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mel-nik [20]
3 years ago
7

An oil refinery can refine about 20 gallons of gasoline from one barrel of crude oil. If an automobile has a fuel economy of 25

miles per gallon, calculate the maximum number of miles an automobile could theoretically travel using the quantity of gasoline that could have been refined from the crude oil that escaped from the well in May of 2010.
Advanced Placement (AP)
1 answer:
Maurinko [17]3 years ago
5 0

Answer:

20(1,800,000)/25= 1,440,00

Explanation:

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Airborne chemical substances travel up the nose to the olfactory_____, which is the receptor got smell
Vlada [557]

Answer:

c) bulb

Explanation:

Airborne chemical substances travel up the nose to the olfactory <u>bulb </u>, which is the receptor got smell

3 0
3 years ago
Regina struggles with reading long passages and feels like no matter how hard she tries she will not understand them, so she ski
tresset_1 [31]

Answer:

uh

Explanation:

6 0
2 years ago
If you were a teacher in such a situation how could you apply the principles of the bank's treat approach to make the experience
vitfil [10]
THE ANSWER IS BECAUSE OF BANKS TREATS APPROACH AND examples being used such as
8 0
2 years ago
Use midpoints to approxiamte the area under the curve y=f(x) 5 sin(xx)!+ 2.5 cos(4xx) on the interval [0,1] using 10 equal subdi
alisha [4.7K]

Subdividing [0, 1] into 10 equally spaced intervals of length \Delta x=\frac{1-0}{10}=\frac1{10} gives the partition

[0,1] = \left[0,\dfrac1{10}\right] \cup \left[\dfrac1{10},\dfrac2{10}\right] \cup \cdots \cup \left[\dfrac9{10},1\right]

The i-th subinterval has left and right endpoints, respectively, given by

\ell_i = \dfrac{i-1}{10} \text{ and } r_i = \dfrac i{10}

where i\in\{1,2,3,\ldots,10\}.

The midpoint of the i-th interval is the average of these,

m_i = \dfrac{\ell_i+r_i}2 = \dfrac{2i-1}{20} \in \left\{\dfrac1{20},\dfrac3{20},\dfrac5{20},\ldots,\dfrac{19}{20}\right\}

We approximate the area under f(x) over [0, 1] by the Riemann sum,

\displaystyle \int_0^1 f(x) \, dx \approx \sum_{i=1}^{10} f(m_i) \Delta x \\\\ ~~~~~~~~ = \frac1{10} \sum_{i=1}^{10} \bigg(5\sin(\pi m_i) + 2.5 \cos(4\pi m_i)\bigg) \\\\ ~~~~~~~~ = \frac{\sin\left(\frac\pi{20}\right) + \sin\left(\frac{3\pi}{20}\right) + \cdots + \sin\left(\frac{19\pi}{20}\right)}2 \\\\ ~~~~~~~~~~~~ + \dfrac{\cos\left(\frac\pi5\right) + \cos\left(\frac{3\pi}5\right) + \cdots + \cos\left(\frac{19\pi}5\right)}4 \\\\ ~~~~~~~~ \approx 3.19623

(D)

6 0
2 years ago
What is the missing area?<br>​
nalin [4]

Answer:

3

Explanation:

if correctly visualising that rectangle is equivalent to the rectangle missing on the upper left side, also half of 7 is 3 1/2. so 3 kinda of fits. im sorry if this is wrong.

3 0
2 years ago
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