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vitfil [10]
3 years ago
7

There are 125 sheets of paper in a pack. A classroom uses 30 packs of paper in a year. How many total sheets of paper does the c

lassroom use?
A.
155

B.
625

C.
1,875

D.
3,750
Mathematics
2 answers:
balu736 [363]3 years ago
5 0
Answer:
D. 3,750
Explaintion:
125 sheets in each pack x 30 packs of paper in a year = 3,750 sheets of paper used in a year
tresset_1 [31]3 years ago
4 0

Answer:

3,750

Step-by-step explanation:

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What is the solution for -3(6x+5)-2(1-17x)=3(5+5x)
Novosadov [1.4K]

-3(6x+5)-2(1-17x)=3(5+5x)

Mutiply the first bracket by -3

(-3)(6x)= -18x

(-3)(5)= -15

Mutiply the second bracket by -2

(-2)(1)= -2

(-2)(-17x)= 34x

Mutiply the third bracket by 3

(3)(5)= 15

(3)(5x)= 15x

-18x-15-2+34x= 15+15x

-18x+34x-15-2= 15+15x

16x-17= 15+15x

move 15 to the other side

sign changes from +15 to -15

16x-17-15= 15-15+15x

16x-32= 15x

move -32 to the other side

sign changes from -32 to +32

16x-32+32=15x+32

16x= 15x+32

move 15x to the other side

sign changes from +15x to -15x

16x-15x=15x-15x+32

x= 32

Answer: x= 32

3 0
3 years ago
Jake cuts out five-pointed stars. They are different sizes, but they all have the same shape. The side lengths within each star
avanturin [10]
60 I think I am not 100%
6 0
3 years ago
The ages of three siblings are consecutive integers. The sum of their ages is 42. Calculate their ages. *
stealth61 [152]

Answer:

13, 14, 15

Step-by-step explanation:

Let x = age of one of the siblings

Since the age of the siblings are consecutive integers, the age of the next sibling would be x + 1 and the age of the third sibling would be x + 2

x +( x +1) + (x +2) = 42

3x + 3 = 42

collect like terms and solve for x

x = 13

The age of the first sibling is 13

The age of the second sibling is x + 1 = 13 + 1 = 14

The age of the third sibling is x + 2 = 13 + 2 = 15

8 0
3 years ago
last year 1/5 of the students in your class played a sport this year 10 students join the class five of the new students play a
tino4ka555 [31]

Answer:

60

Step-by-step explanation:

last year 1/5 of the class played sports, so s = (1/5) c

the next year, s increased by 5, c increased by 10 and the fraction changed to 1/4, so for the next year (s+5) = (1/4)(c+10)

To solve this we can substitute s = (1/5)c from the first equation into the second, so

((1/5)c + 5)=(1/4)(c+10), simplify both sides

(1/5)c + 5 = (1/4)c + 10/4, simplify 10/4 to 5/2

(1/5)c + 5 = (1/4)c + 5/2, multiply both sides by 20 to eliminate fractions (least common multiple of 2, 4 and 5)

4c + 100 = 5c + 50, subtract 4c, subtract 50 from both sides

50 = c, number of students in class last year was 50, this year is 10 more, so this year is 60

5 0
3 years ago
The sum of four<br>consecutive odd number is 8o. Find the number​
yarga [219]

Answer:

<em>The sum of 4 consecutive odd number is 80</em>

<em>Let X be the first of these numbers</em>

<em>Then the next odd number is X+2</em>

<em>The third is X+4The fourth is X+6</em>

<em>All of these add up to 80</em>

<em>(X) + (X+2) + (X+4) + (X+6) = 80</em>

<em>Using the commutative and associative laws, let's transform this equation into</em>

<em>(X + X + X + X) + (2 + 4 + 6) = 804X + 12 = 80</em>

<em>Subtract 12 from both sides of the equation gives4X = 68</em>

<em>Divide both sides by 4 gives</em>

<em>X = 17</em>

<em>Going back to the original question:What are the 4 consecutive odd numbers: 17, 19, 21, 23Checking our answer:17 + 19 + 21 + 23 = 80 Correct!</em>

5 0
2 years ago
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