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Step2247 [10]
3 years ago
6

What is the discriminant in the quadratic equation x2 + 11x + 121 = x + 96

Mathematics
1 answer:
Pani-rosa [81]3 years ago
5 0
Discriminant is the value b^2 -4*a*c once the quadratic is in standard form
x^2 + 11x + 121 = x + 96
subtract x from both sides
x^2 + 10x + 121 = 96
subtract 96 from both sides
x^2 + 10x + 25 = 0
10^2 - 4 * 1 * 25
100 - 100 = 0
Discriminant is 0
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T_3=T_{2+1}=\ ^6C_2(a)^{6-2}(b)^2\\\\

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A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated tha
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To model this situation, we are going to use the decay formula: A=Pe^{rt}
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P is the initial population 
e is the Euler's constant
r is the decay rate 
t is the time in years

A. We know for our problem that the initial population is 1,250, so P=1250; we also know that after a year the population is 1000, so A=1000 and t=1. Lets replace those values in our formula to find r:
A=Pe^{rt}
1000=1250e^{r}
e^{r}= \frac{1000}{1250}
e^{r}= \frac{4}{5}
ln(e^{r})=ln( \frac{4}{5} )
r=ln( \frac{4}{5} )
r=-02231

Now that we have r, we can write a function to model this scenario:
A(t)=1250e^{-0.2231t}.

B. Here we are going to use a graphing utility to graph the function we derived in the previous point. Please check the attached image.

C. 
- The function is decreasing
- The function doe snot have a x-intercept 
- The function has a y-intercept at (0,1250)
- Since the function is decaying, it will have a maximum at t=0: 
A(0)=1250e^{(-0.2231)(0)
A_{0}=1250e^{0}
A_{0}=1250
- Over the interval [0,10], the function will have a minimum at t=10:
A(10)=1250e^{(-0.2231)(10)
A_{10}=134.28

D. To find the rate of change of the function over the interval [0,10], we are going to use the formula: m= \frac{A(0)-A(10)}{10-0}
where 
m is the rate of change 
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A(0) is the function evaluated at 0
We know from previous calculations that A(10)=134.28 and A(0)=1250, so lets replace those values in our formula to find m:
m= \frac{134.28-1250}{10-0}
m= \frac{-1115.72}{10}
m=-111.572
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.

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2 years ago
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