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3 years ago

8

Charlie is mailing packages each small package costs hin $2.06 to send each large package costs him $4.30 how much will it costs

him to send 7 small packages and 5 large packages

1 answer:

vodka [1.7K]3 years ago

4 0

Multiple 7 x 2.06 and 5 x 4.30

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Use the drop-down menus to complete the statements about factoring 14x2 + 6x – 7x – 3 by grouping. The GCF of the group (14x2 –

Yuliya22 [10]

14 x² + 6 x - 7 x - 3 =
= ( 14 x² - 7 x ) + ( 6 x - 3 ) =
= 7 x ( 2 x - 1 ) + 3 ( 2 x - 1 ) =
= ( 2 x - 1 ) ( 7 x + 3 )
Answer:
1. GCF of the group ( 6 x - 3 ) is 3.
2. The common binomial factor is 2 x - 1.
3. The factored expression is: ( 2 x - 1 ) ( 7 x + 3 ).

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3 years ago

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Y(1 - 2x) + xy²(I - 2x)²

Ghella [55]

How do you want this solved?

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3 years ago

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Shortern this expression pls

pogonyaev

Answer:

$c =\frac{8}{3}$

Step-by-step explanation:

Given

$c = \sqrt{\frac{4 + \sqrt 7}{4 - \sqrt 7}} + \sqrt{\frac{4 - \sqrt 7}{4 + \sqrt 7}}$

Required

Shorten

We have:

$c = \sqrt{\frac{4 + \sqrt 7}{4 - \sqrt 7}} + \sqrt{\frac{4 - \sqrt 7}{4 + \sqrt 7}}$

Rationalize

$c = \sqrt{\frac{4 + \sqrt 7}{4 - \sqrt 7} * \frac{4 + \sqrt 7}{4 + \sqrt 7}} + \sqrt{\frac{4 - \sqrt 7}{4 + \sqrt 7}*\frac{4 - \sqrt 7}{4 - \sqrt 7}}$

Expand

$c = \sqrt{\frac{(4 + \sqrt 7)^2}{4^2 - (\sqrt 7)^2}} + \sqrt{\frac{(4 - \sqrt 7)^2}{4^2 - (\sqrt 7)^2}$

$c = \sqrt{\frac{(4 + \sqrt 7)^2}{16 - 7}} + \sqrt{\frac{(4 - \sqrt 7)^2}{16 - 7}$

$c = \sqrt{\frac{(4 + \sqrt 7)^2}{9}} + \sqrt{\frac{(4 - \sqrt 7)^2}{9}$

Take positive square roots

$c =\frac{4 + \sqrt 7}{3} + \frac{4 - \sqrt 7}{3}$

Take LCM

$c =\frac{4 + \sqrt 7 + 4 - \sqrt 7}{3}$

Collect like terms

$c =\frac{4 + 4+ \sqrt 7 - \sqrt 7}{3}$

$c =\frac{8}{3}$

4 0

3 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

A word processing program requires a user to enter a 7-digit registration code made up of the digits 1,2,4,5,6,7 and 9. Each num

ahrayia [7]

For the first digit there are 7 choices. For the second digit there are 6 choices (because we can't use the same one again). And so on, until there is only one choice for the last digit.

The number of possible codes is this:
7*6*5*4*3*2*1 = 7! = 5040

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3 years ago