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Lostsunrise [7]
4 years ago
14

Clara earns $9 per

Mathematics
1 answer:
Monica [59]4 years ago
3 0
18$ for 2 /36$ for 4 /54$ for 6 /72$ for 8
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1/4 10/9 simplify the complex fraction shown
DaniilM [7]

Answer:

Step-by-step explanation:

\frac{1}{4}*\frac{10}{9}=\frac{1*5}{2*9}=\frac{5}{18}

7 0
3 years ago
Jackrabbits are capable of reaching speeds up to 40 miles per hour. How fast is this in feet per second? (Round to the nearest w
pishuonlain [190]

Answer:

59 feet per second

Step-by-step explanation:

  • (40m)(5280ft)=211,200 ft/hr
  • (211,200 ft/hr )=(211,200ft/3600s)
  • 58.7 ft/s
6 0
3 years ago
Read 2 more answers
CALCULUS HELP AGAIN PLS HELP ME GRADUATE
Lesechka [4]
1. Using the trapezoidal rule, you get

\displaystyle\int_0^8R(t)\,\mathrm dt\approx\dfrac{2.5+1.95}2(2-0)+\dfrac{2.8+2.5}2(3-2)+\dfrac{4+2.8}2(7-3)+\dfrac{4.26+4}2(8-7)
\displaystyle\int_0^8R(t)\,\mathrm dt\approx24.830\text{ gal}

R(t) gives the rate at which the water flows, so if you integrate with respect to time, you get the actual amount of water. In the summation above, the differences between successive t's form the heights of the trapezoids, while the successive values of R(t) form the "bases" of the trapezoids for which you take the averages.

2. Possibly... We know that R(t) is differentiable, so R'(t) certainly exists and must be continuous. The intermediate value theorem says that, on the interval [0,8], we can find some 0 such that R'(c) would fall between R(0) and R(8). But from the given data, we can't guarantee that R'(t) is ever 0, because both R(0)=1.95 and R(8)=4.26.

3. The average rate of water flow would be

R_\text{average}=\dfrac{W(8)-W(0)}{8-0}=\dfrac{\ln71-\ln7}8\approx0.290\text{ gal/hr}

4. g(x) is increasing whenever g'(x)>0. By the fundamental theorem of calculus,

g'(x)=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_{-2}^xf(t)\,\mathrm dt=f(x)

and f(x)>0 for -3 only.

5. For 0\le x\le6, we have f(x)=-x+3, so

g(x)=\displaystyle\int_{-2}^xf(t)\,\mathrm dt
g(x)=\displaystyle\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt
g(x)=9+\left(3t-\dfrac{t^2}2\right)\bigg|_{t=0}^{t=x}
g(x)=9+3x-\dfrac{x^2}2
5 0
3 years ago
What was the net change in the elevation of the ocean's surface from 1 p.m. to 7p.m.?​
emmasim [6.3K]

Answer:

4.6

Step-by-step explanation:

The net change is found by subtracting the starting number from the change. 2.2 minus -2.4 is 4.6. If this is wrong I apologise.

8 0
3 years ago
Please answer the picture
quester [9]
B is the correct answer
5 0
3 years ago
Read 2 more answers
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