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when in doub tuse some calculations and make it on desmos.com

lol

i am 313 now :)

use upload when you make the image

i think cphill told you there was no image last time

HOPE THIS HELPED!

No sorry, I'm barely a freshman, otherwise i would so help you

 (2^(2+4)-2(2^n))/(2(2^(n+3)).  =

[ 2^6  - 2(2^n ] / [ 2 * 2^(n + 3) ]  =

[ 2^6 - 2^(n + 1) ] / [ 2^(n + 4) ]  =

[ 64 - 2^(n + 1) ] / [2 ^(n + 4)  ]  =

[ 2 * 2^5 - 2 * 2^n ] / [ 2* 2^(n + 3) ] =

[ 2^5 - 2^n ] / [ 2^(n + 3) ]  =

[32 - 2^n ] / [ 8 * 2^n ]   

This is all we can do

8x^2 + 15x + c = 0

To have two real roots, the discriminant must be > 0

So

15^2 -  4*8*c > 0

225 > 32c

c < 225/32

c< floor [ 225/32]   

c < 7       but     also > 0 since c must be positive

So....the product of all positve c's =   7!  =   5040

x = y^4

x + y^2 = 1        sub the first into the second and we have that

y^4 + y^2  = 1        we can complete the square on y thusly

y^4 + y^2 + 1/4    = 1 + 1/4       factor and simplify

(y^2 + 1/2)^2 =  5/4           take both roots

y^2 + 1/2  =  ± sqrt (5) / 2

y^2 =  ±sqrt (5) / 2 - 1/2

y^2 =      [ -1 ± sqrt (5) ] / 2        take both roots again

y = ± sqrt [ [  -1 ± sqrt (5) ] / 2 ]

Let y  =   a ,  -a

So  ......x = y^4   =   a^4

So   we have the points      (a^4, a )   and ( a^4, - a)

So....the distance between these points  =

sqrt [ (a^4 - a^4) + ( a  -  -a)^2 ]  =  sqrt  [ (2a)^2]  =    2a

So...we have, 2a  =

2 √ [ ( -1  ± √5) / 2 ]      but we cannot take a root of a negative so the distance must be

2 √ [ ( -1  + √5) / 2 ]   =

√ [  4 ( -1  + √5) / 2 ]  =

√ [ - 2 + 2√5 ]

So  (u, v)  =  (-2, 2)

From the top, we have

2(4x+2)=4x−12(x−1)       simplify

8x + 4 =  4x - 12x + 12

8x + 4 = - 8x + 12         add 8x to both sides, subtract 4 from both sides

16x = 8                       divide both sides by 16

x = 8 /16  =  1 / 2      .....just as tertre found.....!!!

Can't draw the diagram.....but

tan 50° = height of statue / shadow length

tan 50° = height of statue / 30      multiply both sides by 30

30 * tan (50°) = height of statue  = 35.8 ft

The shadow represents the adjacent leg to the 50 degree angle of a right triangle....

Opposite leg is the HEIGHT of the statue

Tangent 50 = opposite/30

30 tangent 50 = height = tan(50)*30 = 35.8  ft (rounded)

Hint: Try to simplify the expression.

After simplifying, you should get 8x+4=4x-12x+12, so 16x=8, and x=1/2. 

6x+2=-10

     -2   -2

6x=-12

6      6

x=-2

See my answer here

Thank you!

It comes out to 12.1 when I plugged it into Math Papa calculator.

-(-14.6)-|2.5| ----- substitute 

-(-14.6)-2.5 ------ absolute value of 2.5 is 2.5 

14.6-2.5    opposite of -14.6 is 14.6 then just solve 

12.1

:)

If climbing at same speed, their rates will be equal   (if they start at the same height)

John rate = (x+5) / (x-1)          = Jane rate (x+11) / (x+1)

(x+5)(x+1) = (x+11)(x-1)

x^2 + 6x + 5 = x^2 +10x-11

   6x + 5 = 10x -11

16 =4x

x=4                John speed = (4+5)/(4-1) = 3 mph      Jane speed   (4+11)/(4+1) = 3 mph

I like CPhill's answer best but they are all correct.

Thanks Tertre, that is a good response.

I would word it just a little differently.  I'd say

If you change one side you must do the SAME to the other side  

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Here is a full answer from me:        (of course I know people do not have time to give full answers very often)

I will do another similar question and then you can try your own one:

18=2+6+2t

First you must simplify both sides - if it is possible. 

that gives

18= 8 +2t

you can swap sides if you want to

8+2t=18

2t+8=18

Now you must keep the equation balanced. Just like old fasioned see-saw style scales.

I used a t but the picture uses an x, it makes no difference. 

    2t+8=18

If you change one side you must do the SAME to the other side

You want the 2 lots of t all by itself so you must get rid of 8 on the right hand side.

You can do this by subtracting 8 from BOTH sides.

    2t+8=18

2t+8-8=18-8

        2t=10

Now if you divide 2t by 2 you get t on its own BUT if you divide the LHS by 2 you must also divide the RHS by 2

A fraction line is the same as a divide.

\(\frac{2t}{2}=\frac{10}{2}\\ t=5\)

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Now try your question and let us know how you go.  We can check your answer for you if you want.   

CPhill is right. You can also simplify! -28r-28 becomes, -28(r+1). 

I'll check asad.

Evaluate and Simplify.

f(x) = 2 + 4x − x^2,        f(5 + h) − f(5) / h

I got an answer but I am not sure if I am right or not.  My answer was, "4 - h"

\(\frac{f(5+h)-f(5)}{h}\\ =\frac{[2+4(5+h)-(5+h)^2]-[2+20-25]}{h}\\ =\frac{[2+20+4h-(25+h^2+10h)]-[-3]}{h}\\ =\frac{2+20+4h-25-h^2-10h+3}{h}\\ =\frac{4h-h^2-10h}{h}\\ =\frac{-h^2-6h}{h}\\ =\frac{h(-h-6)}{h}\qquad \text{where } h>0\\ =-h-6\)

So Asad, you were correct.   Good work!   

Hint: Try to group like terms. +5 and +10 change negative. 

Here, I'll solve one and you can solve the rest. Okay, I'll pick the 2nd question.

\(\sqrt{14xy}\cdot \sqrt{12}\cdot \sqrt{30y}=12\sqrt{35} y\sqrt{x}\). 

You do not need to appologize asad.  You did nothing wrong.

If you want to attempt to teach this person something then you can but please do not directly answer his/her questions.  Just give suggestions or smallish hints. That would be fine :)

If he interacts with you then it will be an indication that maybe he actually would like to learn something.