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3 years ago

A system of linear equations is shown below.

Mathematics

1 answer:

Drupady [299]3 years ago

5 0

Answer: D.5 and 2

because (-3;5) is a solution to the system of equations

$\left \{ {{-3p+5q=-5} \atop {-3p-5q=-25}} \right.\\\\\left \{ {{-3p=-15} \atop {5q=10}} \right.\\\\\left \{ {{p=5} \atop {q=2}} \right.$

Step-by-step explanation:

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Determine the decimal equivalent of 19/50

Dovator [93]

Answer:

0.38

Step-by-step explanation:

Simply divide 19 by 50 on a calculator.

5 0

3 years ago

Read 2 more answers

A vertical right circular cylindrical tank has height h=8 feet high and diameter d=6 feet. It is full of kerosene weighing 50 po

Mariana [72]

Answer:

The work done to pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$

Step-by-step explanation:

The work is defined by:

$W=\int dFdx$ (1)

The force here will be the product between the volume and the kerosene weighing, so we have :

$dF=\pi R^{2}dy*50$

This force will be in-lbs.

Where R is the radius (3 feet)

Then using (1), we have:

$W=\int \pi R^{2}dy*50(8-y)$

Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.

$W=\int_{0}^{8} \pi 3^{2}dy*50(8-y)$

$W=450\pi \int_{0}^{8}dy(8-y)$

$W=450\pi(\int_{0}^{8} 8dy-\int_{0}^{8} ydy)$

$W=450\pi(8y|_{0}^{8} -\frac{y^{2}}{2}|_{0}^{8})$

$W=450\pi(8*8 -\frac{8^{2}}{2})$

$W=450\pi(64 -\frac{64}{2})$

Therefore, the work done to pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$

I hope it helps you!

6 0

3 years ago

Find the indefinite integral. (Use C for the constant of integration.) <br> e2x 25 e4x dx.

Sladkaya [172]

Answer:

The solution is $\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] + C$

Step-by-step explanation:

From the question

The function given is $f(x) = \frac{e^{2x}}{ 25 + e^{4x}} dx$

The indefinite integral is mathematically represented as

$\int\limits {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx$

Now let $e^{2x} = u$

=> $\frac{du}{dx} 2e^{2x}$

=> $2 e^{2x}dx = du$

$\int\limits {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx = \int\limits {\frac{1}{ 2(25 + u^2)} } \, du$

$= \frac{1}{2} \int\limits {\frac{1}{ 25 + u^2)} } \, du$

$= \frac{1}{2} \int\limits {\frac{1}{ 5^2 + u^2)} } \, du$

$= \frac{1}{2} \frac{tan^{-1} [\frac{u}{5} ]}{5} + C$

Now substituting for u

$\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] + C$

3 0

3 years ago

Factor the expression by finding the GCF. 16m2 − 12m

ziro4ka [17]

Answer:

4m(4m-3)

Step-by-step explanation:

16m^2 − 12m

16 m^2 = 2*2*2*2*m*m

12m = 2*2*3*m

The greatest factor for 16m^2 and 12m is 2*2*m or 4m

Factor out 4m

16 m^2 = 2*2*2*2*m*m = 4m(2*2m)=4m(4m)

12m = 2*2*3*m = 4m(3)

Factoring out 4m

16m^2 − 12m

4m(4m-3)

3 0

3 years ago

Read 2 more answers

What is 45 X45?////////////////

Vlad1618 [11]

Answer:

2025!!

Step-by-step explanation:

3 0

3 years ago