1. y=3/4x-3 (just from a triangle from one point on the line to anotheron the line where both end at a point.
2. y=5/1x which is also y=5x (basically subtract y^2-y^1/ x^2-x^1 ex. -2--7/5-4
3.y=2/3x-1/2
4.-3/-1 which is two negatives dividing which makes it positive. y=3x
5.cant see
6. Find slope first. y=3x then use point (1,1) move one left and 3 down. and the y intercept would be -2
Answer:
The cost is;
$18.43
Step-by-step explanation:
The general model for a linear equation is;
y = mx + b
let y be the monthly cost in dollars
x is the number of minutes
b will represent the base cost
m is the cost of the call per minute
so let us get m and b
Thus, we have;
18.25 = 31m + b
19.42 = 44m + b
Subtract i from ii
(19.42-18.25) = (44-31)m
1.17 = 13m
m = 1.17/13 = 0.09
To get b, we substitute the value of m in any of the equation
18.25 = 31 m + b
18.25 = 31(0.09) + b
b = 18.25 - 2.79
b = 15.46
so, the equation of the line is;
y = 0.09x + 15.46
Now to get the cost for 33 minutes
simply substitute the value of 33 for x
we have this as;
y = 0.09(33) + 15.46
y = $18.43
Answer:

Step-by-step explanation:
we know that
the formula to calculate the distance between two points is equal to

we have

step 1
Find the distance AB
we have

substitute in the formula



step 2
Find the distance BC
we have

substitute in the formula



step 3
Find the distance DE
we have

substitute in the formula




step 4
Find the distance AE
we have

substitute in the formula




step 5
Find out the area of the four walls
To determine the area sum the length sides and multiply by the height of the walls
so

substitute the given values
The height of the walls is 10 ft



step 6
Determine the number of gallons needed to paint the four walls
we know that
one gallon of paint is enough to cover 400 square feet
using proportion

Round up

Answer:
I hope your question is 4x²+3x+5, then the degree of this polynomial is 2