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irakobra [83]
3 years ago
13

Please help me ASAP! and thank you

Mathematics
1 answer:
FromTheMoon [43]3 years ago
8 0

Answer:

214°

Step-by-step explanation:

Full circle measure = 360°

Measure of arc AB = 146° (central angle has the same measure as measure of intercepted arc)

Therefore:

Measure of arc ACB = 360° - arc AB

Measure of arc ACB = 360° - 146°

= 214°

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In order to get the answer of that, let's understand first that the question can first be presented as a fraction form. So by breaking down the question, we will be able to get the fraction form which is 300/2530. Now in order to get the percent form of that, we simply divide 300 and 2530. The answer to that will result in 0.11858. Now we know for a fact that in order to make a decimal a percentage, we just have to move the decimal point two times to the right. So the answer is 11.858%
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Lateral area of cylinder
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lateral area = 2 X pi X r X H

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What is one benefit of privately issued student loans?
Leona [35]

Answer:

they have lower interest rates and can be paid back with a lower out of pocket cost

Step-by-step explanation:

Student loans are issued as a kind of financial aid that assist students in their quest to acquire higher education. Private student loans are offered by the private-sector lenders. The alternative to this is a Federal loan.

Actually, private student loans are issued at a lower interest rate. Option of a fixed or variable interest rate may be offered on privately issued student loans. This offers a lower out of pocket cost, hence the answer.

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3 years ago
In problem solve the given differential equation by underdetermined coefficients y''-2y+y=xe^x
Alexus [3.1K]

Answer:

Solution is y(t)=C_1e^x+C_2xe^x+\frac{x^3e^x}{6}

Step-by-step explanation:

Given Differential Equation,

y"-2y'+y=xe^x ...............(1)

We need to solve the given differential equations using undetermined coefficients.

Let the solution of the given differential equation is made up of two parts. one complimentary solution and one is particular solution.

\implies\:y(x)=y_c(x)+y_p(x)

For Complimentary solution,

Auxiliary equation is as follows

m² - 2m + 1 = 0

( m - 1 )² = 0

m = 1 , 1

So,

y_c(x)=C_1e^x+c_2xe^x

Now for particular solution,

let y_p(x)=Ax^3e^x

y'=Ax^3e^x+3Ax^2e^x

y"=Ax^3e^x+6Ax^2e^x+6Axe^x

Now putting these values in (1), we get

Ax^3e^x+6Ae^2e^x+6Axe^x-2(Ax^3e^x+3Ax^2e^x)+Ax^3e^x=xe^x

Ax^3e^x+6Ae^2e^x+6Axe^x-2Ax^3e^x-6Ax^2e^x+Ax^3e^x=xe^x

6Axe^x=xe^x

6A=1

A=\frac{1}{6}

\implies\:y_p(x)=\frac{x^3e^x}{6}

Therefore, Solution is y(t)=C_1e^x+C_2xe^x+\frac{x^3e^x}{6}

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4 years ago
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