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tresset_1 [31]
2 years ago
14

Multiply (3a³-6b³) (4a² + 5b²)

Mathematics
1 answer:
GREYUIT [131]2 years ago
3 0

Answer:

12a^5+15a^3b^2-24b^3a^2-30b^5

Step-by-step explanation:

(3a³-6b³) (4a² + 5b²) -> <u>Simplify</u> not Multiply

Expand (3a³-6b³) (4a² + 5b²) using the FOIL Method.

Apply the <u>distributive property</u>.

3a^3(4a^2+5b^2)-6b^3(4a^2+5b^2)

3a^3(4a^2)+3a^3(5b^2)-6b^3(4a^2+5b^2)

3a^3(4a^2)+3a^3(5b^2)-6b^3(4a^2)-6b^3(5b^2)

<h3><u /></h3>

<u>Simplify each term.</u>

Rewrite using the <u>commutative property of multiplication</u>.

3*4a^3a^2+3a^3(5b^2)-6b^3(4a^2)-6b^3(5b^2)

↓

12a^5+15a^3b^2-24b^3a^2-30b^5

<em />

<em>Equation Simplified.</em>

<em>---------------------------------------------------------------------------------------</em>

Hope this helped!

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If the simple interest on $5,000 for 4 years is $1,600, then what is the interest rate?
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If
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Answer:

\frac{s^2-25}{(s^2+25)^2}

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given:\mathcal{L}[t \cos 5t]=(-1)F'(s) with F(s)=\mathcal{L}[\cos 5t].

Now, F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt.

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s).

Solving for F(s) on the last equation, F(s)=\frac{s}{s^2+25}, then the Laplace transform we were searching is -F'(s)=\frac{s^2-25}{(s^2+25)^2}

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3 years ago
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