All Answers 356328 Answers

Call   the  sides  a , b

So

ab = 676

b  =  676/a

Perimeter  =   2 (  a  + b)

P   =  2  (  a   + 676 / a )

P  =  2a  +  1352 / a                     take  the derivative     and set to   0 

P'  =  2    -  1352/a^2    =   0

2 =  1352/a^2

a^2  = 1352 / 2

a^2  =  676

a  =  26

b = 676 /  26    =  26

Min perimeter  =   2  ( 26  + 26)   =   104

                      b 

           x         x

a                    d                 x               c

angle bac  = 70

ab  =bd  ....so.....

angle  adb   =  70

So  angle  abd   =  180 - 2(70)  =  40

And  angle  bdc =   180  - 70   =  110

bd = dc

So  angle  dbc =  (180  -110) / 2  =   35

So....angle   abc  =  angle abd   + angle  dbc =    40  +  35   =  75°   

pi  (7.5 / 2)^2 * 1.4   ⇒  61.85    m^3

61.85   *  1000   =  

61850  L

                   P

           13                     18

Q                      M           7              R

Using the Law of Cosines twice

(PQ^2 - QR^2  - PR^2)  /  (-2 QR * PR)  =  cos  (PRQ)

( 13^2  - 14^2  - 18^2)  / ( - 2 * 14 * 18)  =   cos (PRQ)

              39  / 56   =  cos (PRQ)

And

PM^2  =   MR^2  + PR^2   - 2 ( MR * PR)  cos (PRQ)

PM^2  =  7^2  +  18^2  - 2 ( 7 * 18)   (39/56)

PM^2  =  197.5

PM  = sqrt  (202)   ≈   14.05

Intersecting Chord Theorem

AQ  *  QB  =  CQ *  QD

6  *  14    =     x  *  ( 38 - x)

84   =  38x   -  x^2

x^2   -38x  +  84  =  0

x =      (  38 -   sqrt  [ 38^2  - 4* 84 ] )  /   2   ≈  2.356  =  min for  CQ

Hey there, Guest!

I might've missed a few permutations that I'll find in the future, but for the moment, I'm pretty sure that the answer is 35.

Here's why:

4 0 0 0 x4 boxes

3 1 0 0 x12 permutations amongst the four boxes

2 2 0 0 x6

2 I 1 0 x 12

1 1 1 1

6 + 4 + 12 + 12 + 1 = 35.

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Guest!

It appears that the question that you asked has already been answered here: https://web2.0calc.com/questions/more-hw-please-help.

In case if you're one of the people online of which don't like links, this is what it says:

In triangle PQR, PQ=13, QR=14, and PR=15. Let M be the midpoint of QR. Find PM.

Let angle PRQ be  >β<

1)   Let's find β  first:             cos(β) = (15² + 14² - 13²) / 2*15*14       β = 53.13°

2)   Now we can find  PM      (PM)² = 7² + 15² - 2*7*15 * cos(β)    

                                                 PM = 12.166     

Answerer: Dragan

Hope this helped! :)

( ﾟдﾟ)つ Bye

can u explain where the 200 comes from?

WLOG, let there be $6+3+1=10$ games played.

There are 6 wins, 3 losses, and 1 tie.

a) $\frac{6}{10} = \frac{3}{5}$

b) $\frac{3}{10}$

c) $\frac{1}{10}$

d) $6 \cdot 5 = 30$ wins, $3 \cdot 5 = 15$ losses, and $1 \cdot 5 = 5$ ties.

e) $6 \cdot 8.2 = 49.2$ wins, $3 \cdot 8.2 = 24.6$ losses, and $1 \cdot 8.2 = 8.2$ ties. This is absurd, so I am positive that (e) is wrongly written.

This problem can be rephrased into: What is the x-value of the vertex of the parabola $t^2 - 12t - 36$?

The formula for $x$ is $- \frac{b}{2a},$ for polynomial $ax^2 + bx + c.$

We have: $-\frac{-12}{2} = \boxed{6}$

You can see that the graph shown here: https://www.desmos.com/calculator/fztszalids is least when $x=6.$

First off, we do offer explanations. Most of those answer-only guest answers are wrong and they are misleading the OP, which the OP kind of deserves due to not paying attention in class. Obviously, you attended a class and that is why you have that as your homework. If you played video games or such, they obviously don't know how to do their homework. I daresay they are doing video games or watching videos after they post the question, waiting for a response. I realise that some people take all the work and copy it and mark it as their own, but that's warez. Cheating. Plagiarism. Even if they don't get caught, realize that if they don't learn, it's their problem, and so they won't have as smooth life as others. You don't need to beat a dead horse.

Yep. 

Not sure why it was posted again. 

=^._.^=

This looks a lot like catmg's word problem: https://web2.0calc.com/questions/math_66748#r1

x^2 + 2x + 5 = (x+1)^2 + 3. 

At x = c, the parabola becomes a reflection of itself. 

Where is c?

=^._.^=

You're welcome. :))

=^._.^=

Correct! Thank you!

The total number of ways to get to (5, 7) is 12!/5!/7!. 

The total number of ways to get to (5, 7) while passing through (4, 3) is 7!/3!/3! * 5!/1/!/4!

Any idea how to solve this?

=^._.^=

0 0 0 4

0 0 1 3

0 0 2 2

0 1 1 2

1 1 1 1

I count 5. :))

=^._.^=

-3 + 4 = 1

1 + 4 = 5

6 + 4 = 10

Add 4 to each input. :))

=^._.^=

t^2 -12t - 36 = (t-6)^2 - 72

Do you have an idea on how to solve this now?

=^._.^=

Let's put ax^2 + bx + c into the form a(x-h)^2 + k because I think it's easier to work with. 

(1) The maximum value of y = ax^2 + bx + c is 5, which occurs at x = 3.

That means that a is a negative value, that h = 3, and that k = 5. 

a(x-3)^2 + 5

(2) The graph passes through the point (0,8).

8 = a(0-3)^2 + 5

Next, you want to solve for a, and then find m by plugging in x = 4. 

=^._.^=

After doing some long division, 6/11 = .545454.......... 

So what's in the 100th place?

=^._.^=

I assume you want to expand this?

(-x^2)*(x^3+3/2x+1)

\( (-x^2)*(x^3+\frac{3}{2}x+1)\)

Muliply each term in the second bracket by -x^2

\(=(-x^2)*(x^3)\quad+\quad (-x^2)* (\frac{3}{2}x)\quad +\quad (-x^2)* (1)\)

now simplify each term separately.

That is excellent Krxzy    

Thank you Bginner for your correct answer.

Guest you sound like a smart #@#  and that will not get you far in life.