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we have 19-12=7 so we have that the volume of the small part is 7*5*4=140 cubic feet and there is 14.5*12*5=870 so the area is 870+140=1010 cubic feet so we have \(\boxed{1010\text{ cubic feet }}\)

4^(x+3)  = 4^x*4³ = (2*2)^x.64 = (2^x)²*64  =  9².64 =  5184

common denominator is    x (x+6) = x^2 +6x

[3 (x+6) - x(x) ]  /  x^2+6    = 18 / x^2 + 6      equate the numerators

3x+18 - x^2 = 18

3x = x^2

+-3 = x      (throw out -3)

Thank you! :)

You can divide up the pool into two rectangles to calculate the volume. So, you should do: 12*5*14.5+4*7*5.

(remember,order of operations!)

:)

Can you format your Latex correctly? I am having trouble understanding. 
 

Sorry, but I don't see a diagram. Could you please include it?

:)

y  =  (x - 3) / 2                                 x² + y²  =  5

Step 1)  Substitute  (x - 3)/2  for y into the equation  x² + y²  =  5

              --->     x² + [ (x - 3)/2 ]²  =  5

              Square: 

              --->     x² + (x² - 6x + 9) / 4  =  5

              Multiply each term by 4:

              --->    4x² + (x² - 6x + 9)  =  20

              Simplify:

               --->    5x² - 6x - 11  =  0

Step 2)  Use the quadratic formula to find the two solutions:  x₁  and  x₂.

Step 3)  Replace  x₁  into the equation y  =  (x - 3) / 2  to find the value of  y₁.

              You now have the point  ( x₁, y₁ ).

              Replace  x₂  into the equation y  =  (x - 3) / 2  to find the value of  y₂.

              You now have the point  ( x₂, y₂ ).

Step 4)  Use the midpoint formula to find your answer.

From first equation   x = -3y/(1-k)    sub this in to the second equation

3 (-3y)/(1-k)  + y(1-k) = 0

-9y + y (1-k)^2 = 0

y ( k^2-2k-8) = 0                  already stated y cannot be 0

(k+2)(k-4) = 0       k = -2, 4   (if I did the math correctly !) 

Shape is a rectangle and triangle

Rectangle area =  l x w = 9 x 20

triangle area = 1/2 b h = 1/2  9 x 5       Add them together

Edited reason: wrong spelling :(

so we square the diagonal to get 1600 then we subtract 2(30^2-20^2) to get 600 then add 20^2 to get 1000 then we square root it to get \(\boxed{\sqrt{1000}}\)

https://web2.0calc.com/questions/please-help_4628

220^o since arc(HE) and arc(GF) are both 35^o and arc(HG) is 70^o so arc EPF is (360-35-35-70)^o=220^o and we have that arc(EPF) is \(\boxed{220^\circ}.\)

hi guest, great job, but I am slightly confused why you used smiley faces instead of things that may be more understandable. It may just be that there is something wrong with my brain, but I am a little confused on why you took that approach. :)

We get 

(x^4+2x^3−19x^2−32x+48)/(2x3−3x2−23x+12) when we make a common denominator and we then we factor to get 

((x−1)(x+3)(x+4)(x−4))/((2x−1)(x+3)(x−4)) which simplifies to (x^2+3x−4)/(2x−1) or \(\boxed{\dfrac{x^2+3x−4}{2x−1}}\)

rate  x   time = distance

rate1 = 40

rate 2 = 30 m/hr

t1 = t

t2 = t + 1/3         (1/3 =  20 minutes)

Equate the distances

40 t   =   30 (t+1/3)       solve for t      then distance =   40 t

Here is the pic

The similar triangles are shown.

So the answer is easy to find.

\(angle:~draw~the~2~endpoints~to~the ~center ~and~ find ~that~ angle \\ length:~360/[angle]*[circumfrence]\\ area:~360/[angle]*[area ~of ~circle]\)

or

angle: draw the 2 endpoints to the center and find that angle 

length: 360/[angle]*[circumfrence]

area: 360/[angle]*[area of circle]

@lokiisnotdead, sorry for not seeing your post earlier, as I had to go to sleep. 

My way of doing it was similar to Melody's, so I will let her take the spotlight 

:)

Hi Loki,

Most of your logic looks spot on.

Have a look at mine and see if you can work out what is different (I have not looked that hard)

Thanks heureka, I've no idea why my images have sometimes been rejected and sometimes accepted during the last few days!

Thanks Heureka, 

Here is my contribution. 

It is not much different from yours, but the explanation is maybe a little more different.

If a 7-digit number 13ab45c is divisible by 792, what is b?

