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3 years ago

Determine the horizontal asymptote ..

Mathematics

1 answer:

inessss [21]3 years ago

3 0

If we do the division we get x^2 + 3/x-2
there will be an a vertical asymptote at x = 2 but no horizontal asymptote

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Instruction: use the ratio of a 30-60-90 triangle to solve for the variables. Leave your answers as radicals in simplest form.u=

almond37 [142]

Let's start by identify a 30-60-90 triangle

If we compare those two traingles we can stablish the following relation:

$\begin{gathered} a\sqrt[]{3}=6\sqrt[]{3} \\ \end{gathered}$

Therefore, a=6

u is the opposite side to the 90° angle, therefore

$\begin{gathered} u=2a \\ u=2\cdot6 \\ u=12 \end{gathered}$

on the other hand, v is the opposite side to the 30° angles, so

$\begin{gathered} v=a \\ v=6 \end{gathered}$

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1 year ago

What is the number written in standard notation?

expeople1 [14]

Answer:

option B

Step-by-step explanation:

To find the number which is written in the standard form, $5 \times {10}^{3}$

We expand ${10}^{3}$ and multiply by 5 to get

$= 5 \times (10 \times 10 \times 10)$

$=5\times 1000$

=5000

5 0

3 years ago

Read 2 more answers

What is the surface area of the rectangle pyramid below?

iren [92.7K]

Answer:

Um.. where's the picture?... wheres the rectangle and the measurements.

7 0

3 years ago

What is 15 minutes before 4:35? what time would that be? pls answer today!!!

Andrews [41]

Answer:

4:20

Step-by-step explanation:

35-15= 20

7 0

3 years ago

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(a) Use the definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit. lim n→∞ n i = 1 Correct

Luba_88 [7]

Splitting up the interval of integration into $n$ subintervals gives the partition

$\left[0,\dfrac1n\right],\left[\dfrac1n,\dfrac2n\right],\ldots,\left[\dfrac{n-1}n,1\right]$

Each subinterval has length $\dfrac{1-0}n=\dfrac1n$ . The right endpoints of each subinterval follow the sequence

$r_i=\dfrac in$

with $i=1,2,3,\ldots,n$ . Then the left-endpoint Riemann sum that approximates the definite integral is

$\displaystyle\sum_{i=1}^n\frac{{r_i}^3}n$

and taking the limit as $n\to\infty$ gives the area exactly. We have

$\displaystyle\lim_{n\to\infty}\frac1n\sum_{i=1}^n\left(\frac in\right)^3=\lim_{n\to\infty}\frac{n^2(n+1)^2}{4n^3}=\boxed{\frac14}$

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3 years ago