Answer:
a. t
b. a (or x = a)
c. r
d.
1) c
2) t
3) a
4) p
Step-by-step explanation:
a. Draw vertical line passing through the point (c,0). This line intersects the graph at point L. Point L has coordinates (c,t), so
b. If 
draw the horizontal line passing through the point (0,p). This line intersects the graph at point K with coordinates (a,p), so 
c. Note that 
then
d. Coordinates of point L are (c,t), coordinates of point K are (a,p)
Answer:
y=(3/2)x+-14
First blank: 3
Second blank:2
Last blank:-14
Step-by-step explanation:
The line form being requested is slope-intercept form, y=mx+b where m is slope and b is y-intercept.
Also perpendicular lines have opposite reciprocal slopes so the slope of the line we are looking for is the opposite reciprocal of -2/3 which is 3/2.
So the equation so far is
y=(3/2)x+b.
We know this line goes through (x,y)=(4,-8).
So we can use this point along with our equation to find b.
-8=(3/2)4+b
-8=6+b
-14=b
The line is y=(3/2)x-14.
wheat = $0.96 lb rye = $1.89 lb
12w + 15r = 3987 15w + 10r = 3330 r = (3330 - 15w)/10 12w + 15r = 3987 12w + 15((3330 - 15w)/10) = 3987 12w + (15/10)(3330 - 15w) = 3987 12w + 4995 - 22.5w = 3987 10.5w = 1008 w = 96 cents r = 189 cents.
Hope this helps mate
Hi there! :)
Answer:
Length = 10.5 m, width = 7 m.
Step-by-step explanation:
Given:
Perimeter, or P = 35 m
Ratio of l to w = 3 : 2
Since the ratio is 3 : 2, let l = 3x, and w = 2x.
We know that the formula for the perimeter of a rectangle is P = 2l + 2w. Therefore:
35 = 2(3x) + 2(2x)
35 = 6x + 4x
35 = 10x
x = 3.5
Plug this value of "x" into each expression to solve for the dimensions:
2(3.5) = w
w = 7 m
3(3.5) = l
l = 10.5 m
Therefore, the dimensions are:
Length = 10.5 m, width = 7 m.
I think you meant to say
(as opposed to <em>x</em> approaching 2)
Since both the numerator and denominator are continuous at <em>t</em> = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:
Because these expressions are continuous at <em>t</em> = 2, we can compute the limits by evaluating the limands directly at 2:
