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Olegator [25]
2 years ago
12

The coordinates of the midpoint of gh are m(-13/2,-6) and the coordinates of one endpoint are g(-4,1). what is the other endpoin

t?
Mathematics
1 answer:
Alex777 [14]2 years ago
4 0

Applying the midpoint formula, the coordinates of the other endpoint are: (-9, -13).

<h3>How to Apply the Midpoint Formula?</h3>

The midpoint formula, which can be used in determining the coordinates of the midpoint between two endpoints is usually expressed as: M[(x1 + x2)/2, (y1 + y2)/2].

Given the following coordinates:

Midpoint of GH --> m(-13/2,-6) = (x, y)

One endpoint ---> G(-4, 1) = (x1, y1)

The other endpoint ---> (x2, y2)

Plug in the values into the formula used in calculating the midpoint:

m(-13/2,-6) = [(-4 + x2)/2, (1 + y2)/2]

Solve for x2

-13/2 = (-4 + x2)/2

-13/2 × 2 = -4 + x2

-13 = -4 + x2

-13 + 4 = x2

-9 = x2

x2 = -9

Solve for y2

-6 = (1 + y2)/2

2(-6) = 1 + y2

-12 = 1 + y2

-12 - 1 = y2

-13 = y2

y2 = -13

Thus, using the midpoint formula, the coordinates of the other endpoint are: (-9, -13).

Learn more about the midpoint formula on:

brainly.com/question/13115533

#SPJ1

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Work Shown:

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An old bone contains 80% of its original carbon-14. Use the half-life model to find the age of the bone. Find an equation equiva
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Answer:

An old bone contains 80% of its original carbon-14 in 1844.6479 years

Step-by-step explanation:

We know that

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so, h=5730

now, we can use formula

P(t)=P_0(\frac{1}{2})^{\frac{t}{h} }

we can plug back h

and we get

P(t)=P_0(\frac{1}{2})^{\frac{t}{5730} }

An old bone contains 80% of its original carbon-14

so,

P(t)=0.80Po

we can plug it and then we solve for t

0.80P_0=P_0(\frac{1}{2})^{\frac{t}{5730} }

0.80=(\frac{1}{2})^{\frac{t}{5730} }

\ln \left(0.8\right)=\ln \left(\left(\frac{1}{2}\right)^{\frac{t}{5730}}\right)

t=-\frac{5730\ln \left(0.8\right)}{\ln \left(2\right)}

t=1844.6479

So,

An old bone contains 80% of its original carbon-14 in 1844.6479 years


4 0
3 years ago
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Select the correct answer from each drop-down menu.
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Answer: she invested $2500 at the rate of 10%.

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Step-by-step explanation:

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