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3 years ago

I wasn’t at school when we learned this. Could someone he

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

6 0

Answer:

a= (1,4)

b=(4,2)

c=(2,6)

d=(2,-1)

Step-by-step explanation:

All I did was looked at the regions the points began and ended. I'm not 100% this is correct but I hope I helped!

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3 years ago

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3 years ago

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Andrew [12]

The greatest common factor is the biggest number taken from the values.

Q1. The answer is <span>A. 5y^6
</span>
$20 y^{9} +5 y^{6}= 4*5 y^{9}+5 y^{6}$
Since $x^{a}* x^{b} =x^{a+b}$ , then $x^{9}= x^{3}* x^{6}$

Back to our expression:
$4*5 y^{9}+5 y^{6}=4*5 y^{3}*y^{6}+5 y^{6}=4 y^{3}*5y ^{6}+ 5y ^{6}*1=5 y^{6} (4y ^{3} +1)$
The greatest common factor is thus $5 y^{6}$

Q2. The answer is <span>D. 12xy^2
</span>
$12x y^{5}+60 x^{4} y^{2} -24 x^{3} y^{3}=12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}$
We will use the rule $x^{a}* x^{b} =x^{a+b}$ to factorise the exponents:
$12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}= \\ =12x*y^{2}*y^{3}+5*12*x* x^{3} *y^{2}-2*12x* x^{2} *y*y^{2}= \\ =12xy^{2}*y^{3}+12xy^{2}*5 x^{3} -12xy^{2}*2 x^{2} y= \\ =12xy^{2}(y^{3}+5 x^{3}-2 x^{2} y)$
The greatest common factor is thus $12xy^{2}$

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3 years ago