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2 years ago

a circle with a 12-inch diameter is folding in half and then folded in half again what is the area of the resulting shape

Mathematics

1 answer:

andrew-mc [135]2 years ago

7 0

Answer:

3π in^2

Step-by-step explanation:

Area of the circle before folding = π r^2

= 6^2 π

= 36 π is^2

The area of the resulting shape = 1/2 * 1/2 * 12π

= 3 π in^2.

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Use forward and backward difference approximations of O(h) and a centered difference approximation of O(h2) to estimate the firs

strojnjashka [21]

Answer:

Answer has explained below.

Step-by-step explanation:

Consider the function is:

F(x) = 25x3 – 6x2 +7x -88

Differentiate with respect to x, we get

F’(x) = 25. 3x2 – 6.2x + 7

= 75x2 – 12x +7

At x = 2, we have

F (2) = 25(2)3 – 6(2)2 + 7(2)-88

=102

And f’(2) = 75(2)2 – 12 (2) +7

=283

Now, calculate forward divided difference as:

xi + 1 = xi + h

=2 + 0.25

=2.25

F (xi + 1) = f (2.25) = 25 (2.25)3 – 6(2.25)2 +7(2.25) -88

=182.21

f’(2) = f(2.25) – f(2) / 0.25 = 182.21 – 102 / 0.25

= 320.84

Єt = 283 – 320.8 / 283 = -13.36%

Now calculate backward divided difference:

Xi-1 = xi –h = 2 – 0.25 = 1.75

F(xi-1)= f(1.8) = 25 . (1.8)3 -6 (1.8)2 + 7 (1.8) – 88

= 50.96

F’(2) = f(2) – f(1.8) / 0.25 = 102 – 50.96 / 0.25 = 204.16

Єt = 283 – 204.16 / 283 = 27.86%

Finally, centered divided difference is obtain by inserting f(xi+1) and f (xi-1):

F’(2) = f(2.25) – f(1.8) /2 x 0.25 = 320.84 -50.96 / 0.5 = 539.68

Єt = 283 – 539.68 / 283 = -90.7%

6 0

2 years ago

Fill in the blank. In the triangle below, y=__. Round your answer to two decimal places.

Kobotan [32]

To solve this we use trigonometric functions that would relate the hypotenuse y and the given values. For this case we use cosine function which is expressed as:

cosine theta = adjacent side / hypotenuse
cosine 52 = 35 / y
y = 35 / cos 52
y = 56.85

7 0

2 years ago

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The number of ducks to geese in Miller's Pond last year was 2:3. This year the ratio of ducks to geese is 5:9. The number of gee

My name is Ann [436]

No decreased-

Set up equaivalent ratioo. It started out 2 duck to every 3 geese, continue the pattern and see for a change

ducks 2 4 6
geese 3 6 9

but now its 5 ; 9 instead of 6 to 9, therefore the number of ducks decreased

4 0

2 years ago

At a college, 69% of the courses have final exams and 42% of courses require research papers. Suppose that 29% of courses have a

laila [671]

Answer:

a) 0.82

b) 0.18

Step-by-step explanation:

We are given that

P(F)=0.69

P(R)=0.42

P(F and R)=0.29.

P(course has a final exam or a research paper)=P(F or R)=?

P(F or R)=P(F)+P(R)- P(F and R)

P(F or R)=0.69+0.42-0.29

P(F or R)=1.11-0.29

P(F or R)=0.82.

Thus, the the probability that a course has a final exam or a research paper is 0.82.

P( NEITHER of two requirements)=P(F' and R')=?

According to De Morgan's law

P(A' and B')=[P(A or B)]'

P(A' and B')=1-P(A or B)

P(A' and B')=1-0.82

P(A' and B')=0.18

Thus, the probability that a course has NEITHER of these two requirements is 0.18.

3 0

2 years ago

In one town, 66% of adults have health insurance. What is the probability that 4 adults selected at random from the town all ha

LiRa [457]

Answer:

Probability that randomly selected 4 persons all have insurance is = 0.1897

Step-by-step explanation:

Given that the probability of having an insurance equals 66%=0.66

Thus

Probability that person A has insurance = 0.66

Probability that person B has insurance = 0.66

Probability that person C has insurance = 0.66

Probability that person D has insurance = 0.66

Probability that randomly selected 4 persons all have insurance is

$P(A)\times P(B)\times P(C)\times P(D)\\\\=0.66\times 0.66\times 0.66\times 0.66=0.1897$

3 0

2 years ago

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