Ask question

Ask question

All categories

2 years ago

14

How do I do this question: Six numbers have been left to the final door. Those numbers are: 6,2,3,3,2,2. Fill in the following b

lanks, and use those 6 digits to make the largest number possible. Enter that number (without the comma) to escape!

1 answer:

ASHA 777 [7]2 years ago

5 0

Answer:

633222

Step-by-step explanation:

just put them in order from biggest - smallest

You might be interested in

Lori wants to distribute 35 peaches equally into baskets. she will use more then 1 but fewer than 10 baskets. how many baskets d

Ilya [14]

She needs either 5 baskets or 7 baskets.

3 0

3 years ago

Read 2 more answers

If the diameter of the sphere is 12cm what is the volume of the cylinder

german

$r^{2} \times \pi \times (h = height \: of \: the \: cilinder)$

6 0

3 years ago

Explain how they are different AND the same

True [87]

Proportional and linear functions are almost identical in form. The only difference is the addition of the “b” constant to the linear function. Indeed, a proportional relationship is just a linear relationship where b = 0, or to put it another way, where the line passes through the origin

3 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

The point F(5, 5) is translated 7 units left.

QveST [7]

5 - 7 = -2. Therefore, the answer is (-2, 5).

6 0

3 years ago