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BigorU [14]
3 years ago
15

Pls help. Abcdefghijklmnopqrstuvwxyz?????

Mathematics
2 answers:
Natasha_Volkova [10]3 years ago
8 0

Answer:

72

Step-by-step explanation:

KIM [24]3 years ago
6 0

Answer:

72

Step by Step Explanation:

By definition, the angle labeled as x and the angle labeled as 2x-36, when added, they're equal to 180.

You set up an equation to solve for x:

x+2x-36=180

3x-36=180

3x=216

x=72

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Help I need the answer ASAP!!​
saw5 [17]
Yes I’m going back to the sun yet I have a couple questions and I’m not gonna be able I do that but I’m sorry I don’t have a lot to say about to do this but I’m not going back to the sun today or
3 0
3 years ago
Help Please?<br> thats my image below
kenny6666 [7]

Answer:

6) 7.85 7) 32.97

Step-by-step explanation:

6) p= 3.14×(5/2)

p= 7.85

7) p= 3.14×10×5

p= 32.97

4 0
1 year ago
Pre cal/cal master needed
weqwewe [10]
Take x-2 and insert it into 2x^2 + 3x-2 where the x is located 
2x^2 + 3x-2
2(x-2)^2 + 3(x-2)-2

Now work out 2(x-2)^2 + 3(x-2)-2 also follow PEMDAS
2(x-2)^2 + 3(x-2)-2 
Since (x-2)^2 is an Exponent, lets work with that first and expand (x-2)^2.
(x-2)^2
(x -2)(x-2)
x^2 -4x + 4
Now Multiply that by 2 because we have that in 2(x-2)^2
(x-2)^2  =  x^2 -4x + 4
2(x-2)^2  =  2(x^2 -4x + 4)
2(x^2 -4x + 4) = 2x^2 - 8x + 8
2x^2 - 8x + 8

Now that 2(x-2)^2 is done lets move on to 3(x-2).

Use the distributive property and distribute the 3
3(x-2) = 3x - 6

All that is left is the -2 

Now lets put it all together 
2(x-2)^2 + 3(x-2)-2

2x^2 - 8x + 8 + 3x - 6 - 2

Now combine all our like terms 
2x^2 - 8x + 8 + 3x - 6 - 2
Combine:  2x^2      =  2x^2 
Combine: -8x + 3x  =  -5x
Combine:  8 - 6 - 2 =   0

So all we have left is
2x^2 - 5x





5 0
3 years ago
Find the value of the integral that converges.<br> ∫^-5_-[infinity] x^-2 dx.
Bingel [31]

Answer:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

Step-by-step explanation:

For this case we want to find the following integral:

\int_{-\infty}^{-5} x^{-2}dx

And we can solve the integral on this way:

\int_{-\infty}^{-5} x^{-2}dx= \frac{x^{-2+1}}{-2+1} \Big|_{-\infty}^{-5}

\int_{-\infty}^{-5} x^{-2}dx= -\frac{1}{x} \Big|_{-\infty}^{-5}

And if we evaluate the integral using the fundamental theorem of calculus we got:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

8 0
3 years ago
Solve for j.<br>-13j - 20 = -8j + 20​
aleksley [76]

Answer:

j = -8

Step-by-step explanation:

-13j - 20 = -8j + 20​

Add 13 j to each side

-13j+13j - 20 = -8j+13j + 20

-20 = 5j+20

subtract 20 from each side

-20-20 = 5j +20-20

-40 = 5j

Divide by 5

-40/5 = =5j/5

-8 =j

8 0
3 years ago
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