Not much can be done without knowing what

is, but at the least we can set up the integral.
First parameterize the pieces of the contour:


where

and

. You have


and so the work is given by the integral
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
Answer:
x=23.5
Step-by-step explanation:
90=25-x/2+18
combine like terms
90=43-x/2
subtract 43 from both sides
47= -×/2
47/2= 23.5
x=23.5
or
90-25= 65
65-18=47
47/2=23.5
Answer:
y = 8•6^x
Step-by-step explanation:
Here, we want to write an exponential equation
To fully write this, we need the values of a and b
From the first coordinates given;
8 = a•b^0
a = 8 since b^0 = 1
Furthermore;
288 = 8•b*2
b^2 = 36
b = √36
b = 6
So we have ;
y = 8•6^x
Answer:
30 beads to be exact your welcome
Step-by-step explanation: