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3 years ago

Factor each polynomial 6m^2-15m

1 answer:

6 0

You can do prime factorization then find common factors to factor out.

6m² - 15m

2·3·m·m - 3·5·m

Both terms have 3·m so we can factor it out:

3·m (2·m - 5)

So that would be 3m (2m - 5)

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For a singing contest in which 42,000 votes were cast, the winner received 3/5 of the votes. How many votes did the winner not r

kirza4 [7]

42,000 x .60 (3/5)=25200

42,000-25,200= 16,800

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3 years ago

Choose the best score on a quiz. 24 points out of 42 24 points out of 42 75 points out of 135 75 points out of 135 150 points ou

stellarik [79]

Answer:

150/250

Step-by-step explanation:

24/42, 75/135, 150/250, and 75/150 I believe is what you're asking. We know 75/135 is better than 75/150 so 75/150 is out of the equation. If we multiple 75/135 by 2 we get 150/270 which is more than 150/250, so 75/135 is out of the equation. Now to compare 24/42 and 150/250 we can divide their fractions, if we divide 24/42 we get 0.57, so 57%. If we did 150/250 we get 0.6 or 60%. So 150/250 is the best score.

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2 years ago

Read 2 more answers

A girl cut off 2/5 of a ribbon. The piece she cut off was 5 meters long.How

valentinak56 [21]

Answer:

12.5

Step-by-step explanation:

5/2*5=12.5

7 0

3 years ago

Read 2 more answers

In a rectangle MPKN, the diagonals intersect each other at point O. A line segment OA is an altitude of a triangle MOP, and m∠AO

WINSTONCH [101]

Answer:

Step-by-step explanation:

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3 years ago

Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ

atroni [7]

Answer: $y=Ce^(^3^t^{^9}^)$

Step-by-step explanation:

Beginning with the first differential equation:

$\frac{dy}{dt} =27t^8y$

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

$\frac{1}{y} \frac{dy}{dt} =27t^8$

Multiply both sides by 'dt' to get:

$\frac{1}{y}dy =27t^8dt$

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

$\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt$

$ln(y)=27(\frac{1}{9} t^9)+C$

$ln(y)=3t^9+C$

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

$e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)$

$y=e^(^3^t^{^9} ^+^C^)$

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

$y=e^(^3^t^{^9}^)e^C$

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

$y=Ce^(^3^t^{^9}^)$

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

$\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8$

Now check if the derivative equals the right side of the original differential equation:

$(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)$

$Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)$

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

7 0

2 years ago