Answer:
g(x)=15/20 x^2
Step-by-step explanation:
write 15/20 as a fraction then x^2 comes after it
I took the khan test and aced it
Answer:
The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .
Step-by-step explanation:
Given as :
The mass of liquid water = 50 g
The initial temperature =
= 15°c
The final temperature =
= 100°c
The latent heat of vaporization of water = 2260.0 J/g
Let The amount of heat required to raise temperature = Q Joule
Now, From method
Heat = mass × latent heat × change in temperature
Or, Q = m × s × ΔT
or, Q = m × s × (
-
)
So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )
Or, Q = 50 g × 2260.0 J/g × 85°c
∴ Q = 9,605,000 joule
Or, Q = 9,605 × 10³ joule
Or, Q = 9605 kilo joule
Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer
Û = (-1, -1, -1)
^v = (2, 3, -5)
^v - û = (2 + 1, 3 + 1, -5 + 1) = (3, 4, -4)
Half way from ^v to ^(v - u) = ((3 - 2)/2, (4 - 3)/2, (-4 + 5)/2) = (1/2, 1/2, 1/2)
Halfway from û to ^v = ((2 + 1)/2, (3 + 1)/2, (-5 + 1)/2) = (3/2, 2, -2)
The required vector ^w = ((3/2 - 1/2), (2 - 1/2), (-2 - 1/2)) = (1, 1/2, -5/2)
Check the picture below.
so notice, their perimeter is the same, because the perimeter is just one rod anyway, and all rods are the same length, thus
Answer:
18 servings
Step-by-step explanation:

Hope this helps!