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nope

rate * time = distance

distance/rate = time

105 minutes moving time      Bike time = distance/rate =  20/(2x+1)

                                              Run time  = 5/x

added together the time = 105 min = 1.75 hr

20/(2x+1) + 5/x = 1.75

20x /(2x+1) + 5 = 1.75 x

20 x + 5(2x+1) = 1.75 x (2x+1)

20 x + 10x + 5 = 3.5 x2 + 1.75x

3.5 x^2 -28.25x -5 =0

x = 8.24 m/hr    running speed (via quadratic formula)

Solve for x:
1/12 + 5/x + 20/(2 x + 1) = 11/6

Bring 1/12 + 5/x + 20/(2 x + 1) together using the common denominator 12 x (2 x + 1):
(2 x^2 + 361 x + 60)/(12 x (2 x + 1)) = 11/6

Cross multiply:
6 (2 x^2 + 361 x + 60) = 132 x (2 x + 1)

Expand out terms of the left hand side:
12 x^2 + 2166 x + 360 = 132 x (2 x + 1)

Expand out terms of the right hand side:
12 x^2 + 2166 x + 360 = 264 x^2 + 132 x

Subtract 264 x^2 + 132 x from both sides:
-252 x^2 + 2034 x + 360 = 0

Divide both sides by -252:
x^2 - (113 x)/14 - 10/7 = 0
Add 10/7 to both sides:
x^2 - (113 x)/14 = 10/7

Add 12769/784 to both sides:
x^2 - (113 x)/14 + 12769/784 = 13889/784

Write the left hand side as a square:
(x - 113/28)^2 = 13889/784

Take the square root of both sides:
x - 113/28 = sqrt(13889)/28 or x - 113/28 = -sqrt(13889)/28

Add 113/28 to both sides:
x = 113/28 + sqrt(13889)/28 or x - 113/28 = -sqrt(13889)/28

Add 113/28 to both sides:

x = 113/28 + sqrt(13889)/28 =8.24 - miles per hour.                    

this is trying to teach you about the slope intercept problem
okay so what you wanna do is first gather up your information about the problem, which was that they bought 7 sodas 
so... 
.5*7= 3.5
28.5-3.5= 25

so your equation would be: 
25= 1.25x+2.50y
 
and when you want to graph it
you want to use the formula for the slope intercept which would be y=mx+b 
2.5y= 1.25x+25
so 
y=-0.5x+10

and then youre just gonna locate the y intercept on your graph, so youre gonna have to plot it on 10 on the y axis
and then from there since our slope was 1/2 you want to go down one and to the right by 2 because it is your rise/run so your rise would be 1 and your run would be 2. and yeah! that's how you graph it p much. 

hope this helped you!!
 

We have to find a multiple of 9, that when divided by 25 leaves a remainder of 1.

Therefore, 126 is congruent to 1(mod25)...so x = 126 / 9 = 14.

Thus, 14+11=25 and dividing this result by 25 leaves a remainder of 0.

-tertre

1) \(x^2 - 10x + y^2 + 8y -8 = 0\\ (x^2 - 10x +25 - 25)+ (y^2 +8y + 16-16)-8=0\\ (x-5)^2 - 25 + (y+4)^2-16 - 8 = 0\\ (x-5)^2+ (y+4)^2 = 49= 7^2\\ C=(5,-4),~r=7\)

2) \(\text{The center is at }(-2,3)\\ \text{The radius is }3\\ (x+2)^2 +(y-3)^2 = 9\)

3) \(\text{Vertex at }(2,3)\\ f(x) = a(x-2)^2+3\\ (0,4) \text{ on the parabola}\\ 4a+3=4\\ a=\dfrac 1 4\\ f(x) = \dfrac 1 4 (x-2)^2+3\)

What is the next number  in the sequence 2 3 6 8 8 4 

The next number in that sequence is 2

Multiply each adjacent pair and delete everything except the units digit.

          2 3 6 8 8 4      6                                        6

          2 3 6 8 8 4     18      drop the 1, leave the 8

          2 3 6 8 8 4     48      drop the 4, leave the 8

          2 3 6 8 8 4     64      drop the 6, leave the 4

The next number will be 8 x 4 = 32  drop the three, leave the 2

2 *3 = 6

3*6 =   1  8

8 * 6 =  4  8

8 * 8 =  6  4

8 * 4  = 3  2

100 = 5 * 10^(0.3t), solve for t
Divide both sides by 5:
20 =10^(0.3t)
Take the log of both sides:
0.3t =log(20)
Divide both sides by 0.3:

Multiply the RHS by 10 / 10
t =log(20) / 0.3 = 10*log(20) / log(1,000) - years - when the tree will have 100 branches.
 

Another way:

Let the current teacher's age = T
Let the current student's age = S
T - S = S - 3.............(1)
T - S =63 - T............(2), solve for S, T
S = 23 - Student's age
T = 43 - Teacher's age.

Thank you so much for the help CPhill!!!

\(\text{Let student }k \text{ have baked }n_k \text{ cookies}\\ \text{the }{n_k}'s \text{ might not all be equal}\\ N = \sum \limits_{k=1}^4 ~n_k= \dfrac 1 4 N+81\\ 81 = \dfrac 3 4 N\\ N=108\)

Seriously, GM..... you asked this before....did you bother to look at the answer?  (barely two weeks ago.....) 

https://web2.0calc.com/questions/for-each-problem-below-solve-algebraically-then-use_3

HAHAHA!!!!......sadly, no I'm not.....

l 2x - 1 l   > 5       

We have two equations

2x - 1 > 5             and           - (2x - 1 ) > 5

2x > 6                                     -2x + 1 > 5

x > 3                                       -2x >  4

                                                  x < - 2

I  and II   will make this inequality true

I found the answer. Thanks for trying eveyone

Here you go :

D = (-8, 0)

C = (4,0)

G = (-8/3, 1)

H = (4.3, 1)

Slope  CD  =  [  0 - 0 ]  / [ 4 - - 8 ]  =  0 / 12  =  0

Slope GH  =  [ 1 - 1 ] / [ 4/3 -  - 8/3 ]  =  0 / [ 12/3 ]  = 0 / 4   =   0

What is 0^0? Is 1 or undefined?

Stated as 0⁰ is meaningless.  You have to say it as "the limit of X^X as X approaches zero" 

.

Nice job CPhill and just wondering are you a teacher??