The area of each face is 25 cm
Sid and Libby are planning to sell pies at a local fair.
=> $200 to rent a table at the fair.<span>
=> $3.00 per pie </span>costs for ingredients, other supplies, baking, and packaging<span>
They plan to sell the pies for $8 each.
Let's find out how many pies will they be selling inorder to gain profit
=> 200 + 3.00(n) = 8(n), where n is the number of pie.
Let's try 40
=> 200 + 3(40) = 8 (40)
=> 320 = 320
</span>=> 200 + 3(41) = 8 (41)
=> 323 = 328
<span>Thus, in their 41st pie, they will get a profit
</span>
Y-2=3x+3
Then add two
So y=3x+5
Hope this helps sorry if wrong
Since ![[0,4]=[0,1]\cup(1,4]](https://tex.z-dn.net/?f=%5B0%2C4%5D%3D%5B0%2C1%5D%5Ccup%281%2C4%5D)
, we can rewrite the integral as
Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:
Both integrals are quite immediate: you only need to use the power rule
to get
![\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E11-3t%5E2%5C%3Bdt%20%3D%20%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%2C%5Cquad%20%5Cint_1%5E4%202t%5C%3B%20dt%20%3D%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4)
Now we only need to evaluate the antiderivatives:
![\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15](https://tex.z-dn.net/?f=%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%20%3D%201-1%5E3%3D0%2C%5Cquad%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4%20%3D%204%5E2-1%5E2%3D15)
So, the final answer is 15.
Answer:
$9.00
Step-by-step explanation:
~~~~Hey there~~~~
1. Since it was 15% discount, that means you have to solve 15% of 60, which is equivalent to 0.15 x 60.
2. Since the product of 0.15 and 60 is 9, then 9 is your answer.
3. Your final answer is $9.00
Hope this helped!
~delightedsunrise8
