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1 year ago

6

Please help with the picture below

1 answer:

elena-s [515]1 year ago

4 0

Answer:

Step-by-step explanation:

Remark

The angles are marked as alternate interior angles. To be parallel, they must be equal. In other words x - 20 must equal 52. That's all you have to do. Just equate the 2 givens and solve.

Givens

x - 20 is one angle

52 is the other.

Solution

x - 20 = 52 Add 20 to both sides.

x - 20 + 20 = 52 +20 Combine

x = 72

Answer: C

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You toss a coin three times. What is the probability of getting three heads?

RSB [31]

Answer:1/6

Step-by-step explanation: So you toss it 3 times(1/3) and you have 50% chance(1/2) of choosing heads or tails... so,

1/3 x 1/2 = 1/6

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2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx....%7D%20%7D%20%7D%20%7D%20%20%3D%20%

andrey2020 [161]

First observe that if $a+b>0$ ,

$(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}$

Let $a=0$ and $b=x$ . It follows that

$a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}$

Now let $b=1$ , so $a^2+a=4x$ . Solving for $a$ ,

$a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2$

which means

$a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}$

Now solve for $x$ .

$x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2$

(note that we assume $2x-1\ge0$ )

$4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}$

(we omit $x=0$ since $2\cdot0-1=-1\ge0$ is not true)

3 0

1 year ago

Need help please !!!

ehidna [41]

Using SOH CAH TOA

Sin =opposite/hypotenuse

that means opposite=2

Hypotenuse=3

Cos=Adjacent/Hypotenuse

Using Pythagorean theorem

Hyp^2=Opp^2 +Adj^2

3^2 =2^2+adj^2

9=4+adj^2

9-4=adj^2

5=adj^2

Taking the square root of both sides

√5=adj

:: cos=√5/3

5 0

2 years ago

Rewrite with only sin x and cos x.

Annette [7]

Option A

$\cos 3 x=\cos x-4 \cos x \sin ^{2} x$

<em><u>Solution:</u></em>

Given that we have to rewrite with only sin x and cos x

Given is cos 3x

$cos 3x = cos(x + 2x)$

We know that,

$\cos (a+b)=\cos a \cos b-\sin a \sin b$

Therefore,

$\cos (x+2 x)=\cos x \cos 2 x-\sin x \sin 2 x$ ---- eqn 1

We know that,

$\sin 2 x=2 \sin x \cos x$

$\cos 2 x=\cos ^{2} x-\sin ^{2} x$

Substituting these values in eqn 1

$\cos (x+2 x)=\cos x\left(\cos ^{2} x-\sin ^{2} x\right)-\sin x(2 \sin x \cos x)$ -------- eqn 2

We know that,

$\cos ^{2} x-\sin ^{2} x=1-2 \sin ^{2} x$

Applying this in above eqn 2, we get

$\cos (x+2 x)=\cos x\left(1-2 \sin ^{2} x\right)-\sin x(2 \sin x \cos x)$

$\begin{aligned}&\cos (x+2 x)=\cos x-2 \sin ^{2} x \cos x-2 \sin ^{2} x \cos x\\\\&\cos (x+2 x)=\cos x-4 \sin ^{2} x \cos x\end{aligned}$

$\cos (x+2 x)=\cos x-4 \cos x \sin ^{2} x$

Therefore,

$\cos 3 x=\cos x-4 \cos x \sin ^{2} x$

Option A is correct

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2 years ago

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Irina18 [472]

The answer to this question is C.

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3 years ago