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3^x-3^x-1=54

With what you have written, the first two terms will "cancel" leaving us with

- 1 = 54

Which is impossible.......

I forgot to add how I arrived at that formula......

Draw a diagonal across the bottom of the cube.....this will be the hypotenuse of a right triangle whose legs are both 'a.'

Then, the length of this will be   a√2

Now.....another right triangle is formed by this diagonal and the edge of the cube extending from one end of the diagonal to a top vertex. These are the two legs. The hypotenuse is the distance from the bottom vertex on the other end of this diagonal to this top vertex.  This is what we're looking for. It is given by √[a^2 + ( a√2)^2]  = √[a^2 + 2a^2]  = √[3a^2]  =  a√3  .......!!!!

I assume you have

log16^(⁴√[1/4096])

I admit.....this isn't very intuitive......

Hit the  "  ^x√y  "  key, first

Now key this in just like this....    sqrt(1/4096, 4)  ... and hit the "= " key.....you should get   .125

So now  just key in " log16^.125 " and hit the "=" key again......that should do it....

Ah..thanks, Melody....I forgot that this angle could also be in the 4th quad, so I should have added that " -" to my answer...!!!

Notice that the center of the cube and the center of the sphere will coincide. Now....let's forget the sphere for a moment.

Let us take some cube of edge length, a.

A diagonal drawn from a bottom vertex of the cube to an opposite top vertex will go through the center of the cube and have a length given by  √[a^2 + a^2 + a^2]  = √[3a^2]  = a√3. And 1/2 of this length will be from the center of the cube to either the top vertex or the bottom vertex........whatever your preference. And this equals a√3/2.

But.....from the center of the cube to either vertex  will be where that vertex touches the sphere. And since the center of the cube and the center of the sphere coincide,  this distance will be = to the radius of the sphere. Thus.....the radius = a√3/2....

Does that make sense???

7ba is the same as 7ab, so therefore 3ab and 7ba are like terms.

=10ab +2b

If you are then asked to factorize:

that out the highest common factor (2b) so;

= 2b(5a+1)

I always draw a quick pic for these.

let 

$$\\\theta=sec^{-1}u\qquad so \\
u=sec\;\theta\\
sec\;\theta = \frac{1}{cos\theta}=\frac{hyp}{adj}=\frac{u}{1}$$

Now I can read the positive answer straight off the triangle. 

$$cot(sec^{-1}(u))=cot\;\theta = \frac{adj}{opp}=\frac{1}{\sqrt{u^2-1}}$$

NOW u is positive so sec(theta) and cos(theta) are positive. So theta is in the 1st or 4th quadrant.

So cot(theta) can be positive or negative so

$$cot(sec^{-1}(u))=\pm\;\frac{1}{\sqrt{u^2-1}}$$

Well Chris, you have given lots of choices there  LOL    :))

$$\\24=4*6=4\div \frac{1}{6}\\\\
and\\\\
24=3*8=(4-1)*8\\$$

Brilliant as usual, heureka....!!!!

In a triangle ABC, a,b,c are 3 opposite side of each angle A,B,C, if a^2-c^2=2b, and SinA*CosC=3*CosA*SinC, find b

$$\boxed{ (1)\quad a^2-c^2=2b\qquad (2) \quad SinA*CosC=3*CosA*SinC}$$

$$\small{\text{ sine rule: }} \frac{sinA}{a}=\frac{sinC}{c} \quad \Rightarrow \quad sinA=\frac{a}{c}\cdot sinC \\
(2):\qquad \frac{a}{c}\cdot sinC\cdot cosC = 3cosA\cdot sinC\Rightarrow
\boxed{ a\cdot cosC=3c\cdot cosA }\\
\small{\text{ in every triangle: }}b=a\cdot cosC+c\cdot cosA\Rightarrow\boxed{ c\cdot cosA = b - a\cdot cosC }\\
a\cdot cosC=3 (b - a\cdot cosC)\\
4\cdot a\cdot cosC = 3b\\
\boxed{ a\cdot cosC = \frac{3}
{4}\cdot b }\\\\
\small{\text{ law of cosines: }} c^2=a^2+b^2-2ac\cdot cosC\\
(1): \qquad a^2-c^2=2ab\cdot cosC-b^2 =2b\\
2ab\cdot cosC-b^2 =2b \quad | \quad :b \\
2\cdot a\cdot cosC-b =2 \quad | \quad a\cdot cosC = \frac{3}{4}\cdot b \\
2\cdot \frac{3}{4}\cdot b - b =2\\
\frac{3}{2}\cdot b - b =2 \quad | \quad \cdot 2 \\
3b-2b=4\\
b=4$$

Enter this in the calculator on the home page and you'll get your answer 

4/3 * 134.69

or if you aren't a lazy person then divide the whole number by 3 until you can no more divide and multiply the answer by 4 ! That's easy too! Just I like not to use a lot of energy! ;)

what is Φ6?

$$\!\ \varphi(6) = 2. \qquad 1 \text{ and } 5 \\\\
6 = 2 \cdot 3\\
\!\ \varphi(6)}=6\cdot (1-\frac{1}{2})( 1-\frac{1}{3} )}=2$$

200C8×0.04^8×0.96^192

TOO MANY                  

The ladder's length forms the hypotenuse of a right triangle.....and we're trying to find the opposite side.......the function that relates these is the sine.....so we have

Sin 20 = Opp / 10   rerrange

10* Sin 20  = Opp = height of wall = 3.42m

the answer is 21!!!

3y-x=8

1st.....let x = 0 ...then 3y = 8    .....divide by 3 on each side  ....so y = 8/3

So we have one point on the graph  ....(0, 8/3)

Now....to get a second point, let y= 0   ..... then we have .... - x = 8 ,  ....and it's obvious that x = -8  ..... so, our second point is (-8, 0)

Plot both these points on a graph and draw a kine through them.....that will be your graph

The answer is given by....

50(23) + 15  =   ???

Since this is a parallelogram and KM = LN, this must be a rectangle (at least) with each corner angle = 90 degrees

And consecutive angles in a parallelogram are supplemental - they add to 180 degrees

Therefore, <KLM and <LMN  = 180

 {I think you must mean that <LMN = 90 = 9x.....therefore x = 10}

So we have

2xy + 9x = 180

2(10)y + 90 = 180

20y = 90      

y = 90/20 = 9/2 = 4.5

9/60×100 percent

9^2 is 9 times itself two times.

9^2 = 9 * 9 = 81

4x-4=-3x+45

7x=49

x=7

Yeah, I know :)