Et's give this a go:h(x) = cos(x) / f(x)
derivative (recall the quotient rule)h'(x) = [ f(x) * (-sin(x)) - cos(x)*f'(x) ] / [ f(x) ]^2
simplifyh'(x) = [ -sin(x)*f(x) - cox(x)*f '(x) ] / [ f(x) ]^2h'(π/3) = [ -sin(π/3)*f(π/3) - cox(π/3)*f '(π/3) ] / [ f(π/3) ]^2h'(π/3) = −(3–√/2)∗(3)−(1/2)∗(−7)/(3)2
h'(π/3) = (−33–√/2+7/2)/9
And you can further simplify if you want, I'll stop there.
3 + ___ <---- number is ______
Hope I helped :)
Answer:
555,555 is the number you are rounding.
1001 is rounded to 1000
9999 is rounded to 10,000
130,000 is rounded to 100,000
Step-by-step explanation:
when rounding to the nearest thousand, look at the hundreds number. if it is greater than or equal to 5, round up, less than round down and set everything after it to 0. rounding up increases the thousand by one, rounding down doesn't change it. same applies for ten thousand and hundred thousand.
Since Square Roots must be greater than or equal to three, just add three to the square root to make it be greater than or equal to negative 3
To get the z-value of the scores of the four students, we are going to use the formula for standard score or z-score. It is score minus the mean score, then divided by standard deviation.
z= Score (X)-Mean / SD
To find the z-value of each score, we have to use a Z table. Using the z-score, we are to look first at the y-axis of the table which will highlight the first two digits of the z-score. Then, the x-axis for the second decimal place of the z-score.
You can use this as reference for the z-table: http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf
Mean= 500SD= 100Scores= 560, 450, 640, 530
For the student who scored 560,z= X-Mean / SDz= 560-500 / 100z= 60 / 100z= 0.6
The score is 0.6 standard deviation above the mean. The z-value is 0.7257 or 72.57%.
For the student who scored 450,z= X-Mean / SDz= 450-500 / 100z= -50 / 100z= -0.5
The score is -0.5 standard deviation above the mean. The z-value is 0.3085 or 30.85%.
For the student who scored 640,z= X-Mean / SDz= 640-500 / 100z= 140 / 100z= 1.4
The score is 1.4 standard deviation above the mean. The z-value is 0.9192 or 91.92%.
For the student who scored 530,z= X-Mean / SDz= 530-500 / 100z= 30 / 100z= 0.3
The score is 0.3 standard deviation above the mean. The z-value is 0.6179 or 61.79%.