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KatRina [158]
3 years ago
14

Solve: a.) 5(x - 2) > -15 AND b.) x/1 - 9 < -10

Mathematics
1 answer:
uysha [10]3 years ago
6 0

Answer:

a.) x>-1

b.) x/1 - 9 < -10

Step-by-step explanation:

a.)

5(x - 2) > -15

(divide both sides)

x-2>-3

(move the constant to the right)

x>-3+2

(calculate)

x>-1

(answer)

b.)

x/1 - 9 < -10

(divide)

x-9<-10

(move the constant to the right)

x<-10+9

(calculate)

x/1 - 9 < -10

(answer)

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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

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h =√13

This implies that hypotenuse = √13

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

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

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From the trigonometric identity;

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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}

OR

\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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