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Oxana [17]
3 years ago
5

PLEASE HELP

Mathematics
1 answer:
Paladinen [302]3 years ago
7 0

1. Need distance formula to find how long it is then cut it in half to find the radius....d= 10.....r= 5
Need midpoint formula to find the center....midpoint= (-1, -1)
(h,k) in the equation of a circle is the center, "r" is the radius. When you plug in the center, you have to put the opposite signs...so......
equation of a Circle = (x - h)^2 + (y - k)^2 = r^2 = (x + 1)^2 + (y + 1)^2= 25

2. A= pi(r^2)
200.96= 3.14(r^2)
64 = r^2
r = 8
Then to find the circumference, C= 2(pi)(r) = 2 (3.14) (8) = 50.24

For 3 and 4, is there a diagram?

A quilt design uses parallelogram-shaped pieces that have a base measure of 4 inches and a height of 6 inches. If 15 pieces are used in the quilt, what is the total area of the parallelogram-shaped pieces in the quilt?

1.225 in.

2.360 in.

3.(450 in.)

4.900 in.

<span>   15*4*6 = 360 sorry but I only have the answers for number 1 and 2






</span>


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sattari [20]

Option a is correct. The calculated answer is 0.150

<h3>How to get the value using the cdf</h3>

In order to get  P(0.5 ≤ X ≤ 1.5).

This can be rewritten as

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and P = 1.5

We have the equation as

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This would be written as

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This is approximately 0.1250

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brainly.com/question/19884447

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<h3>complete question</h3>

Use the cdf to determine P(0.5 ≤ X ≤ 1.5).

a) 0.1250

b) 0.0339

c) 0.1406

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Blababa [14]

Answer:

Solving the given formula for v2 gives us:

a(t_2-t_1)+v_1 = v_2

Step-by-step explanation:

Solving an equation for a particular variable means that the variable has to be isolated on one side of the equation.

Given equation is:

a = \frac{v_2-v_1}{t_2-t_1}

Multiplying both sides by t2-t1

(t_2-t_1) . a = \frac{v_2-v_1}{t_2-t_1} . (t_2-t_1)\\a(t_2-t_1) = v_2-v_1

Adding v1 to both sides of the equation

a(t_2-t_1)+v_1 = v_2-v_1+v_1\\a(t_2-t_1)+v_1 = v_2

Hence,

Solving the given formula for v2 gives us:

a(t_2-t_1)+v_1 = v_2

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Answer:

  1.  x^2 +2x -24 = 0

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This knowledge lets you write down the standard form equation with no particular effort.

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