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finlep [7]
3 years ago
15

There are 40 people at Drake’s sweet sixteen party. Drake’s only rule is that everyone in the party must meet and shake each oth

er’s hand. If all40 guest shake each other’s hand once and only once.
1. How many handshakes are there altogether?
2. How did you get your solution?
3. What’s the quickest way to figure this out if Drake invited 1,000 people?
Mathematics
1 answer:
Nonamiya [84]3 years ago
3 0

Answer:

Given:

Total Number of people at the Drake’s sweet sixteen party = 40

Everyone in the party must meet and shake each other’s hand

1) How many handshakes are there altogether?

There was total of 780 handshakes in the party.

2. How did you get your solution:

If you have 7 people, person seven shakes six other hands, person  

six shakes five other hands, person five shakes fours other hands,  

person  ,four shakes three other hands, person three shakes two other hands,  and person two shakes one hand.  Another way to see it is,

Person 7  +person 6 + person 5 + person 4 +person 3 + person 2 +  person 1

= 6+5+ 4+3+2+1

=>21

so similairly in case of 40 persons

= 39+38+37+36+37+36+35+34 .................+3+2+1

=>780

3) What’s the quickest way to figure this out if Drake invited 1,000 people?

People at Party               Number of Handshakes

     2                                                     1

     3                              1 + 2              = 3

     4                            1 + 2 + 3          = 6

     5                          1 + 2 + 3 + 4      = 10

     6                        1 + 2 + 3 + 4 + 5  = 15

     .

     .

     .                                          

     n                              1 + 2 + ...+ (n-1) =  --------     \frac{n(n-1)}{2}

Hence for n persons the number of handshakes will be  \frac{n(n-1)}{2}

So for 1000 persons the number of handshakes will be

=>\frac{1000(1000-1)}{2}

=>\frac{1000(999)}{2}

=>\frac{999000}{2}

=>499,500

                                                     

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4K+44 = 10k- 10 GIVE ME THE ANSWER AS FAST AS YOU CAN!!!!!
Shalnov [3]
4k+44 = 10 k -10
-4k          -4k 

44 = 6k-10
+10        +10

54=6k
54/6 = 6k/6

9=k     hope this helps

3 0
3 years ago
I don’t know how to solve this
Lady_Fox [76]

Answer:

\theta =2\pi k,\ \ k\in Z\ \\\text{or}\ \\\theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z

Step-by-step explanation:

Given:

\cos \theta-\sqrt{3}\sin \theta=1

Divide this equation by 2:

\dfrac{1}{2}\cos \theta-\dfrac{\sqrt{3}}{2}\sin \theta=\dfrac{1}{2}

Note that

\cos \dfrac{\pi }{3}=\dfrac{1}{2}\\ \\\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}

So, the previous equation is

\cos \dfrac{\pi}{3}\cdot \cos \theta-\sin \dfrac{\pi}{3}\cdot \sin \theta=\dfrac{1}{2}

Remind that

\cos x\cos y-\sin x\sin y=\cos (x+y),

then

\cos \left(\dfrac{\pi}{3}+\theta\right)=\dfrac{1}{2}

The solution of this equation is

\dfrac{\pi}{3}+\theta=\pm \arccos \dfrac{1}{2}+2\pi k,\ \ k\in Z\\ \\\dfrac{\pi}{3}+\theta=\pm \dfrac{\pi}{3}+2\pi k,\ \ k\in Z\\ \\\theta =2\pi k,\ \ k\in Z\ \text{or}\ \theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z

8 0
3 years ago
Which graph represents the solution set of and ?
Anit [1.1K]

Answer:

<h2>GRAPH III</h2>

Step-by-step explanation:

Look at the picture.

We have

y = x² - 4 → move the graph of x² 4 units down (the vertex is (0, -4))

x + y + 2 = 0   <em>subtracy x and 2 from both sides</em>

y = -x - 2 → move the graph of -x 2 units down (the y-intercept os -2)

Only graph III and IV meets these conditions.

The solution is the points of intersection of the graphs.

Therefore your answer is GRAPH III

<em />

8 0
3 years ago
How do you solve (7^-1)^-1 and 5^2 * 5^4 * 5^-3
Katyanochek1 [597]

Answer:

Step-by-step explanation:

exponents raised to a - power  is an inverse function probably not the right word

       a^-b    =  1/(a^b)     example       4^-2   = 1/(4^2)     =  1/16

so   (7^-1)^-1  is a double inverse which is back to the original number  7

      inverted 7 twice

      (7^-1)^-1     = (1/7)^-1     =  1/(1/7)     =   7     trust me or use your calculator

since the coeffients are the same  just add the powers...  some exponent rule

     5^2 * 5^4 * 5^-3   = 5^(2+4-3)   = 5³

here is it written out

5^2 * 5^4 * 5^-3    = 5²  ×   5^4   ×  5^-3    =   (5×5) × (5×5×5×5)  × (1/5×1/5×1/5)

    5×5 × 5× (5×5×5  × 1/5×1/5×1/5)           (5×5×5  × 1/5×1/5×1/5)  = 1

    5×5×5    = 5³

   so

7 0
3 years ago
What’s the answer of a multiplying polynomials (3x-1) to the second power and (x+6) to the second power
Vladimir [108]

9x^4+102x^3+253x^2-204x+36 is the result of multiplying (3x-1) to the second power and (x+6) to the second power

<h3><u>Solution:</u></h3>

Given that we have to find the result of multiplying polynomials (3x-1) to the second power and (x+6) to the second power

"Second power" means the term is raised to power of 2

Therefore,

We have to multiply (3x-1)^2 \text{ and }(x+6)^2

\rightarrow (3x-1)^2 \times (x+6)^2

We can use the algebraic identity to expand the above expression

(a+b)^2 = a^2+2ab+b^2\\\\(a-b)^2=a^2-2ab+b^2

Applying these in above expression, we get

\rightarrow ((3x)^2-2(3x)(1)+1^2) \times (x^2+2(x)(6)+6^2)\\\\\rightarrow (9x^2-6x+1) \times (x^2+12x+36)

Multiply each term in first bracket with each term in second bracket

\rightarrow 9x^2(x^2)+(9x^2)(12x)+(9x^2)(36)-6x(x^2)-6x(12x)-6x(36) + x^2+12x+36

Simplify the above expression

\rightarrow 9x^4+108x^3+324x^2-6x^3-72x^2-216x+x^2+12x+36

Combine the like terms

\rightarrow 9x^4+102x^3+253x^2-204x+36

Thus the above expression is the result of multiplying (3x-1) to the second power and (x+6) to the second power

4 0
3 years ago
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