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Ganezh [65]
3 years ago
14

What is another way to write 75 ? 7+7+7+7+7 5+5+5+5+5+5+5 7×7×7×7×7 5×5×5×5×5×5×5

Mathematics
2 answers:
kakasveta [241]3 years ago
6 0
77-2 = 75 so that is another way :) yet, you can do 5*15 :)

Mark as brainliest pleases and thank you!
swat323 years ago
4 0

I think its 7x7x7x7x7. Hope this helps


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Divide 56 in a ratio of 5.3
Varvara68 [4.7K]
In order to find what one part is in a ratio, you have to add the ratio up ( 5+ 3) and divide it by the number you're looking for (56). In this case, you get 56/8, which gives you 7. Therefore, each part is worth 7. You then have to multiply both sides of the ratio (5 and 3) by 7. 5x7= 35. 3x7= 21. Therefore, 56 divided into the ratio of 5:3 is 35:21
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A triangular prism has a height of 9 meters and a triangular base with the following dimensions
nadya68 [22]

Answer:

b 432m

Step-by-step explanation:

A triangular prism has a height of 9 meters and a triangular base with the following dimensions

What is the volume of the prism?

Answers: 420m, 432m, 288m 324m

7 0
3 years ago
Read 2 more answers
PLEASE HELP! I am having trouble with both questions
schepotkina [342]

Answer:

someone had the same exact question i just helped him on it sub 4 for x and 1 for h

Step-by-step explanation:

6 0
2 years ago
Multiply 9· (-1) can u pls help me with this q
kirill115 [55]
A positive times a negative will always be negative
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6 0
4 years ago
Plz need help on 1. 2. and 3 plz help and dont answer if you dont know it plz
sesenic [268]

Answer:

1) 32m^{6}n^{14}

2) 3j^{4}k^{8}

3) 135a^{11}b^{16}

Step-by-step explanation:

to know: when multiplying terms with same base and exponents, multiply the bases and add the exponents

to know: when dividing terms with same base and exponents, divide the bases and subtract the exponents

to know: whenever exponents are contained within a set of parentheses and separated by another set of parentheses, multiply the exponents

1) (4^{4}· m^{12} · n^{20}) ÷ 2³m^{6}n^{6} = 3j^{4}k^{8}

2) (15j^{5}k^{2}· 25j²k^{9}) ÷ 5³j³k³ = (375j^{7}k^{11}) ÷ 125j³k³ = 3j^{4}k^{8}

3) (5 · 3³) a^{11}b^{16} = 135a^{11}b^{16}

5 0
3 years ago
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