Answer:
B
Step-by-step explanation:
B'(7, -4)
Let the pre-image coordinates be (x,y)
Consider x-coordinate,
x + 4 = 7
x = 3
Consider y-coordinates,
y + 5 = -4
y = -9
pre-image coordinates is (3, -9)
5 x 1 = 5
5 x 2 = 10
5 x 3 = 15
5 x 7 = 35
Answer:
10 ,16 AND 8
Step-by-step explanation:
PLEASE MARK ME AS BRAINLIEST
156 and (14y+30) are opposite angles so they are the same. You can use this information to put them into an equation to solve;
14y+30 = 156
You want you the the y's alone, so you subtract 30 from each side.
14y = 126
Then divide each side by 14 to get what just 1 y is.
y = 9
Then to find what the other pair of opposite angle add up to subtract 156x2 from 360
156x2 = 312 360-312 = 48
Halve 48 to get the sum of just one of the opposite angles
48/2 = 24
Now you know that 3x = 24. Solve this equation
x = 8
Answer:
The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.
Step-by-step explanation:
Given that,
The length of fencing of the rectangular plot is = 108 ft.
Let the longer side of the rectangular plot be x which is also the side along the river side and the width of the rectangular plot be y.
Since the fence along the river does not need.
So the total perimeter of the rectangle is =2(x+y) -x
=2x+2y-y
=x+2y
So,
x+2y =108
⇒x=108 -2y
Then the area of the rectangle plot is A = xy
A=xy
⇒A= (108-2y)y
⇒ A = 108y-2y²
A = 108y-2y²
Differentiating with respect to x
A'= 108 -4y
Again differentiating with respect to x
A''= -4
For maximum or minimum, A'=0
108 -4y=0
⇒4y=108

⇒y=27.

Since at y= 27, A''<0
So, at y=27 ft , the area of the rectangular plot maximum.
Then x= (108-2.27)
=54 ft.
The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.