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k0ka [10]
3 years ago
13

Need help Asap what is the approximate area of the circle use π =3.14

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
4 0
You need to include the radius.
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Consider the circle shown with center O and points A , B , C , and D on the circle.
leva [86]

Answer:

the answer is c

Step-by-step explanation:

3 0
2 years ago
What is the vertex of y=-3x^2+6x+15? Help is much appreciated.
DENIUS [597]

Answer:

<u>(1, 18)</u>

Step-by-step explanation:

Rewrite the equation in vertex form by completing the square for -3x^2 + 6x + 15. This = -3(x - 1)^2 + 18.

Set y equal to the new right side.

y = -3(x - 1)

Use the vertex form, y = a(x - h)^2 + k, to determine the values of a, h, and k.

a = -3

h = 1

k = 18

Vertex = (h, k) / (1, 18)

5 0
3 years ago
Is 26 divisible by 2
sergij07 [2.7K]
Yes because 26 is an even number
3 0
3 years ago
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(2a+1) - 4a(2a+1)+4a
maxonik [38]

Answer:

2a (3-4a)

Step-by-step explanation:

(2a+1)-4a(2a+1)+4a

2a+1-8a²-1+4a

= 2a+4a-8a²

=6a-8a²

= 2a(3-4a)

6 0
2 years ago
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The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop
Anettt [7]

Answer:

a. \frac{1}{15}

b. \frac{2}{5}

c. \frac{14}{15}

d. \frac{8}{15}

Step-by-step explanation:

Given that there are two laptop machines and four desktop machines.

On a day, 2 computers to be set up.

To find:

a. probability that both selected setups are for laptop computers?

b. probability that both selected setups are desktop machines?

c. probability that at least one selected setup is for a desktop computer?

d. probability that at least one computer of each type is chosen for setup?

Solution:

Formula for probability of an event E can be observed as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

a. Favorable cases for Both the laptops to be selected = _2C_2 = 1

Total number of cases = 15

Required probability is \frac{1}{15}.

b. Favorable cases for both the desktop machines selected = _4C_2=6

Total number of cases = 15

Required probability is \frac{6}{15} = \frac{2}{5}.

c. At least one desktop:

Two cases:

1. 1 desktop and 1 laptop:

Favorable cases = _2C_1\times _4C_1 = 8

2. Both desktop:

Favorable cases = _4C_2=6

Total number of favorable cases = 8 + 6 = 14

Required probability is \frac{14}{15}.

d. 1 desktop and 1 laptop:

Favorable cases = _2C_1\times _4C_1 = 8

Total number of cases = 15

Required probability is \frac{8}{15}.

8 0
3 years ago
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