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Aliun [14]
3 years ago
8

Find the area of the circle with the given radius or diameter. Use pi = 3.14. r = 6

Mathematics
2 answers:
Step2247 [10]3 years ago
6 0

Answer:

113.04

Step-by-step explanation:


svetoff [14.1K]3 years ago
3 0
The answer is 113.1. Hope this helps
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bonufazy [111]
The correct answer is 427
5 0
3 years ago
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You have found the ages (in years) of 5 gorillas. Those gorillas were randomly selected from the 29 gorillas at your local zoo:
zalisa [80]

Answer:

the answer is 10

Step-by-step explanation:

(8+4+14+16+8)/5

8 0
2 years ago
Which table represents a direct variation function?
sineoko [7]

Answer:

Option C is correct.

Step-by-step explanation:

A direct variation function is

y/x = k

i.e. we can say that the ratio of y and x is equal to a constant value k.

We will check for each Option given.

Option A

7/2 = 7/2

8/3 = 8/3

9/4 = 9/4

10/5 = 2

11/6 = 11/6

Option D is incorrect as y/x ≠ k as ratio of y/x for each value of table doesn't equal to constant

Option B

-3/2 = -3/2

-5/4 = -5/4

-6/6 = -1

-7/8 = -7/8

-8/10 = -4/5

Option B is incorrect as y/x ≠ k as ratio of y/x for each value of table doesn't equal to constant

Option C

10/-5 = -2

8/-4 = -2

6/-3 = -2

4/-2 = -2

2/-1 = -2

Option C is correct as y/x = k as ratio of y/x for each value in table c is equal to constant value -2

Option D

-3/-2 = 3/2

-3/1 = -3

-3/0 = 0

-3/1 = -3

-3/2 = -3/2

Option D is incorrect as y/x ≠ k as ratio of y/x for each value of table doesn't equal to constant .

SO, Option C is correct.

8 0
3 years ago
Read 2 more answers
Use Newton’s Method to find the solution to x^3+1=2x+3 use x_1=2 and find x_4 accurate to six decimal places. Hint use x^3-2x-2=
luda_lava [24]

Let f(x) = x^3 - 2x - 2. Then differentiating, we get

f'(x) = 3x^2 - 2

We approximate f(x) at x_1=2 with the tangent line,

f(x) \approx f(x_1) + f'(x_1) (x - x_1) = 10x - 18

The x-intercept for this approximation will be our next approximation for the root,

10x - 18 = 0 \implies x_2 = \dfrac95

Repeat this process. Approximate f(x) at x_2 = \frac95.

f(x) \approx f(x_2) + f'(x_2) (x-x_2) = \dfrac{193}{25}x - \dfrac{1708}{125}

Then

\dfrac{193}{25}x - \dfrac{1708}{125} = 0 \implies x_3 = \dfrac{1708}{965}

Once more. Approximate f(x) at x_3.

f(x) \approx f(x_3) + f'(x_3) (x - x_3) = \dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125}

Then

\dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125} = 0 \\\\ \implies x_4 = \dfrac{5,881,319,037}{3,324,107,515} \approx 1.769292663 \approx \boxed{1.769293}

Compare this to the actual root of f(x), which is approximately <u>1.76929</u>2354, matching up to the first 5 digits after the decimal place.

4 0
2 years ago
Jill just got her pilot’s license and wants to rent a plane. The Platinum Plane Company charges $180 plus $92 per hour to rent a
Luden [163]

Answer:

5

Step-by-step explanation:

solve equation 180+92x= 250+78x

6 0
3 years ago
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