Answer:
Your answer is dilation.
Step-by-step explanation:
A dilation either shrinks or makes the shape bigger. It changes the size.
You multiply the GCF of the numerical part 3 and the GCF of the variable part x^2y to get 3x^2y
Answer:
Morgan, its a straight line and it goes through the origin
Step-by-step explanation:
Answer:
59 to 66
Step-by-step explanation:
Mean test scores = u = 74.2
Standard Deviation =
= 9.6
According to the given data, following is the range of grades:
Grade A: 85% to 100%
Grade B: 55% to 85%
Grade C: 19% to 55%
Grade D: 6% to 19%
Grade F: 0% to 6%
So, the grade D will be given to the students from 6% to 19% scores. We can convert these percentages to numerical limits using the z scores. First we need to to identify the corresponding z scores of these limits.
6% to 19% in decimal form would be 0.06 to 0.19. Corresponding z score for 0.06 is -1.56 and that for 0.19 is -0.88 (From the z table)
The formula for z score is:

For z = -1.56, we get:

For z = -0.88, we get:

Therefore, a numerical limits for a D grade would be from 59 to 66 (rounded to nearest whole numbers)
I solved this using a scientific calculator and in radians mode since the given x's is between 0 to 2π. After substitution, the correct pairs
are:
cos(x)tan(x) – ½ = 0
→ π/6 and 5π/6
cos(π/6)tan(π/6) – ½ = 0
cos(5π/6)tan(5π/6) – ½ = 0
sec(x)cot(x) + 2 =
0 → 7π/6 and 11π/6
sec(7π/6)cot(7π/6) + 2 = 0
sec(11π/6)cot(11π/6) + 2 = 0
sin(x)cot(x) +
1/sqrt2 = 0 → 3π/4 and 5π/4
sin(3π/4)cot(3π/4) + 1/sqrt2 = 0
sin(5π/4)cot(5π/4) + 1/sqrt2 = 0
csc(x)tan(x) – 2 = 0 → π/3 and 5π/3
csc(π/3)tan(π/3) – 2 = 0
csc(5π/3)tan(5π/3) – 2 = 0