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3 years ago

Determine the vertical asymptote for the rational function f(x) = x - 4 over 2x - 3

Mathematics

1 answer:

k0ka [10]3 years ago

5 0

Answer:

x = $\frac{3}{2}$

Step-by-step explanation:

Given

f(x) = $\frac{x-4}{2x-3}$

The denominator cannot be zero as this would make f(x) undefined.

Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non zero for this value then it is a vertical asymptote.

2x - 3 = 0 ⇒ 2x = 3 ⇒ x = $\frac{3}{2}$

Thus x = $\frac{3}{2}$ is the vertical asymptote

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5. Find the thickness of each layer. Use the data to determine the percentage of the total

zmey [24]

Answer:

Exosphere - 10,000 kilometers thick

Thermosphere - 513 kilometers

Mesosphere - 2200 km

Stratosphere - Depends, but most occurring thickness is 50km

Troposphere - 8-14km

Step-by-step explanation:

Double check and verify this information with your lesson since honestly the numbers could very well differ depending on your lesson and when they were taken. I hope this helps though..have a good one

3 0

1 year ago

A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .