Answer:
see below
Step-by-step explanation:
y = 8x − 9
y = 4x − 1
Since both equations are equal to y, we can set them equal to each other
8x-9 = 4x-1
Subtract 4x from each side
8x-4x-9 = 4x-4x-1
4x-9 = -1
Add 9 to each side
4x-9+9 = -1+9
4x = 8
Divide each side by 4
4x/4 = 8/4
x=2
Now find y
y = 4x-1
y = 4(2)-1
y = 8-1
y = 7
(2,7)
The point of intersection when the 2 lines are graphed is (2,7)
Moooooooosssssssssseeeeeeeeseseses
Answer:
Tn = 17-6n
Step-by-step explanation:
d = 5-11 = -6
Tn = a + (n-1)d = 11 + (n-1)×-6
= 11 -6n +6 = 17-6n
Nine billion, nine hundred thousand, nine.
To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write
0.10
x
. This expression represents a variable cost because it changes according to the number of miles driven.
If a quantity is independent of a variable, we usually just add or subtract it according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost
C
.
C
=
0.10
x
+
50
When dealing with real-world applications, there are certain expressions that we can translate directly into math. The table lists some common verbal expressions and their equivalent mathematical expressions.
