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2 years ago

Help me pls i want to pass my math class

Mathematics

1 answer:

kati45 [8]2 years ago

4 0

Answer:

The slope is -2

The y-intercept is (0,1)

Slope=mx

y-intercept=b

y=mx+b

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The expression representing the distance is
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3 years ago

Find , , and if and terminates in quadrant .

ioda

sin2x =12/13

cos2x = 5/13

tan2x = 12/5

STEP - BY - STEP EXPLANATION

What to find?

• sin2x

• cos2x

• tan2x

Given:

tanx = 2/3 = opposite / adjacent

We need to first make a sketch of the given problem.

Let h be the hypotenuse.

We need to find sinx and cos x, but to find sinx and cosx, first determine the value of h.

Using the Pythagoras theorem;

hypotenuse² = opposite² + adjacent²

h² = 2² + 3²

h² = 4 + 9

h² =13

Take the square root of both-side of the equation.

h =√13

This implies that hypotenuse = √13

We can now proceed to find the values of ainx and cosx.

Using the trigonometric ratio;

$\sin x=\frac{opposite}{\text{hypotenuse}}=\frac{2}{\sqrt[]{13}}$ $\cos x=\frac{adjacent}{\text{hypotenuse}}=\frac{3}{\sqrt[]{13}}$

And we know that tanx =2/3

From the trigonometric identity;

sin 2x = 2sinxcosx

Substitute the value of sinx , cosx and then simplify.

$\sin 2x=2(\frac{2}{\sqrt[]{13}})(\frac{3}{\sqrt[]{13}})$ $=\frac{12}{13}$

Hence, sin2x = 12/13

cos2x = cos²x - sin²x

Substitute the value of cosx, sinx and simplify.

$\begin{gathered} \cos 2x=(\frac{3}{\sqrt[]{13}})^2-(\frac{2}{\sqrt[]{13}})^2 \\ \\ =\frac{9}{13}-\frac{4}{13} \\ =\frac{5}{13} \end{gathered}$

Hence, cos2x = 5/13

tan2x = 2tanx / 1- tan²x

$\tan 2x=\frac{2\tan x}{1-\tan ^2x}$ $=\frac{2(\frac{2}{3})}{1-(\frac{2}{3})^2}$ $=\frac{\frac{4}{3}}{1-\frac{4}{9}}$ $=\frac{\frac{4}{3}}{\frac{9-4}{9}}$ $=\frac{\frac{4}{3}}{\frac{5}{9}}$ $=\frac{4}{3}\times\frac{9}{5}=\frac{4}{1}\times\frac{3}{5}=\frac{12}{5}$

$\tan 2x=\frac{\sin 2x}{\cos 2x}=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5}$

Hence, tan2x = 12/5

Therefore,

sin2x =12/13

cos2x = 5/13

tan2x = 12/5

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