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Andreas93 [3]
3 years ago
5

5,286÷3 how many digits will the classroom number have

Mathematics
1 answer:
docker41 [41]3 years ago
5 0
<span>5,286÷3 how many digits will the classroom number have
Notice that we have 4 digits as dividend with a place value of thousands as the highest and to be divided with our divisor that have only 1 digit with a place value of ones.
Now, let’s see how many digit will our quotient have:
=> 5 286 / 3
=> 1 762 is the quotient, it has still 4 digit with a place value of thousands.
To check simply multiply our quotient and divisor.</span>



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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}
3 0
3 years ago
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