1) A Rectangle has a length of 4x+2 and a width of 7x+6. The

Which algebraic expressions are polynomials? Check all that apply. PLEASE HELP

dexar [7]

Selections 2, 3, 5, 6 are polynomials.

1 and 4 are not. The coefficients don't have to be integers, but the powers of the variables need to be positive integers. In 1, you have x^-1. in 4, you have x^(1/2).

3 0

3 years ago

Solve for x. Help, please. ༼ つ ◕_◕ ༽つ

ahrayia [7]

Answer:

x = -10

Step-by-step explanation:

first you multiply both sides by -3

x + 7 = -3

then you subtract 7 from both sides

x = -10

hope this helps

5 0

3 years ago

Help, please!!! question in picture

lutik1710 [3]

Answer:

20 m^2

Step-by-step explanation:

First multiply the parallelogram part: 3 x 5 = 15

Then multiply the triangle part: (2 x 5)/2 =5

Then add: 15 + 5 = 20

Good Luck!

4 0

3 years ago

The water was pumped out of a backyard pond. What is the domainof this graph?

nalin [4]

I think it's D because you can't have negative minutes and after 8 minutes you can't go any deeper than 0 inches.

8 0

3 years ago

Can someone tell me how

gulaghasi [49]

the way I get the subsequent term, nevermind the exponents, the exponents part is easy, since one is decreasing and another is increasing, but the coefficient, to get it, what I usually do is.

multiply the current coefficient by the exponent of the first-term, and divide that by the exponent of the second-term + 1.

so if my current expanded term is say 7a³b⁴, to get the next coefficient, what I do is (7*3)/5 <----- notice, current coefficient times 3 divided by 4+1.

anyhow, with that out of the way, lemme proceed in this one.

$\bf ~~~~~~~~\textit{binomial theorem expansion} \\\\ \qquad \qquad (1+ax)^n\implies \begin{array}{llll} term&coefficient&value\\ \cline{1-3}&\\ 1&+1&(1)^n(ax)^0\\\\ 2&+\frac{(1)(n)}{1}\to n&(1)^{n-1}(ax)^1\\\\ 3&+\frac{n\cdot (n-1)}{2}&(1)^{n-2}(ax)^2 \end{array}$

so, following that to get the next coefficient, we get those equivalents as you see there for the 2nd and 3rd terms.

so then, we know that the expanded 2nd term is 24x therefore

$\bf n(1)^{n-1}(ax)1 = 24x\implies n(1)(ax)=24x\implies nax=24x\implies n=\cfrac{24}{a}$

we also know that the expanded 3rd term is 240x², therefore we can say that

$\bf \cfrac{n(n-1)}{2}~~(1)^{n-2}(ax)^2 = 240x^2\implies \cfrac{n(n-1)}{2}(1)(a^2x^2) = 240x^2 \\\\\\ \cfrac{(n^2-n)(a^2x^2)}{2}=240x^2\implies \cfrac{(n^2-n)(a^2)}{2}=\cfrac{240x^2}{x^2}\implies \cfrac{a^2n^2-a^2n}{2}=240 \\\\\\ a^2n^2-a^2n=480$

but but but, we know what "n" equals to, recall above, so let's do some quick substitution

$\bf a^2n^2-a^2n=480\qquad \boxed{n=\cfrac{24}{a}}\qquad a^2\left( \cfrac{24}{a} \right)^2-a^2\left( \cfrac{24}{a} \right)=480 \\\\\\ a^2\cdot \cfrac{24^2}{a^2}-24a=480\implies 24^2-24a=480\implies 576-24a=480 \\\\\\ -24a=-96\implies a=\cfrac{-96}{-24}\implies \blacktriangleright a = 4\blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ n=\cfrac{24}{a}\implies n=\cfrac{24}{4}\implies \blacktriangleright n=6 \blacktriangleleft$

7 0

4 years ago