It looks like you want to compute the double integral
over the region <em>D</em> with the unit circle <em>x</em> ² + <em>y</em> ² = 1 as its boundary.
Convert to polar coordinates, in which <em>D</em> is given by the set
<em>D</em> = {(<em>r</em>, <em>θ</em>) : 0 ≤ <em>r</em> ≤ 1 and 0 ≤ <em>θ</em> ≤ 2<em>π</em>}
and
<em>x</em> = <em>r</em> cos(<em>θ</em>)
<em>y</em> = <em>r</em> sin(<em>θ</em>)
d<em>x</em> d<em>y</em> = <em>r</em> d<em>r</em> d<em>θ</em>
Then the integral is
The shaded areas add to 2/5 of the area of the circle, which is given by
A = πr²
Then the shaded area is ...
(2/5)A = (2/5)π(10 in)²
shaded area = 40π in²
Answer:
I wish I know the answer
Step-by-step explanation:
lemme try again
Let us make a list of all the details we have
We are given
The cost of each solid chocolate truffle = s
The cost of each cream centre chocolate truffle = c
The cos to each chocolate truffle with nuts = n
The first type of sweet box that contains 5 each of the three types of chocolate truffle costs $41.25
That is 5s+5c+5n = 41.25 (cost of each type of truffle multiplied by their respective costs and all added together)
The second type of sweet box that contains 10 solid chocolate trufles, 5 cream centre truffles and 10 chocolate truffles with nuts cost $68.75
That is 10s+5c+10n = $68.75
The third type of sweet box that contains 24 truffles evenly divided that is 12 each of solid chocolate truffle and chocolate truffle with nuts cost $66.00
That is 12s+12n=$66.00
Hence option C is the right set of equations that will help us solve the values of each chocolate truffle.
Answer:
8+1/4
Step-by-step explanation:
8+1/4 because 10-2 is 8.
