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pshichka [43]
3 years ago
13

Mr. Lyle made a round trip between two cities that were 450 kilometers apart. On the return trip, he increased his average of ra

te travel by 5 km/h and made the trip in one hour less. Find his rate for each direction.
Mathematics
1 answer:
GREYUIT [131]3 years ago
7 0

Answer:

45 km/h

50 km/h

Step-by-step explanation:

Distance traveled in one way = 450 km

Let x = Velocity when he was going to his destination in km/h

He traveled 5 km/h faster when he returned so his velocity was x+5

It took him 1 hour less when he returned so

\dfrac{450}{x}-\dfrac{450}{x+5}=1\\\Rightarrow x^2+5x-2250=0\\\Rightarrow x=\frac{-5\pm \sqrt{5^2-4\times 1\times \left(-2250\right)}}{2\times 1}\\\Rightarrow x=45,-50

So, velocity while going to his destination is 45 km/h

While returning his velocity was 45+5 = 50 km/h.

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stellarik [79]

Answer:

2062.79 inches cubed

Step-by-step explanation:

Formula- πr2h

=  π×6.82×14.2

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7 0
3 years ago
100 POINTS! :)
nasty-shy [4]

Answer: So we'll use trig to solve this because it's a right triangle. The hypotenuse is the ladder (h) and the two smaller sides are the ground and the vertical wall (w).

That angle ladder makes with the ground = €

SOH: sin € = opposite/hypotenuse

No the ladder won't be safe!!!

Now let's make it safe:

The ladder's length (w) is constant, so stays 12

So now let's ask in an inequality what height will be safe (75° or less)

what does that mean?? well as long as you put the ladder against the wall so that the height from ground to top of the ladder is < 11.6 ft!!

Hope that helps! :-D

Step-by-step explanation:

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3 years ago
t^2+4t+4 t^2-2t+1 Consider the parametric curve a. Find points on the parametric curve where the tangent lines are vertical. b.
kondaur [170]

Answer:

a) t = 1, b)  t = -2, c) Concave upward: (-\infty, 1). No intervals being concave downward.

Step-by-step explanation:

a) Tangent line is vertical if slope is undefined, whose points are associated with discontinuities of the function. Rational functions have discontinuities associated with the denominator:

f(t) = \frac{t^{2}+4\cdot t +4}{t^{2}-2\cdot t + 1}

By factorizing each component, the function is re-arranged as:

f(t) = \frac{(t+2)^{2}}{(t-1)^{2}}

There is no avoidable discontinuities. The only point where tangent line is vertical is t = 1.

b) Tangent line is horizontal if slope is equal to zero, whose point is associated to the points that makes function equal to zero. Rational functions have horizontal tangent lines if numerator is equal to zero and denominator is different to zero. Hence, the only point where tangent line is  horizontal is t = -2.

c) An interval is concave upward if exist an absolute minimum inside, which can be found by the help of the First and Second Derivative Tests.

f'(t) = \frac{2\cdot (t+2)\cdot (t-1)^{2}-2\cdot (t-1)\cdot (t+2)^{2}}{(t-1)^{4}}

f'(t) = 2\cdot (t+2)\cdot (t-1)\cdot \left[\frac{t-1-t-2}{(t-1)^{4}}\right]

f'(t) = -\frac{6\cdot (t+2)\cdot (t-1)}{(t-1)^{4}}

f'(t) = -\frac{6\cdot (t+2)}{(t-1)^{3}}

The only critical point is:

-6\cdot (t+2) = 0

t = -2 (which coincides with the result of point b)

f''(t) = -6\cdot \left[\frac{(t-1)^{3}-3\cdot (t-1)^{2}\cdot (t+2)}{(t-1)^{6}}\right]

f''(t) = -6\cdot \left[\frac{1}{(t-1)^{3}}-\frac{3\cdot (t+2)}{(t-1)^{4}}  \right]

The value associated with the critical point is:

f''(-2) = \frac{2}{9}

Which means that critical point is an absolute minimum, and, consequently, the interval that is concave upward is (-\infty, 1). There is no absolute maximums and, therefore, there is no interval that is concave downward.

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Anon25 [30]
The mean of the combined group is 71.3 --- hope this helps
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Answer:

Less than

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