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3 years ago

CAN SOMEONE ANSWER 2 AND SHOW WORK THANK U LOVES

Mathematics

1 answer:

Rzqust [24]3 years ago

7 0

Answer:

16x^2-24x+9

Step-by-step explanation:

expand equation:

(4x-3)(4x-3)

distribute:

4x(4x-3) - 3(4x-3)

distribute:

16x^2-12x - 12x+9

combine like terms:

16x^2-24x+9

hope this helps!

You might be interested in

An ice-cream store sold 70 sundaes. The ratio of

elixir [45]

Answer:

30 strawberry sundaes.

Step-by-step explanation:

We know there is a total of 70 sundaes.

3 sundaes will be strawberry.

4 sundaes will be choclate.

When you add 3 and 4 together, you get 7. This is your total ratio.

So 3/7 of the sundaes are strawberry.

And 4/7 of sundaes are choclate.

We know there is a total of 70 sundaes.

The total ratio we figured out is 7.

If the total sundaes is 70, and the total ration is 7. Then the total sundaes is 10 times bigger than the ratio.

This means we need to multiply the 3 strawberry sundaes and 4 choclate sundaes by 10.

3*10=30

4*10=40

So there are 30 strawberry sundaes and 40 choclate sundaes.

We need to find the strawberry sundae amount, so the answer is 30.

Hope this helps!

3 0

2 years ago

What is m<1 <br><br> Pls help and show work

insens350 [35]

Answer: 30

Step-by-step explanation: Free math problem solver answers your algebra homework questions with step-by-step explanations.

You visited this page on 12/2/21.

8 0

3 years ago

Read 2 more answers

Leno4ka [110]

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

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