Answer:
sorry if im wrong i think wrong numbers. plz check
Step-by-step explanation:
The center of dilation of the question is (-4,-3) .
let say that
x0=-4
y0=-3
Label the image A'B'C'
The new coordinate would be
A(-4,-1)
x=4
y=-1
x'=x0+ 2(x - x0)
x'= -4+ 2(-4 +4)
x'=-4
y'=y0+ 2(y - y0)
y'= -3+ 2(-1 +3)
y'=-3 +4= 1
______________________________
B(-4,-3)
x=-4
y=-3
x'=x0+ 2(x - x0)
x'= -4+ 2(-4 +4)
x'=-4
y'=y0+ 2(y - y0)
y'= -3+ 2(-3 +3)
y'=-3
______________________________
C(-1,-3)
x=-1
y=-3
x'=x0+ 2(x - x0)
x'= -4+ 2(-1 +4)
x'=-4 +6= 2
y'=y0+ 2(y - y0)
y'= -3+ 2(-3 +3)
y'=-3
A'(-4,1)
B'(-4,-3)
C'(2,-3)
Answer:
10 miles per hour.
Step-by-step explanation:
Let x represent athlete's walking speed.
We have been given that her jogging rate is 5 mph faster than her walking rate, so athlete's jogging speed would be 
miles per hour.
10 minutes = 1/6 hour.
20 minutes = 1/3 hour
While walking, we will get 
While jogging, we will get 
Since athlete is covering same distance while walking and jogging, so we can equate both expressions as:
Cross multiply:
Therefore, athlete's walking speed is 5 miles per hour.
Jogging speed: 
Therefore, athlete's jogging speed is 10 miles per hour.
Answer:
distribute the 2 on the left side to become 2(x+7)-->2x+14=-4x+14, add 4x to both sides to become 2x+14+4x=-4x+14+4x-->6x+14=14, subtract 14 from both sides 6x+14-14=14-14-->6x=0 and divide both sides of the equation by 6 to get x alone, 6x/6=0/6-->x=0 so x would equal 0.
Step-by-step explanation:
I'd say yes. If you use the diagonal as a reference. Take the square and set your compass to the width of the diameter of the square. Now put it on the page and mark a point. Put the point of the compass on that mark and make another mark. Now you can connect the two marks with the straight edge and you have a line that, if you made a square with sides that long, it'd have 2x the area of the first one. That's because the diagonal is the square root of 2 larger than one side. Square the square root of 2 and you've got 2. You lust need to make a perpendicular line to the first one to get the box going.
