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natali 33 [55]
2 years ago
9

Calculate the GPA of a student with the following grades: B (77 hours), D (66 hours), F (2020 hours). Note that an A is equivale

nt to 4.04.0, a B is equivalent to a 3.03.0, a C is equivalent to a 2.02.0, a D is equivalent to a 1.01.0, and an F is equivalent to a 00. Round your answer to two decimal places.
Mathematics
1 answer:
d1i1m1o1n [39]2 years ago
8 0

Answer:

The student's GPA is of 0.82.

Step-by-step explanation:

GPA:

To find the student's GPA, we find his weighed mean.

Grades:

7 hours worth 3(B)

6 hours worth 1(D)

20 hours worth 0(F). So

M = \frac{7*3 + 6*1 + 20*0}{7+6+20} = 0.82

The student's GPA is of 0.82.

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Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth
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Answer:

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

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We first solve the homogeneous differential equation y′′+4y=0.

y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\

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Now, we finding a particular solution.

y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\

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y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

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