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Rama09 [41]
3 years ago
11

Mrs smith has 5 times as many markers as colored pencils. The total is 54. How many markers does she have?

Mathematics
1 answer:
zhenek [66]3 years ago
3 0
5 part marker and 1 part pencil 5+1=6  54/6=9  9 pencils 9 times 5       45 markers
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3. What are the roots of the polynomial y = x³ - 8?
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Step-by-step explanation:

x^3 is a perfect cube, 8 is a perfect cube, so we use difference of cubes.

{a}^{3}  -  {b}^{3}  = (a - b)( {a}^{2}  + ab +  {b}^{2} )

Cube root of x^3 is x.

Cube root of 8 is 2

So

a=x

b= 2.

(x - 2)( {x}^{2}  - 2x + 4)

Set these equations equal to zero

x - 2 = 0

x = 2

{x}^{2}  - 2x + 4 = 0

If we do the discriminant, we get a negative answer so we would have two imaginary solutions,

Thus the only real root is 2.

If you want imaginary solutions, apply the quadratic formula.

1 + i  \sqrt{ 3 }

and

1 - i \sqrt{3}

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3 years ago
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5 0
2 years ago
Read 2 more answers
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
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