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tekilochka [14]
3 years ago
15

HELPPP!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Xelga [282]3 years ago
3 0
First question : A
2cm×1cm×2+2cm×4cm×2+1cm×4cm×2= 28cm^2
Third question
7x+3(x-2)-4x+8
= 7x+3x-6-4x+8
=7x+3x-4x-6+8
=10x-4x+2
=6x+2
Idon't know the second one but i hope i could still help
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F(x)= 13,500 .89x, is this growing or decay
Oksanka [162]
I hope and think it’s 12015
Because this what I did I (.89)x
And x = 13500 and I got 12015
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Factor completely 3x − 12.<br><br> 3(x − 4)<br><br> 3(x + 4)<br><br> 3x(−12)<br><br> Prime
garri49 [273]
3x - 12
3(3x/3 - 12/3)
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What is the solution
gladu [14]

Answer:

  1. 3x-1/4=-5/11
  2. 11 (3x-1)=4 (5)
  3. 33x-11=20
  4. 33x=20+11
  5. 33x=31
  6. x=31/33
  7. x=?
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2 years ago
What is a y-intercept of the graphed function?
Flura [38]
(0,-9) is the y intercept
8 0
3 years ago
How would the factor pairs change if nancy had only 5 posters to arrange?
Romashka [77]

Answer:

Part A) Nancy can arrange the posters in 4 ways

Part B) Nancy can arrange the posters in 2 ways

Step-by-step explanation:

<u><em>The complete question is</em></u>

Nancy has 10 movie posters. She wants to hang them on a wall in equals rows.

A) Find all the ways that she can arrange the posters.

B) How would the factor pairs change if Nancy had only 5 posters to arrange?

Part A) Find all the ways that she can arrange the posters.

we know that

The factors of 10 are

1x10 -----> 1 row of 10 posters

2x5 -----> 2 rows of 5 posters

5x2 -----> 5 rows of 2 posters

10x1 -----> 10 rows of 1 poster

therefore

Nancy can arrange the posters in 4 ways

Part B) How would the factor pairs change if Nancy had only 5 posters to arrange?

we know that

if Nancy had only 5 posters to arrange

then

The factors of 5 are

1x5 -----> 1 row of 5 posters

5x1 -----> 5 rows of 1 poster

therefore

Nancy can arrange the posters in 2 ways

3 0
3 years ago
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