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Serhud [2]
2 years ago
14

Apply the distributive property to factor out the greatest common factor.75+20=

Mathematics
2 answers:
Oksi-84 [34.3K]2 years ago
4 0

Answer:

5(15+4)

Step-by-step explanation:

The greatest common factor is the greatest number that will divide two values. We have two values 75 and 20. Each has numbers which multiply together to give the number. We need to find the highest value or most in common they share. Each has the factors:

75: 1,3,5,15,25, 75

20: 1, 2, 4, 5, 10, 20

Both have 5 and this is the GCF. So we write the expression int he form 5(___+___). We divide each term by 5 to find the corresponding factor.

75/5= 15

20/5=4

5(15+4)


stiks02 [169]2 years ago
4 0

Answer:

5(15+4)

Step-by-step explanation:

5

15+

and (4)

add them all together.

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What is 3.242424 as a mixed number
kicyunya [14]

Answer:

Step-by-step explanation:

The answer is 3 /8/33.

step by step explanation

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Then we can subtract each side of the first equation from each side of the second equation giving:

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100

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(

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(

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+

(

0

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x

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321

99

x

99

=

321

99

99

x

99

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3

×

107

3

×

33

x

=

3

×

107

3

×

33

x

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107

33

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33

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33

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99

33

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8

33

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3

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8

33

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8

33

3

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6 0
1 year ago
Read 2 more answers
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qaws [65]
The answer is true because if you multiply 3•(3) is will give you 9 and if you multiply 10•(3) it will give you 30 which gives you both of the sides for the larger one
7 0
2 years ago
What is 1/999 as a decimal ​
juin [17]

Answer:0.001001001001001

Step-by-step explanation: I have no idea. just put it in a calculator

5 0
1 year ago
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Which of the following events are equal? (a) A = {1,3}; (b) B = {x | x is a number on a die }; (c) C = {x | x2 −4x +3=0 }; (d) D
rosijanka [135]

Answer:

A and C

Step-by-step explanation:

To determine which events are equal, we explicitly define the elements in each set builder.

For event A

A={1.3}

for event B

B={x|x is a number on a die}

The possible numbers on a die are 1,2,3,4,5 and 6. Hence event B is computed as

B={1,2,3,4,5,6}

for event C

C=[x|x^{2}-4x+3]\\solving  x^{2}-4x+3\\x^{2}-4x+3=0\\x^{2}-3x-x+3=0\\x(x-3)-1(x-3)=0\\x=3 or x=1

Hence the set c is C={1,3}

and for the set D {x| x is the number of heads when six coins re tossed }

In the tossing a six coins it is possible not to have any head and it is possible to have head ranging from 1 to 6

Hence the set D can be expressed as

D={0,1,2,3,4,5,6}

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