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Rzqust [24]
3 years ago
9

If I made $20 in 4 hours, what is the rate at which I made money?

Mathematics
1 answer:
slega [8]3 years ago
8 0

Answer:

it would be five each hour.

Step-by-step explanation:

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Work out the area of a rectangle of dimensions 4 cm by 8cm
dedylja [7]
Area = 4 cm * 8 cm = 32 cm²
5 0
3 years ago
Read 2 more answers
Simplify to create an equivalent expression 2-6(-5t+1)
Mariulka [41]

Answer:

30t - 4

Step-by-step explanation:

Remember to follow PEMDAS. First, distribute -6 to all terms within the parenthesis:

2 -6(-5t + 1) = 2 + (-6 * -5t) + (-6 * 1)

2 + (30t) + (-6) = 2 + 30t - 6

Simplify. Combine like terms:

2 + 30t - 6 = 30t + (-6 + 2) = 30t - 4

30t - 4 is your answer.

~

6 0
3 years ago
Read 2 more answers
Topic/what am learning:
zloy xaker [14]
4 7/8 + 11 1/2 = 16 3/8
4 + 7/8 + 11 + 1/2
4 + 11 = 15
7/8 + 1/2 = 7/8 + 4/8 = 11/8
11/8 = 1 3/8
15 + 1 + 3/8 = 16 3/8
3 0
2 years ago
If (a+b) / (x−y) = 3/5 then (10a + 10b) / (9x−9y) = ?
lbvjy [14]

Answer:

2/3

Step-by-step explanation:

Look at what you are getting on the right hand question. 10a +10b has a common factor that can be pulled out. It is 10

10a + 10b = 10(a + b)

Now look at the denominator

9x - 9y = 9(x -y)

There are a lot of possible answers.

Mine would be to multiply 3/5 * 10/9

\dfrac{3}{5}*\dfrac{10}{9} = \dfrac{30}{45} = \dfrac{2}{3}

so

(10a + 10b)/(9x - 9y) = 2/3

5 0
2 years ago
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Two numbers have these properties:
almond37 [142]

Answer:

12\ and\ 30

Step-by-step explanation:

Given both numbers are greater than 6

Their HCF is 6

Their LCM is 60

The product of the HCF and LCM of two numbers is the same as the product of the numbers themselves.

Let us say those number are a and b

a\times b=6\times 60\\a\times b=360

So, the product of those number is 360

Let us factorize 360

360=1\times 2\times 2\times 3\times 5\times 6

It is given that number should be greater than 6

The possible pairs of number are (10,36)\ and\ (12,30)

But only (12,30) has LCM as 60.

So those numbers are 12\ and\ 30

5 0
3 years ago
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