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Vedmedyk [2.9K]
2 years ago
13

Am i smart because i think im not smart IS IT BUSSIN

Mathematics
2 answers:
Harlamova29_29 [7]2 years ago
7 0

Answer:

sure?

Step-by-step explanation:

Harman [31]2 years ago
3 0

Answer:

umm I don't know

jxsbcjbdjcbsub

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Kaynard's first step in solving this rational equation is to simplify it by multiplying both sides by a common denominator. Whic
bulgar [2K]

x^2 - 9 = (x + 3)(x -3)

3x - 9 = 3(x - 3)

and

6

Common denominator would be 6(x+3)(x - 3)

Answer: 6(x+3)(x - 3)

5 0
3 years ago
What type of angles are 1 and 8 ?
wel

Answer:

A. None of the answers are correct

Step-by-step explanation:

∠3 is a verticle angle to ∠1

linear pair means they add up to 180

complementary mean they add up to 90

8 0
2 years ago
Read 2 more answers
The two numbers between the square root of two
KatRina [158]

Answer:

1 and 1.5

Step-by-step explanation:

but im not sure

8 0
3 years ago
Read 2 more answers
Find the area of a square if the perimeter is equal to 38 in.
Misha Larkins [42]
Hello! I would love to help!

Alright! We know that perimeter is basically adding all the sides of a shape together. We also know that a square has 4 sides, so 38 is the result of 4 sides being added. Now, we also know that all sides of a square are even. So, to find one side of a square, we just need to do 38/4


38/4=9.5


Alright. So now we know one side of a square is 9.5 in. We know that in order to find area, we have to multiply length and width. We know that since all sides of a square are even, 9.5 is both our length and width. So, in order to find the area, we just have to multiply 9.5 by 9.5

9.5*9.5=90.25

Alright! So we have determined your answer!!!


90.25 or in fraction form 90 1/4 inches squared!

Hope this helped! Comment if you have questions!!
8 0
2 years ago
Sin^22x - 2cos2x + 2=0
zhuklara [117]

Answer:

x=πn, n∈Z

Step-by-step explanation:

if sin²2x=1-cos²2x, then

1-cos²2x-2cos2x+2=0; ⇒ cos²2x+2cos2x-3=0; ⇔ (cos2x+3)(cos2x-1)=0;

\left[\begin{array}{cc}cos2x+3=0\\cos2x-1=0\\\end{array} \  \ \left[\begin{array}{cc}x \ 'does-not-exist'\\cos2x=1\\\end{array} \  \ 2x=2\pi n, n=Z \  x=\pi n, n=Z

5 0
2 years ago
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