Using V = ⅓[base area×height]
V= ⅓[ (5.5÷4)² × 2.6]
v= ⅓[ 4.92]
v = 1.6m³
X is 13, p is 126, Q is 30, r is 24
1
∫ -------------------- dx
x² + 6x + 10
1
∫ -------------------- dx
(x+3)² + 1
Let t = x + 3
dt = dx
1
∫ -------------------- dt
t²+1
1/1 arctan(t/1)
1/1 arctan (x+3 / 1)
=arctan(x+3) + C
Let x = bags of chips
Let y = jars of dipping sauce
Corinne: 6x + 3y = 25.32
Ginger: 4x + 5y = 24.32
Use the elimination method. I will be cancelling out the x's by multiplying the first equation by -2 and the second equation by 3.
-12x - 6y = -50.64
<u>12x + 15y = 72.96</u>
0x + 9y = 22.32
y = 2.48
Everything below is optional. It would be used to check your work.
Plug this y back into any of the original expressions to find the x. I plugged it into the first equation.
6x + 3(2.48) = 5.32
6x + 7.44 = 25.32
6x = 17.88
x = 2.98
To continue to check your answers, you can plug the x and y into the other (in my case, the second,) equation.
4(2.98) + 5(2.48) = 24.32
11.92 + 12.40 = 24.32
24.32 = 24.32 <--- If the answer is the same on both sides, you will know the x and y are correct.
One jar of dipping sauce is $2.48.
