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kakasveta [241]
3 years ago
10

The Cunninghams are moving across the country. Mr. Cunningham leaves 3 hours before Mrs. Cunningham. If he averages 40mph and sh

e averages 80mph, how long will it take Mrs. Cunningham to overtake Mr. Cunningham?
Mathematics
1 answer:
svp [43]3 years ago
5 0
Recall your d = rt, distance = rate * time

now, if say, by the time they meet, Mr Cunningham has travelled "d" miles, that means Mrs Cunningham must also had travelled "d" miles as well.

However, he left 3 hours earlier, so by the time he travelled "d" miles, and took say "t" hours, for her it took 3 hour less, because she started driving 3 hours later, so, she's been on the road 3 hours less than Mr Cunningham, so by the time they meet, Mrs Cunningham has travelled then "t - 3" hours.

\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
\textit{Mr Cunningham}&d&40&t\\
\textit{Mrs Cunningham}&d&80&t-3
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=40t\\
d=80(t-3)\\
----------\\
\boxed{40t}=80(t-3)
\end{cases}
\\\\\\
\cfrac{40t}{80}=t-3\implies \cfrac{t}{2}=t-3\implies t=2t-6\implies 6=2t-t
\\\\\\
6=t
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2 years ago
A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The​ plate, including the bound
rjkz [21]

Answer:

We have the coldest value of temperature T(\frac{3}{4},0) = -9/16. and the hottest value is T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}.

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x

Let's take the partials derivatives.

T_{x}(x,y)=2x-\frac{3}{2}=0

T_{y}(x,y)=4y=0

So, we can find the critical point (x,y) of T(x,y).

2x-\frac{3}{2}=0

x=\frac{3}{4}

4y=0

y=0

The critical point is (3/4,0) so the temperature at this point is: T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})

T(\frac{3}{4},0)=-\frac{9}{16}    

Now, we need to evaluate the boundary condition.

x^{2}+y^{2}=1

We can solve this equation for y and evaluate this value in the temperature.

y=\pm \sqrt{1-x^{2}}

T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x  

T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2

Now, let's find the critical point again, as we did above.

T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0            

x=-\frac{3}{4}    

Evaluating T(x,y) at this point, we have:

T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2  

T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}

Now, we can see that at point (3/4,0) we have the coldest value of temperature T(\frac{3}{4},0) = -9/16. On the other hand, at the point -(3/4),\frac{\sqrt{7}}{4}) we have the hottest value of temperature, it is T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}.

I hope it helps you!

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The angle of depression from the top of a flag pole to a point on the ground is 30°. If the point on the ground is 75 feet from
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Answer:

Height of flag pole to the nearest foot is 43 feet

Step-by-step explanation:

In the diagram we have flag pole AB and C is the point on the ground.

It is given that angle of depression is 30°

and the distance of the point C from the base B is 75 feet

so we have

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now in right ΔABC

tan C = \frac{opposite \ side }{base} \\tan C=\frac{AB}{BC}

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tan 30= \frac{h}{75}

75 tan30 = h            ( multiply both sides by 75)

h= 75 tan 30 =75 (0.5773)\\ h= 43.3 feet\\

Height of flag pole to the nearest foot is

h= 43 feet

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