Hope someone can.help.me with this
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Answer:
48 kg
Step-by-step explanation:
The given relations can be used to write a system of equations for the two weights. Those can be solved to find Jane's weight.
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<h3>setup</h3>Let x and y represent Jane's and Jessica's original weight, respectively. The ratio of weights was ...
x/y = 8/9
After the changes in weight, they were equal:
x+2 = y-4
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<h3>solution</h3>Adding 4 to the second equation, we have an expression for y that can be substituted into the first equation.
y = x +6 . . . . . . . . . . solve the second equation for y
x/(x+6) = 8/9 . . . . . . substitute for y in the first equation
9x = 8(x +6) . . . . . . . multiply by 9(x+6)
x = 48 . . . . . . . . . . . simplify and subtract 8x
Jane weighed 48 kg at first.
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<em>Alternate solution</em>
The original difference in "ratio units" was 9-8 = 1 ratio unit. We find that this corresponds to 6 kg after the weight changes make the weights equal. Then 8 ratio units will be 8(6 kg) = 48 kg—Jane's original weight.
(This mental solution is virtually the same as the solution using equations shown above.)
x-coordinates for the maximum points in any function f(x) by f'(x) =0 would be x = π/2 and x= 3π/2.
<h3>How to obtain the maximum value of a function?</h3>To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.
we want to find x-coordinates for the maximum points in any function f(x) by f'(x) =0
Given f(x)= 4cos(2x -π)
In general
from x = 0 to x = 2π :
when k =0 then x = π/2
when k =1 then x= π
when k =2 then x= 3π/2
when k =3 then x=2π
Thus, X-coordinates of maximum points are x = π/2 and x= 3π/2
Learn more about maximum of a function here:
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