Which type of function best models the data in the table--linear quadratic, logarithmic, cubic, or exponential?

Mathematics

1 answer:

kobusy [5.1K]3 years ago

7 0

Probably the answer is exponential

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Translate the sentence into an inequality.

stealth61 [152]

Answer:

6 (20 + c) > 16

Step-by-step explanation:

the sum of the number and 20 is 20 + c. so you multiply that equation by 6 giving you 6 (20 + c)

4 0

2 years ago

Can somebody explain how trigonometric form polar equations are divided/multiplied?

koban [17]

Answer:

Attachment 1 : Option C

Attachment 2 : Option A

Step-by-step explanation:

( 1 ) Expressing the product of z1 and z2 would be as follows,

$14\left[\cos \left(\frac{\pi \:}{5}\right)+i\sin \left(\frac{\pi \:\:}{5}\right)\right]\cdot \:2\sqrt{2}\left[\cos \left(\frac{3\pi \:}{2}\right)+i\sin \left(\frac{3\pi \:\:}{2}\right)\right]$

Now to solve such problems, you will need to know what cos(π / 5) is, sin(π / 5) etc. If you don't know their exact value, I would recommend you use a calculator,

cos(π / 5) = $\frac{\sqrt{5}+1}{4}$ ,

sin(π / 5) = $\frac{\sqrt{2}\sqrt{5-\sqrt{5}}}{4}$

cos(3π / 2) = 0,

sin(3π / 2) = - 1

Let's substitute those values in our expression,

$14\left[\frac{\sqrt{5}+1}{4}+i\frac{\sqrt{2}\sqrt{5-\sqrt{5}}}{4}\right]\cdot \:2\sqrt{2}\left[0-i\right]$

And now simplify the expression,

$14\sqrt{5-\sqrt{5}}+i\left(-7\sqrt{10}-7\sqrt{2}\right)$

The exact value of $14\sqrt{5-\sqrt{5}}$ = $23.27510\dots$ and $(-7\sqrt{10}-7\sqrt{2}\right))$ = $-32.03543\dots$ Therefore we have the expression $23.27510 - 32.03543i$ , which is close to option c. As you can see they approximated the solution.

( 2 ) Here we will apply the following trivial identities,

cos(π / 3) = $\frac{1}{2}$ ,

sin(π / 3) = $\frac{\sqrt{3}}{2}$ ,

cos(- π / 6) = $\frac{\sqrt{3}}{2}$ ,

sin(- π / 6) = $-\frac{1}{2}$

Substitute into the following expression, representing the quotient of the given values of z1 and z2,

$15\left[cos\left(\frac{\pi \:}{3}\right)+isin\left(\frac{\pi \:\:}{3}\right)\right] \div \:3\sqrt{2}\left[cos\left(\frac{-\pi \:}{6}\right)+isin\left(\frac{-\pi \:\:}{6}\right)\right]$ ⇒