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g100num [7]
3 years ago
6

Find the greatest common factor: 12ab^2c^4 + 20a^4bc^3 + 16a^3b

Mathematics
1 answer:
nordsb [41]3 years ago
5 0

Option A

The greatest common factor is 4ab

<em><u>Solution:</u></em>

Given that,

We have to find the greatest common factor

12ab^2c^4+20a^4bc^3+16a^3b

<em><u>First find the GCF of numbers</u></em>

GCF of 12, 20, 16

The factors of 12 are: 1, 2, 3, 4, 6, 12

The factors of 16 are: 1, 2, 4, 8, 16

The factors of 20 are: 1, 2, 4, 5, 10, 20

Then the greatest common factor is 4

<em><u>Now find the GCF of variables</u></em>

ab^2c^4\\\\a^4bc^3\\\\a^3b

Here,

ab is the greatest common factor

Therefore, GCF of variables is ab

Thus, GCF of given expression is: 4ab

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46) Which of the following is equivalent to the expression below?
ddd [48]

Answer:

Step-by-step explanation:

Your question doesn't match the picture. OCR is often inaccurate!—not that it matters much in this case. None of the choices is correct.

(20-4)·6² + 2 + 18 = 16·6² + 2 + 18

= 16·36 + 2 + 18

= 576 + 2 + 18

= 578 + 18

= 596

5 0
2 years ago
From a list of ten books, how many groups of 4 books can be selected
andreyandreev [35.5K]

First you have all ten books, but then every time after, you loose one because you can't put the same book in the group twice.

10 * 9 * 8 * 7 = 5040 groups

6 0
3 years ago
12. In a certain cafe, all the sandwiches and drinks are priced the
kakasveta [241]

The cost of one sandwich is $3.90

The cost of one deknk is $1.50

hopefully this is correct

7 0
2 years ago
Alaina needs to sell 225 boxes in order to be the top seller. She sells thin mints and samoas to all of her customers for a tota
Digiron [165]

Answer:

94 boxes of samoas

131 boxes of thin mints

Step-by-step explanation:

m=mints

s=samoas

s+m=225    m=225-s

s(2.5)+m(3)=628

2.5s+3m=628

substitute for m

2.5s+3(225-s)=628

2.5s+675-3s=628

reduce

-0.5s=-47

s=94

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4 0
2 years ago
How do you solve sin (5pi/3) without a calculator?
Savatey [412]

very simple, we use the formula sin(a+b)=sinacosb + sinbcosa and sin(20)=2sinacosa

5pi = 2pi/3+3pi/3,

First, we use sin(a+b)=sinacosb + sinbcosa

sin(5pi/3)=sin(2pi/3+3pi/3)= sin(2pi/3+pi)= sin(2pi/3)cos(pi) +sin(pi)cos(2pi/3)

but we know that sin(pi)= 0, and cos (pi) = -1, so sin(5pi/3)= - sin(2pi/3)

now, use sin(2a)=2sinacosa, sin(5pi/3)= - sin(2pi/3)= -2sin(pi/3)cos(pi/3)

sin<span>(5pi/3)=  -2sin(pi/3)cos(pi/3)</span>

<span>sin(pi/3)= 0.86, cos(pi/3)=0.5, finally we have   </span>sin<span>(5pi/3)=  -0.86 x 0.5= -0.43</span>

5 0
3 years ago
Read 2 more answers
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