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professor190 [17]
3 years ago
15

Mr ogata drove 276 miles from his house to los angeles at an average speed of 62 miles per hour. his trip home took 6.5 hours. h

ow did his peed on the way home compare to his speed on the way to los angeles
Mathematics
1 answer:
Ann [662]3 years ago
8 0
His trip home was at an average speed in miles per hour of
  (276 miles)/(6.5 hours) = 42.46 miles per hour

His speed on the way home was about 19.5 mph less, or about 68% of his speed to LA.
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Answer:

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3. x = 15°

4. x = 30°

5. x = 40°

6. x = 20°

Step-by-step explanation:

All the angles should add up to 90° as they are complementary.

6 0
2 years ago
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If the radius of a circle equals 20 cm find its circumference correct to one decimal place
Brums [2.3K]

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125.66

Step-by-step explanation:

i don't get the one decimal place part but there your answer

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How do I find the area of a rhombus?
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A = pq/2
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Which of the following equations shows how substitution can be used to solve the following system of equations?
klasskru [66]

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Step-by-step explanation:

5 0
2 years ago
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
2 years ago
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