Answer:
A) x² + 2x + 6
B) x² + 2x - 7
C) ¼(x²+2x+1))
D) 6x²+12x+6
E) -x²-2x-1
Step-by-step explanation:
A) f(x) + 5 =x²+2x+1 + 5
= x² + 2x + 6
B) f(x)-8,=x^2+2x+1-8
= x² + 2x - 7
C) ¼f(x) = ¼(x²+2x+1)
D) 6f(x) = 6(x²+2x+1) = 6x²+12x+6
E) -f(x) = -(x²+2x+1) = -x²-2x-1
Answer:
The sample range is 2.61.
Step-by-step explanation:
We are given the following grade-point averages which apply to a random sample of graduating seniors below;
3.81, 2.73, 2.51, 3.09, 3.28, 3.51, 2.86, 1.20, 3.13, and 3.24.
As we know that the range of the data is calculated bu using the following formula;
Range = Highest value in the data - Lowest value in the data
Here, Highest value = 3.81
Lowest value = 1.20
So, the sample data range = 3.81 - 1.20
= 2.61.
Answer:
Step-by-step explanation:
The given relation between length and width can be used to write an expression for area. The equation setting that equal to the given area can be solved to find the shed dimensions.
__
<h3>Given relation</h3>
Let x represent the width of the shed. Then the length is (2x+3), and the area is ...
A = LW
20 = (2x+3)(x) . . . . . area of the shed
__
<h3>Solution</h3>
Completing the square gives ...
2x² +3x +1.125 = 21.125 . . . . . . add 2(9/16) to both sides
2(x +0.75)² = 21.125 . . . . . . . write as a square
x +0.75 = √10.5625 . . . . . divide by 2, take the square root
x = -0.75 +3.25 = 2.50 . . . . . subtract 0.75, keep the positive solution
The width of the shed is 2.5 feet; the length is 2(2.5)+3 = 8 feet.
Answer:
yes
Step-by-step explanation: yes
<h3>
Answer: B) angle 4</h3>
==================================
Explanation:
We can think of lines L and M as sort of train tracks. Inside the train tracks we have the interior angles of: 3, 4, 5, 6
Angles 3 and 6 are one pair of alternate interior angles. They are on alternate sides of the transversal line N.
The other pair of alternate interior angles are 4 and 5
Alternate interior angles are only congruent when L and M are parallel.
