The points are A is (1,1) and B (3,3)
With that 2/2
in it simplified form it is = 1
So your slope is 2/2 or 1
Hope this helps
Sorry, I thought you meant what 1 minus three is, not questions 1 through three
Explanation:
When you perform long division of 50 by 7, you find that the quotient is 7 and the remainder is 1. The remainder expressed over the original divisor is the fraction. (In some cases, that fraction may need to be reduced.)
That is ...
Answer:
0.7 seconds
Step-by-step explanation:
You have to subtract both lap times, but before doing so you need to know what 2/5 is in decimal form. To do so, you must do 2 divided by 5, getting 0.4 seconds as it's a fraction converting to decimal. This means Fam swam 29.4 seconds, which instead of 2/5, you have 0.4 as your converted answer, so you would substitute it after 29. Then, you would do 29.40 subtracted by 28.7 and get 0.7 seconds as your answer to find out the difference of their swim times.
<h3>Given</h3>
- a cone of height 0.4 m and diameter 0.3 m
- filling at the rate 0.004 m³/s
- fill height of 0.2 m at the time of interest
<h3>Find</h3>
- the rate of change of fill height at the time of interest
<h3>Solution</h3>
The cone is filled to half its depth at the time of interest, so the surface area of the filled portion will be (1/2)² times the surface area of the top of the cone. The filled portion has an area of
... A = (1/4)(π/4)d² = (π/16)(0.3 m)² = 0.09π/16 m²
This area multiplied by the rate of change of fill height (dh/dt) will give the rate of change of volume.
... (0.09π/16 m²)×dh/dt = dV/dt = 0.004 m³/s
Dividing by the coefficient of dh/dt, we get
... dh/dt = 0.004·16/(0.09π) m/s
... dh/dt = 32/(45π) m/s ≈ 0.22635 m/s
_____
You can also write an equation for the filled volume in terms of the filled height, then differentiate and solve for dh/dt. When you do, you find the relation between rates of change of height and area are as described above. We have taken a "shortcut" based on the knowledge gained from solving it this way. (No arithmetic operations are saved. We only avoid the process of taking the derivative.)
Note that the cone dimensions mean the radius is 3/8 of the height.
V = (1/3)πr²h = (1/3)π(3/8·h)²·h = 3π/64·h³
dV/dt = 9π/64·h²·dh/dt
.004 = 9π/64·0.2²·dh/dt . . . substitute the given values
dh/dt = .004·64/(.04·9·π) = 32/(45π)
