Solve for x. B) 7 A) 1 C) 8 D) 0

Mathematics

1 answer:

8_murik_8 [283]3 years ago

4 0

Answer:

answer must be letter D

sorry I can't show you my solution but trust me that is the correct answer...

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Which table does not show a direct variation?

stiv31 [10]

Answer:

Table J

Step-by-step explanation:

A direct variation is a simple relationship between x and y.

When you are looking for the direct relationship you should use the formula for it y=kx, or you could just divide y over x ( $\frac{y}{x}$ ).

Analysing Table F:

The relationship between x and y is 4.

$\frac{-8}{-1}=4\\\\\frac{-4}{-1}=4\\\\\frac{0}{0}= origin\\\\\frac{4}{1}=4$

The origin is fine and doesn't ruin the relationship.

Analysing Table G:

The relationship between x and y is 5.

$\frac{5}{1}=5\\\\\frac{10}{2}=5\\\\\frac{15}{3}=5\\\\\frac{20}{4}=5$

Analysing Table H:

The relationship between x and y is $\frac{1}{3}$ .

$\frac{0}{0}=origin\\\\\frac{1}{3}=\frac{1}{3}\\\\\frac{2}{6}=\frac{1}{3}\\\\\frac{3}{9}=\frac{1}{3}$

Like I said before the origin doesn't ruin the relationship so it's fine.

Analysing Table J:

This one doesn't have a direct relationship between x and y.

$\frac{1}{1}=1\\\\\frac{4}{2}=2\\\\\frac{9}{3}=3\\\\\frac{16}{4}=4$

6 0

3 years ago

Find sin(2x), cos(2x), and tan(2x) from the given information.

larisa86 [58]

Since $\cot(x)=\frac{2}{3}$ and $\cot^{2} x+1=\csc^{2} x$ , we know that:

$\left(\frac{2}{3} \right)^{2}+1=\csc^{2} x\\\\\frac{13}{9}=\csc^{2} x\\\\\csc x=\frac{\sqrt{13}}{3}$

If $\csc x=\frac{\sqrt{13}}{3}$ , this means that $\sin x=\frac{3}{\sqrt{13}}$ and by the Pythagorean identity,

$\sin^{2} x+\cos^{2} x=1\\\left(\frac{3}{\sqrt{13}} \right)^{2}+\cos^{2} x=1\\\frac{9}{13}+\cos^{2} x=1\\\cos^{2} x=\frac{4}13}\\\cos x=\frac{2}{\sqrt{13}}$

Using the double angle formula for sine, $\sin(2x)=2\left(\frac{3}{\sqrt{13}} \right)\left(\frac{2}{\sqrt{13}} \right)=\boxed{\frac{12}{13}}$
Using the double angle formula for cosine, $\cos(2x)=1-2\left(\frac{3}{\sqrt{13}} \right)^{2}=\boxed{-\frac{5}{13}}$
So, since tan=sin/cos, $\tan (2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{\frac{12}{13}}{-\frac{5}{13}}=\boxed{-\frac{12}{5}}$