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polet [3.4K]
2 years ago
12

Find the missing segment

Mathematics
1 answer:
patriot [66]2 years ago
7 0
Answer: x=49
Step by step explanation: to find one of the sides just subtract 35 from 60, then you get 35/25=x/35, which is 25x=1225, the last step is to divide both sides by 25 and you get the final result of 49
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Marco needs to buy some cat food. At the nearest​ store, 6 bags of cat food cost ​$28.50. How much would Marco spend on 5 bags o
olga55 [171]

Answer:

$23.75

Step-by-step explanation:

divide $28.50 by 6 to find the price for 1 bag which is $4.75 then multiply $4.75 by 5 and that equals $23.75

3 0
2 years ago
Read 2 more answers
Anyone please could help
Shalnov [3]
The answer would be c.127
5 0
3 years ago
Before beginning voice lessons, Wanda already knew how to sing 19 pieces, and she expects to learn 1 new piece during each week
irakobra [83]

Answer:

19+12 = 31 pieces

Step-by-step explanation:

4 0
2 years ago
2/3[8(5-2)^{2}+3*2] <br> How do I distribute the 2/3 and then solve the rest?
Dafna1 [17]
Before I distribute, I solve the inside first:

2/3[8(3)^2+6]

2/3[8(9)+6]

2/3[72+6]

2/3[78]

2*26= 52 is the answer

another way:

2/3[8(3)^2+6]

2/3[8(9)+6]

2/3[72+6]

(2/3)*72+(2/3)*6

2*24+2*2
48+4= 52
6 0
3 years ago
(Y+3)(y^2-3y+9)<br><br> A. Y^3+27<br> B. Y^3-27<br> C. Y^3-6y^2+27<br> D. Y^3+6y^2+27
KonstantinChe [14]

Answer:

A) y^3+27

Step-by-step explanation:

There are two ways of solving this problem:

1. Recognizing this as the factored form of the sum of perfect cubes

2. Distribute and add the like terms.

1. In order to distribute we must multiply y by y^2-3y+9, and then 3 by y^2-3y+9:

(y+3)(y^2-3y+9)=y(y^2-3y+9)+3(y^2-3y+9)

y(y^2-3y+9)+3(y^2-3y+9)=y^3-3y^2+9y+3y^2-9y+27

After we add the positive and negative 3y^2 and 9y, they will cancel out and be gone entirely:

y^3-3y^2+9y+3y^2-9y+27=y^3+27

2. You know how you can factor the difference of perfect squares?

As an example:

a^2-b^2=(a+b)(a-b)

Well, not many people know this but you can actually factor both the sum and difference of perfect cubes:

a^3+b^3=(a+b)(a^2-ab+b^2)

a^3-b^3=(a-b)(a^2+ab+b^2)

Because we have these identities, we can easily establish here that we have the sum of perfect cubes, and that (y+3)(y^2-3y+9)= y^3+3^3 = y^3+27

6 0
2 years ago
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