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60º 

-3 = -20log10(1-x)      divide both sides by -20

3/20 = log (1-x)

10^(3/20) = 1-x

1.4125 = 1-x

.4125 = -x

x= - .4125

Actually I made a mistake it should be:

(x) + (x + 4) > (20)

2x + 4 > 20

2x > 16

x > 8

So the final answer is  8 < x < 12

I counted 97

Let's put it into   y= mx + b form

2x+4y =1

4y=-2x+1

y=-1/2x + 1/4     -1/2 =m= slope    any line with this slope is parallel

y=-1/2x + n        any 'n' will make a line parallel to the original

30x1.5

The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side. (Think about it, if one side was the same length as the other two combined that would make a straight line. If it was any longer the other two sides couldn't reach the endpoints.)

This means:

(x) + (x + 4) > (20)

2x + 4 > 20

2x > 5

x > 2.5

It also means:

(x) + (20) > (x + 4)

and

(x + 4) + (20) > x

but the last two don't help any because they are true for all values of x, so it doesn't narrow down the answer any more.

Now you also said the triangle is acute. That means that all angles are less than 90°. Since 20 is the longest side length, the angle across from the side with length 20 is the biggest angle. So x has to be less than whatever values make that angle exactly 90. So just find the values of x when the angle opposite the side with length 20 is 90.

Pythagorean theorem says:

x² + (x+4)² = 20²

x² + (x+4)(x+4) = 400

x² + x² + 8x + 16 = 400

2x² + 8x + 16 = 400

Divide everything by 2.

x² + 4x + 8 = 200

x² + 4x - 192 = 0

Use quadratic formula to figure out that:

x = 12 and x = -16

It can't be negative because we are talking about side lengths.

So x has to be less than 12. x < 12

SO the final answer is 2.5 < x < 12

Can you give an example of what you are trying to calculate?

I'm sorry about the unclarity, you got all the assumptions correct. I'll try to explain more details next time. I gave you five points for the answer! :)

Some rather unclear questions in your post, but I'll give my answers with the assumptions:

he has 24 parrots = 12 pairs     12 pairs x 5 hatchlings/pair = 60 hatchlings.

How many pairs?  

   original 12 pairs    + 60 hatchlings  = 42 pairs (assumng equal parts males and females in hatchlings)

" If The birds ate 3 pounds of food a week, and it takes a month to breed, how much food is needed? "

    Is this how much ALL of the 24 birds eat? (...or EACH bird?)

        Assuming ALL 24 birds eat 3 pounds/week and 4 weeks in a month  = 12 pounds food needed.

wow, I've gotta be honest, I did not expet the gcf to be so small!

Set a right-angled triangle with one of the angle = x.

sin x = 5/8 means opposite of x = 5 and hypotenuse = 8.

By means of Pythagoras theorem we get the adjacent of x = \(\sqrt{8^2 - 5^2}\) = \(\sqrt{39}\)

Therefore cos x = adjacent/hypotenuse = \(\dfrac{\sqrt{39}}{8}\)

tan x = sin x / cos x = \(\dfrac{5}{\sqrt{39}}\) = \(\dfrac{5\sqrt{39}}{39}\)

By formula we get:

\(\sin(2x)=2\sin x \cos x = 2\left(\dfrac{5}{8}\right)\left(\dfrac{\sqrt{39}}{8}\right)=\dfrac{5}{32}\sqrt{39}\)

\(\cos(2x)=1-2\sin^2 x = 1- 2\cdot\left(\dfrac{5}{8}\right)^2=\dfrac{7}{32}\)

\(\tan(2x)=\dfrac{\sin (2x)}{\cos (2x)}=\dfrac{\frac{5}{32}\sqrt{39}}{\frac{7}{32}}=\dfrac{5}{7}\sqrt{39}\)

\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\)

Where A and B denotes the angle opposite to a and b respectively.

If you want to find A you have to have the length of sides a and b and the angle B.

So:

\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\\ \dfrac{\sin A}{a}=\dfrac{\sin B}{b}\\ \sin A = \dfrac{a\sin B}{b}\\ A = \arcsin \left(\dfrac{a\sin B}{b}\right)\)

And,

\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\\ b =\dfrac{a\sin B}{\sin A}\\ \dfrac{\sin B}{\sin A}=\dfrac{b}{a}\\ \sin B = \dfrac{b\sin A}{a}\\ B = \arcsin\left(\dfrac{b\sin A}{a}\right)\)

Just exchange all the 'A's into 'B's and all 'a's into 'b's you get the formula for B.

You did this: 

\(\dfrac{3.33^2}{\sqrt5^2}\)

But you should take a square root after squaring both numerator and denominator......

Correct answer:

\(\sqrt{\dfrac{3.33^2}{\sqrt5^2}}\\ =\sqrt{\dfrac{100}{9}\cdot \dfrac{1}{5}}\\ =\sqrt{\dfrac{20}{9}}\\ =\dfrac{\sqrt{20}}{\sqrt9}\\ =\dfrac{\sqrt{20}}{3}\)

And knowing that \(\sqrt{20}=\sqrt{2^2\cdot 5}=2\sqrt 5\)

Final answer = \(\dfrac{2}{3}\sqrt 5\)

I know you did not use Wolfram Alpha..... I just want a really awesome solution......

That means I have agreed with your comment, Question.

I wish I had known that when I had Lucy, I wasted alot of lettuce on her! I tried feeding Pepper mealworms, not sure if he liked them or not,  but the ones from the live bait section at the store die too fast, So I switched back to crickets. :) 

140 pics x $2/pic = $ 280.00 from the sale of the pictures.

Why not just   19 - 3n    ?

Hi Melody!

Perhaps I should have included the following extra step:

 \(\frac{10}{3\sqrt5}\rightarrow\frac{10\sqrt5}{3\times\sqrt5\times\sqrt5}\rightarrow\frac{10\sqrt5}{3\times5}etc\)

i.e. you multiply the expression by 1 in the form \(\frac{\sqrt5}{\sqrt5}\)  Then you multiply the two \(\sqrt5\)s on the bottom together.

Hope this explains it adequately for you.

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How does the 3sqrt5 turn to 3*5?

Hi Alan :))

If the equation is x = (y - 1)/y then Guest #1 is correct.  

If it is  x = y - 1/y  then it is quadratic in y and we have:   \(y = \frac{x\pm \sqrt{x^2+4}}{2}\)

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Like so:  \(\frac{10}{3\sqrt5}\rightarrow\frac{10\sqrt5}{3\times5}\rightarrow\frac{2\sqrt5}{3}\)

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