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3 years ago

PLEASE HELP ME AND DONT JOKE AROUND

Mathematics

2 answers:

Nuetrik [128]3 years ago

6 0

The answer is 12 hope this helped

VladimirAG [237]3 years ago

5 0

Answer:

Step-by-step explanation:

Simplify 2+3 to 5.

7+3× 3 /5

Cancel 3

7+5

Simplify.

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option 1 drop down are: even-odd identity, quotient identity, Pythagorean identity, double-number identity.option 2 drop down ar

motikmotik

Answer:

The equation is given below as

$\frac{\cos2x}{\cos x}=\cos x-\sin x\tan x$

Step 1:

We will work on the left-hand side, we will have

$\begin{gathered} \cos x-\sin x\tan x \\ \text{recall that,} \\ Quoitent\text{ identity is} \\ \tan x=\frac{\sin x}{\cos x} \end{gathered}$

By substituting the identity above, we will have

$\begin{gathered} \cos x-\sin x\tan x=\cos x-\frac{\sin x.\sin x}{\cos x}=\cos x-\frac{\sin^2x}{\cos x} \\ \end{gathered}$

Here, we will make use of the quotient identity

Step 2:

By writings an expression, we will have

$\begin{gathered} \cos x-\sin x\tan x=\cos x-\frac{\sin x.\sin x}{\cos x} \\ \cos x-\sin x\tan x=\frac{\cos^2x-\sin^2x}{\cos x} \end{gathered}$

Here, we will use the definition of subtraction

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

Step 3:

We will apply the double number identity given below

$\begin{gathered} \cos 2\theta=\cos (\theta+\theta)=\cos ^2\theta-\sin ^2\theta \\ \cos 2x=cos(x+x)=\cos ^2x-\sin ^2x \end{gathered}$

By applying this, we will have

$\frac{\cos^2x-\sin^2x}{\cos x}=\frac{\cos2x}{\cos x}$

Here, we will use the double number identity

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

5 0

1 year ago

What is the value of x in the equation below?

11111nata11111 [884]

Answer: B 9.25

Step-by-step explanation:

5 0

3 years ago

One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times sm

sergey [27]

One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle. Then the lines are parallel

<h3>Solution:</h3>

Given that, One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle.

We have to prove that the lines are parallel.

If they are parallel, sum of the described angles should be equal to 180 as they are same side exterior angles.

Now, the 1st angle will be 1/6 of right angle is given as:

$\begin{array}{l}{\rightarrow 1^{\text {st }} \text { angle }=\frac{1}{6} \times 90} \\\\ {\rightarrow 1^{\text {st }} \text { angle }=15 \text { degrees }}\end{array}$