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1 year ago

Simplify the following expression by combining like terms.

Mathematics

1 answer:

Nikitich [7]1 year ago

4 0

Answer:

5d² - 3d

Step-by-step explanation:

Given

4d² + d² - 2d - d ← collect like terms

= (4 + 1)d² + (- 2 - 1)d

= 5d² + (- 3)d

= 5d² - 3d

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Express each ratio as a unit rate. Round to the nearest tenth, if necessary. 19 yards in 2.5 minutes

harkovskaia [24]

19 yards divided by 2.5 minutes.

=19/2.5

= 7.6 yards/minute

4 0

2 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

1 year ago

Circle towns limit form a perfectly circular shape that has a population of 20000 in the population density of 480 people per sq

Alex

Answer:

$D =\frac{P}{\pi X^2}$

And solving for the radius we got:

$X = \sqrt{\frac{P}{\pi D}}$

And replacing the data given we got:

$X = \sqrt{\frac{480 \frac{people}{km^2}}{\pi *20000 people}}= 0.0874Km$

And this value converted to meters is $X = 87.40 m$

Step-by-step explanation:

For this case we know the population size $P = 20000$ and we also know the population density $D = 480 \frac{people}{km^2}$

We can assume that the area is a circle. We also know that the formula for the population density is given by:

$D= \frac{P}{A}$

Where P represent the number of people and A the area. Since we are assuming a circle then the area is given by:

$A = \pi X^2$

With X the radius of the circle

And then the populationd density become:

$D =\frac{P}{\pi X^2}$

And solving for the radius we got:

$X = \sqrt{\frac{P}{\pi D}}$

And replacing the data given we got:

$X = \sqrt{\frac{480 \frac{people}{km^2}}{\pi *20000 people}}= 0.0874Km$

And this value converted to meters is $X = 87.40 m$

6 0

3 years ago

Use the replacement set {3, 4, 9} to find the solution of the equation below.<br> 3n−18=9

Mrac [35]

Answer:

$n=9$

Step-by-step explanation:

$3n-18=9$

$3n+18=$ +18

$3n=27$

$3(9)=27$

$n=9$

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2 years ago

Nate is making a picture frame by cutting a small rectangular prism out of a larger rectangular prism. When the project is compl

dybincka [34]

Surface area= [15*3*2]+[20*3*2]+[10*3*2]+[14*3*2]+[15*3*2]+[14*2.5*2]

surface area=90+120+60+84+90+70=514 in²

the answer is 514 in²

6 0

2 years ago