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Lyrx [107]
2 years ago
10

Which of the following is the quotient of the rational expressions shown here? x/x-1 / 1/x+1.

Mathematics
1 answer:
grandymaker [24]2 years ago
6 0

Answer:

Assignment: 01.07 Laboratory

: 01.07 Laboratory Techniques

Assignment: 01.07 Laboratory

: 01.07 Laboratory Techniques

Assignment: 01.07 Laboratory

: 01.07 Laboratory Techniques

Assignment: 01.07 Laboratory

: 01.07 Laboratory Techniques

Assignment: 01.07 Laboratory

: 01.07 Laboratory Techniques

Step-by-step explanation:

Assignment: 01.07 Laboratory

: 01.07 Laboratory Techniques

Assignment: 01.07 Laboratory

: 01.07 Laboratory Techniques

Assignment: 01.07 Laboratory

: 01.07 Laboratory Techniques

Assignment: 01.07 Laboratory

: 01.07 Laboratory Techniques

Assignment: 01.07 Laboratory

: 01.07 Laboratory Techniques

Assignment: 01.07 Laboratory

: 01.07 Laboratory Techniques

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Translate to an expression. How far have I travelled if I biked at 10mph for hhrs, I walked at 3mph for twice as long as I biked
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Step-by-step explanation:

4 0
3 years ago
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Please help me, i'm trying to graduate!!!
Svet_ta [14]
If you use this equation then you say that the ground is h=0 and solve as a quadratic.
The quadratic formula is (-b±<span>√(b^2-4ac))/2a when an equation is in the form ax^2 + bx + c
So the equation you have been given would be -16t^2-15t-151 = 0
This equation has no real roots which leads me to believe it is incorrect.
This is probably where your difficulty is coming from, it's a mistake.

The equation is formed from S=ut+(1/2)at^2+(So) where (So) is the initial height and S is the height that you want to find.
In this case you want S = 0.
If the initial height is +151 and the initial velocity and acceleration are downwards (negative) and the initial velocity (u) is -15 and the initial acceleration is -32 then you get the equation S=-15t-16t^2+151

Solving this using the quadratic formula gives you t = 2.64 or t = -3.58
Obviously -3.58s can't be the answer so you're left with 2.64 seconds.

Hope this makes sense.
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3 years ago
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