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1 year ago

Lenny bought x shares of stock for Sy per share last month. He paid

Mathematics

1 answer:

gulaghasi [49]1 year ago

8 0

Lenny's net proceeds algebraically are found as; $\rm |xy + 20 - \frac{98}{100} |$

<h3>What is the equation?</h3>

A mathematical statement consisting of an equal symbol between two algebraic expressions with the same value is known as an equation.

Let the bought share will be $ x, and $y is the price per share

Paid fees to the broker = $ 20

The total amount spent = $(xy+20)

The price at which each share sells = $ A

Broker commission = 2 %

The total amount obtained after selling the shares;

⇒ ax - 2% of ax

⇒ 98 % of ax

Hence, Lenny's net proceeds algebraically are found as; $\rm |xy + 20 - \frac{98}{100} |$

To learn more, about equations, refer;

brainly.com/question/10413253

#SPJ1

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There are two numbers, and one is 7 more than twice the other. The sum of the numbers is 43. What are the numbers? If x is less

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Answer:

43-7=x

Step-by-step explanation:

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2 years ago

Write the ratio as a fraction in simplest form:

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The answer is 4/5 I hope that is right

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3 years ago

I need help ;^; pleeeeease

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2 years ago

if a cylinder has a volume of approximately 125.6 cubic cm and has a height of 10cm what is the length of the radius

Elina [12.6K]

Answer:

radius = 4 cm

Step-by-step explanation:

To find the length of the radius, we will follow the step below;

First, write down the formula for finding the volume of a cylinder

v=πr²h

where v is the volume of a cylinder

r is the radius and

h is the height of the cylinder

from the question given,

v=125.6 and h = 10 cm

we can now proceed to insert the values into the formula and solve for r

note that π is a constant which is equal to 3.14

v=πr²h

125.6 =3.14×r×10

125.6 =31.4 r

Divide both-side of the equation by 31.4

125.6/31.4 =31.4 r/31.4

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2 years ago

Use the Limit Comparison Test to determine whether the series converges.

lianna [129]

Answer:

The infinite series $\displaystyle \sum\limits_{k = 1}^{\infty} \frac{8/k}{\sqrt{k + 7}}$ indeed converges.

Step-by-step explanation:

The limit comparison test for infinite series of positive terms compares the convergence of an infinite sequence (where all terms are greater than zero) to that of a similar-looking and better-known sequence (for example, a power series.)

For example, assume that it is known whether $\displaystyle \sum\limits_{k = 1}^{\infty} b_k$ converges or not. Compute the following limit to study whether $\displaystyle \sum\limits_{k = 1}^{\infty} a_k$ converges:

$\displaystyle \lim\limits_{k \to \infty} \frac{a_k}{b_k}\; \begin{tabular}{l}\\ $\leftarrow$ Series whose convergence is known\end{tabular}$ .

If that limit is a finite positive number, then the convergence of the these two series are supposed to be the same.
If that limit is equal to zero while $a_k$ converges, then $b_k$ is supposed to converge, as well.
If that limit approaches infinity while $a_k$ does not converge, then $b_k$ won't converge, either.

Let $a_k$ denote each term of this infinite Rewrite the infinite sequence in this question:

$\begin{aligned}a_k &= \frac{8/k}{\sqrt{k + 7}}\\ &= \frac{8}{k\cdot \sqrt{k + 7}} = \frac{8}{\sqrt{k^2\, (k + 7)}} = \frac{8}{\sqrt{k^3 + 7\, k^2}} \end{aligned}$ .

Compare that to the power series $\displaystyle \sum\limits_{k = 1}^{\infty} b_k$ where $\displaystyle b_k = \frac{1}{\sqrt{k^3}} = \frac{1}{k^{3/2}} = k^{-3/2}$ . Note that this

Verify that all terms of $a_k$ are indeed greater than zero. Apply the limit comparison test:

$\begin{aligned}& \lim\limits_{k \to \infty} \frac{a_k}{b_k}\; \begin{tabular}{l}\\ $\leftarrow$ Series whose convergence is known\end{tabular}\\ &= \lim\limits_{k \to \infty} \frac{\displaystyle \frac{8}{\sqrt{k^3 + 7\, k^2}}}{\displaystyle \frac{1}{{\sqrt{k^3}}}}\\ &= 8\left(\lim\limits_{k \to \infty} \sqrt{\frac{k^3}{k^3 + 7\, k^2}}\right) = 8\left(\lim\limits_{k \to \infty} \sqrt{\frac{1}{\displaystyle 1 + (7/k)}}\right)\end{aligned}$ .

Note, that both the square root function and fractions are continuous over all real numbers. Therefore, it is possible to move the limit inside these two functions. That is:

$\begin{aligned}& \lim\limits_{k \to \infty} \frac{a_k}{b_k}\\ &= \cdots \\ &= 8\left(\lim\limits_{k \to \infty} \sqrt{\frac{1}{\displaystyle 1 + (7/k)}}\right)\\ &= 8\left(\sqrt{\frac{1}{\displaystyle 1 + \lim\limits_{k \to \infty} (7/k)}}\right) \\ &= 8\left(\sqrt{\frac{1}{1 + 0}}\right) \\ &= 8 \end{aligned}$ .

Because the limit of this ratio is a finite positive number, it can be concluded that the convergence of $\displaystyle a_k &= \frac{8/k}{\sqrt{k + 7}}$ and $\displaystyle b_k = \frac{1}{\sqrt{k^3}}$ are the same. Because the power series $\displaystyle \sum\limits_{k = 1}^{\infty} b_k$ converges, (by the limit comparison test) the infinite series $\displaystyle \sum\limits_{k = 1}^{\infty} a_k$ should also converge.

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3 years ago