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natima [27]
2 years ago
10

If a bridge will hold only 4500 kilograms and an empty truck weighs 26500 hectograms, whats the heaviest load the truck can safe

ly carry across the bridge? Give the answer in kilograms.
Mathematics
1 answer:
Klio2033 [76]2 years ago
4 0

1850 kilograms

26500x0.1 (conversion rate from hg to kg) = 2.650 kg

4500-2650 = 1850

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Answer:

C.

Step-by-step explanation:

Theoretical probability and Experimental probability are related in that theoretical probability is based on your reasoning (e.g. two sides; therefore, there is a 50% chance of heads), but experimental probability is where your numbers are based off of actual results (e.g. you landed head 1/4th of the time; therefore, there is a 25% chance of getting heads.)

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Kimberly has $120 to spend at the bookstore. Kimberly buys a hardcover book for $36, as well as some gift cards for her family a
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she can buy 5 more

Step-by-step explanation:

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A rectangular deck has an area of 320ft^2. the length of the deck is 4 feet longer than the width. find the dimensions of the de
Digiron [165]
A = L * W
A = 320
L = W + 4

320 = W(W + 4)
320 = W^2 + 4W
W^2 + 4W = 320
W^2 + 4W + 4 = 320 + 4
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W = -2 + 18 = 16 ft <== this is the width
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in summary...the width is 16 ft and the length is 20 ft 

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3 years ago
f(x) = 3 cos(x) 0 ≤ x ≤ 3π/4 evaluate the Riemann sum with n = 6, taking the sample points to be left endpoints. (Round your ans
Kruka [31]

Answer:

\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558

Step-by-step explanation:

We want to find the Riemann sum for \int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx with n = 6, using left endpoints.

The Left Riemann Sum uses the left endpoints of a sub-interval:

\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_0)+f(x_1)+2f(x_2)+...+f(x_{n-2})+f(x_{n-1})\right)

where \Delta{x}=\frac{b-a}{n}.

Step 1: Find \Delta{x}

We have that a=0, b=\frac{3\pi }{4}, n=6

Therefore, \Delta{x}=\frac{\frac{3 \pi}{4}-0}{6}=\frac{\pi}{8}

Step 2: Divide the interval \left[0,\frac{3 \pi}{4}\right] into n = 6 sub-intervals of length \Delta{x}=\frac{\pi}{8}

a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b

Step 3: Evaluate the function at the left endpoints

f\left(x_{0}\right)=f(a)=f\left(0\right)=3=3

f\left(x_{1}\right)=f\left(\frac{\pi}{8}\right)=3 \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}=2.77163859753386

f\left(x_{2}\right)=f\left(\frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{2}=2.12132034355964

f\left(x_{3}\right)=f\left(\frac{3 \pi}{8}\right)=3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=1.14805029709527

f\left(x_{4}\right)=f\left(\frac{\pi}{2}\right)=0=0

f\left(x_{5}\right)=f\left(\frac{5 \pi}{8}\right)=- 3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=-1.14805029709527

Step 4: Apply the Left Riemann Sum formula

\frac{\pi}{8}(3+2.77163859753386+2.12132034355964+1.14805029709527+0-1.14805029709527)=3.09955772805315

\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558

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3 years ago
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