Answer:252in. squared
Step-by-step explanation:
16+2=18
12+2=14
14×18=252 ^2
look ar the photo, x is 4
Given:
Slope = -3/5
y - intercept = (0 , 5)
The answer is y = -3/5 + 5
Note:
The x = 0 and the y = 5 . We are looking for y- intercept so we use 5 instead of 0.
Since the formula for slope intercept form is y = mx + b, we use option number 4 as our equation. for option number 1, it did not include the negative sign that was given in the slope so we opt to choose option 1.
Answer: (4, -9)
<u>Step-by-step explanation:</u>
Use elimination method. Manipulate one (or both) equations to eliminate one of the variables and solve for the remaining variable. <em>I will be eliminating y</em>
6x + y = 15 → 2(6x + y = 15) → 12x + 2y = 30
-7x - 2y = -10 → 1(-7x + 2y = -10) → <u> -7x - 2y = -10</u>
5x = 20
x = 4
Next, replace "x" with "4" into either equation and solve for y.
6(4) + y = 15
24 + y = 15
y = -9
<u>Check:</u>
Plug in x = 4 and y = -9 into the other equation to verify it makes a true statement.
-7x - 2y = -10
-7(4) - 2(-9) = -10
-28 - -18 = -10
-28 + 18 = -10
-10 = -10 
Answer:
I guess that you want to model the elevation of Lake Sam Rayburn.
During the summer, it is 165 ft above the sea level (the sea level is our position 0ft).
If it does not rain, the elevation of the lake decreases by 0.5ft each week.
So if we assume that there is no rain, we can write the elevation fo the lake as a linear relationship with slope equal to -0.5ft and y-intercept equal to 165ft.
L(w) = 165ft - 0.5ft*w
Where w is the number of weeks without rain, if we have 0 weeks without rain, then the level of the lake remains constant at 165ft above sea level,
L(0) = 165ft - 0.
