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2 years ago

15

Use the net to find the surface area of the regular pyramid

1 answer:

Alja [10]2 years ago

8 0

Answer:

151.9 if the side length of the pentagon/ base of each triangle is 6 m

Step-by-step explanation:

if the side length of the pentagon/ base of each triangle is 6 m then it is a pretty simple question, we just need to add the surface area of the base pentagon and each triangle.

We have the area of the base, so we just need the triangles. The area of a triangle is .5bh, where the base of a triangle here is one side of the pentagon and the height is that indicated red 6. so that means one triangle has an area of .5*6*6 or 18. There are 5 triangles total so that means that with all the triangles there is an area of 90. Adding that to 61.9 gets us 151.9. let me know if you need any more help.

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Step-by-step explanation:

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Let a equal measure of a angle a. The equation 360ﾟ= a + 90ﾟ+ 135ﾟ+ 75ﾟ represents the sum of the angles in the quadrilateral. F

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A = 60°

Step-by-step explanation:

First to find the missing angle you add the three known angles,

90 + 135 + 75 = 300

Then, you subtract, 360 - 300 = 60

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2 years ago

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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.

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3 years ago