Answer:(x^2+y^2)^2=(x^2+y^2)(x^2+y^2)
Step-by-step explanation:
We can rewrite left side into right side form
(x^2+y^2)^2=(x^2+y^2)(x^2+y^2)
we can expand it
(x^2+y^2)^2=x^4+x^2y^2+x^2y^2+y^4
(x^2+y^2)^2=x^4+y^4+2x^2y^2
we can add and subtract 2x^2y^2
(x^2+y^2)^2=x^4+y^4+2x^2y^2+2x^2y^2-2x^2y^2
(x^2+y^2)^2=x^4-2x^2y^2+y^4+2x^2y^2+2x^2y^2
(x^2+y^2)^2=x^4-2x^2y^2+y^4+4x^2y^2
(x^2+y^2)^2=x^4-2x^2y^2+y^4+(2xy)^2
(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2
Answer:
Retracing your steps
Step-by-step explanation:
The computer will finish in 2 seconds
2/3 = 4/6
1 = 6/6
6/6 - 4/6 = 2/6
solve this answer by using substitution. set the first equation to x=-y+4 then substitute -y+4 for x in the second equation.
2(-y+4) + 3y=0
-2y+8 + 3y=0
y+8=0
y=-8
now use -8 and solve for x
x+(-8)=4
x=12
Answer:
The minimum score required for an A grade is 88.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

Find the minimum score required for an A grade.
Top 12%, which is at least the 100-12 = 88th percentile, which is the value of X when Z has a pvalue of 0.88. So it is X when Z = 1.175.




Rounding to the nearest whole number
The minimum score required for an A grade is 88.