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4^x -1

(2²) ^x - 1

(2^x)² - 1         sub in 2^x =7

7² -1 = 48

slope of this line = - dying rate/60          if they are dying at rate of 120 bacteria per hour slope would be - 120/60 = - 2

See https://web2.0calc.com/questions/pls-help-me-smart-people

The answer is 1/4.

See https://web2.0calc.com/questions/graphing-question_40

The distance between the circumcenters is 12*sqrt(3).

4^x-1 = 13

Thank you so much for explaining!

Distance between P&Q=26

So, you have to find the midpoint of P&Q

(15, 24)

Hello Guest,

use the double summation sign \(\displaystyle\sum\) to get the answer ... 

\(\displaystyle\sum_{n=1}^{52} 2n = 2756\)

\({\color{red}7},{\color{red}14},{\color{red}21},...,{\color{red}2751},{\color{blue}(2756)},{\color{red}2758}\)

So the remainder is \((2756-2751=)\) \({\color{blue}5}\) .

Straight

How many isosceles triangles with whole-number length sides have a perimeter of 30 units?     

Well, first, for the equal sides to be whole numbers, their total has to be even.  That narrows it down.  

Let's make a little table.    

size of base      size of each of the other two sides  

      2                                     14      2 + 14 + 14 = 30   

      4                                     13      4 + 13 + 13 = 30  

      6                                     12  

      8                                     11   

    10                                     10  

    12                                       9  

    14                                       8  

    16                                       7      This one can't work, nor any subsequent sizes in this progression, because  

                                                       the sum of any two sides of a triangle has to be larger than the third side.

So it looks like there are seven (7) such isoceles triangles that satisfy the criteria of the problem.  

_.

x = 2|y|− 3

Range is all real y

Domain is x>=3

So for the inverse:

Domain is all real x     (-infinty, infinty)

Range is y>=3            [-3,infinity)

check by graphing

The green one is the original and the purple one is the inverse  (reflection across y=x which is the red one)

Use the Euclidean algorithm to find integers x and y  such that  164x + 39y = 1.  State your answer as a list with x first and y second, separated by a comma.

\(164x+39y=1\\~\\ 164=4*\boxed{39}+8\\ 39=4*\boxed{8}+7\\ 8=1*\boxed{7}+1\\~\\ \text{Now the extended euclidean algorithm}\\ 1=8-1*\boxed{7}\\ 1=8-1*\left[39-4*\boxed{8}\right]\\ 1=8+1\left[-39+4*\boxed{8}\right]\\ 1=5*\boxed{8}-1*39\\ 1=5*\left[ 164-4*\boxed{39} \right]-1*39\\ 1=5*164-21*39\\ 1=164(5)+39(-21)\\ 1=164(5-39s)+39(-21+164a)\\~\\ x=5-39a\qquad y=-21+164a \qquad \text{Where a is any integer.} \)

You can finish it.

LaTex

164x+39y=1\\~\\
164=4*\boxed{39}+8\\
39=4*\boxed{8}+7\\
8=1*\boxed{7}+1\\~\\
\text{Now the extended euclidean algorithm}\\
1=8-1*\boxed{7}\\
1=8-1*\left[39-4*\boxed{8}\right]\\
1=8+1\left[-39+4*\boxed{8}\right]\\
1=5*\boxed{8}-1*39\\
1=5*\left[  164-4*\boxed{39} \right]-1*39\\
1=5*164-21*39\\
1=164(5)+39(-21)\\
1=164(5-39s)+39(-21+164a)\\~\\
x=5-39a\qquad y=-21+164a \qquad \text{Where a is any integer.}

angle CBD = 25 degrees

Median length = 60/6

[WXYZ] = 18√50

6 meters

6 meters, sir

For how many seconds is the height of the cannonball at least  meters?

Hello Guest!

How high should the cannonball at least fly? 6 meters. Thanks!

\(h(t) = -4.9t^2 + 14t - 0.4=6\)

\(h(t) = -4.9t^2 + 14t - 6.4=0\)

                  a            b          c

\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(t = {-14 \pm \sqrt{14^2-4\cdot (-4.9)\cdot (-6.4)} \over 2\cdot (-4.9)}\\ t=\dfrac{-14\pm 8.4}{-9.8}\\ \color{blue}t\in \{0.571,2.286\}\)

The cannonball is located at a height of 6 meters and above

from 0.57 seconds to 2.29 seconds after take-off, i.e. for 1.71 seconds.

  !

