Answer:
1. y=(x+3)^3. Zero: x=-3 multiplicity 3.
2. y=(x-2)^2 (x-1). Zeros: x=2 multiplicity 2; x=1 multiplicity 1.
3. y=(2x+3)(x-1)^2. Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.
Step-by-step explanation:
1. y=(x+3)^3
![y=0\\ (x+3)^3=0\\ \sqrt[3]{(x+3)^3}=\sqrt[3]{0}\\ x+3=0\\ x+3-3=0-3\\ x=-3](https://tex.z-dn.net/?f=y%3D0%5C%5C%20%28x%2B3%29%5E3%3D0%5C%5C%20%5Csqrt%5B3%5D%7B%28x%2B3%29%5E3%7D%3D%5Csqrt%5B3%5D%7B0%7D%5C%5C%20x%2B3%3D0%5C%5C%20x%2B3-3%3D0-3%5C%5C%20x%3D-3)
Zero: x=-3 multiplicity 3.
2. y=(x-2)^2 (x-1)
Zeros: x=2 multiplicity 2; x=1 multiplicity 1
3. y=(2x+3)(x-1)^2
Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.
Step-by-step explanation:
umm what grade is u im only in 6th i wannna help u
The horizontal compression of a graph means that the graph has come closer to the y-axis. Horizontal stretching and compression occur when a factor is multiplied to the original equation. If the factor being multiplied is greater than 1, then the graph is compressed. If the factor being multiplied is less than 1, the graph is stretched. Therefore, in this case, a number greater than 1 must be multiplied to -x + 5.
The answer is B. F(x) = -2(x-5)
Answer:
Yes
Step-by-step explanation:
Since the circle is split into 6 pieces, we can split the probability of landing on each section into fractions
There is one piece with the number 1 on it, so there is a 1/6 chance of landing on that.
There are 3 pieces with the number 2 on it, so there is a 3/6 chance of landing on it, simplifies that would be 1/2
There is one piece with the number 3 on it, so there is a 1/6 chance of landing on that.
There is one piece with the number 4 on it, so there is a 1/6 chance of landing on that.
Since the pieces with a 1 and 4 on the both have a 1/6 probabilit of landing, there is an equal chance on landing on either tile.
Answer:
Step-by-step explanation:
So you just need:
Number here = (3 * variable ^ number greater than 2) - 2
