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Anika [276]
3 years ago
5

What is c•c•c•c in algebra

Mathematics
1 answer:
irakobra [83]3 years ago
7 0

Answer:

c·c·c·c is the same as saying 4c

You might be interested in
What is the slope of the line of (6,22) and (15,55) please simplify​
Dennis_Churaev [7]

Answer:

11/3

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(55-22)/(15-6)

m=33/9

simplify

m=11/3

5 0
4 years ago
Which expression is a sum of cubes?
erik [133]

we know that

A polynomial in the form a^{3} +b^{3} is called a sum of cubes

so

Let's verify each case to determine the solution

<u>case A)</u> -64x^{6} y^{12} +125x^{16} y^{3}

we know that

-64=-4^{3}

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{16}=x^{15} *x=x*(x^{5})^{3} -------> is not a perfect cube

y^{3}= (y)^{3}

therefore

the case A) is not a sum of cubes

<u>case B)</u> -32x^{6} y^{12} +125x^{16} y^{3}

we know that

-32=-2^{5} -------> is not a perfect cube

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{16}=x^{15} *x=x*(x^{5})^{3} -------> is not a perfect cube

y^{3}= (y)^{3}

therefore

the case B) is not a sum of cubes

<u>case C)</u> 32x^{6} y^{12} +125x^{9} y^{3}

we know that

32=2^{5} -------> is not a perfect cube

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{9}=(x^{3})^{3}

y^{3}= (y)^{3}

therefore

the case C) is not a sum of cubes

<u>case A)</u> 64x^{6} y^{12} +125x^{9} y^{3}

we know that

64=4^{3}

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{9}=(x^{3})^{3}

y^{3}= (y)^{3}

Substitute

4^{3}(x^{2})^{3}(y^{4})^{3} +5^{3}(x^{3})^{3}(y)^{3}

(4x^{2}y^{4})^{3} +(5x^{3}y)^{3}

therefore

<u>the answer is</u>

64x^{6} y^{12} +125x^{9} y^{3} is a sum of cubes

6 0
3 years ago
Read 2 more answers
I=PRT; if P=10, R=0.1 and I=200,
brilliants [131]
10 × 0.1 × t = 200
1 × t = 200
t = 200
4 0
3 years ago
Sorry for the messy laptop but can someone help me qwq
Troyanec [42]

I believe the correct answer is H. 1,-5

5 0
3 years ago
What is the answer to 4+2(x-3)=-8
Afina-wow [57]

Answer: x= -3

Step By Step Explanation:

3 0
3 years ago
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