Ratio of marbles of Peter by Sam, P : S = 9 : 5

Now,  J : P : S = 36 : 63 : 35

(a)    Ratio of John by Sam : J : S = 36 : 35

(b)    Let the number of marbles with John = 36x

                                                           Peter = 63x

                                                           Sam = 35x

        When John Gabe 1/12 of his marble to Sam and

        Peter gave 1/9 of his marbles to Sam. Then

        
        marbles with Sam = 35x + (36x * 1/12)

                                                + (63 * 1/9)

                                      = 35x + 3x + 7x

                                      = 45x

        According to Question

           45x = 135

          => x = 135/45 = 3

  
      Total marbles + (36x + 63x + 35x) = 134x = 134 * 3

                                                                         = 402

Let total money Joe have = $x

Let Cost of can of drink = $y

     Hence cost of plate of chicken = $(y+3)

He spent 3/5 of his money on a can of drink and

 plate of chicken

Now money left = x - 3/5x = 2/5x

He spent rest of money to buy another 2 similar

cans of drink and has $1 left.

So Cost of 2 cans = y + y = $2y

Now money left = $1

     =>  2/5x - 2y = 1

     =>  2x - 10y = 5                 -- (A)

Also Cost of one can and a plate of chicken = 3/5x

            =>       y + (y + 3) = 3/5x

            =>     5 (2y + 3) = 3x

            =>       3x - 10y = 15                -- (B)

     Subtract (A) from (B) we get

                x = 10

      from (A)       y = 1/10 (2x-5) = 1/10 (20-5) = 1.5

  => Cost of Can of drink = $y = $1.5

this is required solution , may this help you.

let x represents the dollars owned by Tom

and y represents the dollars owned by Peter.

Given that, the price of the game cartridge is 

$45.

 If Tom buys the game cartridge, then the

money left to Tom is x−45. and the ratio of

his money and the Peter's is 1:3.

That is, \( {x-45 \over y}={1 \over 3}.\)
\( {x-45 \over y}={1 \over 3}.\) 
3x-135=y

3x-y=135

If Peter buys the game cartridge, then the

money left to Peter is y - 45. and the ratio of

his money and the Peter's is 3:5.

\( {y-45\over x}={3 \over 5}\)

3x=5y-225

x=5/3y-223/3

Substitute (2) in (1),
3x-y=135

3(5/3y-223/5)-y=135

y=90

Substitute y=90 in (2),

x=5/3(90)-225/3

x=75

Therefore, Tom has $75 and Peter has $90 at first.

Let total number of marbles in the bag were = x

Given, then jerry has marbles = 1/4 of the total number of marbles

                                         
                                                 = 1/4^x = x/4

       
          Lucas had marbles = x/4 + 36 = (36 more than Jerry)

          Kumar had marbles = 1/2 of the number of marbles

                                                was had

                                           = 1/2 (x/4 + 36)

          As, there were total x marbles in the bag

           So;

                    x = x/4 + (x/4 + 36) + 1/2(x/4 + 36)

                     
                       = 2x/4 + x/8 + 36 + 18

                    x = \( {4x+x \over 8}\) + 36 + 18

                   ^x- 5x/8 + 54

                      3x/8 = 54

                           x = \( {54 * 8 \over 3} = {432 \over 3} = 144\)

x = 144

a) 

   Lucas had marbles = (x/4 + 36)

                                   = 144/4 + 36

                                     = 36 + 36

                                      = 72

      So;     Lucas had 1/2 of the total marbles.

    b)     Total marbles were x = 144.

Zachary kept 290 balls in two boxes.

A as B

Let A ---> x balls => x+y=290 

      B ---> y

Now, 2/5A ---> B        B has total

                                     =y+2x/5

A has        x-2x/5=3x/5

Still A has 10 balls more than box B

        3x/5 = y+2x/5+10

   => 3x/5 - 2x/5 = y+10

   =>    x/5 = y+10

   =>    x - 5y = 50

   x+y=290

   x-5y=50

-   +      -

---------------                   |

      6y=240                   |    x = 290 - 40

        y=240/6=40          |        = 250

                                     |

There were 250 balls in A 

                     40 balls in B at first

Create Equations:

• B = C&D = 8
• C = 16
• D = 9

Create Venn Diagram:

Solve Question:

• How many people own a cat or a dog?
• 16 + 1 = 17 people

Define Variables:

C = Total Chocolate Chips

S = Small Christmas Cake/Tree

B = Big Christmas Cake

Set the 3 Equations:

1. 1/3 C = 12 S

2. 3/4 (2/3 C) = 10 B

3. B = 16 + S

Begin Solving for Variables:

• 1/2 C = 10 (16 + S)

• 1/2 C = 160 + 10S

• C = 320 + 20S

• 1/3 (320 + 20S) = 12 S

• 320 + 20S = 36S

• S = 20

Solve for Rest of Variables:

• B = 16 + (20) = 36

• C = 320 + 20 x (20) = 720

Check:

• 1/3 (720) = 240 | 12 (20) = 240
• 3/4 (2/3 (720) = 360 | 10 (36) = 360
• (36) = 16 + (20) = 36