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3 years ago

How to Factor 3x^2-x-10

Mathematics

1 answer:

adoni [48]3 years ago

5 0

Solution

( − 2)(3 + 5)

Step-by-step solution:

Use the sum-product pattern

3x^2−−10

3^2+5−6x−10

Common factor from the two pairs

3^2+5−6−10

(3+5)−2(3+5)

Rewrite in factored form

(3+5)−2(3+5)

(−2)(3+5

hope this was good! <3

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Which is the value of the expression (StartFraction (10 Superscript 4 Baseline) (5 squared) Over (10 cubed) (5 cubed)) cubed?

Flura [38]

Answer:

The value to the given expression is 8

Therefore $\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=8$

Step-by-step explanation:

Given expression is (StartFraction (10 Superscript 4 Baseline) (5 squared) Over (10 cubed) (5 cubed)) cubed

Given expression can be written as below

$\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3$

To find the value of the given expression:

$\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=\frac{((10^4)(5^2))^3}{((10^3)(5^3))^3}$

( By using the property ( $(\frac{a}{b})^m=\frac{a^m}{b^m}$ )

$=\frac{(10^4)^3(5^2)^3}{(10^3)^3(5^3)^3}$

( By using the property $(ab)^m=a^mb^m$ )

$=\frac{(10^{12})(5^6)}{(10^9)(5^9)}$

( By using the property $(a^m)^n=a^{mn}$ )

$=(10^{12})(5^6)(10^{-9})(5^{-9})$

( By using the property $\frac{1}{a^m}=a^{-m}$ )

$=(10^{12-9})(5^{6-9})$ (By using the property $a^m.b^n=a^{m+n}$ )

$=(10^3)(5^{-3})$

$=\frac{10^3}{5^3}$ ( By using the property $a^{-m}=\frac{1}{a^m}$ )

$=\frac{1000}{125}$

$=8$

Therefore $\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=8$

Therefore the value to the given expression is 8

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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:

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

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

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:

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

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

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

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

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

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

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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.

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