All categories

4 years ago

Please help! 50 points and i'll give brainliest!

Mathematics

2 answers:

igomit [66]4 years ago

8 0

Answer:

It's 70.

Step-by-step explanation:

I realllllllly hope this helps and I realllllly hope you give me brainliest.

Nitella [24]4 years ago

3 0

if im correct its 70

You might be interested in

I need help really bad help me pls

olasank [31]

Answer: w= 13

Step-by-step explanation: 8+5 is 13

3 0

2 years ago

Read 2 more answers

Find the area of each figure. Round to the nearest hundredth where necessary. Only 1 and 5 please!

Alex787 [66]

Answer:

See below

Step-by-step explanation:

To begin, it is volume, not area

1. 8 x 12 x 17 = 1632 in ^3

5. 9 x 9 x 13.5 = 1093.5 in^3

If you want to find SURFACE AREA:

1. 2(8 x 12) + 2(8 x 17) + 2(17 x 12) = 872 in^2

5. 2(9 x 9) + 4(9 x 17) = 774 in^2

6 0

3 years ago

A Bag contains eight titles labeled A,B,C,D,E,F,G, and H. One title will be randomly picked. What is the probability of picking

Lynna [10]

Answer:

1/6

Step-by-step explanation:

A and E are vowels so you minus it by 8, which makes it 6.

4 0

3 years ago

Read 2 more answers

Please Help <br><br> help please I hope you help me thanks

velikii [3]

Answer:

Every number of years 365 and I have been trying to get a hold of

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