The choices are found elsewhere and the figures would be:
a. rectangleb. trianglec. squared. trapezoid
From the choices, the answer would be option B. A triangle figure would be the cross section when <span>a rectangular pyramid that has been intersected by a plane perpendicular to its base and through</span>its vertex.
The easiest way is to graph it based upon the slope (m) and y-intercept (b), in the standard slope-intercept form: y = m (x) + b.
The line above intercepts the y-axis at y = -2, which is b. The slope (m) = rise/run = (y2-y1)/(x2-x1 ); so for the point (-4, 2) to (-6, 4) is:
(4-2)/(-6--4) = 2/(-6+4) = 2/-2 = -1.
So one form of the equation would be:
y = -1x - 2
Now the other form of an equation is point-slope: y-k = m (x-h), where the point is at (h, k)
and if we pick -5 for x (bc 5 it listed in 3 of the answers), the y at x=-5 looks like around +3
so we get: y-k = -1 (x--5)...
y-3 = -(x+5)... therefore D) is the correct answer:
Answer:
5 parts are shaded and 4 parts are white so:
There are 9 parts all together.
We can then form ratio's of the white areas and the shaded areas:
White Area Ratio =

Shaded Area Ratio =

Let the area of sqaure be equated to x, which means let the entire area of the square equal to x:
x = Area of whole square
Now we can form an equation :

So now we just need to solve for x:


The area of the square is:

Answer:
142.2 meters.
Step-by-step explanation:
We have been given that a box measures 70 cm X 36 cm X 12 cm is to be covered by a canvas.
Let us find total surface area of box using surface area formula of cuboid.
, where,
= Length of cuboid,
= Breadth of cuboid,
= Width of cuboid.




Therefore, the total surface area of box will be 7584 square cm.
To find the length of canvas that will cover 150 boxes, we will divide total surface area of 150 such boxes by width of canvass as total surface area of canvas will also be the same.





Let us convert the length of canvas into meters by dividing 14220 by 100 as 1 meter equals to 100 cm.




Therefore, 142.2 meters of canvas of width 80 cm required to cover 150 such boxes.