\(792=8*9*11\)

For this number to be dividable by 8, the last 3 digits must be divisible by 8

456/8=57  no other digits will work so  c=6

\(13ab45c\\ becomes\\ 13ab456\)

Now a number is divisible by 11 if

| (sum of even digits) - (sum of odd digits) | = multiple of 11

Sum of odd digits = 1+a+4+6 = 11+a

Sum of even digits = 3+b+5 = 8+b

absolute value of difference = | 11+a - 8-b | = |3+a-b |  

so 3+a+b can be negative but it must be a multiple of 11

3+a-b = 11k    where k is an integer.

\(-9\le a-b\le9\\ -6\le 3+a-b \le12\\ which\;\; means\\ 3+a-b=11\;\;or\;\;0\\ a-b=8\;\;or\;\;-3\\ a=8+b\;\;or \;\; a=-3+b\)

Now for a number to be divisible by 9, the sum of the digits must be a multiple of 9

so

\(1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=8+b\;\;then\\ \\\quad =19+b+8+b\\\quad =27+2b\\ \quad b=0\;\;or\;\;9\\ b=9 \;\;doesn't \;\;work\\ so\;\; b=0,\;\;a=8 \; \text{is a possible solution}\\ ~\\ 1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=-3+b\;\;then\\ \\\quad =19+b-3+b\\\quad =16+2b\\ \quad b=1,\;\;a=-2 \;\text{which is invalid}\)

If b=9 then a=17 or 6     17 is not good so

If b=9  a=-3+9=6

If b=0  a=8+0=8

So the number could be

1380456   or

1369456

So the number is

\(13ab45c= 13ab456 =1380456\\ a=8\\ b=0\\ c=6\)

LaTex:

-9\le a-b\le9\\ -6\le 3+a-b \le12\\ which\;\; means\\ 3+a-b=11\;\;or\;\;0\\ a-b=8\;\;or\;\;-3\\ 
a=8+b\;\;or \;\; a=-3+b

1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=8+b\;\;then\\ \\\quad =19+b+8+b\\\quad  =27+2b\\ \quad b=0\;\;or\;\;9\\
b=9 \;\;doesn't \;\;work\\
so\;\;   b=0,\;\;a=8 \; \text{is a possible solution}\\
~\\

1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=-3+b\;\;then\\ \\\quad =19+b-3+b\\\quad  =16+2b\\
 \quad b=1,\;\;a=-2 \;\text{which is invalid}

A, B and C each represent a single digit with no two digits being the same.
Find the values of A, B and C so that the following is true: \(AB + C=38\) and \(BC+A=29\)

My attempt:

\(\begin{array}{|rcll|} \hline \mathbf{AB + C} &=& \mathbf{38} \\ 10A+B+C &=& 38 \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{BC+A} &=& \mathbf{29} \\ 10B+C+A &=& 29 \qquad (2) \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)-(2): & 10A+B+C -(10B+C+A) &=& 38-29 \\ & 10A+B+C - 10B -C -A &=& 38-29 \\ & 10A+B - 10B -A &=& 9 \\ & 9A-9B &=& 9 \quad | \quad :9 \\ & A-B &=& 1 \\ & \mathbf{B} &=& \mathbf{A-1} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1): & \mathbf{10A+B+C} &=& \mathbf{38} \quad | \quad \mathbf{B=A-1} \\ & 10A+A-1+C &=& 38 \\ & 11A-1+C &=& 38 \\ & 11A +C &=& 39 \\ & \mathbf{ 11A } &=& \mathbf{39-C} \\ \hline \end{array}\)

\(\begin{array}{|c|c|c|} \hline C & 39-C & A = \dfrac{39-C}{11} \\ \hline 0 & 39 & \text{not integer} \\ \hline 1 & 38 & \text{not integer} \\ \hline 2 & 37 & \text{not integer} \\ \hline 3 & 36 & \text{not integer} \\ \hline 4 & 35 & \text{not integer} \\ \hline 5 & 34 & \text{not integer} \\ \hline \color{red}6 & 33 & \mathbf{A =3} \\ \hline 7 & 32 & \text{not integer} \\ \hline 8 & 31 & \text{not integer} \\ \hline 9 & 30 & \text{not integer} \\ \hline \end{array}\), so \(\mathbf{C=6}\)

\(\begin{array}{|rcll|} \hline \mathbf{B} &=& \mathbf{A-1} \quad | \quad \mathbf{A =3} \\ B &=& 3-1 \\ \mathbf{B} &=& \mathbf{2} \\ \hline \end{array}\)

check:

\(\begin{array}{|rcll|} \hline 32+6 &=& 38 \qquad ( AB+C = 38) \\ 26+3 &=& 29 \qquad ( BC+A = 29) \\ \hline \end{array}\)