Answer:
Step-by-step explanation:
x-intercepts exist where y is equal to 0. Where y is equal to 0 is where the graph goes through the x-axis. Our x-intercepts are (2x-3), (x + 3), and (x-4). Again, since x-intercepts exist where y = 0, then by the Zero Product Property, 2x - 3 = 0, x - 4 = 0, and x + 3 = 0. In the first x-intercept:
2x - 3 = 0 and
2x = 3 so
x = 3/2
In the second:
x - 4 = 0 so
x = 4
In the third:
x + 3 = 0 so
x = -3
So the x-intercepts in the correct order are x = 3/2, 4, -3
Answer:
Time of murder = 10:39 am
Step-by-step explanation:
Let the equation of exponential function representing the final temperature of the body after time 't' is,
f(t) = 
Here, a = Initial temperature
n = Constant for the change in temperature
t = Duration
At 11:30 am temperature of the body was 91.8°F.
91.8 =
--------(1)
Time to reach the body to the morgue = 12:30 pm
Duration to reach = 12:30 p.m. - 11:30 a.m.
= 1 hour
Therefore, equation will be,
84.4 = 
eⁿ = 
ln(eⁿ) = ln(0.9194)
n = -0.08403
From equation (1),
91.8 = 

![ln[(e)^{0.08403t}]=ln[\frac{98.6}{91.8}]](https://tex.z-dn.net/?f=ln%5B%28e%29%5E%7B0.08403t%7D%5D%3Dln%5B%5Cfrac%7B98.6%7D%7B91.8%7D%5D)
0.08403t = 0.07146
t = 0.85 hours
t ≈ 51 minutes
Therefore, murder was done 51 minutes before the detectives arrival.
Time of murder = 11:30 - 00:51
= 10:90 - 00:51
= 10:39 am
A,C, and D! x factors should always have a number next to them to show through tape diagrams , those without them are automatically eliminated , the rest have number answers and are correct
x^2 -12 x = -28
(-12/2) ^ 2 =36
x^2 -12 x + 36 = -28 + 36
x^2 -12x + 36 = 8
(x-6)^2 = 8
take the square root of each side
x-6 = sqrt (8) x-6 = -sqrt (8)
you get a positive and a negative when taking the square root
x= 6 + sqrt (8)
x = 6 - sqrt (8)
sqrt (8) = 2 sqrt (2)
x= 6 + 2 sqrt (2)
x = 6 - 2 sqrt (2)
Answer:
x= 6 + 2 sqrt (2)
x = 6 - 2 sqrt (2)
Answer:
a. V = 4/3
b. See attachment
Step-by-step explanation:
a.
Given
Z = x² + y²
V = ∫ ∫ (x² + y²) dxdy {0,1}{0,y} + ∫ ∫ (x² + y²) dxdy {0,1}{x,2-x}
V = ∫ ∫ (x² + y²) dxdy {0,1}{x,2-x}
Integrate with respect to y
V = ∫ x²y+ y³/3 dx {0,1}{x,2-x}
V = ∫ x²(2-x) + (2-x)³/3 - x²(2) - (2)³/3 dx {0,1}
V = ∫ 2x² -7x³/3 + (2-x)³/3 dx {0,1}
V = 2x³/3 - 7x⁴/12 + (2-x)⁴/12 {0,1}
V = (⅔ - 7/4 + 2/12) - (0-0+16/12)
V = 4/3