Answer: IX - 4I ≤ 4
Step-by-step explanation:
In the numer line we can see that our possible values of x are in the range:
0 ≤ x ≤ 8
And we want to find an absolute value equation such that this set is the set of possible solutions.
An example can be:
IX - 4I ≤ 4
To construct this, we first find the midpoint M of our set, in this case is 4.
Then we write:
Ix - MI ≤ IMI
Notice that i am using the minor and equal sign, this is because the black dots means that the values x = 0 and x = 8 are included, if the dots were empty dots, it would be an open set and we should use the < > signs.
Basically this app helps you with any question you need you got to ask a question and people will answer it for you just like i am doing now also you can answer other peoples questions to get points.
2x+2y=38
y=x+3
First you would substitute the y for x+3
2x+2(x+3)=38
You would multiply the 2 by the x+3
2x+2x+6=38
Add the 2x and the 2x
4x+6=38
Subtract the 6 on both sides
4x=32
Divide the 4 on both sides
x=8
Now substitute the x in the second problum for 8
y=8+3
Add the 8 and the 3 to solve for y
y=11
Your answers
x=8
y=11
the correct answer would be A:(8,11)
To solve this, you need to isolate/get the variable "m" by itself in the equation:
1. 2m - 1 = 3m Subtract 2m on both sides to get "m" on one side of the equation
2m - 2m - 1 = 3m - 2m
-1 = m
2. 2m = 1 + m Subtract m on both sides to get "m" on one side of the equation
2m - m = 1 + m - m
m = 1
3. m - 1 = 2 Add 1 on both sides to get "m" by itself
m - 1 + 1 = 2 + 1
m = 3
4. 2 + m = 3 Subtract 2 on both sides to get "m" by itself
m = 1
5. -2 + m = 1 Add 2 on both sides to get "m" by itself
m = 3
6. 3 = 1 + m Subtract 1 on both sides to get "m" by itself
2 = m
Answer:
The standard form of the equation for the conic section represented by
is:
![4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2](https://tex.z-dn.net/?f=4%5Cleft%28-%5Cfrac%7B3%7D%7B2%7D%5Cright%29%5Cleft%28y-12%5Cright%29%3D%5Cleft%28x-%5Cleft%28-5%5Cright%29%5Cright%29%5E2)
Step-by-step explanation:
We know that:
is the standard equation for an up-down facing Parabola with vertex at (h, k), and focal length |p|.
Given the equation
![x^2\:+\:10x\:+\:6y\:=\:47](https://tex.z-dn.net/?f=x%5E2%5C%3A%2B%5C%3A10x%5C%3A%2B%5C%3A6y%5C%3A%3D%5C%3A47)
Rewriting the equation in the standard form
![4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2](https://tex.z-dn.net/?f=4%5Cleft%28-%5Cfrac%7B3%7D%7B2%7D%5Cright%29%5Cleft%28y-12%5Cright%29%3D%5Cleft%28x-%5Cleft%28-5%5Cright%29%5Cright%29%5E2)
Thus,
The vertex (h, k) = (-5, 12)
Please also check the attached graph.
Therefore, the standard form of the equation for the conic section represented by
is:
![4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2](https://tex.z-dn.net/?f=4%5Cleft%28-%5Cfrac%7B3%7D%7B2%7D%5Cright%29%5Cleft%28y-12%5Cright%29%3D%5Cleft%28x-%5Cleft%28-5%5Cright%29%5Cright%29%5E2)
where
vertex (h, k) = (-5, 12)