Okay so, togo flies 30 miles in 3/5 of an hour with the wind.togo flies 30 miles in 5/6 of an hour against the wind. his rate of speed with the wind was 30 / (3/5) = 30*5/3 = 150/3 = 50 mph.his rate of speed against the wind was 30 / (5/6) = 30*6/5 = 6*6 = 36 mph.3/5 of an hour times 50 mph = 30 miles.
5/6 of an hour times 36 mph = 30 miles.mph looks good.
going he was with the wind.
his rate of speed then was the airplane speed plus the wind speed.coming back he was against the wind.his rate of speed then was the airplane speed minus the wind speed.-----50 mph = A + W36 mph = A - W-----looks like A + W - (A - W) should equal 50 - 36.removing parentheses, equation becomesA + W - A + W = 14.
combining like terms, the equation becomes2*W = 14.dividing both sides by 2, equation becomesW = 7.if W = 7, then A can be found as follows:50 = A + 7A = 43.likewise,36 = A - 7A = 43.since the airplane speed is the same, it looks like the answer is good.solving in the original equations, we get3/5 * (43 + 7) = 303/5 * 50- = 3030 = 30good.-----5/6 * (43 - 7) = 305/6 * 36 = 3030 = 30good.
answer is W = 7 and A = 43.
Answer:
<h3>22 ÷ 0 = 0 </h3>
<h3>So ur Answer is 0 .</h3>
Answer:
(4,0)
Step-by-step explanation:
The object is first at (0,0)
It is reflected across line x=-2, this means you draw the mirror line at x=-2 and count 2 equal units backwards to get the image.The image will be at ;
The image (-4,0) is then reflected on the y-axis
You know reflection on the y-axis, the y-coordinate remains the same but the x-coordinate is changed to its opposite sign.
Hence;
(- -4,0)= (4,0)
The image will move 8 units towards positive x-axis.This is the same as moving 4 units from the mirror line at (0,0) and land at (4,0)
Yes they can! Just because two plots have the same IQR and range, does not mean that the rest of the data has to be the exact same. For example, the median.
5.B
6.D
7.A
8.B
9.A
10.C hsvs whehsg
