Answer:
in the 1st row, one x^2 is positive while the other is negitive, making the total zero. also in the top row 8x+(-6x) = 2x, but in the 2nd row is says -2x, showing they are not the same. Also there is no -1/2 in the 2nd row like there is in the 1st row
Step-by-step explanation:
Answer:
19.) r = -4 20.) r = 1
Step-by-step explanation:
19.) Because we know the slope and have some points, we can use point-slope form to find the equation.
Point-slope form is:
y - y1 = m(x - x1)
Substitute the numbers into the equation:
y - 3 = 7/6(x - 1)
y - 3 = 7/6x -7/6
y = 7/6x + 11/6
Now that we know the equation, we can put in the x value to find the y (In this case, y = r):
r = 7/6(-5) + 11/6
r = -35/6 + 11/6
r = -4
20.) Because the slope is undefined, that means on the graph, the line will appear straight and vertical. This means the equation will be something like x = a number.
We have the point (1,4). Since we know that the x value will never change, we know that r will have to equal 1 as well.
r = 1
You can plot a linear equation, if you are given the coordinates and the slope for THE SAME line.
Assuming you have those, the first step is to find the y-intercept.
That is done by replacing 
and 
in the general formula by their coordinated counterparts, and replacing 
by the slope that is given to you.
Let's assume the coordinates you are given are 
, and the slope you are given is 1.
By substituting into the general formula...
...we can see that it is possible to find the y-intercept (
).
Knowing the y-intercept, it is possible to plot the line and solve any question.
I'll leave a picture of the example used above when graphed.
9514 1404 393
Answer:
the difference of 2 or 3 rectangles
Step-by-step explanation:
In every case, the "shaded" area can be computed by finding the area of a "bounding" rectangle, and subtracting the areas of the rectangular cutouts that give the figure its shape.
(a) The cutout is the white space at upper right. (Insufficient dimensions are given.)
(b) The cutout is the white space at lower left. The bounding rectangle is 8×7, and the cutout is 4×3.
(c) The cutout is the rectangle in the middle. The bounding rectangle is 13×7, and the cutout is 4×1.
(d) The cutouts are the rectangles on either side. They could be considered as a single unit. The bounding rectangle is 20×25; the cutouts have a total width of 16 and a height of 20, so total 16×20.
(e) Similar to (d), the cutouts are the white spaces on either side. The bounding rectangle is 14×12. The cutouts total 12 in width and 3 in height, so total 12×3.
__
You will note in (d) and (e) that the dimensions of the cutouts have something in common with the dimensions of the bounding rectangle. This means the problem can be simplified a little bit by factoring out that common factor. In (e), for example, 14×12 -(12×3) = 12(14 -3) = 12×11
