All Answers 358455 Answers

the answer is -13.5

the answer is 10

No solution!

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(However, 8ⁿ - 1 = 2ⁿ does have a solution.)

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

1.  Write the circle equation in the form (x-xc)² + (y-yc)² = r²,  where (xc, yc) are the coordinates of the centre and r is the radius.  

2. Your circle equation can be written as (x-5)² + (y-11)² = 10², so the centre is at (5, 11).

3. Your straight line can be written in the form y1 = -(12/5)x - 54/5, so it has slope -12/5.

4. The slope of the straight line perpendicular to this is y2 = (5/12)x + k, where k is a constant.

5. For this second line to go through (5, 11) we must have 11 = (5/12)*5 + k so that k = 107/12

6. The equation of the second line is therefore y2 = (5/12)x + 107/12

7. This hits the first line when y1=y2.  The value of x at which this occurs is given by equating the two straight line equations:  -(12/5)x - 54/5 = (5/12)x + 107/12

8.  Rearrange as;  (5/12 + 12/5)x = -54/5 - 107/12  or (169/60)x = -1183/60 so that x = -1183/169 = -7

9. When x = -7 then y1 = y2 = -(5/12)*7 +107/12 = 72/12 = 6

10. The coordinates of P are therefore (-7, 6)

11. The distance between P and Q is given by √( (5 - (-7))² + (11 - 6)² ) = 25√2 ≈ 35.355

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Oops!  Ignore my result for 11.  It should be √( (5 - (-7))² + (11 - 6)² ) =√( (12²+5²) = √169 = 13. (thanks Chris!).

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1+2+3+4+5+6+7+8+9+10=55

$${\mathtt{60}}{\mathtt{\,\times\,}}{\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{15.95}} = {\mathtt{735.95}}$$

its D, $60

hi,your question :

x=2y+12

4x+4y=0

find x and y?

first let x=2y+12 be (1) and 4x+4y=0 be (2)

Sub (1) into (2), you will get the following equation:

4(2y+12) + 4y=0

then solve the equation:

8y+48+4y=0

12y+48=0

y=-4 

then sub y=-4 into (1) , you will get x=2(-4)+12,x=4

so y=-4 and x=4

hope I can help u :) 

$$\frac{12-x}{1+x}=0.3$$

Multiply both sides by (1 + x)

$$12-x = 0.3(1+x)$$

Multpliy each term in brackets on the right-hand side by the 0.3

$$12-x=0.3+0.3x$$

Subtract 0.3 from both sides and add x to both sides

$$12-0.3=x+0.3x$$

Collect like terms

$$11.7=1.3x$$

Divide both sides by 1.3

$$x=\frac{11.7}{1.3}$$

or $$x=9$$

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generallly i used google

2x + 4 = -24

subtract 4 from both sides,

2x = -28

divide both sides by 2,

x = -14

To find 3x, multiply both sides by 3

It seems she drove from 5:30am to 1:00pm  

5:30 to 6 is a half hour

7,8,9,10,11,12,1   that is 7 hours.  So the total time travelling is  7.5 hours

She drove 355 miles

so her speed was   355/7.5    miles per hour

The diameter is all the way acros the coin.  that is 23mm.

the radius is only half way across the coin - so how much would that be?

the area of a circle is    A= pi*r²  = pi*r*r

pi is about 3.14  

so you can work that out .  the units will be mm²

now volume is just A*thinkness and the units will be  mm³

x=2y+12

4x+4y=0 

find x and y 

Let 

This answer is quite similar to mine Nauseated.

You may have made it a smidge easier by getting a common denominator in the first place but otherwise our answers are the same.    

SEE great minds do think alike!  

Thanks Rosala - It is Australia Day too !!

Our Great nations have much in common :))

@@ End of Day Wrap    Mon 26 /1/15     Sydney, Australia      Time 9:25 pm

How did you guess  -   It was a Public Holiday in Australia today.     

Australia Day marks the anniversary of the 1788 arrival of the First Fleet in Sydney Cove.

(Well the first boat anyway)  

Now down to business.

There were a lot of questions and some very busy  answerers today.  Thank you Heureka, Sasini, CPhill, MsTeya, Saseflower, Newbie, Alan, SevenUP, and I think I just saw a pop-up for Nauseated.  

And the cheer goes out.   

Now for the interest Posts

Another Solution

$$Prove \\\\
\frac{\tan \left(a\right)}{1-\cot \left(a\right)}+\frac{\cot \left(a\right)}{1-\tan \left(a\right)}=\sec \left(a\right)\csc \left(a\right)+1 \\$
\\
LHS
\\\\
\hspace*{0.7cm}\ $=\frac{-\tan ^2\left(a\right)+\tan \left(a\right)-\cot ^2\left(a\right)+\cot \left(a\right)}{\left(\tan \left(a\right)-1\right)\left(\cot \left(a\right)-1\right)}$ \\

$=\frac{-\left(\frac{\cos \left(a\right)}{\sin \left(a\right)}\right)^2-\left(\frac{\sin \left(a\right)}{\cos \left(a\right)}\right)^2+\frac{\cos \left(a\right)}{\sin \left(a\right)}+\frac{\sin \left(a\right)}{\cos \left(a\right)}}{\left(-1+\frac{\cos \left(a\right)}{\sin \left(a\right)}\right)\left(-1+\frac{\sin \left(a\right)}{\cos \left(a\right)}\right)} \\\\

$=\frac{\cos ^2\left(a\right)+\sin ^2\left(a\right)+\cos \left(a\right)\sin \left(a\right)}{\cos \left(a\right)\sin \left(a\right)}\\\

\hspace*{3.0cm}$\mathrm{Using \; identity}: \cos ^2\left(x\right)+\sin ^2\left(x\right)=1 \\$

$=\frac{1+\cos \left(a\right)\sin \left(a\right)}{\sin \left(a\right)\cos \left(a\right)}$
\\\\$$

$$RHS \\\\
\sec \left(a\right)\csc \left(a\right)+1 \\$

\ Transform using Sin and Cos \\

$=1+\frac{1}{\cos \left(a\right)}\frac{1}{\sin \left(a\right)}\\\

\ Simplify \\

$=\frac{1+\cos \left(a\right)\sin \left(a\right)}{\cos \left(a\right)\sin \left(a\right)}\\$$

Snarky Comments pending . . ..

Tues 27/1/15

1) Chris's retort was great - I had not heard this saying before.  Thanks Chris

How about the following:

Perhaps I should clarify a step:  Note that n*rⁿ = r*n*r^n-1= r*d(rⁿ)/dr  (or perhaps this just muddies the waters - in which case ignore it!)

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