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My name is Ann [436]
3 years ago
10

Identify each sequence below as geometric, arithmetic, or neither.

Mathematics
1 answer:
vitfil [10]3 years ago
8 0
1) arithmetic
2) neither
3) geometric
4) geometric
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Divide each polynomials by the monomials. Choose all that are correct. (9xy2 + 12x3y4 − 6x) ÷ 3x = 4x2y4 + 3y2 − 2 (25x4y2 + 10x
UkoKoshka [18]

Answer:

(21x^4y + 7x^3y^2 − 28x^2y^2) ÷ 7xy = 3x^3 + x^2y − 4xy

Step-by-step explanation:

(9xy^2 + 12x^3y^4 − 6x) ÷ 3x = 4x^2y^4 + 3y^2 − 2 (False: 9xy^2:3x=3y^2)

25x^4y^2 + 10x^2y^4 − 15y) ÷ 5y = 5x^4y + 2x^3y^2 − 3 (False: 10x^2y^4:5y=2x^2y^3)

(16x^4y^2 + 24x^2y^2 − 8xy^2) ÷ 4xy = 4x^4y + 6xy− 2y(False: 16x^4y^2:4xy=4x^3y)

(21x^4y + 7x^3y^2 − 28x^2y^2) ÷ 7xy = 3x^3 + x^2y − 4xy (True)

7 0
3 years ago
Find the commission.<br><br>Sales: $530 Commission Rate: 8% Commissions: ?
kipiarov [429]

Answer:

42.40

Step-by-step explanation:

8 0
3 years ago
Can 3.65909090909 be expressed as a fraction whose denominator is a power of 10? Explain.
GuDViN [60]
\bf 3.659\textit{ can also be written as }\cfrac{3659}{1000}\textit{ therefore }3.6590909\overline{09}\\\\&#10;\textit{can be written as }\cfrac{3659.0909\overline{09}}{1000}

notice above, all we did, was isolate the "recurring part" to the right of the decimal point, so the repeating 09, ended up on the right of it.

now, let's say, "x" is a variable whose value is the recurring part, therefore then

\bf \cfrac{3659.0909\overline{09}}{1000}\qquad \boxed{x=0.0909\overline{09}} \qquad \cfrac{3659+0.0909\overline{09}}{1000}\implies \cfrac{3659+x}{1000}

now, the idea behind the recurring part is that, we then, once we have it all to the right of the dot, we multiply it by some power of 10, so that it moves it "once" to the left of it, well, the recurring part is 09, is two digits, so let's multiply it by 100 then, 

\bf \begin{array}{llllllll}&#10;100x&=&09.0909\overline{09}\\&#10;&&9+0.0909\overline{09}\\&#10;&&9+x&#10;\end{array}\quad \implies 100x=9+x\implies 99x=9&#10;\\\\\\&#10;x=\cfrac{9}{99}\implies \boxed{x=\cfrac{1}{11}}\\\\&#10;-------------------------------\\\\&#10;\cfrac{3659.0909\overline{09}}{1000}\qquad \boxed{x=0.0909\overline{09}} \quad \cfrac{3659+0.0909\overline{09}}{1000}\implies \cfrac{3659+x}{1000}&#10;\\\\\\&#10;\cfrac{3659+\frac{1}{11}}{1000}

and you can check that in your calculator.
8 0
3 years ago
First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x
e-lub [12.9K]

Answer:

(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

\int x ln(5+x)dx

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

\int (U-5) ln(U)dU

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

\int (pq')=pq-\int qp'

so we must define p, q, p' and q':

p=ln U

p'=\frac{1}{U}dU

q=\frac{U^{2}}{2}-5U

q'=U-5

and now we plug these into the formula:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU

Which simplifies to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU

Which solves to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C

so we can substitute U back, so we get:

\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C

and now we can simplify:

\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

notice how all the constants were combined into one big constant C.

7 0
3 years ago
12xy-44y+5 determine if it is prime or factorable
Elza [17]

PrimeAnswer:

Step-by-step explanation:

8 0
3 years ago
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