(4y-3)(2y²+3y-5)
First , let's start with "4y"
4y*2y² = 8y³
4y*3y = 12y²
4y*-5 = -20y
Next, let's multiply by "-3"
-3*2y² = -6y²
-3*3y = -9y
-3*-5 = 15
Now, let's combine all of our values.
8y³+12y²-6y²-20y-9y+15 = 8y³+6y²-29y+15
Answer:
a) vertical translation: up five units
b) horizontal translation: right 2 units
c) vertically stretched by 3 units
Step-by-step explanation:
https://www.humbleisd.net/cms/lib2/TX01001414/Centricity/Domain/3611/Square%20Root%20Graphs%202015.pdf
Answer:
<h2> Combination</h2>
Step-by-step explanation:
In this case the order of selection does not matter since we are concerned in the number of ways possible a set of students (5) can be grouped for a project, we are going to be using combination technique
In permutation the order of selection matters hence will not give the desired result
Answer:
∠2 and ∠5
Step-by-step explanation:
we know that
<u>Alternate Exterior Angles</u> are a pair of angles on the outer side of each of those two lines but on opposite sides of the transversal
In this problem
∠12 and ∠2 are alternate exterior angles
∠12 and ∠5 are alternate exterior angles
therefore
∠2 and ∠5 are each separately alternate exterior angles with ∠12
First you need to make both bases the same:
Lets remove the ^p and ^4
To make the base of 42 equal to 41, you would have 41^x = 41
X - ln(42) / ln(41) = 1.00648904
Now you have 41^1.00648904(p) = 41^4
Now the bases are equal so we need to set the exponents to equal:
1.00648904(p) = 4
Divide both sides by 1.00648904 to solve for P
P = 4 / 1.00648904
P = 3.97421114
