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3 years ago

Mary travelled y meters from point A to point B at an average speed of 4 m/s. She continued

Mathematics

1 answer:

Rus_ich [418]3 years ago

6 0

Answer: dont know

Step-by-step explanation:

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2. Tom bought 500 shares of a company’s stock for $11.06/share. He pays a broker a commission of $12 to buy and sell stock. Afte

goldfiish [28.3K]

Step-by-step explanation:

a.)Buying: total cost

Total cost= commission + (price per share× Number of of shares ) ;

Total cost= 10 + (11.06×500)=$5540

b.)Net gain or loss;

First, find cash received from sale of stock and deduct commission;

Cash from sale =10×500= $ 5000

deduct commission= 5000-10= $4990

Gain or loss= sale-cost = 4990-5540 = -$550, meaning there is a loss.

c.) Annual rate of return= (net gain or loss/amount paid)×100%

return= -550/(5540)×100 = -9.92%

3 0

3 years ago

Pls help its a grade

Allushta [10]

Answer:

Step-by-step explanation:

3 divided by 9 is 3

3+2=5

6 0

3 years ago

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Answer please. If you answer, you get 44 points. I don't use the points here. So answer please. I want full work.

ra1l [238]

Answer: you simplfy the two equations already there then plug them in to find factors

Step-by-step explanation:

5 0

3 years ago

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The Hyperbolic Sine (sinh(x)) and Hyperbolic Cosine (cosh(x)) functions are defined as such: sin h(x) = e^x - e^-x/2 cosh(x) = e

labwork [276]

Answer:

y-incercepts:

sinh(x):0, cosh(x)=1

Limits:

positive infinity: sinh(x): infinity, cosh(x): infinity

negative infinity: sinh(x): - infinity, cosh(x): infinity

Step-by-step explanation:

We are given that

$\sinh(x)=\frac{e^{x}-e^{-x}}{2}$

$\cosh(x)=\frac{e^{x}+e^{-x}}{2}$

To find out the y-incerpt of a function, we just need to replace x by 0. Recall that $e^{0}=1$ . Then,

$\sinh(0) = \frac{1-1}{2}=0$

$\cosh(0) = \frac{1+1}{2}=1$

For the end behavior, recall the following:

$\lim_{x\to \infty}e^{x} = \infty, \lim_{x\to \infty}e^{-x} = 0$

$\lim_{x\to -\infty}e^{x} = 0, \lim_{x\to -\infty}e^{-x} = \infty$

Using the properties of limits, we have that

$\lim_{x\to \infty} \sinh(x) =\frac{1}{2}(\lim_{x\to \infty}e^{x}-\lim_{x\to \infty}e^{-x})=(\infty -0) = \infty$

$\lim_{x\to \infty} \cosh(x) =\frac{1}{2}(\lim_{x\to \infty}e^{x}+\lim_{x\to \infty}e^{-x}) =(\infty -0)= \infty$

$\lim_{x\to -\infty} \sinh(x) =\frac{1}{2}(\lim_{x\to -\infty}e^{x}-\lim_{x\to -\infty}e^{-x}) = (0-\infty)=-\infty$

$\lim_{x\to -\infty} \cosh(x) =\frac{1}{2}(\lim_{x\to -\infty}e^{x}+\lim_{x\to -\infty}e^{-x}) =(0+\infty)= \infty$

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3 years ago

Who thinks the wap song is weird

Anna35 [415]

Answer:

It a bit vulgar, but Cardi B is still a good artsist

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4 years ago

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