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True [87]
2 years ago
5

Directions: Solve for X Round your answer to the nearest tenth. Make sure you calculator

Mathematics
2 answers:
ale4655 [162]2 years ago
6 0

Answer:

36.87

Step-by-step explanation:

Using tan inverse of 9 over 12

ipn [44]2 years ago
5 0

Answer:

x ≈ 36.9°

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan x = \frac{opposite}{adjacent} = \frac{9}{12} , then

x = tan^{-1} (\frac{9}{12} ) ≈ 36.9° ( to the nearest tenth )

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eimsori [14]

Answer:

Your solution would be -1x^4 - 1x^3 + x^2 - 6x - 6.

Step-by-step explanation:

5 0
2 years ago
What is the answer for 1,296.5 × 10.2 ?
Elden [556K]
 the answer for 1296.5 times 10.2 is
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3 years ago
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Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!
LUCKY_DIMON [66]

Answer:

=8\sqrt{15}b^{\frac{7}{2}}

Step-by-step explanation:

\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}

=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}

=\sqrt{40}b^2\sqrt{24b^3}

\sqrt{24b^3}

=\sqrt{24}\sqrt{b^3}

=\sqrt{24}b^{\frac{3}{2}}

=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

\sqrt{2^3}

=2^{3\cdot \frac{1}{2}

=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}

2^{\frac{3}{2}+\frac{3}{2}}

=2^3

=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}

b^{\frac{3}{2}+2}

=b^{\frac{7}{2}}

=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}

=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}

=8\sqrt{15}b^{\frac{7}{2}}

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Answer:

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yuradex [85]

Answer:

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Step-by-step explanation:

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