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valentina_108 [34]

3 years ago

8

Find all 2-digit natural numbers that have exactly three factors. Thanks!

2 answers:

Tasya [4]3 years ago

8 0

If the number has 3 factors, the factors would be 1, the number itself, and a a number that could square into it... meaning that you are finding numbers that are 2-digit, postive, perfect squares with no other factors besides 1 and itself. 49 is the only number that will work.

yKpoI14uk [10]3 years ago

5 0

It would be 25 & 49.

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How to figure out 5 - 0.53

sveta [45]

the answer would be 4.47

because every whole number is made out of 100 small parts.

So that would be like 400-53 which is 447 and then add the decimal point. That would make it 4.47

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3 years ago

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vfiekz [6]

Answer. the answer for the problem you asked is A

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2 years ago

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grigory [225]

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3 years ago

Divide. (18x^3 + 12x^2 - 3x) ÷ 6x^2

nlexa [21]

$\bold{[ \ Answer \ ]}$

$\boxed{\bold{\frac{x^3\left(6x^2+4x-1\right)}{2}}}$

$\bold{[ \ Explanation \ ]}$

$\bold{Divide: \ \left(18x^3\:+\:12x^2\:-\:3x\right)\:\div \:6x^2}$

$\bold{-------------------}$

$\bold{Rewrite}$

$\bold{18x^3+12x^2-3x \ = \ x^2\frac{x\left(6x^2+4x-1\right)}{2}}$

$\bold{Rewrite}$

$\bold{x^2\frac{x\left(6x^2+4x-1\right)}{2}}$

$\bold{Multiply \ Fractions \ (a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c})}$

$\bold{\frac{x\left(6x^2+4x-1\right)x^2}{2}}$

$\bold{Rewrite}$

$\bold{x\left(6x^2+4x-1\right)x^2 \ = \ x^3\left(6x^2+4x-1\right)}$

$\bold{Simplify}$

$\bold{\frac{x^3\left(6x^2+4x-1\right)}{2}}$

$\boxed{\bold{[] \ Eclipsed \ []}}$

3 0

3 years ago

Read 2 more answers

It says 7/6x= 140 I am lost can somebody please help me

kobusy [5.1K]

7/6x=140

since variable x is multiplied by 7/6, find x by multiplying by the inverse of 7/6 which is 6/7.

7/6x × 6/7 = 140 × 6/7
x= 120

8 0

3 years ago