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erma4kov [3.2K]
3 years ago
12

3x - 8 < -2x +22 solve the inequality

Mathematics
2 answers:
Salsk061 [2.6K]3 years ago
5 0
<span>x < 8/5
22
Hope this helps you</span>
kotegsom [21]3 years ago
4 0
X < 6
Hope it helped!
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5. Amy deposited $7,000 in a bank account earning 3.5% interest, compounded
Semenov [28]

Answer:

Step-by-step explanation:

Given the following data;

Principal = $7,000

Interest = 3.5% = 3.5/100 = 0.035

To find the future value, we would use the compound interest formula;

A = P(1 + \frac{r}{n})^{nt}

Where;

A is the future value.

P is the principal or starting amount.

r is annual interest rate.

n is the number of times the interest is compounded in a year.

t is the number of years for the compound interest.

Substituting into the equation, we have;

A = 7000(1 + \frac{0.035}{n})^{nt}

7 0
3 years ago
Use the drop-down menus to compare −2.25 and −3.
Kisachek [45]
-2.5>-3 because the bigger a negative number is like -15,-18, etc the smaller it’s value actually gets so -2.5 is greater because it is closer to 0 on a number line.
5 0
2 years ago
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Masteriza [31]
I think the answer to your question might be 6
4 0
2 years ago
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The school band is selling raffle tickets for $2. If Kyle spent p dollars on tickets this week and q dollars on tickets the week
Alex

Answer:

Tickets = p/2 + q/2

Step-by-step explanation:

I don't understand the options, as written.

The answer should be Tickets = p/2 + q/2

If q = 10 and p = 20, Kyle would have bought 5 tickets the week before, and 10 tickets this week.

5 0
2 years ago
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What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)
Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\

Now to solve the integral on the right hand side we shall use Integration by parts Taking 7t^{3} as first function thus we have

\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\

Again repeating the same procedure we get

=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\

Again repeating the same procedure we get

\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\

Now solving this integral we have

\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}

Thus we have

L[7t^{3}]=\frac{7\times 3\times 2}{s^4}

where s is any complex parameter

5 0
3 years ago
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