V = 20.88 km / h ( velocity )
d = 145 km ( distance )
Velocity formula: v = d / t
t = d / v ( time )
t = 145 km : 20.88 km/h
t = 6.94444 h
We have to multiply 0.94444 by 60 minutes.
0.94444 · 60 = 56.67 ≈ 56 minutes 40 seconds
Answer: It would take 6 hours 56 minutes and 40 seconds.
Answer:
(35x+56)-6
Step-by-step explanation:
distrubuting
The Pythagorean's Theorem for our situation would look like this:

So let's call the short leg s, the long leg l and the hypotenuse h. It appears that all our measurements are based on the measurement of the short leg. The long leg is 4 more than twice the short leg, so that expression is l=2s+4; the hypotenuse measure is 6 more than twice the short leg, so that expression is h=2s+6. And the short leg is just s. Now we can rewrite our formula accordingly:

And of course we have to expand. Doing that will leave us with

Combining like terms we have

Our job now is to get everything on one side of the equals sign and solve for s

That is now a second degree polynomial, a quadratic to be exact, and it can be factored several different ways. The easiest is to figure what 2 numbers add to be -8 and multiply to be -20. Those numbers would be 10 and -2. Since we are figuring out the length of the sides, AND we know that the two things in math that will never EVER be negative are time and distance/length, -2 is not an option. That means that the short side, s, measures 10. The longer side, 2s+4, measures 2(10)+4 which is 24, and the hypotenuse, 2s+6, measures 2(10)+6 which is 26. So there you go!
Answer:
sin⁴x - sin²x
sin²x(sin²x - 1)
(1 - cos²x)(1 - cos²x - 1)
(1 - cos²x)(-cos²x)
-cos²x + cos⁴x
cos⁴x - cos²x
To determine the effect of changing the diameter of a sphere to its volume, we would need to know the formula for the volume of a sphere which is V = 4πr³/3 where r is half of the diameter. As you can see, there is a direct relationship.
r1 = d1/2
r2 = d2/2 = 2d1/2 = d1
V2/V1 = (4πd1³/3) / (4π(d1/2)³/3)
V2/V1 = 8
Therefore, the volume of the sphere would be 8 times the original volume.