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3 years ago

PLZ HELP Find the SURFACE AREA of the figure below.

Mathematics

2 answers:

LenKa [72]3 years ago

5 0

Itsssss 35280 have a nice dayyyyyyyyt

Nat2105 [25]3 years ago

3 0

Answer:

what grade are you in

Step-by-step explanation:

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The radio of 2,787 to 1,750 is ____

pishuonlain [190]

Answer:

1.59

Step-by-step explanation:

Divide 2,787 by 1,750

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3 years ago

I really need help. What would be the slope-intercept equation?

Naddik [55]

Answer:

y = -3.71x + 147.25

Step-by-step explanation:

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3 years ago

The standard normal curve shown here is a probability density curve for a continuous random variable. This means that the area u

faust18 [17]

the answer is 0.7068

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3 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

2 years ago

P(t) = 250 * (3.04)^t/1.98,

Alex17521 [72]

Answer:

The concept which best describes the change of the population is the derivative of $p(t)=250(3.04)^{\frac{t}{1.98}}$ .

Step-by-step explanation:

Observe that the function $p(t)=250(3.04)^{\frac{t}{1.98}}$ describes the amount of rabbits at the time t (in years) but no the rate of change of the population at a given instant. So you have to use the derivative of $p(t)$ to obtain that rate of change at any instant. For example, if we derivate the function $p(t)=250(3.04)^{\frac{t}{1.98}}$ we obtain:

$p'(t)=250\cdot \log(\frac{1}{1.98})(3.04)^{\frac{t}{1.98}}$

And if we want to find the rate of change at $t=5$ years we evaluate

$p'(5)=250\cdot \log(\frac{1}{1.98})(3.04)^{\frac{5}{1.98}}=2326$ rabbits/year

5 0

3 years ago