expressions here: brainly.com/question/11688560

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2 years ago