$(7^2)^3=7^5\qquad\bold{FALSE}\\\\\text{because}\\\\7^2=7\cdot7\\\\(7^2)^3=(7\cdot7)^3=(7\cdot7)\cdot(7\cdot7)\cdot(7\cdot7)=\underbrace{7\cdot7\cdot7\cdot7\cdot7\cdot7}_6=7^6\\\\\text{or use}\ (a^n)^m=a^{nm}\\\\(7^2)^3=7^{2\cdot3}=7^6$

4 0

2 years ago

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I need help with my pre-calculus homework, please show me how to solve them step by step if possible. The image of the problem i

Anna [14]

We know that the law of sines states that:

$\frac{\sin\alpha}{a}=\frac{\sin\beta}{b}$

For simplicity, let:

$\beta=m\angle A_1BC$

In triangle A1BC this leads to:

$\begin{gathered} \frac{\sin31}{5}=\frac{\sin\beta}{6} \\ \sin\beta=\frac{6}{5}\sin31 \\ \beta=\sin^{-1}(\frac{6}{5}\sin31) \\ \beta=38.174 \end{gathered}$

Therefore:

$m\angle A_1BC=38.174$

Now, triangle A2BC is isosceles which means that both the base angles are equal, since angle CA1B and A2A1B are supplementary we have:

$m\angle CA_2B=180-38.174=141.826$

8 0

1 year ago