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gizmo_the_mogwai [7]
3 years ago
15

Solve by Elimination x-8y=18 -16x +16y=-8

Mathematics
1 answer:
nadya68 [22]3 years ago
8 0
<h3><u>The value of y is -2.5</u></h3><h3><u>The value of x is -2.</u></h3>

x - 8y = 18

-16x + 16y = -8

We can multiply each term in the first equation by 2 to better assist this elimination.

2x - 16y = 36

Now, we can use elimination.

Add equations together.

-14x = 28

Divide both sides by -14

x = -2

Now that we have a value for x, we can solve for y.

-2 - 8y = 18

Add 2 to both sides.

-8y = 20

Divide both sides by -8

y = -2.5


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This is the answer. 20% of 890 =178
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Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua
goblinko [34]
<h2>Hello!</h2>

The answer is:  There is a total of 5.797 gallons pumped during the given period.

<h2>Why?</h2>

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

D(t)=\frac{5t}{1+3t}

So, the integral will be:

\int\limits^4_0 {\frac{5t}{1+3t}} \ dx

So, integrating we have:

\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx

Performing a change of variable, we have:

1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}

Then, substituting, we have:

\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du

\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )

\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du

\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]

Reverting the change of variable, we have:

\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]

Then, evaluating we have:

\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0
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Martha os taking the bus into work 3 times a week (12 times a month) for $4.50 a day. Would it be better to continue to pay dail
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Answer:

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$4.50 x 12 = 54

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A.)

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NEED HELP FAST!!!
mina [271]
Remember

we can do anything to an equation as long as we do it to both sides

try to isolate the variable

you have 2 types
x+b=c
x/b=c

fior the first type, minus b from both sides to get
x=c-b

for the second, multiply both sides by b to get rid of the fraction to get
x=cb

also remember that -x times -1=x



b.add 25 to both sides
-a=20
multiply -1
a=-20


c.
-t/8=-4
multiply both sides by 8
-t=-32
mutiply -1
t=32

d. -n/-5=-30
mulitply both sides by -5
-n=150
multiply both sides by -1
n=-150

e. multiply both sides by 12
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A number is multiplied by 3,and 7 is added to it to get the result 25.Find the number.
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Answer:

6

Step-by-step explanation:

25-7=18 18/3= 6.

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THE ANSWER IS 6

4 0
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