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Phantasy [73]
3 years ago
11

Colin drove 45 minutes to the airport. He arrived 90 minutes before his flight departed, and then he spent 70 minutes in the air

. Once he landed, Colin spent 20 minutes gathering his luggage, and then he drove 35 minutes to his hotel. What must be true of any expression that represents the total time that Colin spent traveling from his house to the hotel?
Mathematics
1 answer:
shtirl [24]3 years ago
8 0
The answer is 50 minutes bro
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At the beginning of year 1, Gracie invest $350 at annual simple interest rate of 4%. She makes no deposits to or withdrawals fro
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I guess you want to know the total amount after one year. If so, the interest yielded after one year is 350 x 4% = $14 and her new capital becomes
350 (initial capital)  + 14 (interest over 1 year) = 350+14 = $364

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3 years ago
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X – y = 3 (line a)
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Answer. You must first put the equation into slope-intercept form (y = mx + b). To do this, subtract x from both sides to isolate y. Then, since y will be negative, divide each side by -1, resulting in y = x - 3.

3 0
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A byte is used to measure a computer's memory. There are 2^30 bytes in a gigabyte. A computer has an 8 gigabyte memory. Which ex
aksik [14]

Answer:

8x2^30

Step-by-step explanation:

5 0
3 years ago
Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)
USPshnik [31]
See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) +  f(\frac{x_{1} + x_{2} }{2}) + ... +  f(\frac{x_{n-1} + x_{n} }{2})]

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
M_{4} = \frac{1 - 0}{4} [ f(\frac{0 +  \frac{1}{4} }{2}) +  f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) +  f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]
M_{4} = \frac{1}{4} [ f(\frac{1}{8}) +  f(\frac{3}{8}) +  f(\frac{5}{8}) + f(\frac{7}{8})]

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]

M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]

M₄ = 1/4 · (2 - 478/512)
     = 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

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