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Papessa [141]
3 years ago
6

What is 2.5 (x - 1) + 4.5 (-x - 2)

Mathematics
2 answers:
yaroslaw [1]3 years ago
8 0

Answer:

2.5(x-1)+4.5(-x-2)

2.5x-2.5-4.5x-9

-2x-11.5

Step-by-step explanation:

Stels [109]3 years ago
3 0

Answer:

2x-11.5

Step-by-step explanation:

2.5(x-1)+4.5(-x-2)\\2.5x-2.5+4.5x-9\\2.5+2x-9\\2x-11.5

hope this helps

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Steven sold two less boxes of candy for the school fundraiser than Rodney with some of the boxes they sold with 8 how many boxes
wel
If Steven sold two boxes of candy less than Rodney and they sold 8 boxes, you can calculate how many boxes did each sell using the following steps:

Steven ... s = r - 2
Rodney ... r

s + r = 8
r - 2 + r = 8
2 * r = 8 + 2
2 * r = 10    /2
r = 5

s = r - 2 = 5 - 2 = 3

Result: Steven sold 3 boxes of candy and Rodney sold 5 boxes of candy.
7 0
3 years ago
What is 6 as a percent?
mr Goodwill [35]

6 as a percent is 600%.

6 0
3 years ago
What number is 40% of 720
Zarrin [17]
You would have to set you problem up like
   x         40
____=_____
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you would multiply 40 by 720 then you would divide 28,800 by 100 your answer will be 288.
3 0
3 years ago
What is the difference of the polynomials?<br><br> (–2x3y2 + 4x2y3 – 3xy4) – (6x4y – 5x2y3 – y5)
ANEK [815]
For the answer to the question above, I'll provide my solutions to my answers for the problem below.

(–2x3y2 + 4x2y3 – 3xy4) – (6x4y – 5x2y3 – y5)

(−2x3)(y2)+4x2y3+−3xy4+−1(6x4y)+−1(−5x2y3)+−1(−y5)

(−2x3)(y2)+4x2y3+−3xy4+−6x4y+5x2y3+y5

−2x3y2+4x2y3+−3xy4+−6x4y+5x2y3+y5

−2x3y2+4x2y3+−3xy4+−6x4y+5x2y3+y5

(−6x4y)+(−2x3y2)+(4x2y3+5x2y3)+(−3xy4)+(y5)

−6x4y+−2x3y2+9x2y3+−3xy4+y5
So the answer is,
= <span><span><span><span><span>−<span><span>6x4</span>y</span></span>−<span><span>2x3</span>y2</span></span>+<span><span>9x2</span>y3</span></span>−<span>3xy4</span></span>+y5</span> 

I hope this helps
4 0
3 years ago
Read 2 more answers
10. Write a number sentence for the model. Let one white tile equal +1 and one black tile equal –1.
Darina [25.2K]
The Correct answer to it is -14 - (-6) = - 8
5 0
3 years ago
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