Givens
Slower plane
r = r
t = 2.5 hours
d = d
Faster Plane
r_faster = 1.5 * r
t = 2.5 hours
d1 = d + 127.5
The time is the same for both
t = d/r
d1/r1 = d/r
(d+ 127.5)/1.5r = d/r Multiply each side by r
(d + 127.5)/1.5 = d Multiply both sides by 1.5
d + 127.5 = 1.5d Subtract d from both sides.
127.5 = 1.5d - d
127.5 = 0.5d Divide by 0.5
127.5 / 0.5 = d
255 = d
Now you can go back and figure out the rates.
First find d1
d1 = d + 127.5
d1 = 255 + 127.5
d1 = 382.5
<em><u>Rate of the slower plane</u></em>
d = 255
t = 2.5 hours
r = d/t
r = 255/2.5
r = 102 miles per hour.
<em><u>Faster plane</u></em>
d1 = 382.5 miles
t = 2.5 hours
r1 = d/t
r1 = 382.5/2.5 = 153 miles per hour.
You need to use some trig... but I will try to explain it as best as possible. If you imagine a unit circle, we describe the coordinate points that line on the circle in terms of cosine and sine. A certain coordinate point on the circle can be written like (cos(x),sin(x)). "X" is going to be the angle from the center of the circle starting at point (1,0). Now imagine this triangle being fitted into a circle. The leg represents the x and y length that lead up to a point on the circle. The hypotenuse of the triangle represents the radius of the circle.
We need to find the hypotenuse first.
cos(37 degrees) x hypotenuse = 2.1
hypotenuse = 2.63 approx.
So the "x" length or the sine length is...
sin(37 degrees) x 2.63 = 1.6 approx.
answer: 1.6
Answer:
x=21
x= -37
Step-by-step explanation:
is this the questions
3x-63=0
x+37=0 then the answer is given above
Answer:
Sorry its not clear there is no attachments or anything
Step-by-step explanation:
Using the point on the graph (4, 1) solve for a
1 = a(4-3)² - 1
2 = a