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3 years ago

7

Área de Cuadriláteros ¿Me pueden ayudar con esta tarea?

1 answer:

IRISSAK [1]3 years ago

7 0

Answer:

I can't understand this language sorry

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Save me the headache

maxonik [38]

$(9\sin2x+9\cos2x)^2=81$

Taking the square root of both sides gives two possible cases,

$9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1$

or

$9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1$

Recall that

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

If $\alpha=2x$ and $\beta=\dfrac\pi4$ , we have

$\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}$

so in the equations above, we can write

$\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1$

Then in the first case,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi$

(where $n$ is any integer)

$\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi$

$\implies x=n\pi\text{ or }\dfrac\pi4+n\pi$

and in the second,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi$

$\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi$

$\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi$

Then the solutions that fall in the interval $[0,2\pi)$ are

$x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4$

5 0

3 years ago

Read 2 more answers

QUICK!!!!!!!!help help help

Luba_88 [7]

I think its "Yes it can because -3.5 lies to the left of -1.5"

5 0

4 years ago

Help me on this please.

frosja888 [35]

it’s system of substitutions ?
so try

a = 2
b = 5

8 0

2 years ago

Planes X and Y are perpendicular. Points A, E, F, and G are points only in plane X. Points R and S are points in both planes X a

NemiM [27]

Given:

Planes X and Y are perpendicular to each other
Points A, E, F, and G are points only in plane X
Points R and S are points in both planes X and Y
Lines EA and FG are parallel

The lines which could be perpendicular to RS are EA and FG.

6 0

3 years ago

Read 2 more answers

Logan is a land surveyor. He wants to determine d, the distance across a lake that cannot be directly measured. He draws this di

Mice21 [21]

Answer:

The first option (5.7km)

Step-by-step explanation:

Since we have a right angle triangle and we know two of its sides we can easily find out the thirds side (the value of d) by using the Pathagoras Theorem. In our case 7 and d are our legs and the hypotenuse is equal to 9, so...

(Based on the Pathagoras Theorem)

$c^{2} = a^{2} + b^{2}$

$9^{2} = 7^{2} + d^{2}$

$d^{2} = 9^{2} - 7^{2} \\d^{2} = 81 - 49\\d^{2} = 32\\d = \sqrt{32}\\$

d ≈ 5.7km

There for aproximate distance across the lake is equal to 5.7km

7 0

3 years ago

Read 2 more answers