All categories

3 years ago

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

1 answer:

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

You might be interested in

What is the coefficient of the third term in the binomial expansion of (a + b)6? 1 15 20 90

Sholpan [36]

Answer: 15

Step-by-step explanation:

(r+1)th term of $(a+b)^n$ is given by:-

$T_{r+1}=\ ^nC_r(a)^{n-r}b^r$

For $(a+b)^6$ , n= 6

$T_3=T_{2+1}=\ ^6C_2(a)^{6-2}(b)^2\\\\$

$=\ \dfrac{6!}{4!2!}a^4b^2\ \ \ [^nC_r=\dfrac{n!}{r!(n-r)!}]\\\\=\dfrac{6\times5\times4!}{4!\times2}a^4b^2\\\\=3\times5a^4b^2\\\\ =15a^4b^2$

Hence, the coefficient of the third term in the binomial expansion of $(a+b)^6$ is 15.

3 0

3 years ago

Read 2 more answers

The rule is Y=2x Pls help me

Alenkinab [10]

8 to 16

2 to 4

13 to 26

5 to 10

6 to 12

8 0

3 years ago

Read 2 more answers

Plz help I’m getting timed

Fed [463]

id say its a rhombus. all the other shapes can have right angles

6 0

2 years ago

Read 2 more answers

The ___ Property of Addition states that for any numbers a and b,a+b=b+a

Nina [5.8K]

Answer:

Commutative

Step-by-step explanation:

6 0

2 years ago

A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated tha

8_murik_8 [283]

To model this situation, we are going to use the decay formula: $A=Pe^{rt}$
where
$A$ is the final pupolation
$P$ 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
$A(10)$ is the function evaluated at 10
$A(0)$ 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.

7 0

2 years ago