All categories

3 years ago

What is the GCF of 28ab and 42b? 6b 14b 4b 7b NEED ASAP

Mathematics

1 answer:

grigory [225]3 years ago

3 0

Answer:

14b

Step-by-step explanation:

Here, we want to get the greatest common factor of the two terms

The greatest common factor can be obtained by finding the expression that factor the given expression

We have this as;

14b(2a + 3)

So the greatest common factor is 14b

You might be interested in

if the sum of two numbers is four and the sum of their squares minus three times their product is 76, find the numbers

Sauron [17]

Answer:

Numbers =(6,-2 )

Step-by step explanation:

Given

the sum of two numbers is four and the sum of their squares minus three times their product is 76,

Let the first term =X

Second term will be=4-x

X^2+(4-x)^2-3x(4-x)=76

x^2+16+x^2-8x-12x+3x^2=76

5x^2-20x+16=76

5x^2-20x-60=0

5(x^2-4x-12)=0

So x^2-4x-12=0

x^2-(6-2)x-12=0

x^2-6x+2x-12=0

X(x-6)+2(x-6)=0

(X-6)(X+2)=0

X=6,-2

Let X=6 and -2 then numbers will be 6 and (4-6)=-2

6 0

3 years ago

Express 16 1/4in simplest radical form

elena-14-01-66 [18.8K]

Answer:

Step-by-step explanation:

${16}^{ \frac{1}{4} } \: can \: also \: be \: written \: as \: ( {{2}^{4})}^{ \frac{1}{4} }$

because ${2}^{4} = 16$

So,

${( {2}^{4} )}^{ \frac{1}{4} } = {(2)}^{4 \times \frac{1}{4} } = 2$

7 0

3 years ago

Determine whether there is a difference between the two data sets that is not apparent from a comparison of the measures of cent

Iteru [2.4K]

Answer:

It seems the question is incomplete as the two data sets implied from the question are not visible.

However, two data sets having identical measures of center have a difference.

Step-by-step explanation:

Let's consider these two data sets:

2, 2, 3, 5, 6, 6 and 1, 1, 1, 1, 1, 19

Both were contrived.

The most reliable among the measures of center is the mean. The mean of each is calculated by summing the data and dividing by the number of elements. In both sets, the mean is 4.

The data in the first set seem to be concentrated around the mean indeed. But for the second data set, one of them, 19, is obviously very far from the others and the mean. We need another measure to describe this: standard deviation.

Its formula is:

$\rho=\sqrt{\dfrac{\sum (x-\overline{x})^2}{n}}$

$x$ represents each data, $\overline{x}$ is the mean and $n$ is the number of items. $x-\overline{x}$ is the deviation from the mean of each data item.