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coldgirl [10]
2 years ago
13

What are a, b, c and d​

Mathematics
1 answer:
GuDViN [60]2 years ago
7 0

Answer:

Step-by-step explanation:

4 g.

c=42

a=b=d

a+b+42=180

a+a=180-42

2a=138

a=138/2=69

a=69

b=69

c=42

d=69

5a.

a=30

angle O=180-(a+30)=180-(30+30)=180-60=120°

d=80

angle O=180-(d+80))=180-(80+80)=180-160=20°

central angle O of middle triangle=180-(120+20)=180-140=40°

b=c

b+c+40=180

b+b=180-40

2b=140

b=140/2=70

b=70°

c=70°

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Step-by-step explanation:

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A rectangular tank measures 20cm x 30cm x 40cm. You want to fill it by using a cubic container that is 20cm on each side to carr
umka2103 [35]

Answer:

4 trips

Step-by-step explanation:

Given that:

Dimension of rectangular tank :

(20cm * 30cm * 40cm)

Dimension of cubic container used to fill tank = 20cm per side

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Number of trips it would take :

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What is the surface area?????
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5 0
3 years ago
“I think of a number and divide it by 5. The answer is 20.”
muminat

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Step-by-step explanation:

7 0
3 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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