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(x+1)/(2-x) < x/(3+x)   

This might be solved easiest by some analysis

Let's divide the inequality into three iintervals

(-∞, -3), (-3, 2) and (2, ∞)

Notice that, on the first interval, the function on the left is always less than the function on the right

We can't say what happens at x = -3 because the function on the right isn't defined there

On the second interval, the function on the left is always greater than the function on the right

And at x = 2 the function on the left is undefined

On the third interval, the function on the left is again always less than the function on the right

So, the solution to this problem is  that the intervals (-∞, -3) and (2, ∞)  make the inequality true and the interval (-3, 2) makes it false

Well you have been given lots of information here.

What do you think the answer might be?  How will you try to do it?

It is very similar to this question - just a little easier.

We are not here just to provide answers so that you can plug the answer in and get it right.

If you want more input on this question you should work with us  :)

$${\frac{\left({{\mathtt{Y}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{\left(-{\mathtt{6}}\right)}{\mathtt{\,\times\,}}{{\mathtt{Y}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{\left(-{\mathtt{3}}\right)}\right)}{\left({{\mathtt{Y}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{\left(-{\mathtt{8}}\right)}\right)}}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{{\mathtt{2}}}$$

$${\frac{\left({{\mathtt{Y}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{\left(-{\mathtt{9}}\right)}\right)}{\left({{\mathtt{Y}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{\left(-{\mathtt{8}}\right)}\right)}}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{{\mathtt{2}}}$$

$${{\mathtt{Y}}}^{{\mathtt{0}}}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{\left(-{\mathtt{1}}\right)}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{{\mathtt{2}}}$$

$${\mathtt{1}}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{\left(-{\mathtt{1}}\right)}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{{\mathtt{2}}}$$

$${\mathtt{1}}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{{\mathtt{1}}}$$

$${{\mathtt{X}}}^{{\mathtt{1}}}$$

$${\mathtt{X}}$$

This is a geometric series.....the sum of the first n terms is given by

S_n = a(1 - r^n) / ( 1 - r)

Where a is the first term, r is the common ratio,  and n is the number of terms we wish to sum

So, here  

n = 12  a = 1 and r = 1.5

So we have

S₁₂ = 1(1 - 1.5^12) /(1 - 1.5)   = 257.49267578125 

Y=U Here

get 

afteer billions, are trillions,quadrillions, quintillions, sextillions,septillions,octillions,googls, googolplex, googleplexians etc.

The accumulated amount after 30 years is given by.....

A = 75000(1+ .125)^30  =  $ 2,568,247.87

S^2    where S is the side length

Type in 

log(a, b)     where a is the anti-log {the number you want to take the log of}  and b is the base

solve for D

B = L/(4$${\mathtt{\pi}}$$$${{\mathtt{D}}}^{{\mathtt{2}}}$$)

B(4$${\mathtt{\pi}}$$$${{\mathtt{D}}}^{{\mathtt{2}}}$$) = L

(4$${\mathtt{\pi}}$$$${{\mathtt{D}}}^{{\mathtt{2}}}$$) = L/B

$${{\mathtt{D}}}^{{\mathtt{2}}}$$ = L/(B*4pi)

D = $${\sqrt{{\frac{{\mathtt{L}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{piB}}\right)}}}}$$

The equation for a line is

y = mx + b

In this case the slope is 3/4, so we can fill some parts in

y = 3x/4 + b

Now we find b by plugging in the point that the line passes through or (2,1)

substituting the numbers in to their respective places gives you

1 = 3(2)/4 + b

then simplifying

-1/2 = b

then returning to the original equation we get

y = 3x/4 - 1/2

then getting rid of any fractions (multiply both sides by 4)

4y = 3x - 2

E = energy in joules

M = mass in Kg

C = speed of light in meters = 3.0 * 10^8 meters/second

For a right triangle.

The tangent of an angle = opposite side length/adjaecent side length. (sin(x)/cos(x)) or (y/x) etc.

 - also note that sin(A) = a/c and cos(A) = b/c, and tan(A) = sin(A) / cos(A) = (a/c) / (b/c) which = (a/b)

In the case of angle A˚ tan(A) = a/b

In the case of angle B˚ tan(B) = b/a

In the case of angle C˚(or any right angle), tan(90˚) = undefined

$${\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}^{{\mathtt{5}}}$$ = 12

If you are solving for x then you must isolate it onto one side. 

The first step is to get rid of that pesky 5th power. 

In order to do that, you must take the 5th root of both sides. 

When you do this you remove the 5th power from the x side.

The next step is easy, you isolate x by adding 3 to both sides. 

In the end you get x = 3 + $${\sqrt[{{\mathtt{{\mathtt{5}}}}}]{{\mathtt{12}}}}$$

cool cool :)

you need to meat einstein in the space time continum

millions comes after billions

You need to use the compound interest formula.

do you know if?

Why are you having problems?

 A.   7C2×5C2

B.     7C3*5C1+  7C4 =  5×7C3+7C4

$${\mathtt{6}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{x}}}{{\mathtt{2}}}}\right) = {\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}$$

$${\mathtt{6}} = \left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{x}}}}\right)$$

$${\mathtt{6}} = \left({\frac{{\mathtt{2}}}{{\mathtt{x}}}}\right){\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{x}}}}\right){\mathtt{\,\times\,}}{\mathtt{3}}$$

$${\mathtt{6}} = {\mathtt{2}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{6}}}{{\mathtt{x}}}}\right)$$

$${\mathtt{4}} = \left({\frac{{\mathtt{6}}}{{\mathtt{x}}}}\right)$$

$${\mathtt{4}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{x}}}{{\mathtt{6}}}}\right) = {\mathtt{1}}$$

$${\frac{{\mathtt{X}}}{{\mathtt{6}}}} = {\frac{{\mathtt{1}}}{{\mathtt{4}}}}$$

$${\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{6}}$$

$${\mathtt{X}} = {\frac{{\mathtt{3}}}{{\mathtt{2}}}}$$

20, 37, 19, 43, 26, 29

There are three ways to find the average of these numbers

1.  Add up all the numbers and divide by the number of numbers.

$${\mathtt{20}}{\mathtt{\,\small\textbf+\,}}{\mathtt{37}}{\mathtt{\,\small\textbf+\,}}{\mathtt{19}}{\mathtt{\,\small\textbf+\,}}{\mathtt{43}}{\mathtt{\,\small\textbf+\,}}{\mathtt{26}}{\mathtt{\,\small\textbf+\,}}{\mathtt{29}} = {\mathtt{174}}$$

 $${\frac{{\mathtt{174}}}{{\mathtt{6}}}} = {\mathtt{29}}$$

2.  Put the numbers in order from lowest to highest and the number in the middle is the average.  If there are two numbers in the middle, add both numbers and then divide by two to get your average.

19, 20, 26, 29, 37, 43

$${\mathtt{26}}{\mathtt{\,\small\textbf+\,}}{\mathtt{29}} = {\mathtt{55}}$$

$${\frac{{\mathtt{55}}}{{\mathtt{2}}}} = {\mathtt{27.5}}$$

3.  Look at the numbers and whatever number you have the most of is your average.

Since no numbers repeat, this option will not work.

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