All Answers 356325 Answers

1) Hard to see where you went wrong without seeing what you actually did.

2) Isn't the area of a circle a multiple of pi ?

Like so:

Two solutions:

Not sure which you meant by the third (including or excluding the one I did!), so:

The period looks to be 6....the (wacky) wave completes a cycle every 6 units

Amplitude appears to be 2  

The diagonal of a square always = 

side length *  √2

So

7 √2

Radius of semi-circle  = 1.5 cm

Area of   semi-circle  =  pi (1.5)^2  / 2      (1)

Area of rectangle  = 3 * (6.5 - 1.5)  = 3 * 5  = 15      (2)

Area of triangle

One leg = 5cm

The other   =  √[5.5^2 - 5^2 ]   = √5.25  cm    

Area of  triangle = (1/2) product of the legs  = (1/2)(5)√5.25    (3)

Add (1) + (2) + (3)   ≈   24.12  cm^2 

is that f''(x) (2 primes) or f'''(x) (3 primes)?

\(f''(x) = 5+\cos(x)\\ f'(x) = 5x + \sin(x)+ c_1\\ f(x) = \dfrac 5 2 x^2 - \cos(x) + c_1 x + c_2\)

\(f(0) = -1 + c_2 = -1\\ c_2=0\\ f(x) = \dfrac 5 2 x^2 - \cos(x) + c_1 x\)

\(f\left(\dfrac{9\pi}{2}\right) = 0\\ \dfrac 5 2 \cdot \dfrac{81\pi^2}{4} - 0 + c_1 \dfrac{9\pi}{2} = 0\\ \dfrac{405\pi^2}{8} + c_1 \dfrac{9\pi}{2} = 0\\ c_1 = -\dfrac{45\pi}{4}\\ f(x) = \dfrac 5 2 x^2 - \cos(x) -\dfrac{45\pi}{4}x \)

Maybe it is f'''(x)

2(a + bi)  - 3i ( a - bi)  = -7 + 3i

2a + 2bi - 3ai - 3b  = -7 +3i

Equate real and imaginary parts

(2a - 3b)  = 7    ⇒  6a - 9b  = 21

(-3a + 2b) = 3   ⇒  -6a + 4b = 6         

Add the last two equations   and we have

-5b = 27

b = -27/5

And  

2a - (3)(-27/5) =  7

2a + 81/5  = 35/5

2a = [ 35 -81] / 5

2a = -46 / 5

a = -46/10  = -23/5

So

z  =  -23/5  - (27/5) i

a*b   = (a + b)b  =  ab + b^2

So

(3 * 5) - (5 * 3)  =

[3*5 + 5^2]  - [ 5*3 + 3^2]  =

[ 15 + 25 ] - [ 24 ]  =

40 - 24  =

16

Thanks to guest for catching my errors    !!!!

We want

a(2) + 3  = (2) - 5           and        ( -2) - 5  = 2(-2) - b

2a +3 = -3                                     -7   = -4 - b

2a = -6                                             b = 3

a = -3       

a + b   =   -3 + 3   = 0

See the graph, here : https://www.desmos.com/calculator/fsg2n87hro

This is always confusing....UNLESS we are solving an equation...by convention.....we always take the positive root

So

√ [ (-8)^2 + (-6)^2 ]  =  √100  = 10

And

l -10  l   =   10

So.....

"A" is correct

I could find only one such prime: 101

1/101 =0.00990099009900............etc.

Thank you! really appreciate it

3. There are 16 teams in a basketball league. Eight teams are in the East conference, and the other eight teams are in the West conference. If a team plays each other team from the same conference 4 times and from the difference conference twice, how many games are there in total?

Never sure about these....but....here's my best attempt

Suppose that each team plays every other team in its conference once  and every other team in the other conference once

Within one conference......it will be similar to the "handshake" problem

There will be  (8*7)/2  = 28 games withiin each conference  if each team plays each other once

So....the total number of intra-coference games will be four times this = 4 (28 + 28) = 4(56)  = 224 games

If each team plays each team from the other conference once there will be  (8)(8)  = 64 games

But....we must double this  = 128 games

So....the total number of games  =  224 + 128  =  352 games

2. Kathy tossed a coin 8 times and got 3 heads and 5 tails. How many different sequences of results could she have gotten?

This is the same as finding the number of identifiable "words" that can be formed from the "word"  HHHTTTTT

We have

   8!

____  =    56  identifiable "words"  or sequences

 3! 5!

anyone?

1. Everyone at the party shook hands with everyone else exactly once. If there were a total of  21 handshakes, how many people were at the party?

There were 7 people

The 7th person shakes 6 hands

The 6th ,5 hands

The 5th, 4 hands

.

.

.

The 2nd,1 hand

The last...0 hands  (everyone else has shaken this person's hand once)

So

6 + 5 + 4 + 3 + 2 + 1  + 0  =     21

It doesn't say....I showed both cases

Which is correct.....I don't know.....

7                        1  way

6, 1                    2 ways

5,2                     2 ways

4,3                     2 ways

5,1,1                  3 ways

4,2,1                  6 ways

3, 3, 1                3 ways

3,2,2                  3 ways

4,1,1,1               4 ways

3,2,1,1              12 ways

2,2,2,1               4 ways

3,1,1,1,1            5 ways

1,1,2,2,1           10 ways

1,1,1,1,1,2         6 ways

1,1,1,1,1,1,1      1 way

So

64 expressions

7 = 7...........................................1
6 + 1 = 7.....................................2
5 + 2 = 7.....................................2
5 + 1 + 1 = 7...............................3
4 + 3 = 7.....................................2
4 + 2 + 1 = 7...............................6
4 + 1 + 1 + 1 = 7.........................4
3 + 3 + 1 = 7...............................3
3 + 2 + 2 = 7...............................3
3 + 2 + 1 + 1 = 7........................12
3 + 1 + 1 + 1 + 1 = 7...................5
2 + 2 + 2 + 1 = 7.........................4.
2 + 2 + 1 + 1 + 1 = 7..................10
2 + 1 + 1 + 1 + 1 + 1 = 7.............6
1 + 1 + 1 + 1 + 1 + 1 + 1 = 7.......1
                                           Total =64 arrangements.