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3 years ago

Need help on #6 (I will mark brainliest)

Mathematics

1 answer:

tatuchka [14]3 years ago

4 0

a. See the attachment for a graph.

b. The value of y changes by -1.5 for each increase in x of 1. That means the relationship is linear and the slope is -1.5. If we "back up" one in the value of x to make it 0, we add 1.5 to the corresponding value of y to make it 15.00. This tells us the y-intercept is 15 and the equation can be written in slope-intercept form as

... y = -1.5x + 15

c. The slope means each bus ride costs $1.50. The intercept is the cost of the ticket: $15.00.

d. The x-intercept of the graph is 10. This means the ticket has no value left after 10 rides. Barry can use his card to ride the bus 10 times.

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Find e^cos(2+3i) as a complex number expressed in Cartesian form.

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Answer:

The complex number $e^{\cos(2+31)} = \exp(\cos(2+3i))$ has Cartesian form

$\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)$ .

Step-by-step explanation:

First, we need to recall the definition of $\cos z$ when $z$ is a complex number:

$\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}$ .

Then,

$\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}$ . (I)

Now, recall the definition of the complex exponential:

$e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)$ .

So,

$e^{2i-3} = e^{-3}(\cos 2+i\sin 2)$

$e^{-2i+3} = e^{3}(\cos 2-i\sin 2)$ (we use that $\sin(-y)=-\sin y)$ .

Thus,

$e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)$

Now we group conveniently in the above expression:

$e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2$ .

Now, substituting this equality in (I) we get

$\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2$ .

Thus,

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)$

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right]$ .

5 0

3 years ago

THIS QUESTION IS WORTH 33 POINTS!! PLEASE PLEASE HELP ME

3241004551 [841]

It's less than a whole number like 2/1 times 1/2 is 2/2 or 1

6 0

3 years ago

Read 2 more answers

Find the coordinates of the midpoint of the segment whose endpoints are H(5, 13) and K(7, 5). (12, 18) (9, 7) (2, 8) (6, 9)

____ [38]

Answer: D. (6, 9)

<u>Step-by-step explanation:</u>

Midpoint is the "average" of the x's and y's:

Given: (5, 13) and (7, 5)

Midpoint: $(\dfrac{5+7}{2},\dfrac{13+5}{2})$

= $(\dfrac{12}{2},\dfrac{18}{2})$

= (6, 9)

4 0

4 years ago

Suppose f(x)=3x^2-2x <br> Find f (-4)

Lerok [7]

f(-4) = 3(-4)^2 - 2(-4)

f(-4) = 3(16) + 8

f(-4) = 48 + 8

f(-4) = 56

I think the answer is 56 :)

Please give BRAINIEST :)

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3 years ago

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If the ratio is 50g to 300 ml how many ml would be needed for 83g

Natali [406]

Answer:

498 ml would be needed for 83g.

Step-by-step explanation:

♧We can solve using proportion.

$\frac{50 \: g}{300 \: ml} = \frac{83 \: g}{x} \\ \\ \frac{24900}{50} = \frac{50x}{50} \\ \\ 498 = x$

▪Happy To Help <3

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3 years ago