Answer:
the horizontal position for the left edge of the image is -4.46 cm.
Step-by-step explanation:
The horizontal midpoint of the page is:
21.59 / 2 = 10.79 cm
The horizontal midpoint of the image is given as:
30.51 / 2 = 15.25 cm
These midpoints must be on the same point in the axis.
By taking the leftmost edge of the paper to be point zero on the axis, then, the distance accommodated by the paper is 10.795 cm.
The distance that goes beyond this leftmost edge is computed as;
This is on the negative side on the axis.
Thus the horizontal position for the left edge of the image is -4.46 cm.
Amelio have 18 Nickels and 12 dimes!!
2.10−(6×0.05)−(12×0.10)−(12×0.05)= 0
Hello!
I've attached a photo for reference.
Lines A and B form straight angles, which measure 180 degrees. That means that -
m∠x + m∠y = 180°
m∠y + m∠z = 180°
m∠z + 43° = 180°
43° + m∠x = 180°
Since you're trying to find z, use the solvable equation with z in it:
m∠z + 43° = 180°
180 = z + 43
137 = z
Answer:
m∠z = 137°
This problem can be readily solved if we are familiar with the point-slope form of straight lines:
y-y0=m(x-x0) ...................................(1)
where
m=slope of line
(x0,y0) is a point through which the line passes.
We know that the line passes through A(3,-6), B(1,2)
All options have a slope of -4, so that should not be a problem. In fact, if we check the slope=(yb-ya)/(xb-xa), we do find that the slope m=-4.
So we can check which line passes through which point:
a. y+6=-4(x-3)
Rearrange to the form of equation (1) above,
y-(-6)=-4(x-3) means that line passes through A(3,-6) => ok
b. y-1=-4(x-2) means line passes through (2,1), which is neither A nor B
****** this equation is not the line passing through A & B *****
c. y=-4x+6 subtract 2 from both sides (to make the y-coordinate 2)
y-2 = -4x+4, rearrange
y-2 = -4(x-1)
which means that it passes through B(1,2), so ok
d. y-2=-4(x-1)
this is the same as the previous equation, so it passes through B(1,2),
this equation is ok.
Answer: the equation y-1=-4(x-2) does NOT pass through both A and B.
