For this case we have the following number:
96
We can rewrite this number in an equivalent way.
For example, we can use words to rewrite the number.
We have then:
96 = ninety six
Answer:
the name of an equivalent name for 96 is:
Ninety-six
The width is half the length, so is
width = (1/2)*length
width = (1/2)*(<span>3.2a + 0.18b) cm
width = (1.6a +0.09b) cm
The perimeter of the rectangle is twice the sum of length and width.
perimeter = 2*(length + width)
perimeter = 2*((3.2a +0.18b) cm + (1.6a +0.09b) cm)
perimeter = 2*(4.8a +0.27b) cm)
perimeter = (9.6a +0.54b) cm
Sasha did not get this answer, so apparently ...
her reasoning was not correct.</span>
Solution:
Matte Satin Glossy Total
Homeowners 0.08 0.20 0.24 0.52
Contractors 0.04 0.26 0.18 0.48
Total 0.12 0.46 0.42 1
Approximately what percentage of contractors prefer the glossy finish?
Answer: Percentage of contractors who prefer the glossy finish is:
or 
Therefore, the option D. 37.5% is correct
Answer:
Step-by-step explanation:
If these 3 points are collinear, then we can find the slope of the linear function using any 2 of those points. Suppose we use (-4, 3) and (0, 1):
As we move from (-4, 3) to (0, 1), x increases by 4 and y decreases by 2. Hence, the slope of this lilne is m = rise/run = -2/4, or m = -1/2.
Using the slope-intercept formula y = mx + b and replacing y with 1, x with 0 and m with -1/2, we get:
1 = (-1/2)(0) + b, or b = 1. Then the desired equation is y = f(x) = (-1/2)x + 1
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