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0.8 zero point eight

cos2A = cos^2(A) - sin^2(A)

So, if cos(A) = -3/5......then

1 - cos^2(A) = sin^(A)

1 - (9/25) = sin^2(A)

16/25 = sin^2(A)    

So

cos(2A) = (9/25) - (16/25) =  -7/25

x = x^1

So

x^1*x^2 = x^3       .......remember a^m * a^n = a^(m+n) .....

1.8 = 1 + 8/10 = 1 + 4/5 = 9/5

the answer is 85695064

yes but your answer will always be zero ex. 4 times 0 is 0

the answer is 0.0180

Thank you -Isaac

Call "x" the total....then...

.60x = 1449     .....divide 1449 by .60

x = 2415

8ft. x 5.6ft = 44.8 sq. ft.

I don't see any "m" in the equation to substitute anything into.......????

Thanks, heureka......that's a lot of work for just 3 points !!!!

@@ End of Day Wrap       Mon20/10/14        Sydney, Australia Time        12:10pm (Really Tues morn)   ♬

Hello everyone,

Great answers today from SevenUP, CPhill, Alan, Geno3141, Landry, Dragons, Heureka and DavidQD.

You are all fantastic.       

Now for the interest posts:

1) These inequalities look interesting.    Thanks Chris and Alan.

Express in the form a+bi where a&b are real numbers.

1. (2+5i)-(3-4i): 2 + 5i - 3 +4i = -1 +9i

2. (4-3i)(2+i): 4*2 +4i -6i -3i*i = 8 -2i -3(-1) = 11 - 2i       | i*i = -1 !

3. i (3-2i)^2: i(3*3-2*3*2i + 4i*i) = i(9-12i-4) = i(5-12i) = 5i -12i*i = 5i + 12 = 12 + 5i

4. i^52: (i*i)^26 = (-1)^26 = 1

5. 1+5i / 1-2i = $$\small{\text{
$
\frac{1+5i} {1-2i} = \frac{(1+5i)(1+2i)} {(1-2i)(1+2i)} = \frac{1*1+7i+10i*i}{1-4i*i}=\frac{1+7i-10}{1+4} =\frac{-9+7i}{5} =-\frac{9}{5}+\frac{7}{5}i
$
}}$$

6. 7+3i / 4i = $$\small{\text{
$
\frac{7+3i} {4i} = \frac{(7+3i)(-4i)} {(4i)(-4i)} = \frac{-28i-12i*i}{-16i*i}=\frac{-28i+12}{16} =\frac{ -7i+3 }{4} =\frac{3}{4}-\frac{7}{4}i
$
}}$$

7. 1 / a+2i = $$\small{\text{
$
\frac{1} {a+2i} = \frac{1*(a-2i)} {(a+2i)(a-2i)} = \frac{ a-2i }{a^2-4i*i}=\frac{a-2i}{a^2+4} =\frac{a}{a^2+4}-\frac{2}{a^2+4}i
$
}}$$

Tues 21-10-14

                    ♫♪  ♪ ♫                                ♬ ♬ MELODY ♬ ♬                           ♫♪  ♪ ♫

It is the question that is inaccurate.  GoldenLeaf's working appears to be good.      

Re-post this again......nothing is showing up....

Thanks for explaining that Alan.  :)

Thanks Alan.   

(1.6755133009522 x 10^13) / (7.122^3)

now you can do it on the calc on the home page. 

(4x+12)-(3x+10)   distribute the "-" across the terms in the second set of parentheses

4x + 12 - 3x - 10    combine like terms

x + 2

OK...that's not as bad as I thought it would be......a couple of G & T's later on would probably help "reinforce" the concept!!!!

The volume of the cube = (7cm)^3 = 343cm^3

Therefore, the cylinder has the same volume (assuming no loss due to melting)...so we have

343 = pi*r^2 *h  →

343 =pi*(1.5)^2*h →

343 / [pi *(1.5)^2] = h =  about 48.52cm

I was already doing it before you answered LOL.  :)

 Prove L'Hopital's rule for the following special case.

$$\\$Prove $\displaystyle\lim_{x\rightarrow a}\frac{f(x)}{g(x)}\; = \;\displaystyle\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\quad\\\\\\ $When $ g(a)=f(a)=0 $ and $ f'(a) $ and $ g'(a)$ exist $ \\\\\\ \displaystyle\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\;=\;\frac{\displaystyle\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}} {\displaystyle\lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a}}\\\\\\\
\displaystyle\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\;=\displaystyle\lim_{x \rightarrow a}\;\left(\frac{\frac{f(x)-f(a)}{x-a}} {\frac{g(x)-g(a)}{x-a}}\right)\\\\\\ \\\displaystyle\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\;=\displaystyle\lim_{x \rightarrow a}\;\left(\frac{f(x)-f(a)}{ g(x)-g(a)}\right)\\\\\\
\displaystyle\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\;=\displaystyle\lim_{x \rightarrow a}\;\left(\frac{f(x)-0}{ g(x)-0}\right)\\\\\\
\displaystyle\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\;=\displaystyle\lim_{x \rightarrow a}\;\frac{f(x)}{ g(x)}$$