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4 years ago

How to find the positive divisors of 372?

1 answer:

8 0

The factors are 1*2*2*3*31
So you can see that all of the divisors listed in your answer are are different products of these factors.

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Anyone good at Ks3 maths

MaRussiya [10]

Answer:

${ \tt{distance = speed \times time}} \\ { \tt{ = 56 \times 3 \frac{1}{2} }} \\ = { \tt{196 \: miles}}$

5 0

3 years ago

Read 2 more answers

Sam has 3 1/3 books left to read for school. Sam read 2 7/8 books on Tuesday. Sam reads 2 7/12 times faster on weekday. How many

oksano4ka [1.4K]

3 1/3 it looks like a trick question cause he stater that same has that many books left to read

5 0

3 years ago

If a positive integer that is not a multiple of 5 is divided by 5, what is the least possible remainder

solniwko [45]

I would say 1 would be the least possible remainder...

11/5 = 2 remainder 1
21/5 = 4 remainder 1
31/5 = 3 remainder 1

and since ur dealing with positive integers (whole numbers), 1 is gonna be the smallest

8 0

4 years ago

Can some one help me out with this answer 1/2 x 3/4

ad-work [718]

Here is your answer

$\huge\frac{3}{8}$

EXPLANATION

$\frac{1}{2} \times \frac{3}{4}$

Here, we multiply numerator of one fraction to the numerator of the other fraction and similarly the denominators of both fractions are multiplied.

$\frac{1 \times 3}{2 \times 4}$

$\frac{3}{8}$

HOPE IT IS USEFUL

6 0

3 years ago

Write the equation of a polynomial of degree 3, with zeros 1, 2 and -1 where f(0)=2

drek231 [11]

Answer:

The equation of a polynomial of degree 3, with zeros 1, 2 and -1 is $x^{3}-2 x^{2}-x+2=0$

Solution:

Given, the polynomial has degree 3 and roots as 1, 2, and -1. And f(0) = 2.

We have to find the equation of the above polynomial.

We know that, general equation of 3rd degree polynomial is

$F(x)=x^{3}-(a+b+c) x^{2}+(a b+b c+a c) x-a b c=0$

where a, b, c are roots of the polynomial.

Here in our problem, a = 1, b = 2, c = -1.

Substitute the above values in f(x)

$F(x)=x^{3}-(1+2+(-1)) x^{2}+(1 \times 2+2(-1)+1(-1)) x-1 \times 2 \times(-1)=0$

$\begin{array}{l}{\rightarrow x^{3}-(3-1) x^{2}+(2-2-1) x-(-2)=0} \\ {\rightarrow x^{3}-(2) x^{2}+(-1) x-(-2)=0} \\ {\rightarrow x^{3}-2 x^{2}-x+2=0}\end{array}$

So, the equation is $x^{3}-2 x^{2}-x+2=0$

Let us put x = 0 in f(x) to check whether our answer is correct or not.

$\mathrm{F}(0) \rightarrow 0^{3}-2(0)^{2}-0+2=2$

Hence, the equation of the polynomial is $x^{3}-2 x^{2}-x+2=0$

3 0

3 years ago