All Answers 358187 Answers

All my armpits from the bottom adore you Gracias! 

In this problem I see two big triangles so there total height is 11.9 and there base is 16.8 and we can plugh that in the formula for the area of a triangle but we need to divide the height by two....  

11.9/2 = 5.95 

So  \(\frac{1}{2}* 5.95 * 16.8 = 49.98\)

(Note) I with that answer I completed the necessary division... already so 49.98 * 2 = 99.96ft²

If you replace the varaibles you will get:

\(\frac{1}{3}(3^2)+5.2(2)\)

First complete: \(\frac{1}{3}(3^2)\)

\(\frac{1*3^2}{3}\)

\(\frac{3^2}{3}\)

\(3\)

Then complete: \(5.2(2)\)

\(5.2(2)\)

\(10.4\)

Now you have:

\(3+10.4\)

\(13.4\)

Your answer is 13.4

Hope this Helps ;P

Your do know semester test are based on your personal knowledge right, are your proctors aware of you looking for help?

=  10x² + 13x - 30

=  10x² + 25x - 12x  - 30

                                             Factor  5x  out of the first two terms.

=  5x(2x + 5) - 12x - 30

                                             Factor  -6  out of the last two terms.

=  5x(2x + 5) - 6(2x + 5)

                                             Factor  (2x + 5)  out of both remaining terms.

=  (2x + 5)(5x - 6)

Does that help answer your question?

\(\displaystyle \Re{(z)} = a,\text{ so let }z = a +ib.\)

\(\displaystyle |z| =\sqrt{ a^{2}+b^{2}}=1, \\ \text{ so }a^{2}+b^{2}=1,\\ b^{2}=1-a^{2}.\)

\(\displaystyle z^{2}=(a+ib)(a+ib)=a^{2}-b^{2}+2iab,\\ \text{so }\Re{(z^{2})}=a^{2}-b^{2}=a^{2}-(1-a^{2})=2a^{2}-1.\)

Oh my god, i am so sorry i did not mean to reply to your comment Hiylin Link, I mean to post a new answer and sorry if i offend you in anyway :( 


Tip for the person who posted this question:

The form of standard form is ax+by=c 
so you should just plug in values from the slope intercept equation, and put it in a way that would fit the standard form. 

Slope intercept is y=mx+B
and thus if you have a line that contains a slope of 1/2 and passes through the point of (-2,3), you would need to find your equation first, and put it into slope intercept, and then change it and put it as standard form! 
 

\(\ h(x)\ =\ f(x)\cdot g(x)\)

                                          Divide both sides of the equation by  f(x)

\(\dfrac{h(x)}{f(x)}\ =\ g(x)\)

                                          Substitute  6x + 15  in for  h(x)  and substitute  2x + 5  in for  f(x)

\(\frac{6x+15}{2x+5}\ =\ g(x)\)

                                          That is the same as....

\(g(x)\ =\ \frac{6x+15}{2x+5}\)

                                          We can factor  3  out of  6x + 15  because  3(2x + 5)  =  6x + 15

\(g(x)\ =\ \frac{3(2x+5)}{2x+5}\)

                                          Cancel the common factor of  (2x + 5)  in the numerator and denominator.

\(g(x)\ =\ 3\)          and note that   2x + 5  ≠  0   so   x  ≠  -5/2

So we can say     g(x)  =  3     for all possible values of  x  except  x = -5/2

I forgot to remove the square at the end. I redited my work to put the correct end result, thank you for correcting me, I see my error

;P

\(y\ =\ \frac{15\sqrt{24}}{\sqrt{x}}\)

What is  x  when  y = 3 ?  Plug in  3  for  y  and solve for  x .

\( 3\ =\ \frac{15\sqrt{24}}{\sqrt{x}}\)

To solve for  x ,

divide both sides of the equation by  15

then square both sides of the equation

then multiply both sides of the equation by  x

then multiply both sides of the equation by  \(\frac{225}{9}\)

Can you do those things sinclairdragon?  If you get stuck let us know!

Thanks!

Thank you!

Thanks!!

