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4 years ago

Help me with this fast!!!!

Mathematics

1 answer:

pashok25 [27]4 years ago

5 0

Answer:

show in cleary please I donsent see it

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VikaD [51]

Your doctor wants you to take 12 tablets per day.

Each tablet is 250 mg, so 4 tablets make 1g.

4 tablets x 3g = 12 tablets a day

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3 years ago

Read 2 more answers

To solve the equation −x − 3 = x2 − 2x− 15, you could graph of the following system. y = −x − 3 y = x2 − 2x − 15 Use the graph o

soldier1979 [14.2K]

Answer:

I think option (D) 4 is correct

7 0

3 years ago

a pair of shoes costs $45 at joe's Emporium, and the same pair costs $56 at Fran's Shoes. what is the difference in price betwee

Ratling [72]

Answer:

$11

Step-by-step explanation:

to find the difference you subtract

56 - 45 = 11

6 0

3 years ago

please help! 20 points for answering. if you guess or just write something to just get points i will make sure your account gets

Rudik [331]

It's the one with the greater than or less than sign (the second option) because Logan wants to make money greater than or less than $1000

6 0

3 years ago

Express the product of z1 and z2 in standard form given that <img src="https://tex.z-dn.net/?f=z_%7B1%7D%20%3D%20-3%5Bcos%28%5Cf

Marta_Voda [28]

Answer:

Solution : 6 + 6i

Step-by-step explanation:

$-3\left[\cos \left(\frac{-\pi }{4})\right+i\sin \left(\frac{-\pi }{4}\right)\right]\cdot \:2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi }{2}\right)\right]$

This is the expression we have to solve for. Now normally we could directly apply trivial identities and convert this into standard complex form, but as the expression is too large, it would be easier to convert into trigonometric form first ----- ( 1 )

( Multiply both expressions )

$-6\sqrt{2}\left[\cos \left(\frac{-\pi }{4}+\frac{-\pi \:\:\:}{2}\right)+i\sin \left(\frac{-\pi \:}{4}+\frac{-\pi \:\:}{2}\right)\right]$

( Simplify $\left(\frac{-\pi }{4}+\frac{-\pi }{2}\right)$ for both $\cos \left(\frac{-\pi }{4}+\frac{-\pi }{2}\right)$ and $i\sin \left(\frac{-\pi }{4}+\frac{-\pi }{2}\right)$ )

$\left(\frac{-\pi }{4}+\frac{-\pi }{2}\right)$ = $\left(-\frac{3\pi }{4}\right)$

( Substitute )

$-6\sqrt{2}\left(\cos \left(-\frac{3\pi }{4}\right)+i\sin \left(-\frac{3\pi }{4}\right)\right)$

Now that we have this in trigonometric form, let's convert into standard form by applying the following identities ----- ( 2 )

sin(π / 4) = √2 / 2 = cos(π / 4)

( Substitute )

$-6\sqrt{2}\left(-\sqrt{2} / 2 -i\sqrt{2} / 2 )$

= $-6\sqrt{2}\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)$ = $-\frac{\left(-\sqrt{2}-\sqrt{2}i\right)\cdot \:6\sqrt{2}}{2}$

= $-3\sqrt{2}\left(-\sqrt{2}-\sqrt{2}i\right)$ = $-3\sqrt{2}\left(-\sqrt{2}\right)-\left(-3\sqrt{2}\right)\sqrt{2}i$

= $3\sqrt{2}\sqrt{2}+3\sqrt{2}\sqrt{2}i:\quad 6+6i$ - Therefore our solution is option a.

4 0

4 years ago