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ddd [48]
3 years ago
10

Which number is equivalent to the fraction 15 over 7?

Mathematics
1 answer:
USPshnik [31]3 years ago
7 0
15/7 is approximately 2.143
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1) Using Order of Operations, solve the following problem: (5²- 6 · 3) + 2³
tino4ka555 [31]

Answer:

1. 24 2. 70

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
In the expression 3x2y + 4x – 5x2y, are the terms 3x2y and –5x2y like terms? Explain by providing support for your answer.
loris [4]

Answer:

yes

Step-by-step explanation:

These are like terms because they  have the same variables and powers. The coefficients do not need to match for it to be considered a like term.

6 0
3 years ago
Write the explicit rule of the arithmetic sequence that has a first term of 3 and a fourth term of 18
mr Goodwill [35]

Answer:

a_{n} = 5n - 2

Step-by-step explanation:

The nth term of an arithmetic sequence is

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₄ = 18 , then

a₁ + 3d = 18 , that is

3 + 3d = 18 ( subtract 3 from both sides )

3d = 15 (divide both sides by 3 )

d = 5

Then

a_{n} = 3 + 5(n - 1) = 3 + 5n - 5 = 5n - 2 ← explicit rule

5 0
3 years ago
What is the value of 8/36 in lowest terms ?
Illusion [34]

Simplify:

8/36=4/18=2/9=0.22222(repeating)

Your answer would be 2/9 or 0.22222(etc.).

I hope this helps :)

4 0
4 years ago
Read 2 more answers
The bed of a truck is stacked with with boxes of paper. The boxes are stacked 5 boxes deep by 4 boxes high by 4 boxes across
marysya [2.9K]

Answer:

3

Step-by-step explanation:

The boxes are stacked 5 boxes deep by 4 boxes high by 4 boxes across, then there are 5\cdot 4\cdot 4=80 boxes in total.

The mass of 1 box of paper is 22.5 kilograms, so 80 boxes weigh 22.5\cdot 80=1800 kilograms.

When the driver is in the truck, the mass is 2948.35 kilograms, then the total mass is

2948.35+1800=4748.35\ kg.

Let n be the number of boxes of paper the driver must deliver at the first stop. Their weigth is 22.5n kg and the weight of the truck without n boxes is

4748.35-22.5n\ kg.

Trucks with a mass greater than 4700 kilograms are not allowed over the bridge, thus

4748.35-22.5n48.35,\\ \\n>\dfrac{967}{450}=2\dfrac{67}{450}.

Hence, the driver must deliver at least 3 boxes at the first shop.

8 0
3 years ago
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