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Feliz [49]
3 years ago
15

Which of the following statements are never true? Select all that apply. Three points are coplanar Two planes meet in exactly on

e point Opposite rays form a line Two lines that lie in a parallel planes are parallel Two lines meet at exactly two points
Mathematics
1 answer:
statuscvo [17]3 years ago
7 0

Answer: Two planes meet in exactly one point

Two lines meet at exactly two points


Step-by-step explanation:

From the given statements there are two statements which are never true :-

1) Two planes meet in exactly one point .

Since when two line meets , they either meet at one point or infinite points (coincidence) , thus its impossible that they will meet at exactly two points.

2) Two lines meet at exactly two points

Since when two planes meet , the intersection of two plane always make a line not a point. Thus its impossible.


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Anarel [89]
Unit rate is miles per 1 gallon
50mi/2 and 1/2 gallon
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times 10/10
500mi/25 gallon
20mi/1gallon

answer is 20mi/gal
6 0
3 years ago
A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so
Debora [2.8K]

Let A(t) = amount of salt (in pounds) in the tank at time t (in minutes). Then A(0) = 11.

Salt flows in at a rate

\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}

and flows out at a rate

\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}

which I'll solve with the integrating factor method.

\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95

-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}

\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}

Integrate both sides. By the fundamental theorem of calculus,

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)

\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}

\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}

\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}

After 1 hour = 60 minutes, the tank will contain

A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131

pounds of salt.

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2 years ago
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6 0
3 years ago
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Answer:

c) The car dealer's business is expanding rapidly.

Step-by-step explanation:

You can tell because they're selling more and more every month.

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What is the range of f(x) = 3x + 9?
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There is no restriction upon the domain, the x values, and is all real numbers or (-oo,+oo)

From such it is clear that the domain also contains all real numbers.

The range of f(x) is (-oo,+oo)
8 0
2 years ago
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