What is the area of the figure?

Mathematics

1 answer:

nirvana33 [79]2 years ago

5 0

We can break it into two parts.

-> First, the rectangle on the bottom is made up of 2 feet x 16 feet. The formula for area of a rectangle is length times width, so in this case it is equal to 32 feet

-> Next, the triangle has sides of 10 feet, 10 feet, and 16 feet. We can calculate that the height is 6. The formula for area of a triangle is base times height divided by two, so, $\frac{16*6}{2}= \frac{96}{2} =$ 48 feet

-> Last, we can add the two totals together. 32 feet + 48 feet = 80 feet giving us our final answer of 80 feet². As this answer is not pictured I am going to assume they are not including Heron's formula or other more complex formulas, leading us to the answer of "can't be determined"

Answer:

O can't be determined

Have a nice day! - <em>Note: It depends on the level of your math class for whether the answer "cannot be determined" is correct as it can be determined, just takes more effort than simple one - two-step formulas.</em>

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

★ Also please leave the rating you think I deserve (It helps other users as well as myself)

- Heather

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Source

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3 years ago

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

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