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Mathematics

1 answer:

Shtirlitz [24]2 years ago

8 0

The coordinates of the pre-image of point F' is (-2, 4)

<h3>How to determine the coordinates of the pre-image of point F'?</h3>

On the given graph, the location of point F' is given as:

F' = (4, -2)

The rule of reflection is given as

Reflection across line y = x

Mathematically, this is represented as

(x, y) = (y, x)

So, we have

F = (-2, 4)

Hence, the coordinates of the pre-image of point F' is (-2, 4)

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VWXY is a parallelogram. Find x.

natulia [17]

Answer: 10

Step-by-step explanation: 6 divided by 120 = 60 ÷ 6 = 10

5 0

3 years ago

A man starts walking north at 5 ft/s from a point P. Five minutes later a woman starts walking south at 6 ft/s from a point 500

aev [14]

Check the picture below.

the man is going North, and he has a rate of 5 f/s, she's going south, 500 miles from P, the green distance, and she's going 6 f/s.

by the time she started walking, he already had been walking for 5 minutes, or 300 seconds, since he's going 5 f/s, then he had already covered 1500 feet by then.

after that, they both walked for 15 minutes more or 900 seconds, and for her that means 5400 feet, whilst for him going slower means 4500 feet, as you see in the picture.

since he's going North and she's going South, dy/dt is the combined rates, as you see there.

now, let's use the pythagorean theorem, bearing in mind that the green 500 feet are constant all the while, that matters because the derivative of a constant is 0.

$\bf r^2=x^2+y^2\stackrel{chain~rule}{2r\cfrac{dr}{dt}}=0+\stackrel{chain~rule}{2y\cfrac{dy}{dt}}\implies \cfrac{dr}{dt}=\cfrac{2y}{2r}\cdot \cfrac{dy}{dt}
\\\\\\
\cfrac{dr}{dt}=\cfrac{y}{r}\cdot \cfrac{dy}{dt}$

so, what is "r" value 15 minutes later anyway?

$\bf r=\sqrt{(1500+4500+5400)^2+500^2}\implies r=\sqrt{11400^2+500^2}
\\\\\\
r=\sqrt{130210000}\implies r=100\sqrt{13021}$

then we just plug those folks in the derivative,

$\bf \cfrac{dr}{dt}=\cfrac{y}{r}\cdot \cfrac{dy}{dt}\qquad 
\begin{cases}
r=100\sqrt{13021}\\
\frac{dy}{dt}=11\\
y=11400
\end{cases}\implies \cfrac{dr}{dt}=\cfrac{11400}{100\sqrt{13021}}\cdot 11
\\\\\\
 \cfrac{dr}{dt}=\cfrac{114\cdot 11}{\sqrt{13021}}\implies \cfrac{dr}{dt}=\cfrac{1254}{\sqrt{13021}}\implies \cfrac{dr}{dt}\approx 10.99~\frac{f}{s}$