List three inequalities

List the potential rational zeros of the polynomial function. Do not find the zeros.

Archy [21]

$f(x)=\boxed{1}x^5-3x^2+3x+\boxed{14}$

The potential rational zeros are divisors of 14:

$\pm1;\ \pm2;\ \pm7;\ \pm14$

and the quotient p/q where p is a divisor of 1, and q is a divisor of 14:

$\pm\dfrac{1}{2};\ \pm\dfrac{1}{7};\ \pm\dfrac{1}{14}$

4 0

3 years ago

Can someone pls help me with number 5 pls show your work so i can understand thanks

Natali [406]

Step-by-step explanation:

Let's represent Aaron's age with x and

Charlene's age with x+8

x+x+8=44

2x+8=44

2x=44-8

2x=36

x=36÷2

x=18

Aaron is 18 years old.

While Charlene is 18+8=26 years old.

Hope this helps!

7 0

3 years ago

Jeanne has many nickels, dimes, and quarters in her wallet. She chooses 3 coins at random. What is the probability that all thre

Whitepunk [10]

Answer:

Step-by-step explanation:

There isn't enough said about the distribution of coins in her wallet, but we'll just assume that the number is so large that any coin is equally likely to be drawn.

Stated another way, there are 27 possible outcomes of the three draws (3 x 3 x 3) and we'll assume each is equally likely.

PROBLEM 1:

This is a conditional probability question. We only have to consider the cases where she could have drawn 2 quarters and another coin. The possible draws are:

DQQ, NQQ, QDQ, QNQ, QQD, QQN or QQQ*.

That's 7 possible draws (with equal probability) and only 1* of them is a draw with 3 quarters.

Answer:

P(three quarters given two are quarters) = 1/7

PROBLEM 2:

Again, this is conditional probability. To help count the ways, let's instead count the ways to *not* draw any dimes. That means you have 2 choices for the first coin, 2 choices for the second coin and 2 choices for the third coin.

So 8 out of the 27 draws would *not* contain a dime. By subtracting, we can see that 19 of the draws *would* contain at least one dime.

Now think of the ways to create a draw consisting of one of each coin. We have the 3 different coins and they can be drawn in any order. That would be 3! or 6 ways.

If that isn't clear, let's list them all out:

DDD, DDN, DDQ, DND, DNN, DNQ*, DQD, DQN*, DQQ, NDD, NDN, NDQ*, NND, NQD*, QDD, QDN*, QDQ, QND*, QQD

There are 19 possible outcomes with at least one dime and exactly 6 of them have one of each type.

P(all different given at least one is a dime) = 6/19

3 0

3 years ago

Which choice is equivalent to the expression below?

morpeh [17]

Answer:

C

Step-by-step explanation:

As we can see, there are perfect squares in the first two expressions

thus, we have it that;

√(25x) = 5√x

√(4x) = 2√x

so we have the expression as;

5 √x - 2 √(x + 2 √x

= 5√x

3 0

2 years ago

20 Points!! Please help

Setler79 [48]

Answer:

The graph of the triangle LMN and the image L'M'N' created by rotation of ΔLMN through 180° made with MS Excel is attached

The transformation involved in the 180° rotation of ΔLMN to create ΔL'M'N' includes;

The changing of the signs of the x and y-values of the coordinates of the vertices ΔLMN to get the corresponding x and y-values of the coordinates of the vertices of ΔL'M'N' as follows;

Before, rotation

The coordinates of the vertices of ΔLMN are; L(3, 3), M(9, 9), and N(9, 3)

After 180° rotation

The coordinates of the vertices of the image of ΔLMN, which is ΔL'M'N' are; L'(-3, -3), M'(-9, -9), and N'(-9, -3)

Therefore, the image of ΔLMN, ΔL'M'N' is located in the third quadrant, while the preimage ΔLMN is in the first quadrant

Part B

The lines drawn through L and L' and through M and M' are colinear lines

Part C

The lines dawn through points N and N' will not pass through the same line as the lines through L and L' and through M and M', because the three points are not colinear

Step-by-step explanation:

8 0

2 years ago

List three inequalities ​

List three inequalities