All categories

3 years ago

What is the value of the discriminant for the equation 3a^2-8a-4=0?

Mathematics

1 answer:

MrMuchimi3 years ago

5 0

Hello,

Δ=8²+4*3*4=64+48=112=2^4 * 7

You might be interested in

Solve for x. 3x + 5 = 23

Lunna [17]

[|] Answer [|]

$\boxed{X \ = \ 6}$

[|] Explanation [|]

3x + 5 = 23

_________

Subtract 5 From Both Sides:

3x + 5 - 5 = 23 - 5

Simplify:

3x = 18

Divide Both Sides By 3:

$\frac{3x}{3} \ = \ \frac{18}{3}$

Simplify:

X = 6

_________

- Check Your Work -

Substitute 6 For X:

3 * 6 + 5 = 23

Parenthesis

Exponents

Multiply

Divide

Add

Subtract

3 * 6 = 18

18 + 5 = 23

$\boxed{[|] \ Eclipsed \ [|]}$

8 0

3 years ago

Read 2 more answers

Given that −4i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable. f

GalinKa [24]

Answer:

$\large \boxed{\sf \bf \ \ f(x)=(x-4i)(x+4i)(x+3)(x-5) \ \ }$

Step-by-step explanation:

Hello, the Conjugate Roots Theorem states that if a complex number is a zero of real polynomial its conjugate is a zero too. It means that (x-4i)(x+4i) are factors of f(x).

$\text{Meaning that } (x-4i)(x+4i) =x^2-(4i)^2=x^2+16 \text{ is a factor of f(x).}$

The coefficient of the leading term is 1 and the constant term is -240 = 16 * (-15), so we a re looking for a real number such that.

$f(x)=x^4-2x^3+x^2-32x-240\\\\ =(x^2+16)(x^2+ax-15)\\\\ =x^4+ax^3-15x^2+16x^2+16ax-240$

We identify the coefficients for the like terms, it comes

a = -2 and 16a = -32 (which is equivalent). So, we can write in $\mathbb{R}$ .

$\\f(x)=(x^2+16)(x^2-2x-15)$

The sum of the zeroes is 2=5-3 and their product is -15=-3*5, so we can factorise by (x-5)(x+3), which gives.

$f(x)=(x^2+16)(x^2-2x-15)\\\\=(x^2+16)(x^2+3x-5x-15)\\\\=(x^2+16)(x(x+3)-5(x+3))\\\\=\boxed{(x^2+16)(x+3)(x-5)}$

And we can write in $\mathbb{C}$

$f(x)=\boxed{(x-4i)(x+4i)(x+3)(x-5)}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

7 0

3 years ago

What is 18/40 in simplest form?

zaharov [31]

9/20 because you can divide them in half but you can divide them any further.

8 0

2 years ago

Read 2 more answers

In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year. Let X be the number of quakes in a given

hichkok12 [17]

Answer:

a) Earthquakes are random and independent events.

b) There is an 85.71% probability of fewer than three quakes.

c) There is a 0.51% probability of more than five quakes.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

In this problem, we have that:

In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year, so $\mu = 1.3$

(a) Justify the use of the Poisson model.

Earthquakes are random and independent events.

You can't predict when a earthquake is going to happen, or how many are going to have in a year. It is an estimative.

(b) What is the probability of fewer than three quakes?

This is $P(X < 3)$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725$

$P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543$

$P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2725 + 0.3543 + 0.2303 = 0.8571$

There is an 85.71% probability of fewer than three quakes.

(c) What is the probability of more than five quakes?

This is $P(X > 5)$

We know that either there are 5 or less earthquakes, or there are more than 5 earthquakes. The sum of the probabilities of these events is decimal 1.

$P(X \leq 5) + P(X > 5) = 1$

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725$

$P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543$

$P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303$

$P(X = 3) = \frac{e^{-1.3}*1.3^{3}}{(3)!} = 0.0980$

$P(X = 4) = \frac{e^{-1.3}*1.3^{4}}{(4)!} = 0.0324$

$P(X = 5) = \frac{e^{-1.3}*1.3^{5}}{(5)!} = 0.0084$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2725 + 0.3543 + 0.2303 + 0.0980 + 0.0324 + 0.0084 = 0.9949$

Finally

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9949 = 0.0051$

There is a 0.51% probability of more than five quakes.

5 0

3 years ago

What is thestandard form of 000 000 000 08

GuDViN [60]

80 million would be the answer

5 0

3 years ago