Someone please help??

Mathematics

1 answer:

Xelga [282]3 years ago

6 0

Answer:

x-axis

Step-by-step explanation:

The asymptote is a straight line that the curve gets closer and closer to but never touches it.

The given exponential function is $f(x)=3^x$ .

The given graph has a horizontal asymptote,

The equation of this horizontal asymptote is y=0.

This is also refers to as the x-axis.

Therefore the asymptote is the x-axis.

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Explain the meaning of continuity as it pertains to a function. Identify and explain the three conditions that must exist in ord

vesna_86 [32]

When there are no gaps or breaks on the graph, a function is continuous. When a function fails to be continuous, it is said to be discontinuous at that location.

<h3>What is Continuity in Calculus?</h3>

A function is continuous when there are no gaps or breaks in the graph. These gaps or breaks can be easily seen in a graph. They are also easily stated as holes, jumps, or vertical asymptotes. However, in calculus, you must be more specific in your definition of continuity.

There are three conditions that must be met in order to state a function is continuous at a certain point.

A function f(x) is continuous at a point a if only if the following three conditions are satisfied:

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1 year ago

Triangle pqr has vertices p(-3 -1) q(-1,-7) and r(3,3) and points a and b are midpoints of segment pq and segment rq respectivel

Serhud [2]

Answer:

View graph

Step-by-step explanation:

Be: p(-3,-1) q(-1,-7) r(3,3)

we must find the value of the middle point a and b

$a=\frac{pq}{2}\\b=\frac{rq}{2}$

remember that distance between two points is:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$

pq=distance between p and q

qr =distance between q and r $pq=\sqrt{(-1+3)^{2}+(-7+1)^{2}}=\sqrt{(2)^{2}+(-6)^{2}}=\sqrt{40}\\ pq=2\sqrt{10}\\ a=\frac{pq}{2}=\frac{2\sqrt{10}}{2}=\sqrt{10}\\ coordinates\\a=\frac{pq}{2}=\frac{(-3-1,-7-1)}{2}=(-2,-4)\\\\qr=\sqrt{(3+1)^{2}+(3+7)^{2}}=\sqrt{(4)^{2}+(10)^{2}}=\sqrt{116}\\qr=2\sqrt{29}\\ b=\frac{qr}{2}=\frac{2\sqrt{29}}{2}=\sqrt{29}\\ coordinates\\b=\frac{pq}{2}=\frac{(-1+3,-7+3)}{2}=(1,-2)\\$