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Mathematics

2 answers:

NemiM [27]3 years ago

4 0

Answer:

A- 3,000 = 50(60) + b

B- 2,100 = 50(42) + b

Step-by-step explanation:

i just did it

zlopas [31]3 years ago

3 0

Answer:

The correct answers are A and E. You're welcome :)

Step-by-step explanation:

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The graph shows the cost of renting skates. Sadie rents a pair of skates at 10 A.M. How much will she pay if she returns the ska

stiv31 [10]

Answer: 1 PM = $8

1:15 PM = $9.0

Step-by-step explanation:

a. Since Sadie rents a pair of skates at 10 A.M and returned the skates by 1 P.M, this means the hours spent was 3 hours. Therefore, based on the graph, for 3 hours rented, he will pay $8

The amount that she will pay if she returns them at 1:15 P.M. will be equal to the value of 3 hours 15 minutes which will be:

= $8 + $1.0

= $9.00

5 0

2 years ago

A circular sector has a central angle of π6 radians and a radius of 9 ft.

ohaa [14]

Answer:

$\approx \: 21.2 \: {ft}^{2}$

Step-by-step explanation:

$central \: \angle = \frac{\pi ^{c} }{6} = 30 \degree \\ \\ area \: of \: the \: sector = \frac{30 \degree}{360 \degree} \times \pi {(9)}^{2} \\ \\ = \frac{1}{12} \times 3.14 \times 81 \\ \\ = \frac{1}{12} \times 254.34 \\ \\ = 21.195 \: {ft}^{2} \\ \\ \approx \: 21.2 \: {ft}^{2}$

8 0

2 years ago

Read 2 more answers

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$