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Thank you so much! They would look like this then, right?

We are using the exponent law  that  says that  (a^m)^n  =  a^(m*n)

(-3a^2b)^3   =   (-3)^3 * (a^2)^3 * (b)^3   =  -27a^6b^3

(5 - 2x)^2  =  (5 - 2x) (5-2x)  =  25 - 10x -10x + 4x^2  =  4x^2 -20x + 25

(5p^2q^3)^4  =  (5)^4 * (p^2)^4 * (q^3)^4  =  625p^8q^12

(-2xy^3)^4  =  (-2)^4 * (x)^4 * (y^3)^4  =  16x^4y^12

(a)

Rewrite l    as

-3y  = -6x + 20          divide through by -3

y = 2x - 20/3

m :  y = 2x - 6

Note that in the form  y = mx + b       m is the slope of the line

So   l and  m    have the same slope...so....they are parallel

(b)   5x + 2 1/2 y -9  = 0

5x + 5/2 y - 9  = 0               add 9 to both sides......subtract  5 x from both sides

(5/2)y = - 5x + 9                 multiply through by (2/5)

y = -2x + 18/5

m and n   are neither parallel  or perpendicular

Here's a graph : https://www.desmos.com/calculator/r0tsvzvdjo

(2x-3)(2x+5) > -7   simplify

4x^2 - 6x + 10x - 15 > -7       

4x^2 + 4x - 15 > -7        add 7 to both sides

4x^2 + 4x - 8 > 0         divide through by 4

x^2 + x - 2 >  0           factor

(x + 2) ( x - 1) > 0

As before.....set each factor to 0 and solve

x + 2  = 0             x - 1  = 0

x = -2                      x  = 1

We have three intervals ..... ( -inf, -2) or (-2, 1)  or  (1, inf)

Test x = 0  in the original inequality  and we get that

(2(0)-3)(2(0)+5) > -7

(-3) (5) > -7

-15 > -7

This is not true

So....the intervals that solve  this   are   (-inf, -2)  and ( 1, inf)

Here's the graph : https://www.desmos.com/calculator/7dgbgkrmaw

x^2 < 25

x^2 - 25 < 0

Factor

(x + 5) ( x - 5) < 0

Set each factor to 0   and solve

x + 5  =  0            x  -  5  = 0

x = - 5                  x   = 5

We have three possible intervals that solve this...... (-inf, -5)  or (-5, 5)  or (5, 0)

What happens is that either the two outside intervals solve the original equation or the middle interval does

Test x =  0,  [which is a point in the middle interval ] in the original inequality

0^2 < 25    which is true

So....the interval  that solves the inequality  is  (-5, 5).....we don't include the endpoints because of the "<" sign

Here's a graph : https://www.desmos.com/calculator/fr1w7y9y6f

\(\ \phantom{=\quad}\frac43x^2+4x+1\)

                                                   Factor  \(\frac43\)  out of all the terms.

\(=\quad\frac43(x^2+3x+\frac34)\)

                                                          Add  \(\frac94\)  and subtract  \(\frac94\)  to complete the square inside the parenthesees.

\(=\quad\frac43(x^2+3x+\frac{9}{4}-\frac{9}{4}+\frac34)\)

                                                          Factor   \(x^2+3x+\frac94\)   as a perfect square trinomial.

\(=\quad\frac43(\ (x+\frac32)^2-\frac{9}{4}+\frac34)\)

                                                          Combine  \(-\frac94\)  and  \(+\frac34\)

\(=\quad\frac43(\ (x+\frac32)^2-\frac{6}{4})\)

                                                   Distribute the  \(\frac43\)

\(=\quad\frac43(x+\frac32)^2-\frac43(\frac{6}{4})\)

                                                   And  \(\frac43(\frac64)\ =\ 2\)

\(=\quad\frac43(x+\frac32)^2-2\)

Now the expression is in the form  a(x + b)² + c  where  a,  b,  and  c  are constants.

