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krok68 [10]
3 years ago
5

Given a triangle with verticies A(-3, 1), B(3,5), and C(5,-5). Vertify the midsegment theorem.

Mathematics
1 answer:
natima [27]3 years ago
3 0
And-6-4 abd (563-55 NDA fib
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A person who weighs 100 pounds on Earth weighs 16.6 lb. on the moon.
mario62 [17]

Answer:

See explanation below.

Step-by-step explanation:

Given: 100 lbs on Earth is 16.6 lbs on the moon.

a. The independent variable is weight. The gravity of the Moon and the gravity of the Earth are constant. Weight can change, but gravity is a constant.

b. An equation that relates the weight of someone on the Moon who travels to the Earth:

100 / 16.6 = 6.02. Take the Moon weight and multiply by 6.02:

Moon Weight * 6.02 = Earth Weight.

Proof:

16.6 * 6.024 = 99.99 - approximately 100 lbs Earth weight.

c. A 185 lb astronaut on Earth would weigh:

16.6 / 100 = .166. Take the Earth weight and multiply by .166:

185 * .166 = 30 lbs on the Moon.

d. A person who weighs 50 lbs on the Moon:

50 * 6.024 = 301.2 lbs on Earth.

Hope this helps! Have an Awesome Day! :-)

6 0
3 years ago
Use the law of sines to find what Angle R is:
stich3 [128]

Answer:

Step-by-step explanation:

So, according to the law of Sines, we have

\frac{Sin(A)}{a} = \frac{Sin(b)}{b}

In our case, we have S and R, so...

\frac{Sin(S)}{s} = \frac{Sin(R)}{r}

And we want to isolate R...

r = sin^{-1} (\frac{Sin(S) * r}{s} ) = sin^{-1} (\frac{Sin(52) * 32}{26} ) = 75.90

the angle of R is then 75.90 degrees.

5 0
4 years ago
Please help me someone ASAP
Artyom0805 [142]
The Answer is b. Hope that helps
5 0
3 years ago
Is y=x^3 a solution of the differential equation yy'=x^5+y
shepuryov [24]

No; we have y=x^3\implies y'=3x^2. Substituting these into the DE gives

3x^5=x^5+x^3

which reduces to x^3=0, true only for x=0.

3 0
3 years ago
The lengths of a certain species of fish are approximately normally distributed with a given mean (look at the picture) and stan
tatyana61 [14]
The empirical rule states that approximately 68/95/99.7% of a normal distribution lies within 1/2/3 standard deviations. So the answer is 68%.
6 0
3 years ago
Read 2 more answers
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