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solution is incorrect , sir.

Try these steps and see how you go :)

You will learn better if you do quite a bit of it yourself.

You can post your answers (or troubles and someone will look at them for you :)

1) find the midpoint of AB

2) Find the gradient of AB

3)Determine the gradient of the perpendicular to AB

4) use the point gradient formula to determine the equation of the line.

My problem as I said is in my english level

not in solutions itselves

Guest , also

I know that the up side is always true , because a>0 , and b^2-4ac<0

2x^2+x-6 can be expressed as (2x-3)(x+2)

x^2+x+3>0 is always true

(2x-3)(x+2)>=0 solution is (-inf;-2] U [3/2;+inf)

but (2x-3)(x+2) is down so we will eliminate -2 and 3/2

the final solution (-inf;-2) U (3/2;+inf)

this is the final answer 

yeah , my problem is that I can't express words in english .

so in writings problems I use their sentences lol

my level in english is not good enough

First set h = 0, then divide each term by -8 to get:

2t^2 + 3t - 20 = 0. This factors as:

(2t - 5)(t + 4) = 0

The only positive solution is given by 2t - 5 = 0 or t = 5/2

So t = 2.5 seconds

I think I figured it out. Please let if I figured it out right or not please someone.

if a=4, then c=19(4)-4

c=19(4)-4

c=19*4=76

76-4=72

c=72

Did I do this problem right?????????????

\(\quad\frac{4}{3}\times\frac{22}7\times8 ^3\\ = \frac{4}{3}\times\frac{22}7\times512 \\ = \frac{45056}{21} \space or \space 2145\frac{11}{21}\)

400 x 100  x 2  x  4  x  10  x  4000  x  50 = 640000000000

Left hand side = 2

Right hand side =\(-\frac{1}{8(-1-2)^3}-1= -\frac{1}{-216}-1=-\frac{215}{216}\)

The 2 sides isn't the same

\(h=\frac{-b}{2a}\\ 2ah=-b\\ b=-2ah\)

f(x) =  (1/4) x + 3      write f(x)  as y

y = (1/4)x + 3      the object is to get x by itself and then  "exchange" x and y

Subtract 3 from both sides

y - 3  = (1/4)x     multiply both sides by 4

4y - 12 = x        "exchange"   x and y

4x - 12   = y

For y, write f^-1 (x)

4x - 12  =  f^-1(x)   = 4(x - 3)    =   "C"

h= -b/2a cross multiply

2ah = -b multiply both sides by -1

b= -2ah

6^(x+8)=12      take the log of both sides

log 6^(x + 8)  = log 12       and by a  log property we can write

(x + 8) log 6   = log 12       divide both sides by log 6

(x + 8)    = [ log 12 / log 6]       subtract 8 from each side

x  = [ log12 / log 6 ]  -   8    =  about    -6.613

Solve for x over the real numbers:
6^(x+8) = 12

Take the logarithm base 6 of both sides:
x+8 = (log(12))/(log(6))

Subtract 8 from both sides:
Answer: |  x = (log(12))/(log(6))-8

\(1. \quad |F_E|=k|\frac{q_1q_2}{r^2}|\\|\frac{F_E}{k}|=|\frac{q_1q_2}{r^2}|\\\frac{F_E}{k}=\frac{q_1q_2}{r^2}\\ q_1=\frac{(F_E)(r^2)}{kq_2}\\ Correct.\)

2nd question. Too lazy to type in LaTeX. Correct.

3rd question Correct.

Yes.

0.0000000062715585 corrected to 1 dp = 0

0.0000000062715585 corrected to 2 dp = 0

......

0.0000000062715585 corrected to 7 dp= 0

0.0000000062715585 corrected to 8 dp= 0.00000001

0.01 corrected to 1dp = 0

\(\quad log(\infty)\\=log(\infty ^{\infty}) \space<--\color{red}\space Note\space that\space infinity\space to\space the\space power\space of\space infinity\space still\space equals\space infinity\space \\\color{red}because\space infinity\space is\space the\space biggest\space number.....\\= \infty \times log({\infty})\\ = \infty\\ \color{red}(It\space is \space because\space everything\space times\space infinity\space equals\space infinity)\)

I did it on the graphing calculator.

https://www.desmos.com/calculator/hntv9qimie

If it's a maths question, it's 19

If it's a joke, it's 21

Btw I don't understand why 9+10 = 21 is a joke. And I am not from America. This is not my fault right? :P

\(2\times \dfrac{2}{5}= \frac{2\times2}{5}=\frac{4}{5}\)

This is not an equation.

Hello guest #2 :D

The name is not important. The more important is what they do inside the baby changing stations......

They don't change the baby inside so they will always come back with the same baby......

\(\frac{x^2+x+3}{2x^2+x-6}\geqslant 0\)

\(x^2+x+3\geqslant 0\)

\(x \geqslant {-1 \pm \sqrt{1^2-4(1)(3)} \over 2(1)}\)

And the right hand side is imaginary.

The parabola:  y  =  ax² + bx + c  has a vertical line of symmetry at  x = 1.

It goes through the point (-1,3)   --->   Since the graph has a vertical line of symmetry at  x = 1  and since the point (-1,3) is two units to the left of this line, there will be a point with the same y-value two units to the right of this line:  (3,3).

Since it goes through the point (2,-2), which is one unit to the right of the vertical line of symmetry, there will be a point with the same y-value one unit to the left of this line:  (0,-2).

We only need to use three of these four points in the equation  y  =  ax² + bx + c:

(-1,3)   --->    3  =  a(-1)² +b(-1) + c     --->      3  =  a - b + c                      (equation #1)

(2,-2)   --->   -2  =  a(2)² + b(2) + c      --->     -2  =  4a + 2b + c                 (equation #2)

(3,3)   --->     3  =  a(3)² + b(3) + c      --->      3  =  9a + 3b + c                  (equation #3)

Combining equations #2 and #1:      -2  =  4a + 2b + c   --->                       -2  =  4a + 2b + c          

                                                           3  =  a - b + c        --->  x -1   --->      -3  =  -a +   b  - c

     Adding down the columns:                                                                       -5  =  3a + 3b

Combining equations #3 and #1:       3  =  9a + 3b + c   --->                    3  =  9a + 3b + c   

                                                           3  =    a -   b + c   --->  x -1   --->   -3  =  -a +   b -  c

    Adding down the columns:                                                                    0  =  8a + 2b     --->     0  =  2a + b

Combing these two new equations:      -5  =  3a + 3b   --->                   -5  =  3a + 3b

                                                               0  =  2a +   b   --->  x -3   --->   0  =  -6a - 3b

     Adding down the columns:                                                                -5  =  -3a     --->     a  =  5/3

Substituting this value into  0  =  2a + b   --->   0  =  2(5/3) + b   --->   0  =  10/3 + b   --->   b  =  -10/3

Substituting this value into  3  =  a - b + c   --->   3  =  (5/3) - (-10/3) + c   --->   c  =  -2

Equation:  y  =  ax² + bx + c   --->   y  =  (5/3)x² - (10/3)x - 2

This equation has two roots:  Using the quadratic formula:  x = [ 10 + sqrt(220) ] / 10  and  x  = [ 10 - sqrt(220) ] / 10

The larger answer is:  x = [ 10 + sqrt(220) ] / 10     =     10/10 + sqrt(220)/10     =     1 + sqrt(220) / 10

     =    1 + sqrt(220) / sqrt(100)     =     1 + sqrt(220/100)     =     1 + sqrt(2.20)     --->     n = 2.20