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3 years ago

6

30 points if answered ASAP one day jasmine earned 25.20 for 6 hours of work. another day she earned 16.80 for 4 hours of work. w

hich function shows how much jasmine earns (c) for working (h) hours?

1 answer:

lapo4ka [179]3 years ago

4 0

Answer:

d)c=4.20h

Step-by-step explanation:

the rate is 16.80 /4

4.20 per hour

earn = hours * rate

c = 4.20*h

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Ali, Ben, and Cliff plant seedlings in their neighborhood park. Ali plants 40% of the total number of seedlings, Ben plants 45%

Juli2301 [7.4K]

Answer:

580 seedlings altogether.

Step-by-step explanation:

Given that,

Ali plants 40% of the total number of seedlings, Ben plants 45% of the total number of seedlings, and Cliff plants the rest of the seedlings. Cliff plants 87 seedlings.

The percentage of the seedling planted by Cliff = 100% - 45% - 40% = 15%

<em>Let x = the total number of seedlings</em>

<em /> $15\%\ of\ x=87\\\\x=\dfrac{87\times 100}{15}\\\\x=580$

Hence, three boys plant 580 seedlings altogether.

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2 years ago

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

ozzi

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,

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

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

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3 years ago

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solniwko [45]

Answer:

Step-by-step explanation:

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