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2 years ago

What is the missing term? -4.-12.x 12.-4 . 2x ? 2.2 3х +1

Mathematics

1 answer:

Murrr4er [49]2 years ago

5 0

Answer:

Step-by-step explanation:

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4x·3x = 12x²

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Four times a number was 22 greater than twice the same number. What was the number?

Setler [38]

Answer: 11

Step-by-step explanation:

Set up an equation

Based on the question, we can determine that 4x = 22 + 2x where x in is the number

"Four times a number" -> 4x

"was" -> =

"22 greater than" -> 22 +

"twice the same number" -> 2x

Put the underlined parts together to get 4x = 22 + 2x

Solve equation

Rearrange/collect like terms to solve for x by subtracting 2x from both sides

4x = 22 + 2x

2x = 22

Solve for x by dividing each side by 2

2x = 22

x = 11

Final Answer

x = 11

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3 years ago

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Answer:

5x+8y

Step-by-step explanation:

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3 years ago

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AVprozaik [17]

It is c because i believe so

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1 year ago

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Professor Blanchard is packing up the books in he office to move to a new location. A small box can hold 44 books and a large bo

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Answer:

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Step-by-step explanation:

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2 years ago

A large tank is filled to capacity with 600 gallons of pure water. brine containing 4 pounds of salt per gallon is pumped into t

nignag [31]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which the amount of salt in the tank changes is given by

$\dfrac{\mathrm dA(t)}{\mathrm dt}=\dfrac{4\text{ lbs}}{1\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}-\dfrac{A(t)\text{ lbs}}{600\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}$
$\dfrac{\mathrm dA}{\mathrm dt}=24\dfrac{\text{lb}}{\text{min}}-\dfrac{A(t)}{100}\dfrac{\text{lb}}{\text{min}}$

Let's drop the units for now. We have

$\dfrac{\mathrm dA(t)}{\mathrm dt}+\dfrac{A(t)}{100}=24$
$e^{t/100}\dfrac{\mathrm dA(t)}{\mathrm dt}+e^{t/100}\dfrac{A(t)}{100}=24e^{t/100}$
$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/100}A(t)\right]=24e^{t/100}$
$e^{t/100}A(t)=\displaystyle24\int e^{t/100}\,\mathrm dt$
$e^{t/100}A(t)=2400e^{t/100}+C$
$A(t)=2400+Ce^{-t/100}$

We're given that the water is pure at the start, so $A(0)=0$ , giving

$A(0)=0=2400+Ce^{-0/100}\implies C=-2400$

So the amount of salt in the tank (in lbs) at time $t$ is

$A(t)=2400\left(1-e^{-t/100}\right)$

4 0

2 years ago