3k(k+10)=0 factor show steps please
1 answer:
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Answer:
T = 3.967 C
Step-by-step explanation:
Density = mass / volume
Use the mass = 1kg and volume as the equation given V, we will come up with the following equation
D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3
= (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1
Find the first derivative of D with respect to temperature T
dD/dT =
Let dD/dT = 0 to find the critical value we will get
= 0
Using formula of quadratic, we get the roots:
T = 79.53 and T = 3.967
Since the temperature is only between 0 and 30, pick T = 3.967
Find 2nd derivative to check whether the equation will have maximum value:
Substituting the value with T=3.967,
d2D/dT2 = -1.54 x 10^(-8) a negative value. Hence It is a maximum value
Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of
D = 1/0.001 = 1000 kg/m3
Therefore T = 3.967 C
Answer:
a. y-intercept = 23350 and x-intercept = 13
b.
c.
Step-by-step explanation:
Given
Solving (a): The x and y intercepts
The y intercept is the initial depreciation value
i.e. when x = 0
This value is the value of the car when it was initially purchased.
Hence, the y-intercept = 23350
The x intercept is the year it takes to finish depreciating
i.e. when y = 0
From the question, we understand that it takes 13 years for the car to totally get depreciated.
Hence, the x-intercept = 13
Solving (b): The slope
The slope (m) is the rate of depreciation per year
This is calculated by dividing the total depreciation by the duration.
So:
Because it is depreciation, it means the slope represents a deduction.
So,
Solving (c): The straight line equation
The general format of an equation is:
Where
In (a), we have that:
In (b), we have that:
Substitute these values in
<em>Hence, the depreciation equation is: </em><em></em>