EmeraldWonder, there are a couple of small errors in your work at the end because  \(\frac{8\,*\,5t^2}{5t\,*\,3}\ \neq\ \frac{8t^2}{3}\)

Here is another way to work this problem:

\(\ \phantom{=\quad}\dfrac{24t^3}{15t^4}\cdot \dfrac{5t^8}{3t^6}\)

\(=\quad\dfrac{24\,\cdot\,t^3\,\cdot\,5\,\cdot\,t^8}{15\,\cdot\,t^4\,\cdot\,3\,\cdot\,t^6}\)

\(=\quad\dfrac{120\,\cdot\,t^3\,\cdot\,t^8}{45\,\cdot\,t^4\,\cdot\,t^6}\)          because we can multiply numbers in any order and   24 · 5 = 120   and   15 · 3 = 45

\(=\quad\dfrac{120\,\cdot\,t^{11}}{45\,\cdot\,t^{10}}\)          because   t³ · t⁸  =  t t t · t t t t t t t t  =  t^{(3 + 8)}  =  t¹¹   and   t⁴ · t⁶  =  t^{(4 + 6)}  =  t¹⁰

\(=\quad\dfrac{120\,\cdot\,t^{10}\cdot\,t}{45\,\cdot\,t^{10}}\)          because   t¹¹  =  t t t t t t t t t t t  =  t¹⁰ · t

\(=\quad\dfrac{120\,\cdot\,t}{45}\)          when  t ≠ 0  because we can cancel the common factor of  t¹⁰  in the numerator and denominator.

\(=\quad\dfrac{8t}{3}\)          because   \(\frac{120}{45}\)   reduces to   \(\frac83\)_

can you please tell me what what is 8 ex 2 + 7 ex 2 -6 ex 2 - 3 ex 2 

By "8 ex 2" etc., does ex mean exponent?  Is this trying to say 8² ? 

IF that is what it means, then

8²  +  7²  –  6²  –  3² 

64  +  49  –  36  –  9    =   68 

.

\(\frac{24t^3}{15t^4}*\frac{5t^8}{3t^6}\)

Simplify \(\frac{24t^3}{15t^4}\)

Both 24 and 15 have a common factor of 3

\(\frac{24t^3}{15t^4} = \frac{8t^3}{5t^4}\)

\(\frac{8t^3}{5t^4}*\frac{5t^8}{3t^6}\)

Apply exponent rule: \(\frac{x^a}{x^b} = \) 1/x^b-a

\(\frac{8}{5t}\)

Continue

\(\frac{8}{5t}*\frac{5t^8}{3t^6}\)

Apply exponent rule: \(\frac{x^a}{x^b} = \) x^a-b

\( \frac{5t^2}{3}\)

Continue

\(\frac{8}{5t} * \frac{5t^2}{3}\)

\(\frac{8*5t^2}{5t*3}\)

\(\frac{8t^2}{t*3}\)

\(\frac{8t}{3}\)

Your answer would be \(\frac{8t}{3}\)

Hope this helps ;P

3. Suppose a is jointly proportional to b and c. If a=4 when b=8 and c=9, then what is a when b=2 and c=18.

a = kbc

4 = k(8)(9)

4 = k (72)

4/72  = k  = 1/18

a = (1/18)(2)(18)  =  2

Thanks, hectictar....I didn't even realize we had that.....LOL!!!!

********

\(\sqrt[3]{\frac{8}{\sqrt{27}}}\ =\ \Bigg(\frac{8}{27^{\frac12}}\Bigg)^{\frac13}\ =\ \frac{8^{\frac13}}{27^{\frac12\cdot\frac13}}\ =\ \frac{8^{\frac13}}{27^{\frac13\cdot\frac12}}\ =\ \frac{8^{\frac13}}{(27^{\frac13})^{\frac12}}\ =\ \frac{2}{3^{\frac12}}\ =\ \frac{2}{\sqrt3}\ =\ \frac{2}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}\ =\ \frac{2\sqrt3}{3}\)

Now it is simplified in the form  \(\frac{a\sqrt{b}}{c}\) .

a + b + c  =  2 + 3 + 3  =  8