Answer:
$1000 ; 500 ; 1000 ; y = 500x + 1000
Step-by-step explanation:
From the graph, the initial deposit which started the savings account is $1000 ; this is the value on the y - intercept, the value in the account at time or period = 0.
The slope :
x1 = 0 ; y1 = 1000
x2 = 6 ; y2 = 4000
Slope = Rise / Run
Rise = y2 - y1 = 4000 - 1000 = 3000
Run = x2 - x1 = 6 - 0 = 6
Hence,
Slope = 3000 / 6
Slope = 500
y - intercept = 1000 (from. Graph)
Equation in slope intercept form:
General form : y = mx + c
m = slope ; c = intercept
Equation is written as ;
y = 500x + 1000
C = 3b+2d is the same as 3b+2d = C
Let's isolate d. To do this, we first need to subtract 3b from both sides
3b+2d = C
3b+2d-3b = C-3b
2d = C-3b
Then divide both sides by 2
2d = C-3b
2d/2 = (C-3b)/2
d = (C-3b)/2
Take note of the parenthesis as they are very important. We want to divide ALL of C-3b over 2. We don't want to just divide -3b over 2.
The answer choices you have aren't 100% clear but I have a feeling your teacher meant to say d = (C-3b)/2 instead of d = C-3b/2 for choice A
If that assumption is correct, then the answer is choice A.
Answer:
The answer is 1
Step-by-step explanation:
-50+50=1
First of all, when I do all the math on this, I get the coordinates for the max point to be (1/3, 14/27). But anyway, we need to find the derivative to see where those values fall in a table of intervals where the function is increasing or decreasing. The first derivative of the function is
. Set the derivative equal to 0 and factor to find the critical numbers.
, so x = -3 and x = 1/3. We set up a table of intervals using those critical numbers, test a value within each interval, and the resulting sign, positive or negative, tells us where the function is increasing or decreasing. From there we will look at our points to determine which fall into the "decreasing" category. Our intervals will be -∞<x<-3, -3<x<1/3, 1/3<x<∞. In the first interval test -4. f'(-4)=-13; therefore, the function is decreasing on this interval. In the second interval test 0. f'(0)=3; therefore, the function is increasing on this interval. In the third interval test 1. f'(1)=-8; therefore, the function is decreasing on this interval. In order to determine where our points in question fall, look to the x value. The ones that fall into the "decreasing" category are (2, -18), (1, -2), and (-4, -12). The point (-3, -18) is already a min value.