Ivy and Andrey were asked to find an explicit formula for the sequence -100,-50,0,50,...−100,−50,0,50,...minus, 100, comma, minu
Arturiano [62]
Answer:
C. Both Ivy and Andrey is the correct answer.
Step-by-step explanation:
Answer:

![g(x)=\sqrt[3]{x}-5](https://tex.z-dn.net/?f=g%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D-5)
Step-by-step explanation:
Consider the given function is
![f(x)=\sqrt[3]{-2x+4}-5](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7B-2x%2B4%7D-5)
It is given that
and neither g(x) nor h(x) is solely x.
![f(x)=\sqrt[3]{(-2x+4)}-5](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7B%28-2x%2B4%29%7D-5)
Let
, then we get
![f(x)=g(h(x))=\sqrt[3]{h(x)}-5](https://tex.z-dn.net/?f=f%28x%29%3Dg%28h%28x%29%29%3D%5Csqrt%5B3%5D%7Bh%28x%29%7D-5)
Substitute h(x)=x in the above function.
![g(x)=\sqrt[3]{x}-5](https://tex.z-dn.net/?f=g%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D-5)
Therefore, the required functions are
and
.
Check the solutions.
![[\because h(x)=-2x+4]](https://tex.z-dn.net/?f=%5B%5Cbecause%20h%28x%29%3D-2x%2B4%5D)
![[\because g(x)=\sqrt[3]{x}-5]](https://tex.z-dn.net/?f=%5B%5Cbecause%20g%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D-5%5D)

Therefore, our solution is correct.
It equals 276 but about 300
Answer:
it is 4
Step-by-step explanation:
dxgfchv
Answer:
B. 16 hrs
Step-by-step explanation:
Distance = rate × time
The best way to do this is to make a table with the info. We are concerned with the trip There and the Return trip. Set it up accordingly:
d = r × t
There
Return
The train made a trip from A to B and then back to A again, so the distances are both the same. We don't know what the distance is, but it doesn't matter. Just go with it for now. It'll be important later.
d = r × t
There d
Return d
We are also told the rates. There is 70 km/hr and return is 80 km/hr
d = r × t
There d = 70
Return d = 80
All that's left is the time column now. We don't know how long it took to get there or back, but if it took 2 hours longer to get There than on the Return, the Return trip took t and the There trip took t + 2:
d = r × t
There d = 70 × t+2
Return d = 80 × t
The distances, remember, are the same for both trips, so that means that by the transitive property of equality, their equations can be set equal to each other:
70(t + 2) = 80t
70t + 140 = 80t
140 = 10t
14 = t
That t represents the Return trip's time. Add 2 hours to it since the There trip's time is t+2. So 14 + 2 = 16.
B. 16 hours