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3 years ago

Help please !!!! my brain hurts

Mathematics

1 answer:

fiasKO [112]3 years ago

8 0

Answer:

the answer id d.

Step-by-step explanation:

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Using the common denominator of 56, what is an equivalent fraction for 3/8 and 3/7?

anyanavicka [17]

Answer: $\frac{3}{8}$ = $\frac{21}{56}$ $\frac{3}{7}$ = $\frac{24}{56}$

Step-by-step explanation:

In order to get from 8 to 56, we have to multiply by 7.

We always do the same to the top! 3 multiplied by 7 = 21.

In order to get from 7 to 56, we have to multiply by 8.

We always do the same to the top! 3 multiplied by 8 = 24.

5 0

3 years ago

TRUE OR FALSE: F(x)=-3(x+2)(x-5)^3>0 when x<-2 or x>5<br> PLEASE HELP!!!!

Hitman42 [59]

F(x) = -3(x + 2)(x - 5)^3 > 0
(x + 2)(x - 5) > 0
x + 2 > 0 or x - 5 < 0
x > -2 or x < 5
-2 < x < 5

Therefore, the stated interval is false.

6 0

3 years ago

Two plumbers charge different rates for service. The first plumber charges $100 per service call plus $25 per hour of service. T

Bond [772]

Answer: 2

Step-by-step explanation:

6 0

3 years ago

Sam bought an electric iron as a price of 25 less than the original price if the original price of it is $2500 at what price did

wlad13 [49]

Answer:

$2475 or $1875

Step-by-step explanation:

Assuming the question meant to say $25 less than the original price, it would be $2475 because $2500-$25= $2475

However, if the question meant to say that Sam bought the iron for 25% less than the original price, it would be $1875 because 2500*0.25= $625

$2500-$625= $1875

4 0

3 years ago

Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).

Firdavs [7]

Answer:

A) Particular solution:

$2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

B) Homogeneous solution:

$y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))$

C) The most general solution is

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

Step-by-step explanation:

Given non homogeneous ODE is

$y''+4y'+5y=10x+e^{-x}---(1)$

To find homogeneous solution:

$D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)$

To find particular solution:

$y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\$

Substituting $y_{p},y'_{p},y''_{p}$ in (1)

$y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\$

Equating the coefficients

$5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\$

The general solution is

$y=y_{h}+y_{p}$

from (2) ad (3)

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

6 0

3 years ago

Help please !!!! my brain hurts ​

Help please !!!! my brain hurts