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Math is life.

IF    y is math 

             x = the answer

cpilllllll

Volume of tank =   (1.75 x 12 )in    x   (1.25 x 12) in   x   11 in =   3465  in³

3465 in^3   /  ( 231 in³ /gal)    x   $ 3 /gal  =  $ 45 to fill ENTIRE tank....but you only need 80% of this as tank is 20% full

      80% x $45 = .8 * 45 =  $ 36   to fill the already 20% full tank

I don't know if this is the easiest way, but it's the way that I saw.

I'm going to find the area of the rhombus and then use the formula:  Area  =  ½ · d₁ · d₂ 

Since a rhombus has 4 equal sides, each side has value:  68 / 4  =  17.

The area of the triangle formed by two sides of the rhombus and the diagonal can be found by using Heron's

formula:  Area  =  sqrt[ s(s - a)(s - b)(s - c) ]  where a, b, and c are the lengths of the three sides and 

s is the semiperimeter:  s  =  (a + b + c) / 2.

s  =  (17 + 17 + 16) / 2  =  25

Area(triangle)  =  sqrt[ 25(25 - 17)(25 - 17)(25 - 16) ]  =  120

Therefore, the area of the rhombus  =  2 x 120  =  240

Now using the formula:  Area  =  ½ · d₁ · d₂    --->   240  =  ½ · 16 · d₂   --->   d₂  =  30

1)

(a)

y  =  k *  2x  / z^3

So

8  = k *  2(6) / 2^3

8 = k  *  12  / 8

8 = k * 3 / 2

(2/3)*8  =  k

16/3  = k

(b)

So  when   y = 9   and z  = 1    we  have that

9 = (16/3) * 2x  / 1^3

9 = (32/3)x

9  (3/32)  = x

27 /32  =  x

The consecutive integers are:  x,  x +1,  x + 2,  x + 3, x + 4

The mean of these integers is:  x + 2

Since the mean is 21:  x + 2  =  21

which means that x (which is the smallest) is 19.

4 ( x + 7) ( 2 - x)      expand

4 ( -x^2  - 7x + 2x  + 14)

4 ( -x^2  - 5x  + 14)

-4x^2  - 20x  + 56

The  max value occurs  at  x  = - (-20)  / [ 2 (-4) ]  =  20 / -8   =  - 5/2  = -2.5

Plugging this into the original function we find the  max  as

4( -2.5 + 7) ( 2  -  -2.5)  =

4 ( 4.5) ( 4.5)   = 

81

I got it wrong it was 12

Hey, GM!!!!!.....good to see you  !!!

Let  x   = 3

Then  3(3)^2 - 3  =   3 *9 - 3   =  27 -3  =    24

And     7(3) + 3  =  21 + 3  =    24

Note  that  the  output  is the same....so.....this piece-wise  function is  continuous at x   = 3

We have  that

( x + 6)  + (6x + 2)  + (2x + 7)

________________________  =    2x +  7        simplify

                  3

9x  + 15  =   3(2x + 7)

9x +  15  =  6x  + 21       subtract  15, 6x  from both sides

3x  = 6

x   = 2

This was posted here: https://web2.0calc.com/questions/helpp_29

Take a look at it! Hope this helped~

If you don't get at least one head, you must get all tails.

To find the probability of getting at least one head, subtract the probability of getting all tails from 1.000.

One toss:         1.000 - all heads  = 1.000 - 0.80  =  0.20

Two tosses:     1.000 - all heads  = 1.000 - 0.80x0.80  =  0.36

Three tosses:  1.000 - all heads  = 1.000 - 0.80x0.80x 0.80  =  0.488

Four tosses:    1.000 - all heads  = 1.000 - 0.80x0.80x.80x0.80  =  0.5904

It takes four tosses.

just a way cooler and explaining ver. of mine 

We  have the form

At^2 + Bt  +  C

The  max height  will be reached  at   t  =  -B  / [2A ]

So....we  have

t   =   -75  / [ 2 * -25 ]   =   -75 / -50   =    3 / 2   sec    =   1.5  sec

t=1.5

1.5 sec it reaches 80.25 ft as a max

Cross-multiply

[ √(3x) - 4√3]  [ √(6x) - 2√3]   = [ √x  - √2] [ 2√(2x)  + √2 ]     simplify

√ (3x * 6x)  - 4√(3 * 6x) - 2√(3 * 3x) + 8√(3 * 3)  =  2√(2x * x)  - 2√(2* 2x)  + √(2x)  - √(2*2)

√(18x^2)  - 4√(18x) - 2√(9x) + 8√9  =  2x√2 - 4√x + √(2x) - 2 

3x√2  - 12√(2x) - 6√x + 24  =   2x√2 - 4√x + √(2x) - 2

x√2 - 13√(2x) - 2√x + 26  =  0 

√(2x) [ √x - √2 ]  =   13√(2x) - 26

√(2x)  [ √x - √2) ]=  13√(2x) - 13(2) 

√(2x) [ √x - √2)  =  13√2  [ √x - √2]

[ √x - √2]  [ √(2x) - 13√2 ]   =   0

[ √x - √2] (√2) [ √(x) - 13 ]   =   0      divide out  √2

[ √x - √2]  [ √(x) - 13 ]   =   0

So  we  have  that

√x - √2  =  0        or       √x - 13  = 0

    x   =  2                       √x   = 13             square both sides

    x  = 2          or          x   =169     

Reject  x  = 2   since it makes  an original denominator    =    0

So.....the solution is that   x   =169

if tails prob=.8 then heads =.2

2*n = 5+

n is greater then 2.5

but you cant toss a coin .5 times

so n=3 tosses

i think this is right i could be wrong

oops 72

72 bags

30 after 50% sub 

30*2=60

60= 5/6 of the box

60/5=12

12=1/6 of the box

12*6=76

the box had 76 bags to begin with

Yeah, I messed up my answer. 

Getting the AwesomeBall is still 0.10 and, if we get that ball, we have won.

Now, what happens when we don't get that ball (which is a probability of 0.90).

B x W x W x W  =  (0.90) x (3/10) x (2/9) x (1/8)  =  0.0075

B x W x W x L   =  (0.90) x (3/10) x (2/9) x (7/8)  =

B x W x L x W   =  (0.90) x (3/10) x (7/9) x (2/8)  =

B x L x W x W   =  (0.90) x (7/10) x (3/9) x (2/8)  =

Now, add  0.10 + 0.0075 + ...

I couldn't upload an image so I tried to get it linked!

https://free-picload.com/image/qJ0Mu3 

a = √25 = 5

b = √32 = 5.656854249

c = AB = ?

∠C = 81.86989765º

The law of cosines:          c² = a² + b² - 2ab * cos(C)

                                         c² = 25 + 32 - 56.56854249 * 0.141421356

                                         c² = 57 - 8

                                         c = 7