How to find zeros of a quadratic function <br> example<br> f(x)=x^3+2x^2-25x-50

I need help Plzz & Thank You .

Afina-wow [57]

${x50506}^{?}$

6 0

3 years ago

The function f(x)=2x-6 and g(x) is shown in the table

lesya692 [45]

Answer:

Multiply then add them and them that's your answer.

Step-by-step explanation:

5 0

2 years ago

50% of babies are born female.Olive wants to find the probability 20 or more of the 50 babies born today were female. If we use

Anna [14]

Answer:

D. Keep track of the number of heads flipped in 50 flips. These represent the number of females born today. Repeat this 200 times.

3 0

3 years ago

A certain store sells small, medium, and large toy trucks in each of the colors red, blue,green, and yellow. The store has an eq

salantis [7]

Answer:

The probability that the truck she selects will have at least one of the two features that Ella Stella wants is 50%.

Step-by-step explanation:

Since a certain store sells small, medium, and large toy trucks in each of the colors red, blue, green, and yellow, and the store has an equal number of trucks of each possible color / size combination, if Stella wants a medium red truck and her mother will randomly select one of the trucks in the store, to determine what is the probability that the truck she selects will have at least one of the two features that Stella wants, the following calculation must be performed:

3 sizes

4 colors

4 x 3 = 12

4 midsize cars

3 red cars

1 medium red car

4 + 3 - 1 = 6

6/12 = 0.5

Therefore, the probability that the truck she selects will have at least one of the two features that Ella Stella wants is 50%.

6 0

3 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago

How to find zeros of a quadratic function example f(x)=x^3+2x^2-25x-50