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Vlad1618 [11]
3 years ago
11

Can someone please explain to me how to do this

Mathematics
2 answers:
Nataly [62]3 years ago
6 0

Answer:

-7

Step-by-step explanation:

-9(x+3)=36

-9x-27=36

-9x=36+27

-9x=63

-9x/9=63/9

x=-7

DerKrebs [107]3 years ago
3 0

Answer:

just  ultiply 9x3

Step-by-step explanation: if you do 9(x+3) that = to 9x or 9*x and 27 or 9*3 so the answer should be -27

also please give brainlest

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Factorize: a^2+10ab -75b^2
soldier1979 [14.2K]

Answer:

9

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
In a regular polygon each side has the same length. find the perimeter of the regular polygon
VLD [36.1K]
First, set these expressions equal in an equation. solve for x, then multiply x by 3 because there are 3 sides. the perimeter is 12

5 0
4 years ago
Need someone with a big brain to help lol<br> Solve for x
AfilCa [17]

Answer:

x=4

Step-by-step explanation:

Hey there!

I am going to use this image as an example so you can understand the formula

The formula for problems like these is

B(BA)=D(DC)

So all we have to do is plug in the numeric values

so now we have 5(5+x+3)=6(6+x)

now we just solve for x

5(5+x+3)=6(6+x)

step 1 distribute the 5 to what is in the parenthesis (5, x and 3)

5x5=25

5*x=5x

5x3=15

so now we have

25+5x+15=6(6+x)

step 2 distribute the 6 to what is in the parenthesis (6 and x)

6x6=36

6*x=6x

now we have

25+5x+15=6x=36

step 3 combine like terms, the only like terms are 25 and 15

15+25=40

now we have

40+5x=6x+36

step 4 subtract 5x from each side

5x-5x cancels out

6x-5x=x

now we have

40=x+36

step 5 subtract 36 from each side

40-36=4

36-36 cancels out

now we have

x=4 so we can conclude that x=4

hope this helps :)

5 0
3 years ago
Cos^2 x+4sin^2 x/2=1
lana [24]

Let\ \dfrac{x}{2}=a,\ therefore\ x=2a.\\\\\cos^2x+4\sin^2\dfrac{x}{2}=\cos^22a+4\sin^2a\\\\\text{use}\ \cos2x=\sin^2x-\cos^2x\\\\=(\sin^2a-\cos^2a)^2+4\sin^2a\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=(\sin^2a)^2-2(\sin^2a)(\cos^2a)+(\cos^2a)^2+4\sin^2a\\\\=\sin^4a-2\sin^2a\cos^2a+\cos^4a+4\sin^2a\\\\=\underbrace{\sin^4a+2\sin^2a\cos^2a+\cos^4a}_{(*)}-4\sin^2a\cos^2a+4\sin^2a\\\\\text{use}\ (*)\qquad(a+b)^2=a^2+2ab+b^2

=\underbrace{(\sin^2a)^2+2\sin^2a\cos^2a+(\cos^2a)^2}_{(*)}-4\sin^2a(\cos^2a-1)\\\\=(\sin^2a+\cos^2a)^2-4\sin^2a(\cos^2a-1)\\\\\text{use}\ \sin^2a+\cos^2a=1\to\sin^2a=\cos^2a-1\\\\=1^2-4\sin^2a(\sin^2a)=1-4\sin^4a=1-(2\sin^2a)^2

\cos^22a+4\sin^2a=1\\\\1-(2\sin^2a)^2=1\qquad\text{subtract 1 from both sides}\\\\-(2\sin^2a)^2=0\to2\sin^2a=0\qquad\text{divide both sides by 2}\\\\\sin^2a=0\to\sin a=0\\\\a=k\pi\ for\ k\in\mathbb{Z}\\\\\dfrac{x}{2}=k\pi\qquad\text{multiply both sides by 2}\\\\\boxed{x=2k\pi\ for\ k\in\mathbb{Z}}

6 0
3 years ago
HELP I NEED HELP ASAP
Vanyuwa [196]
It’s A 18.7 good luck hope you ace it
3 0
3 years ago
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