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podryga [215]
3 years ago
11

Help would be good i dont know how to do this

Mathematics
1 answer:
Mila [183]3 years ago
5 0

\frac{x^2-4x-32}{x^2+7x+12}=\frac{x^2+4x-8x-32}{x^2+4x+3x+12}=\frac{x(x+4)-8(x+4)}{x(x+4)+3(x+4)}=\\\frac{(x+4)(x-8)}{(x+4)(x+3)}=\frac{x-8}{x+3}

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In a classroom of 33 students, the ratio of boys to girls is 3 : 8.
kirill [66]

Answer:

the answer is 9 :)

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Mr moore has saved the liscense plates from all his cars. He has nailed them to a wall in his garage. The wall is 5 meters long
QveST [7]

Answer: 270

Step-by-step explanation:

I might be wrong on this one, but the average license plate is 12 in by 6 in. One meter is around 3 feet. The dimensions of the wall are 9 feet by 15 feet.

Length wise, 15 license plates can fit on the wall, and bright wise, 18 can fit on the wall.

15 times 18 = 270

270 license plates is your answer

4 0
4 years ago
Which of the following is not an identity?
Shkiper50 [21]
A ) cos² x · 1 / sin x - 1 / sin x = - sin x
cos² x - 1 / sin x = - sin x 
- sin² x / sin x = - sin x
- sin x = - sin x     ( correct )
B ) sin x ( cos x / sin x + sin x / cos x ) = 1 / cos x
sin x · ( cos² x + sin²x ) / sin x cos x = 1 / cos x
sin x · 1 / sin x cos x = 1 / cos x
1 / cos x = 1 / cos x  ( correct )
C ) cos² x - sin² x = 1 - 2 sin² x
1 - sin² x - sin² x = 1 - 2 sin² x
1 - 2 sin² x = 1 - 2 sin² x   ( correct )
D ) 1/sin²x + 1/ cos²x = 1
cos²x + sin² x / sin² x cos² x = 1
1 / cos² x sin² x = 1
cos²x sin² x ≠ 1
Answer: D ) is not an identity. 
8 0
3 years ago
What’s is the range of the data below
koban [17]

Answer:

The answer is 28

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
If x&gt;7, then |x|&gt;7. |y|&gt;7, so y=7<br><br><br> Valid or invalid?
Alex

\text{if }x>7\text{, then }|x|>7 is a valid argument

|y|>7\text{, so }y=7 is not a valid argument

For the first argument: \text{if }x>7\text{, then }|x|>7

From the definition of absolute value function

|x|=x   if x\ge0

That is every positive number is its own absolute value. Since

x>7\implies x\ge0,

we can argue that

x>7\implies |x|>7

so the first argument is valid

For the second argument: |y|>7\text{, so }y=7

From the definition of absolute value function

|y|:=\left \{ {y\text{  if }y\ge0}\atop{-y\text{  if }y

This means that

|y|>7:=\left \{ {y>7\text{  if }y\ge0}\atop{-y>7\text{  if }y

or

|y|>7:=\left \{ {y>7\text{  if }y\ge0}\atop{y

no part of the definition allow for the option y=7. So the second argument is not valid.

Learn more here: brainly.com/question/11897796

7 0
3 years ago
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