The answer is 43j you just have to add them together
(3x²y⁶)⁷
(3x²y⁶)(3x²y⁶)(3x²y⁶)(3x²y⁶)(3x²y⁶)(3x²y⁶)(3x²y⁶)
2187x¹⁴y⁴²
Answer:
A) 34.13%
B) 15.87%
C) 95.44%
D) 97.72%
E) 49.87%
F) 0.13%
Step-by-step explanation:
To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:
P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)
= 0.5 - 0.1587 = 0.3413
It means that the PERCENT of scores that are between 90 and 100 is 34.13%
At the same way, we can calculated the percentages of B, C, D, E and F as:
B) Over 110

C) Between 80 and 120

D) less than 80

E) Between 70 and 100

F) More than 130

Dividing the parallelogram in half and applying trigonometry is the easiest solution for me.
Quotient=(dividend-remainder)/divisor