\(abc\ =\ (\frac43)(\frac32)(-2)\ =\ (2)(-2)\ =\ -4\) _

\(f(x)\ =\ 6\sqrt{3x}\\~\\ g(x)\ =\ (2x)^4\ =\ 2^4x^4\ =\ 16x^4\\~\\ h(x)\ =\ \big(2^{x+2}\big)^2\ =\ \big(2^{2}\big)^{x+2}\ =\ 4^{x+2}\ =\ 4^2\cdot4^x\ =\ 16\cdot4^x\\~\\ j(x)\ =\ 3\sqrt{12x}\ =\ 3\sqrt{4\cdot3x}\ =\ 3\sqrt{4}\sqrt{3x}\ =\ 6\sqrt{3 x}\\~\\ m(x)\ =\ 16x^4\)

So we can see that   \(f(x)\ =\ j(x) \)   and   \(g(x)\ =\ m(x)\)

( And  \(h(x)\)  is not equal to any of the other functions. )

\(f(x)\ =\ 2^{x+3}\ =\ 2^x\cdot2^3\ =\ 2^3\cdot2^x\ =\ 8\cdot2^x\ =\ 8\cdot g(x)\\~\\ f(x)\ =\ 8\cdot g(x)\)_

25x² + 20x - 10  =  0

                                       Rewrite the equation as...

(5x)² + 4(5x) - 10  =  0

                                       To make it clearer, let  u  =  5x   and then substitute  u  in for  5x

u² + 4u - 10  =  0

                                       Add  10  to both sides of the equation.

u² + 4u  =  10

                                       Add  4  to both sides of the equation to complete the square on the left side.

u² + 4u + 4  =  14

                                       Factor the perfect square trinomial on the left side.

(u + 2)²  =  14

                                       And since  u  =  5x  we can substitute  5x  in for  u

(5x + 2)²  =  14

Now the equation is in the form  (ax + b)²  =  c   where  a,  b,  and  c  are integers and  a > 0 .

a + b + c  =  5 + 2 + 14  =  21

thanks!

If the quadratic has exactly one real root then the discriminant must be zero.

(4m)² - 4(1)(m)  =  0

16m² - 4m  =  0

m(16m - 4)  =  0

16m - 4  =  0          or          m  =  0

16m  =  4

m  =  1/4

The positive value of  m  is  1/4

Let's say it starts wih  x  fish.

The smallest positive integer value of  x  that makes  \(\frac{8}{27}x-\frac{38}{27}\)  a positive integer is  25 .

Hey this is from a AOPS thingy. I go there so I know

3465

The number cannot end in 5

The number will be divisible by 11  if the  [  sum of the first and third digit ] - [ sum of second and fourth digit ] = 0

5346    ⇒ divisible by 11              3546                                       3456 

5364                                             3564 ⇒  divisible by 11          3654                                       

5436                                             4536                                       4356 ⇒  divisible by 11                                     

5463                                             4563                                       4653  ⇒  divisible by 11

5634                                             6534 ⇒ divisible by 11            6354

5643  ⇒ divisible by 11                6543                                        6453 

6 numbers

    1                   √102  + √98                √102   +  √98           √102   +  √98

__________   *__________       =     ____________  =      ___________

√102  - √98        √102 + √98                 102     -   98                       4

Note  that  √102  and √98  ≈   10   .....so.....

≈10 + ≈ 10               ≈ 20

_________    =        ____        ≈      5

        4                          4

friendly bump to this problem

Oh, that's what I should have done! I messed up. Thanks, hectictar.

1  2  3  4  5    6   7   8  9   10

    D     D       D        D       D

       D           U            D

           U                 U

                D                       U

                     D

                           D

                                D

                                    U

The cards 1, 4, 9 , 10   are face-up at the end

Not sure about this one...but....

We can weigh any positive integer amount  up to

7 grams with weights of 1, 2 , 4   grams

15 grams with weights of 1, 2, 4, 8 grams

.

.

.

63 grams with weights of 1, 2, 4, 8, 16, 32 grams

127 grams with weights of  1, 2, 4, 8, 16, 32 ,64  grams

So...7  weights

90 - 96   ......string length  = 7

3  5   7      11  13        17  19       23              29  31            37       41  43       47           53            59   

            9               15              21       25   27            33  35      39              45     49  51      55  57

61              67          71   73               79         83               89

     63   65         69                 75  77         81        85   87         91   93

94  seems correct

Hahaha!!

OK....but.....I suspect hectictar might have gotten the same answer  !!!

I just beat her to the punch   ....LOL!!!!

Strictly by a little trial and error :

2^5 * 9^2  =  2592

So

x * y  =   5 * 2  =  10

THANK YOU SO MUCH BROHAM