A). To go from ABCD to A'B'C'D' takes two transformations.
Step-1 : Reflect the original figure around the x-axis.
Step-2 : Slide the new figure 7 units to the right.
See the attached drawing. It shows the whole process
including the intermediate status and coordinates of
the four points after Step-1. You'll love it.
B). I looked up a definition of 'congruent'. It said that two figures
are congruent if they have have the same shape and size after
being moved, rotated,
or reflected. (I was worried about the
'reflected' part.)
Answer:
a) OA = 1 unit
b) BC = 3 units
c) OD = 2 units
d) AC = 3√2 units
Step-by-step explanation:
Given function:
![f(x)=\dfrac{2}{x}-2](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7B2%7D%7Bx%7D-2)
<h3><u>Part (a)</u></h3>
Point A is the x-intercept of the curve.
To find the <u>x-intercept</u> of the curve (when y = 0), set the function to zero and solve for x:
![\begin{aligned}f(x) & = 0\\\implies \dfrac{2}{x}-2 & = 0\\\dfrac{2}{x} & = 2\\2 & = 2x\\\implies x & = 1\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%28x%29%20%26%20%3D%200%5C%5C%5Cimplies%20%5Cdfrac%7B2%7D%7Bx%7D-2%20%26%20%3D%200%5C%5C%5Cdfrac%7B2%7D%7Bx%7D%20%26%20%3D%202%5C%5C2%20%26%20%3D%202x%5C%5C%5Cimplies%20x%20%26%20%3D%201%5Cend%7Baligned%7D)
Therefore, A (1, 0) and so OA = 1 unit.
<h3><u>Part (b)</u></h3>
If OB = 2 units then B (-2, 0). Therefore, the x-value of Point C is x = -2.
To find the y-value of Point C, substitute x = -2 into the function:
![\implies f(-2)=\dfrac{2}{-2}-2=-3](https://tex.z-dn.net/?f=%5Cimplies%20f%28-2%29%3D%5Cdfrac%7B2%7D%7B-2%7D-2%3D-3)
Therefore, C (-2, -3) and so BC = 3 units.
<h3><u>Part (c)</u></h3>
<u>Asymptote</u>: a line that the curve gets infinitely close to, but never touches.
The y-value of Point D is the horizontal asymptote of the function.
The function is undefined when x = 0 and therefore when y = -2.
Therefore, D (0, -2) and so OD = 2 units.
<h3><u>Part (d)</u></h3>
From parts (a) and (c):
To find the length of AC, use the distance between two points formula:
![d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
![\textsf{where }(x_1,y_1) \textsf{ and }(x_2,y_2)\:\textsf{are the two points.}](https://tex.z-dn.net/?f=%5Ctextsf%7Bwhere%20%7D%28x_1%2Cy_1%29%20%5Ctextsf%7B%20and%20%7D%28x_2%2Cy_2%29%5C%3A%5Ctextsf%7Bare%20the%20two%20points.%7D)
Therefore:
![\sf \implies AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20AC%3D%5Csqrt%7B%28x_C-x_A%29%5E2%2B%28y_C-y_A%29%5E2%7D)
![\sf \implies AC=\sqrt{(-2-1)^2+(-3-0)^2}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20AC%3D%5Csqrt%7B%28-2-1%29%5E2%2B%28-3-0%29%5E2%7D)
![\sf \implies AC=\sqrt{(-3)^2+(-3)^2}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20AC%3D%5Csqrt%7B%28-3%29%5E2%2B%28-3%29%5E2%7D)
![\sf \implies AC=\sqrt{9+9}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20AC%3D%5Csqrt%7B9%2B9%7D)
![\sf \implies AC=\sqrt{18}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20AC%3D%5Csqrt%7B18%7D)
![\sf \implies AC=\sqrt{9 \cdot 2}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20AC%3D%5Csqrt%7B9%20%5Ccdot%202%7D)
![\sf \implies AC=\sqrt{9}\sqrt{2}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20AC%3D%5Csqrt%7B9%7D%5Csqrt%7B2%7D)
![\sf \implies AC=3\sqrt{2}\:\:units](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20AC%3D3%5Csqrt%7B2%7D%5C%3A%5C%3Aunits)
The answer is the third one aka C.
There must not be two points on the same y-axis because that doesn't make a function but a relation instead.
C. doesn't have 2 points on same y-axis and therefore the third picture is the relation that's a function.
It takes 107.692 minutes approximately to fill the pond
<em><u>Solution:</u></em>
Given that inlet pipe can fill an artificial pond in 70 min
So the inlet pipe can fill
of the pond in 1 min
An outlet pipe can empty the same pond in 200 min
So the outlet pipe can empty
of pond in 1 min
Here negative sign denotes "empties the pond"
To find: With both pipe open, time taken to fill the pond
Let "x" be the time taken by pipes to fill the pond
With both pipes are left on, so together their rate is:
![\frac{1}{x} = \frac{1}{70} - \frac{1}{200}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%7D%20%3D%20%5Cfrac%7B1%7D%7B70%7D%20-%20%5Cfrac%7B1%7D%7B200%7D)
On solving,
![\frac{1}{x} = \frac{200 - 70}{14000}\\\\\frac{1}{x} = \frac{130}{14000}\\\\x = \frac{14000}{130} = 107.692](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%7D%20%3D%20%5Cfrac%7B200%20-%2070%7D%7B14000%7D%5C%5C%5C%5C%5Cfrac%7B1%7D%7Bx%7D%20%3D%20%5Cfrac%7B130%7D%7B14000%7D%5C%5C%5C%5Cx%20%3D%20%5Cfrac%7B14000%7D%7B130%7D%20%3D%20107.692)
Thus it takes 107.692 minutes approximately to fill the pond