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1 year ago

11

Graph a line with a slope of 2/5 that contains the ping (-2,4)

1 answer:

user100 [1]1 year ago

4 0

Hello there! A point that would verify the slope to be 2/5 would be (0, 9).

Given the point (-2,4) and the slope 2/5, we can follow the slope formula of rise/run, where 2 is our rise (y) and 5 is our run (x). Moving up two points on the y-axis will give us our rise, 0, because -2 + 2 = 0. Moving to the right 5 units from 4 on the x-axis will give us our run, 9, because 5 + 4 = 9.

Your final answer: One of the points on the graphed line will be (0,9). When progressing two points up and 5 points to the right, we will obtain more points. Hope this helps!

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Answer:

Part 1) $A=60\ ft^2$

Part 2) $A=80\ cm^2$

Part 3) $A=96\ m^2$

Part 4) $A=144\ cm^2$

Part 5) $A=13\ m^2$

Part 6) $A=(49\pi -33)\ in^2$

Step-by-step explanation:

Part 1) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior rectangle

The area of rectangle is equal to

$A=bh$

where

b is the base of rectangle

h is the height of rectangle

so

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Part 2) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior square

The area of square is equal to

$A=b^2$

where

b is the length side of the square

so

$A=(12)(8)-(4^2)\\A=96-16\\A=80\ cm^2$

Part 3) we know that

The area of the shaded region is equal to the area of four rectangles plus the area of one square

so

$A=4(4)(5)+(4^2)\\A=80+16\\A=96\ m^2$

Part 4) we know that

The shaded region is equal to the area of the complete square minus the area of the interior square

so

$A=(15^2)-(9^2)\\A=225-81\\A=144\ cm^2$

Part 5) we know that

The area of the shaded region is equal to the area of triangle minus the area of rectangle

The area of triangle is equal to

$A=\frac{1}{2}(b)(h)$

where

b is the base of triangle

h is the height of triangle

so

$A=\frac{1}{2}(6)(7)-(4)(2)\\A=21-8\\A=13\ m^2$

Part 6) we know that

The area of the shaded region is equal to the area of the circle minus the area of rectangle

The area of the circle is equal to

$A=\pi r^{2}$

where

r is the radius of the circle

so

$A=\pi (7^2)-(3)(11)\\A=(49\pi -33)\ in^2$

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