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yuradex [85]
2 years ago
14

Round each amount to the nearest hundred to find a reasonable estimate of 6207 divided by 214.

Mathematics
2 answers:
Bas_tet [7]2 years ago
6 0
Someone answered it
sergey [27]2 years ago
3 0
D is the answer I think
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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
How are sequences whose consecutive terms differ by a constant modeled, and what can those models tell us?
alex41 [277]

Answer:

Linear functions

Step-by-step explanation:

3 0
2 years ago
Find the minimum and maximum possible areas of a rectangle measuring 6 km by 11 km.
ExtremeBDS [4]
Assuming your numbers are rounded to the nearest km, the minimum area will be ...
  (5.5 km)·(10.5 km) = 57.75 km² . . . minimum

And the maximum area will be ...
  (6.5 km)·(11.5 km) = 74.75 km² . . . maximum

_____
When a number is rounded to 6 km as the nearest km, its value may actually be anywhere in the range 6 km ± 0.5 km. If you really want to get technical about it, the ranges of possible dimensions are [5.5, 6.5) km and [10.5, 11.5) km, and the range of possible areas is [57.75, 74.75) km².
3 0
3 years ago
What is 569993885.5908653<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20587" id="TexFormula1" title="x^{2} 587" alt="x^{2} 587"
Yanka [14]
99 forgive me if i’m wrong
8 0
2 years ago
Read 2 more answers
Find the three inequalities which define the shaded region on the prid.<br> Answer<br> [5]
nevsk [136]

Answer:

Step-by-step explanation:

<em>1) y ≥ x/2</em>

<em>2) y ≤ -x</em>

<em>3) x ≥ 0</em>

5 0
2 years ago
Read 2 more answers
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