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2 years ago

How to evaluate 4(x+2−5x) when x=2

Mathematics

2 answers:

Crank2 years ago

8 0

Answer:

-24

Step-by-step explanation:

4(x+2−5x)

Combine like terms

4(2 -4x)

Distribute

8 -16x

Let x=2

8 - 16(2)

8 - 32

-24

spayn [35]2 years ago

4 0

<h3>Given :</h3>

how to evaluate 4(x+2−5x) when x=2

The value after evaluating

<h3>Solution :</h3>

We are provided that x equals 2 so have to put 2 instead of x to the desired result

4(x + 2 - 5x)

Putting the value of x we get

4(2 + 2 - 5 × 2)

According to BODMAS multiplication comes first then addition

So 5 × 2 will be solved after that we will simplify 2 added with 2

4(2 + 2 - 10)

4(4 - 10)

4(-6)

- 24

<u>Henceforth, the required answer is -24</u>

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Question 1

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2 years ago

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2 years ago

Give an example of an addition problem in which you would and would not group the addends differently to add

madam [21]

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8 0

2 years ago

Simplify the expression csc(-x)/1+tan^2x)

Charra [1.4K]

Assuming ya meant $\frac{csc(-x)}{1+tan^2(x)}$

to slimplify, we use a variation of the pythagorean identity and a decomposition into the sin and cos

for the pythaogreaon identity
$cos^2(x)+sin^2(x)=1$
divide both sides by $cos^2(x)$
$1+tan^2(x)=sec^2(x)$ since $\frac{sin(x)}{cos(x)}=tan(x)$
subsitute
$\frac{csc(x)}{sec^2(x)}$

recall that $csc(x)=\frac{1}{cos(x)}$
also that cos(x) is an even function and thus cos(-x)=cos(x)
therfore $csc(-x)=\frac{1}{cos(-x)}=\frac{1}{cos(x)}=csc(x)$
so we get

$\frac{csc(x)}{sec^2(x)}$
decompose them into $\frac{1}{cos(x)}$ and $\frac{1}{sin^2(x)}$ to get $\frac{\frac{1}{cos(x)}}{\frac{1}{sin^2(x)}}$
multiply by $\frac{sin^2(x)}{sin^2(x)}$ to get
$\frac{sin^2(x)}{cos(x)}$
we can furthur simlify to get
$(\frac{sin(x)}{cos(x)})(sin(x))=tan(x)sin(x)$
the expression simplifies to tan(x)sin(x)

5 0

3 years ago

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