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Arisa [49]
3 years ago
5

If you placed $1,000.00 in a savings account with an interest rate of 4.5% month versus an account with 3.75% month, how much in

terest could you earn in that month?
A: $1.25 more
B: $12.50 more
C: $7.50 more
D: $75.00 more
Mathematics
1 answer:
Setler [38]3 years ago
7 0

The answer would be C. Hope this helps!

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Logan is going to use a computer at an internet cafe. The cafe charges an initial fee to use the computer and then an additional
aalyn [17]

Answer:

$6

Step-by-step explanation:

Hourly rate is $1.5

6 0
2 years ago
The value of a car decreases $500 for every 1,000 miles it is driven. The current value of the car is $23,000, and the car is dr
Hoochie [10]

Answer: C

Step-by-step explanation:

If we know the value of the car decreases $500 for every 1,000 miles, and that the car is driven about 10,000 miles every year, that means that you need to take the total value of the car (23,000) and subtract it from the amount of money it is losing per year. Again, the car is driven about 10,000 miles per year, so that means that the car will most likely continue to be driven 10,000 miles per year. If you do the math, for one year, the value of the car will drop $5,000 ($500 x 10, because it is $500 per every 1,000 miles) So, for each year, you can just multiply the number of years by $5,000 to find out how much the vehicle has depreciated over time.

Hope this helped you and made sense! Feel free to ask me any questions you have!

4 0
3 years ago
Help!!!Can someone help me with my math ttas fast!!
MAVERICK [17]
The answer to the question is boneless
4 0
3 years ago
HELP I WILL GIVE BRAINLIEST!
Hatshy [7]

Answer:

C. 8.1 Inches

Step-by-step explanation:

Take 100 Ib. and multiply it by 0.9. You get 90 lb. Now referring to the fraction rule, do the same to the "denominator": 9 In. x 0.9, and you get 8.1 Inches.

5 0
3 years ago
Solve y" + y = tet, y(0) = 0, y'(0) = 0 using Laplace transforms.
irina1246 [14]

Answer:

The solution of the diferential equation is:

y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}

Step-by-step explanation:

Given y" + y = te^{t}; y(0) = 0 ; y'(0) = 0

We need to use the Laplace transform to solve it.

ℒ[y" + y]=ℒ[te^{t}]

ℒ[y"]+ℒ[y]=ℒ[te^{t}]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]+s·y(0)-y'(0)=s²·Y(s)

ℒ[y]=Y(s)

ℒ[te^{t}]=\frac{1}{(s-1)^{2}}

So, the transformation is equal to:

s²·Y(s)+Y(s)=\frac{1}{(s-1)^{2}}

(s²+1)·Y(s)=\frac{1}{(s-1)^{2}}

Y(s)=\frac{1}{(s^{2}+1)(s-1)^{2}}

To be able to separate in terms, we use the partial fraction method:

\frac{1}{(s^{2}+1)(s-1)^{2}}=\frac{As+B}{s^{2}+1} +\frac{C}{s-1}+\frac{D}{(s-1)^2}

1=(As+B)(s-1)² + C(s-1)(s²+1)+ D(s²+1)

The equation is reduced to:

1=s³(A+C)+s²(B-2A-C+D)+s(A-2B+C)+(B+D-C)

With the previous equation we can make an equation system of 4 variables.

The system is given by:

A+C=0

B-2A-C+D=0

A-2B+C=0

B+D-C=1

The solution of the system is:

A=1/2 ; B=0 ; C=-1/2 ; D=1/2

Therefore, Y(s) is equal to:

Y(s)=\frac{s}{2(s^{2} +1)} -\frac{1}{2(s-1)} +\frac{1}{2(s-1)^{2}}

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[\frac{s}{2(s^{2} +1)}]-ℒ⁻¹[\frac{1}{2(s-1)}]+ℒ⁻¹[\frac{1}{2(s-1)^{2}}]

y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}

8 0
3 years ago
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