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4 years ago

8

Two Questions please help and explain

2 answers:

ozzi4 years ago

4 0

Question #2
Erica will have to work for 14 hours in order to buy her sneakers.
How to solve:
You subtract $31 from $150 which equals to $119
You multiply 8.50 (which is the amount of money she gets paid every hour) by the answers they gave you.
8.50 multiplied by 14 is 199
Erica had to work 14 hours to make $199

Your answer is D) 14

Scilla [17]4 years ago

4 0

The first one is going to be C because when you like multiplying and them and then you're going to find your answer

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Please help please please help

Inessa05 [86]

2x + 2 = 3x - 52

-2x -2x

2 = 1x - 52

+52 +52

54 = 1x

---- ----

1 1

54 = x

Check:

2(54) + 2 = 3(54) - 52

108 + 2 = 162 - 52

110 = 110

3 0

3 years ago

Read 2 more answers

6 dogs are black, 5 dogs are white, and the rest are blue. If there are 15 dogs in total, how many are blue?

Bezzdna [24]

Answer:

4 blue dogs

Step-by-step explanation:

Since there are 6 black dogs and 5 white dogs, there is 11 black and white dogs. There is 15 dogs in total, so you subtract 11 from 15, and that is 4. So, 4 dogs are blue.

6 0

3 years ago

Zachary invested $130 in an account paying an interest rate of 3.6% compounded daily. Assuming no deposits or withdrawals are ma

TiliK225 [7]

Answer:

$193

Step-by-step explanation:

8 0

3 years ago

Which strategy can be used to solve this problem?

USPshnik [31]

The anser is b guess and test

4 0

3 years ago

Read 2 more answers

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago