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3 years ago

PLEASE ANSWER ASAP! 25 POINTS!

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

3 0

E.) 15/53
F.) 15/53
G.) 18/53
H.) 12/53

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Chuck and Dana agree to meet in Chicago for the weekend. Chuck travels 104 miles in the same time that Dana travels 96 miles. If

Yanka [14]

Answer: Chuck's travel at a rate of 52mph

Step-by-step explanation:

For Chuck's trip:

D=RT

104= (R+4)T

T= 104 / (R+4)

For Dana's trip:

96 = RT

T= 96/R

Set both equation for Chuck's and Dana together

104/(R+4) =96/R

Then we cross multiply

96(R+4) = 104R

96R + 384 = 104R

104R - 96R = 384

8R = 384

To get R, divide both side by 8

8R/8 = 384/8

R= 48mph

This means Dana's speed is 48mph

Chuck's speed will be: 48mph+4mph = 52mph

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3 years ago

You score a 95% or your mouth quiz the quiz was out of 60 points how many points did you get

Genrish500 [490]

Answer:

It will be 57 points.

Step-by-step explanation:

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2 years ago

Solve the equation | z - 12 = 9

MariettaO [177]

Answer:

Step-by-step explanation:

21-12=9

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2 years ago

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Margot is sewing a ribbon on a seam along the perimeter of a square pillow. The side length of the pillow is 2x2+1 inches. She p

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3 years ago

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Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

3 years ago