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2 years ago

The sum of two numbers is 39. One number is 7 more than the other. Find the two numbers.

Mathematics

1 answer:

Charra [1.4K]2 years ago

5 0

Answer:

32 is the other number

think about it if you add

32+7 that is = to 39

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Change Y - 4X = 0 to the slope-intercept form of the equation of a line.

Bingel [31]

Answer:

y=4x

Step-by-step explanation:

Add 4x to both sides to get y=mx+b

0 is y-intercept.

4x is the slope.

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3 years ago

Draw a box and whiskers plot of the numbers 7,4,7,5,5,3,7,6,7

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Answer

3,4,5,5,6,7,7,7,7

Step-by-step explanation:

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2 years ago

A carton has a length of fraction 2 and 2 over 3 feet, width of fraction 1 and 1 over 3 feet, and height of fraction 1 and 1 ove

astraxan [27]

Writing this problem in symbols instead of in words greatly simplifies it:

(2 2/3) * (1 1/5) * (1 1/2)

Write each quantity (inside each set of parentheses) as an improper fraction:

(8/3) * (6/5) * (3/2)

Now multiply the numerators thru: 8*6*3 / 3*5*2

Notice that we can reduce this by dividing the 8 by the 2 and dividing the 6 by the 3: 4*3*3 36
---------- = -----------
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3 years ago

What is 12 1/2 x 4 2/3

egoroff_w [7]

Answer:

174 1/3

Step-by-step explanation:

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3 years ago

Read 2 more answers

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