Step-by-step explanation:
10 red 6 blue so 16 altogether.
apologies if I'm wrong
Answer:
The answer is A.
Step-by-step explanation:
So we have the two equations:
![3x+y=1\\y+4=5x](https://tex.z-dn.net/?f=3x%2By%3D1%5C%5Cy%2B4%3D5x)
To make a substitution of the second equation into the first equation, we need to isolate the <em>y </em>variable in the second equation. Thus:
![y+4=5x\\y=5x-4](https://tex.z-dn.net/?f=y%2B4%3D5x%5C%5Cy%3D5x-4)
Now, we can substitute this into the first equation. Therefore:
![3x+y=1\\3x+(5x-4)=1\\3x+5x-4=1](https://tex.z-dn.net/?f=3x%2By%3D1%5C%5C3x%2B%285x-4%29%3D1%5C%5C3x%2B5x-4%3D1)
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)
Does it show the figures it’s talking about?
Long way: try to divide. But we are lazy, so it's no go. Option number one: all rational numbers that could give you a division can be taken from "THE List": you build it by taking the constant term, that is 6, and all its divisors (1,2,3,6), the highest term coefficent, 3 in this case, its divisors (1,3) , and you divide every one in the first list by the second, and adding a
in front of them. Our list is:
.
Since -3 is part of THE List, it might be a divisor - that system would garantee you that, for example, x-4 would NOT be a factor.
Cheap way. if (x+3) is a factor of g(x), then g(-3) = 0. Let's replace.
![3(-3)^3+ 2(3)^2-17(3)+6= \\ 3(-27) +3(18)-17(-3)+6 = -81+54+51+6=-30](https://tex.z-dn.net/?f=%203%28-3%29%5E3%2B%202%283%29%5E2-17%283%29%2B6%3D%20%5C%5C%203%28-27%29%20%2B3%2818%29-17%28-3%29%2B6%20%3D%20-81%2B54%2B51%2B6%3D-30)
Since we didn't get zero,
is not a factor of g(x)