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3 years ago

why does the method for finding the volume of an irregular-shaped object work for regular-shaped objects as well?

Mathematics

1 answer:

kumpel [21]3 years ago

5 0

Answer:

Its the same formula that your using to find a regular object as an irregular one. The only difference is that your adding the formula of multiple shapes together since an irregular shape consists of multiple regular ones

hope this helps!

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WILL GIVE BRINALIEST!

Liono4ka [1.6K]

-21, -23, -27, -28, -28, -29 order them least to greatest and then find the middle if there is not a middle then put -27.5 as your answer for the median. For the mean add up all the number then you should get -156 then divide it by 6 which should give you -26 for the mean. and for the mode you should get -28 as the one that occurs the most often.

mean: -26

median: -27.5

mode: -28

4 0

3 years ago

Find the area of each sector.

svetlana [45]

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3 0

3 years ago

An ectopic pregnancy is twice as likely to develop when the pregnant woman is a smoker as it iswhen she is a nonsmoker. If 32 pe

Dmitriy789 [7]

Answer:

48.48%

Step-by-step explanation:

Let's assume that there is a number N of women.

32% of these are smokers, then there are 0.32*N smokers

then 68% of these are non-smokers, then there are 0.68*N non-smokers.

Let's assume that the probability of having a ectopic pregnancy for a non-smoker is p (and the probability for a smoker will be 2*p)

Then the number of women with an ectopic pregnancy that are non-smokers is:

p*0.68*N

The number of women with an ectopic pregnancy that are smokers is:

2*p*0.32*N

Then the total number of women with an ectopic pregnancy will be:

p*0.68*N + 2*p*0.32*N

The percentage of women having an ectopic pregnancy that are smokers is equal to the quotient between the number of women with an ectopic pregnancy that are smokers and the total number of women with an ectopic pregnancy, all that times 100%.

The percentage is:

$P = \frac{2*p*0.32*N}{p*0.68*N + 2*p*0.32*N} *100\%$

Taking p and N as common factors, we get:

$P = \frac{2*p*0.32*N}{p*0.68*N + 2*p*0.32*N} *100\% = \frac{N*p}{N*p} \frac{2*0.32}{0.68 + 2*0.32} *100\%$

Then we get:

$\frac{2*0.32}{0.68 + 2*0.32} *100\% = 48.48\%$

8 0

3 years ago

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2

Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

(a) The velocity of the top of the ladder = 1.5 m/s downwards

(b) The rate the area formed by the ladder is changing is approximately -75.29 ft.²/sec

Reasons:

The given parameter are;

Length of the ladder, l = 25 feet

Rate at which the base of the ladder is pulled, $\displaystyle \frac{dx}{dt}$ = 2 feet per second

(a) Let y represent the height of the ladder on the wall, by chain rule of differentiation, we have;

$\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}$

25² = x² + y²

y = √(25² - x²)

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}$

Which gives;

$\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times \frac{dx}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2$

$\displaystyle \frac{dy}{dt} = \mathbf{ \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2}$

When x = 15, we get;

$\displaystyle \frac{dy}{dt} = \frac{15 \times \sqrt{625-15^2} }{15^2- 625}\times2 = \mathbf{-1.5}$

The velocity of the top of the ladder = 1.5 m/s downwards

When x = 20, we get;

$\displaystyle \frac{dy}{dt} = \frac{20 \times \sqrt{625-20^2} }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6$

The velocity of the top of the ladder = $\underline{-2.\overline{6} \ m/s \ downwards}$

When x = 24, we get;

$\displaystyle \frac{dy}{dt} = \frac{24 \times \sqrt{625-24^2} }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}} \approx -6.86$

The velocity of the top of the ladder ≈ -6.86 m/s downwards

(b) $\displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}$

Therefore;

$\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}$

$\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}$

$\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}$

Therefore;

$\displaystyle \frac{dA}{dt} = \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2$

When the ladder is 24 feet from the wall, we have;

x = 24

$\displaystyle \frac{dA}{dt} = \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}$

The rate the area formed by the ladder is changing, $\displaystyle \frac{dA}{dt}$ ≈ -75.29 ft.²/sec

$\displaystyle sin(\theta) = \frac{x}{25}$

$\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}$

$\displaystyle \frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt}$

$\displaystyle\frac{d \theta}{dx} = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}$

Which gives;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}$

When x = 24 feet, we have;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}$

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is $\displaystyle \frac{d \theta}{dt}$ ≈ 0.286 rad/sec