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3 years ago

Encontrar el area de la figura

Mathematics

1 answer:

andrey2020 [161]3 years ago

5 0

2 x 2 = 4
2 x 2 = 4/2 = 2
4 + 2 = 6

area de la figura = 6

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А square tile has a width of foot. How many tiles will fit end-to-end along a<br> 4-foot wall?

irina1246 [14]

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4 tiles

Step-by-step explanation:

4/1 = 4

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2 years ago

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25. A wheelchair access ramp has an angle of elevation of 16°. If the ramp

Elanso [62]

Answer:

C. 76.81

Step-by-step explanation:

We must use one of the trigonometric ratios to find the length, L, of the ramp. If we use 16 as the reference angle, the 22 in would be our opposite side and the ramp length is the adjacent.

Therefore, we must use tan, which is $\frac{opposite}{adjacent}$ :

$tan (16)=\frac{22}{L} \\\\L*(tan(16)=22)\\\\L*tan(16)=22\\$

$\frac{L*tan(16)}{tan(16)}=\frac{22}{tan(16} \\\\L=79.815=79.82 in.$

8 0

2 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

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