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3 years ago

A bulletin board is 3 ft high. the area is 15 sq ft. what is the perimeter?

1 answer:

5 0

First step: Find how wide the bulletin board is
15 ft² ÷ 3 ft = 5 ft
Step the second: Use the perimeter formula P = 2l + 2w
P = 2(5) + 2(3)
<u>P = 16 ft</u>

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Step-by-step explanation:

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3 years ago

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astraxan [27]

Answer:

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Step-by-step explanation:

Step 1: Distribute

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3 years ago

Subtract these polynomials.

Elina [12.6K]

Answer:

D) x^2+2

Step-by-step explanation:

(2x^2+2x+3)-(x^2+2x+1)

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3 years ago

The city council plans to build a new water storage tank for the community. The tank is cylindrical with a base diameter of 25 f

Stella [2.4K]

Answer:

a. $14726.216ft^{3}$

b. $110446.62gal$

Step-by-step explanation:

I attach the diagram of the new water tank.

a. In order to find the volume of the tank in cubic feet, we will need to calculate the volume of a cylinder. We know that its base measures $25ft$ and its height is $30ft$ .

The volume ''V'' of a cylinder is :

$V=(\frac{\pi D^{2}}{4}).(H)$

Where '' $D$ '' is the diameter of the base and '' $H$ '' is the height of the cylinder.

If we replace all the data in the equation :

$V=\frac{\pi .(25ft)^{2}}{4}.(30ft)=14726.216ft^{3}$

We found out that the volume of the tank in cubic feet is $14726.216ft^{3}$

b. We know that 1 cubic foot of water is equal to 7.5 gallons. To find the amount of gallons that the new tank can hold, we need to do an unit conversion.

This equivalence is given to us :

$1ft^{3}=7.5gal$ ⇒ Using the equivalence ⇒

$(14726.216ft^{3}).(\frac{7.5gal}{1ft^{3}})=110446.62gal$

Finally, the new tank will hold $110446.62gal$ of water.

7 0

3 years ago

From Euler’s relation eiθ = cosθ + isinθ,

adelina 88 [10]

Answer:

(a) The graphic representation is in the attached figure.

(b) $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$ .

(c) $\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ .

Step-by-step explanation:

(a) Given a complex number $e^{i\theta}$ we know, from Euler's formula that $e^{i\theta} = \cos(\theta)+i\sin(\theta)$ . So, it is not difficult to notice that

$|e^{i\theta}|^2 = \cos^2(\theta)+\sin^2(\theta) =1$

so it is on the unit circumference. Also, notice that the Cartesian representation of the complex number is $(\cos(\theta), \sin(\theta))$ .

Now,

$e^{-i\theta} = \cos(\theta)+i\sin(-\theta) = \cos(\theta)-i\sin(\theta)$ .

Notice that $e^{-i\theta}$ has the same modulus that $e^{i\theta}$ , so it is on the unit circumference. Beside, its Cartesian representation is $(\cos(\theta), -\sin(\theta))$ .

So, the points $(\cos(\theta), \sin(\theta))$ and $(\cos(\theta), -\sin(\theta))$ are symmetric with respect to the X-axis. All this can be checked in the attached figure.

(b) Notice that

$e^{i\theta} + e^{-i\theta} = \cos(\theta)+i\sin(\theta) + \cos(\theta)-i\sin(\theta) = 2\cos(\theta)$

Then,

$\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$ .

(c) Notice that

$e^{i\theta} - e^{-i\theta} = \cos(\theta)+i\sin(\theta) - \cos(\theta)+i\sin(\theta) = 2i\sin(\theta)$

Then,

$\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ .

7 0

3 years ago