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4 years ago

⦁ The completed construction of a regular hexagon is shown below. Explain why ACF is a 30º-60º-90º triangle. (10 points)

Mathematics

1 answer:

Pani-rosa [81]4 years ago

3 0

Answer:

Step-by-step explanation:

You can prove that angle C is 30°. This is because we know that angle B is 120° also, and co-interior angles add to 180°, which makes angle BCF 60°. Then we can divide this by 2 as angle BCF is bisected by AC, so therefore angle ACF is 30° as 60 ÷ 2 = 30.

You can then prove angle A is 90°. First we can find angle BAC because it is part of a triangle, and angles in a triangle add to 180°. This means we can do 30 + 120 = 150, and 180 - 150 = 30° which is angle BAC. Then we do 120 - 30 = 90, as the whole of angle BAF is 120°.

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3 years ago

The monthly charges for Jims phone is

solmaris [256]

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Explanation:

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3 years ago

What is the true solution to the equation below?

Marina CMI [18]

Answer:

The solution is:

$x=4$

Step-by-step explanation:

Considering the expression

$lne^{lnx}+lne^{lnx}^{^2}=2ln8$

$\ln \left(e^{\ln \left(x\right)}\right)+\ln \left(e^{\ln \left(x\right)\cdot \:2}\right)=2\ln \left(8\right)$

$\mathrm{Apply\:log\:rule}:\quad \:log_a\left(a^b\right)=b$

$\ln \left(e^{\ln \left(x\right)}\right)=\ln \left(x\right),\:\space\ln \left(e^{\ln \left(x\right)2}\right)=\ln \left(x\right)2$

$\ln \left(x\right)+\ln \left(x\right)\cdot \:2=2\ln \left(8\right)$

$\mathrm{Add\:similar\:elements:}\:\ln \left(x\right)+2\ln \left(x\right)=3\ln \left(x\right)$

$3\ln \left(x\right)=2\ln \left(8\right)$

$\mathrm{Divide\:both\:sides\:by\:}3$

$\frac{3\ln \left(x\right)}{3}=\frac{2\ln \left(8\right)}{3}$

$\ln \left(x\right)=\frac{2\ln \left(8\right)}{3}.....A$

Solving the right side of the equation A.

$\frac{2\ln \left(8\right)}{3}$

$\ln \left(8\right):\quad 3\ln \left(2\right)$

Because

$\ln \left(8\right)$

$\mathrm{Rewrite\:}8\mathrm{\:in\:power-base\:form:}\quad 8=2^3$

⇒ $\ln \left(2^3\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$\ln \left(2^3\right)=3\ln \left(2\right)$

$\frac{2\ln \left(8\right)}{3}=\frac{2\cdot \:3\ln \left(2\right)}{3}$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:3=6$

$=\frac{6\ln \left(2\right)}{3}$

$\mathrm{Divide\:the\:numbers:}\:\frac{6}{3}=2$

$=2\ln \left(2\right)$

So, equation A becomes

$\ln \left(x\right)=2\ln \left(2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$=\ln \left(2^2\right)$

$=\ln \left(4\right)$

$\ln \left(x\right)=\ln \left(4\right)$

$\mathrm{Apply\:log\:rule:\:\:If}\:\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\:\mathrm{then}\:f\left(x\right)=g\left(x\right)$