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3 years ago

1/4 / (17 + 4) What’s 1/4 divided by (17+4)

Mathematics

2 answers:

blondinia [14]3 years ago

6 0

Answer:

21/4

Step-by-step explanation:

1/4 ÷ (17 + 4)

1/4 ÷ (21)

21 × 1/4 = 21/4

<u>-TheUnknownScientist</u>

Marrrta [24]3 years ago

5 0

Answer:

0.0119047619

Step-by-step explanation:

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In a group of 60 students, 20%are girls. How many are boys?

Firlakuza [10]

Answer: About 12 of those 60 students are boys.

To find the percent, you'll convert 20% to a decimal.

20 / 100 = 0.2

Then, multiply 60 by 0.2

0.2 × 20 = 12

That's one way, another way is:

Convert 20% to a fraction.

20% = 1/5

Then multiply 60 by 1/5.

60 × 1/5 = 60/5

Lastly, divide 60 by 5

60 / 5 = 12.

Hence, 12 is the answer.

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What is the value of x in the equation 25x2 = 16. A. 5/4. B. 4. C. 4/5. D. 16/25

Sergeeva-Olga [200]

Answer:

x = ⅘

Step-by-step explanation:

For clarity, exponents should be marked with a ^. 25x² can be written as 25x^2.

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x² = 16/25

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3 years ago

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

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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

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

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