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SSSSS [86.1K]
3 years ago
5

Factor the polynomial by it’s greatest common monomial factor. 9b^8 + 24b^3

Mathematics
2 answers:
docker41 [41]3 years ago
8 0

Answer:

9b^8 + 24b^3 = (3b^3) [ 3b^5  +  (8)]

greatest common monomial is  3*b^{3}

Step-by-step explanation:

we see  9b^8 + 24b^3

there is a common factor here.

3b^3  so

9b^8 + 24b^3 = (3b^3) * 3b^5  +  (3b^3) *(8)

9b^8 + 24b^3 = (3b^3) [ 3b^5  +  (8)]

so...

monitta3 years ago
8 0

Answer:

3b^3(3b^5+8)

Step-by-step explanation:

that is the khan academy answer

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Which one is bigger 2/3 or 4/7
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Answer:

\frac{2}{3}

Step-by-step explanation:

To compare the fractions we require them to have a common denominator.

The lowest common multiple of 3 and 7 is 21

Change both fractions into fractions with a denominator of 21

\frac{2(7)}{3(7)} and \frac{4(3)}{7(3)}

= \frac{14}{21} and \frac{12}{21}

The numerator 14 is greater than 12, hence

\frac{2}{3} is the larger fraction

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(a) Find an equation for the family of linear functions with slope 2 and sketch several members of the family.
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Answer:

\textbf{(\text{a})} f(x) = 2x+n

\textbf{(\text{b})}\\ f(x) = mx + 1 -2m = m(x-2) + 1

\textbf{(\text{c})}f(x) = 2x - 3

Step-by-step explanation:

First, we know that family of functions represents a set of functions whose equations have a similar form. In our case, a family of linear functions can be represented as

                                   \{ ax + b | a, b \in \mathbb{R} \}.

Now, we can take an arbitrary member of that family, a function

                                     f(x) = mx + n

for some real constants m and n.

\textbf{(\text{a})}  

In this part of the problem, we know that m = 2\\, so we consider

                                    f(x) = 2x +n.

To graph several members of the family, you can plug in any real number in the equation above instead of n, since \forall n \in \mathbb{R} satisfy the equation.

For n = 0, we have f(x) = 2x.

For n = 15, we have f(x) = 2x + 15.

For n = -15, we have f(x) = 2x - 15.

The graphs for the values n = 0, n = 15 and n = -15 are presented on the first graph below.

\textbf{(\text{b})}

We need to find the member of the family of linear functions such that

                                   f(2) = 1.

Substituting 2 for x in f(x) = mx + n gives

                                 f(2) = 2m+n.

Now, since we have that f(2) = 1, we can equate 1 with 2m +n and express one of them in terms of the other.  

                        2m + n = 1 \implies n = 1 - 2m

Substituting 1-2m for n in f(x) = mx + n gives the equation

                    f(x) = mx + 1 -2m = m(x-2) + 1

which represents the wanted family. To sketch several member, we can choose any real value for m, since \forall m \in \mathbb{R} satisfy the equation.  

For m = 0, we have f(x) = 1.

For m = 5, we have f(x) = 5x - 9.

The graph is presented below.

\textbf{(\text{c})}

A function belongs to both families if it satisfies both conditions; Its slope must be equal to 2 and f(2) = 1.

Let's consider a function

                                f(x) = mx + n

for some real constants m and n.

The objective is to find the numeric value of the constants m and n. Since the slope must be equal to 2, we obtain that m = 2 and

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To find the numeric value of n, we use the fact that f(2) = 1.

Substituting 2 for x gives

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        4 + n = f(2) = 1 \implies 4 + n = 1 \implies n = 1 - 4 \implies n = -3

Therefore, a function that belongs to both families is

                                      f(x) = 2x - 3 .

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