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cuase its stupid

The answer will be long.

$${\mathtt{16.666\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

The answer is $${\mathtt{841.2}}{\mathtt{\,-\,}}\left({\mathtt{1.688\: \!83}}\right) = {\mathtt{839.511\: \!17}}$$. Hope it helps

your welcome!

Thanks rosala! Thank you! Hope it helps!

Is it 

$$\mbox{e=mc^2}$$

Thank you DS and Alan!im sure that will help!points for both of u!

This ones awesome reinout , thank u so much for it!i am a big fan of your lunchboxes so id give u a point for this!

That's great, reinout-g !!!!!

You're welcome CPhill...

Thanks, D.S.......

3^5=243

243 is your answer.

Ok. This happened to me. First, reboot your device. Then, wait for a few minutes to turn it in again. Then, go back there. If it doesn't work, delete the app and and download the app again. That will help.

They probably mean writing a number as a product of it's prime factors.

Remember that prime numbers are numbers that can only be divided by itself and 1.

Now every number that is not a prime number can therefore be written as a product of two or more prime numbers. (Since it is the opposite of a prime number). For example 9 = 3*3, 14 = 7*2 and 18 = 3*3*2

Hi there!

I'm not sure where you're having trouble finding the answer to this question so let me help you.

f(8) means you have a function f(x) where you want to enter x = 8.

In this case your function is $$f(x) = \frac{1}{4}x^4+9x^2-3$$

So if we enter x = 8 into this function we have

$$f(8) = \frac{1}{4} \times 8^4+9 \times 8^2-3$$

Now you probably know the rules of operations in math, so first you would need to calculate

$$8^4 = 4096$$ and $$8^2 = 64$$

This would give us

$$f(8) = \frac{1}{4} \times 4096 + 9 \times 64 - 3$$

second we would need to do the multiplications so

$$\frac{1}{4} \times 4096 = \frac{4096}{4} = 1024$$ and $$9 \times 64 = 576$$

Now we have

$$f(8) = 1024 + 576 - 3$$

This gives

$$f(8) = 1597$$

Reinout-g 

-5.6 repeating

7*10^-27*(3*10⁴)² = 7*10^-27*3²*10^4*2 = 63*10^-19 = 6.3*10^-18 

If you were standing at (x,y) = (1.3,4.8), in which horizontal direction is the steepest slope DOWNWARDS  ?

In a flat plane:  The gradient is independent from the place ( co-ordinates ).

The gradient points in every point of the space in the direction of the strongest increase.

f(x,y) z = 2.73x - 1.51y - 1.99

 $$\dfrac{\partial f(x,y)}{\partial x} = 2.73 \quad and \quad
\dfrac{\partial f(x,y)}{\partial y} = -1.51\\
Gradient =
\left(
\begin{array}{r}
2.73 \\
-1.51
\end{array}
\right)\\$$

horizontal direction is the steepest slope UPWARDS  =

$$\tan^{-1}(\frac{-1.51}{2.73}) = -28\ensurement{^{\circ}}.9475759928 + 360 \ensurement{^{\circ}} =331\ensurement{^{\circ}}.052424007$$

horizontal direction is the steepest slope DOWNWARDS  = $$331\ensurement{^{\circ}}
.052424007 - 180\ensurement{^{\circ}} = 151\ensurement{^{\circ}}.052424007$$

Consider the positive x-axis to be zero degrees, the positive y-axis to be 90 degrees, the negative x-axis to be 180 degrees etc.

In that direction, what is the absolute angle between the surface and the horizontal plane?

slope: $$\tan^{-1}( \sqrt{1.51^2+2.73^2 } ) = \tan^{-1}(3.11977563296)
= 72\ensurement{^{\circ}}.2274815207$$

So

x² + y² = 256  

And

(x + y )² = x² + 2xy + y² =  784

So

x² + 2xy + y² =  784

And

x² + y² = 256

And subtracting the second equation from the first, we have

2xy = 528     and dividing by 2 on both sides, we have

xy = 264   ....and that's the product of the two numbers

a² + b² = 256  

(a+b)² = 784  

Now (a+b)² = a² + b² +2ab  so 784 = 256 +2ab

2ab = 784 - 256 = 528

ab = 528/2 = 264

There are 9 possibilities for the person to the right of the host.  For each of those there are 8 possibilities for the person to their right.  For each of those ... etc.  This makes 9! arrangements in all.

$${\mathtt{9}}{!} = {\mathtt{362\,880}}$$ 

One thing you could try rosala, is going to the Customise and control icon (the three horizontal bars at the top right), click on that, select Tools from the pop-up menu, select Clear browsing data, and then choose to clear several weeks worth of browsing data.

I don't know if this will solve your problem, but it's worth a try.

Thanks for the advice CPhill!but is there anyother way to cure Google!

btw i'll give u some points!