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2 years ago

Which graph represents y as a function of x?

Mathematics

1 answer:

Vilka [71]2 years ago

8 0

Answer:

Bottom left.

Step-by-step explanation:

In order for any of them to be the graph of a function you would need that given a value of x, you get no more than one value of y. Zero is fine, you just picked an x out of the domain.

In layman's terms it means that if you grab a ruler, a piece of paper, a pen, and you scan the graph moving the edge parallel to the y axis (ie, vertically) you are allowed to touch the graph only once. Top right is out. Bottom left is ok, bottom right is not: you have infinite points of contact in there.

Top left would be fine, if it wasn't for x = 1. In there you have two values marked with a solid dot. That makes it not the graph of a function.

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A baseball is thrown into the air from the top of a 224-foot tall building. The baseball's approximate height over time can be r

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3 years ago

Read 2 more answers

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

Thus, we are given the time interval [0,12] for t.

We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

$\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})$

Equating the first derivative to zero, we get,

$\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0$

Solving, we get,

$8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2$