Ask question

Ask question

All categories

4 years ago

8

40 equals 1/2 times P what is p

1 answer:

Sauron [17]4 years ago

5 0

Well this equation should look like 40=1/2p this means we have to get p by itself inorder to do that we have to divide p and 40 by 1/2 or 0.5. why divide well because in order to get rid of a multplication you have to divide and why do it to both sides because everything you do to one side you have to do to the other when you divide 40 by 0.5 you get 80, no if you want a fraction you will have to set 40 over 1 to get a fraction as well to be able to simplifyit this gives you 40/1 divided by 1/2 find a common dinominator for 1 and 2 well it can be 2 so lets multiply 1 times 2 to get 2 and 20 times 2 to get 80 now we have 80/2 and 1/2 now multiply them to get 80/2. now when you divide this does it give you 40?? it does so we have found the right answer

Hope this helps

You might be interested in

a cash register has $1.04 in coins. all the coins are pennies and dimes.there are 24 pennies in the cash register. Which represe

VladimirAG [237]

3-1 finds for she did to is to is to to fix l Fick

8 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,

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

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

5 0

3 years ago

What is K if your equation is 49X-84X+K

Elan Coil [88]

Answer:

1

Step-by-step explanation:

if there is just k, the number behind it is always 1

5 0

3 years ago

What is the width of a rectangle with length 25 in. and area 375 in. 2?

Rufina [12.5K]

The width of the rectangle is 15in

6 0

3 years ago

ANSWER ASAPPPP PLEASE BEING TIMEDDDDDDD

eimsori [14]

The second answer is the correct answer

4 0

3 years ago