Answer:
1 because you get 70 and 80 a day NORMALLY and the 2 indicates double so the 2 multiplies the 70 and 80.
Step-by-step explanation:because you get 70 and 80 a day NORMALLY and the 2 indicates double so the 2 multiplies the 70 and 80.
Answer:
8. yes, ASA
9. yes, SAS
10. yes, AAS
11. no
12. no
13. yes, SSS
Step-by-step explanation:
Answer: See below
Step-by-step explanation:
<u>To find the factor,</u>
f(x) = 0

<u>So,</u>
x = 0 multiplicity 2
x = -6 multiplicity 1
x = 6 multiplicity 1
For an even multiplicty, the graph touches the x-axis, and for an odd multiplicty, the graph crosses the x-axis
<u>Therefore,</u>
x = 0 multiplicty 2, Touch
x = -6 multiplicty 1, Cross
x = 6 multiplicty 1, Cross
Answer:
y=-2x-2
Step-by-step explanation:
y+4=-2(x-1)
y+4=-2x+2
y=-2x+2-4
y=-2x-2
Answer:
The dimensions that minimize the surface are:
Wide: 1.65 yd
Long: 3.30 yd
Height: 2.20 yd
Step-by-step explanation:
We have a rectangular base, that its twice as long as it is wide.
It must hold 12 yd^3 of debris.
We have to minimize the surface area, subjet to the restriction of volume (12 yd^3).
The surface is equal to:

The volume restriction is:

If we replace h in the surface equation, we have:

To optimize, we derive and equal to zero:
![dS/dw=36(-1)w^{-2} + 8w=0\\\\36w^{-2}=8w\\\\w^3=36/8=4.5\\\\w=\sqrt[3]{4.5} =1.65](https://tex.z-dn.net/?f=dS%2Fdw%3D36%28-1%29w%5E%7B-2%7D%20%2B%208w%3D0%5C%5C%5C%5C36w%5E%7B-2%7D%3D8w%5C%5C%5C%5Cw%5E3%3D36%2F8%3D4.5%5C%5C%5C%5Cw%3D%5Csqrt%5B3%5D%7B4.5%7D%20%3D1.65)
Then, the height h is:

The dimensions that minimize the surface are:
Wide: 1.65 yd
Long: 3.30 yd
Height: 2.20 yd