<u>Answer:</u>
The equation of a line, in point slope form that passes through (-2, -6) and has a slope of 1/3 is 
<u>Solution:</u>
Given that line passes through (-2, -6) and has slope of 1/3
We have to find the equation of the line
The point slope form is given as
where m is the slope of the line and a, b are the x, y coordinates of the given point through which the line passes.
Here in this question, m = 1/3 and a = -2 and b = -6
By substituting in point slope form we get,
Hence the equation of a line, in point slope form that passes through (-2, -6) and has a slope of 1/3 is
Answer:
x = 12
y = 22
Total amount = $2596
Step-by-step explanation:
First let's find the value of each deposit until the 10th in relation to x and y:
1st: x
2nd: y
3rd: x + y
4th: x + 2y
5th: 2x + 3y
6th: 3x + 5y
7th: 5x + 8y
8th: 8x + 13y
9th: 13x + 21y
10th: 21x + 34y
Now, we can write a system with two equations and two variables:
2x + 3y = 90
21x + 34y = 1000
From the first equation: x = (90 - 3y)/2
Using this value of x in the second equation, we have:
21*(90 - 3y)/2 + 34y = 1000
945 - 31.5y + 34y = 1000
2.5y = 55
y = 22
Now we can find x:
x = (90 - 3*22)/2 = 12
Now, summing all the deposits, we have a total of 55x + 88y, which is equal to 55*12 + 88*22 = $2596
Answer:
b
Step-by-step explanation:
centroid
x=(x1+x2+x3)/3=(0+3+0)/3=3/3=1
y=(y1+y2+y3)/3=(0+0-3)/3=-3/3=-1
centroid =(1,-1)
Answer:
g(-4) = -1
g(-1) = -1
g(1) = 3
Explanation:
If you are given a function that is defined by a system of equations associated with certain intervals of x, just find which interval makes x true, and then substitute x into the equation of that interval.
For example, given g(-4), this is an expression which is asking for the value of the equation when x = -4. So -4 is not ≥ 2, so ¼x - 1 will not be used. -4 is also not ≤ -1 and ≤ 2, so -(x - 1)² + 3 will not be used either. So in turn, we will just use -1 which is always -1 so g(-4) will just be -1, right because there is no x variable in -1 so it will always be the same.
Using the same idea as before g(-1) is g(x) when x = -1 so -1 will not be a solution because -1 is not less than -1 (< -1). -1 is not ≥ 2 either so we will be using the second equation because -1 is part of the interval -1≤x≤2 (it is a solution to this inequality), therefore -(x - 1)² + 3 will be used.
As x = -1, -(x - 1)² + 3 = -(-1 - 1)² + 3 = -(-2)² + 3 = -4 + 3 = -1.
It is a coincidence that g(-1) = -1.
Now for g(1), where g(x) has an input of 1 or the value of the function where x = 1, we will not use the first equation because x = 1 → x < -1 → 1 < -1 [this is false because 1 is never less than -1], so we will not use -1.
We will use -(x - 1)² + 3 again because 1 is not ≥ 2, 1≥2 [this is also false]. And -1 ≤ 1 < 2 [This is a true statement]. Therefore g(1) = -(1 - 1)² + 3 = -(0)² + 3 = 3
