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4 years ago

What is (8m^7 - 10m^5) / 2m^3 ?

2 answers:

Colt1911 [192]4 years ago

7 0

4m^4-5m^2

Basically you divide each term by the denominator
this is akin to splittting fraction. Just as you can add 2/5 and 4/5 to get (2+4)/5, like wise you can split apart fractions. By doing this, you get two fractions and you can evaluate them individually

kipiarov [429]4 years ago

4 0

$\frac{8m^{7} - 10m^{5}}{2m^{3}} = \frac{8m^{7}}{2m^{3}} - \frac{10m^{5}}{2m^{3}} = 4m^{4} - 5m^{2}$

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How can exponential and logarithmic functions be combined

natulia [17]

Well.. .not exactly what you're asking, but the relationship of them is
${
log_{{ a}}{{ b}}=y \iff {{ a}}^y={{ b}}\qquad\qquad 
% exponential notation 2nd form
{{ a}}^y={{ b}}\iff log_{{ a}}{{ b}}=y 
}$

so... hmm to add some
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3 years ago

Solve the system of equations by substitution 5x +y = 36 y = 4x

natali 33 [55]

Step-by-step explanation:

The answer is mentioned above.

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3 years ago

Read 2 more answers

Suppose a tank contains 400 gallons of salt water. If pure water flows into the tank at the rate of 7 gallons per minute and the

Strike441 [17]

Answer:

Step-by-step explanation:

This is a differential equation problem most easily solved with an exponential decay equation of the form

$y=Ce^{kt}$ . We know that the initial amount of salt in the tank is 28 pounds, so

C = 28. Now we just need to find k.

The concentration of salt changes as the pure water flows in and the salt water flows out. So the change in concentration, where y is the concentration of salt in the tank, is $\frac{dy}{dt}$ . Thus, the change in the concentration of salt is found in

$\frac{dy}{dt}=$ inflow of salt - outflow of salt

Pure water, what is flowing into the tank, has no salt in it at all; and since we don't know how much salt is leaving (our unknown, basically), the outflow at 3 gal/min is 3 times the amount of salt leaving out of the 400 gallons of salt water at time t:

$3(\frac{y}{400})$

Therefore,

$\frac{dy}{dt}=0-3(\frac{y}{400})$ or just

$\frac{dy}{dt}=-\frac{3y}{400}$ and in terms of time,

$-\frac{3t}{400}$

Thus, our equation is

$y=28e^{-\frac{3t}{400}$ and filling in 16 for the number of minutes in t:

y = 24.834 pounds of salt

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3 years ago

5 square root 13^3. A 13^2. B 13^15 C 13^5/3. D 13^3/5

arlik [135]

Answer is D.

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3 years ago

What is the solution to the quadratic equation? y = x^2-4

Anika [276]

(-2,2)

i just did my homework with the same question

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3 years ago