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LOL!!!!....no....some of this stuff CAN  be  hard  !!!!

\(x\&=7-x\)

\(\&x = x -7\)                         

What  is the  value of       \(\&(15\&)\)       ????

15&  =  7 - 15   =   - 8

And

&(15&)   = &(-8)   =  &x =  -8 - 7   =    

-15

ha... even you found it difficult, this is my year 7 homework and you found it hard :), not to be mean but yea and srry if you find this rude

but still thx for the help

Not the  best.....I just know a few things.....LOL!!!

Let  team A's points  = A

Let team B's  points  = B

So....we have this system :

A = B + 12

A = 5B  

To  solve....just set the  A's equal

B + 12  =  5B

I'll let you complete this.....!!!!!

Thanks CPhill, you are the best!

Note  that    the  discriminant of  x^2  + x  + 3 =    1^2  -  4(1) (3)  =  1  - 12 =   -11

This means that   this   function   has no real roots.....and since it's a parabola that turns upward.....then it  never  intersects the  x axis, so it is   never   <  0   for  any  value  of  x

So.....we  only  need to  find  the  values   that   make   2x^2  + x   - 6   <  0

So

2x^2  + x  - 6    <  0   factor   the  left side

(2x  - 3) ( x + 2)     <   0

Note  that      on   (inf, -2)   this  function  is  >  0

And on  ( 3/2 , inf)   the same is true

So....it's  only  <  0    on   (  -2,  3/2)

So.....going back to the original  function...it  must  be  ≥  0    on      (-inf , -2)   U ( 3/2 , inf)

See the graph here :  https://www.desmos.com/calculator/agjohh32v6

f(x) =  ax  + b

y = ax  + b               get x by itself

y - b  = ax

[y -b ] / a  = x         "swap"  x  and  y

[x - b ]  / a = y    =  f^-1(x)

So

f^-1(x)  - 3   =    g (x)

[ x - b ]  /  a   -  3 =   5x  - 4         add 3 to both sides

[ x - b ] / a  =  5x    -  1           simplify 

(1/a)x  -  b/a   =  5x   - 1

Equating  terms

(1/a)  =  5    ⇒   5a  = 1   ⇒    a  = 1/5

And 

b/a =  1

b /(1/5)   = 1

b  = 1 * (1/5)

b = 1/5

So

5a  +  5b   =

5(1/5)  + 5 (1/5)   =

1  +   1     =

2

Perimeter =  2 (W + L)

And we  know that

112 =   2 ( W  +  L )    divide  both sides  by  2

56  = W  +  L

56  -   W   =   L

And

L             4

__  =     ___    so

W            3

56 - W            4  

______   =  _____       cross-multiply

     W               3 

3 ( 56 - W )   =  4W

168  - 3W  = 4W        add W to  both sides

168  =   7 W               divide  both sides  by 6

24  = W

And  L =   56 -  24  =   32

A little  difficult, TB....!!!!

Let   a  be the amount  pumped   from  tank  B   into C

And since they all  hold the  same amount

Amount in B =  Amount in A

(150 + n)  -  a =   n -  500

150 - a  =  -500     add 500, A  to both sides

a = 650

And the  amount in  C =   amount in A

500 + 650 =  n - 500      add 500  to  both sides

1000 + 650 =   n

1650 = n

Proof

A  contains  1650  - 500  =  1150  after the pumping its amount into C

B contains 150  + 1650 - 650 =   1150   after pumping its amount into C

C contains   500 + 650 =  1150  after receiving  oil from A  and B

Cos B  = x / r

r =    sqrt  (x^2 + y^2 ) =    sqrt  [ (-4)^2  +  (3sqrt(5))^2  ] =    sqrt [ 16  + 45  ]=   sqrt (61)

So

Cos B =     -4 / sqrt (61)   =   -4[sqrt (61) ]  / [ sqrt (61)  * sqrt (61) ] =   [-4 sqrt (61) ]  / 61 

First  one  - this is odd because  if    ( -x, y)  is on the  graph, then so is  (x , -y)

Second  one....

If even  then  m(x)  = m(-x)

If odd then m(-x)  = - m(x)

So

g(x) = x^4 - x^2  =  (-x)^4 - (-x)^2 =  x^4 - x^2      so....even

f(x)  = x^3  - 3x^2  = (-x)^3  - 3(-x)^2  =   -x^3  - 3x^2     not even

f(x)  = x^3  - x^2     -f(x)  = - [ x^3 - 3x^2 ]=   -x^2 + 3x^2     not odd, either

So....neither

h(x)   = x^3  - x

h(-x)=   (-x)^3  - (-x) =  -x^3 + x    not even

-h(x)  =  -[ x^2  -x ]  =   -x^3  +  x    not odd , either

So...neither 

No prob, TB  !!!!

OK.....!!!!

Thank you very much!!

Let  the  number of red  marbles  = R

Then the number of blue marbles  =  4R + 7

So

R  + ( 4R + 7) =   107    simplify

5R  + 7  = 107     subtract  7 from  both sides

5R  = 100   divide  both sides  by  5

R =   20  = red marbles

4 (20) + 7 =   87   = blue marbles

Second  one

csc is  positive  in Q1   [all trig  functions  are positive in Q1 ]

tan  =  y / x  = is  negative in Q2  [  y is positive   but  x  is negative ]

cot   =  x / y   =  is positive in Q3  [ x , y  are  both negative here, but   -x / -y  =  x / y ]

F

G               E

FG  and EG  are  the legs  and  EF is the  hypotenuse

EG  =   sqrt   [ EF^2  -  FG^2 ]=  sqrt [ 8^2  - 6^2  ] = sqrt [ 64 - 36 ] = sqrt [ 28 ]  =   2sqrt (7)

So

cos F  =    FG / EF  =  6 / 8  =   3 / 4

tan F =  EG  / FG  =   2sqrt (7)  / 6  =     sqrt (7)  / 3

sec E    =  EF / GE  =  8 / [ 2 sqrt (7) ]  =     4 / sqrt [7]  =    4sqrt (7)  /  [ sqrt (7) * sqrt (7) ] = 4sqrt (7)  / 7

Thank you!!

x = y + 8

x  = y*3

Set the x's equal

y*3  = y + 8   simplify

3y   =  y  +  8      subtract y from  both sides

2y  = 8            divide both sides  by  2

y =  4

And

x  = y +  8

x   = 4  +  8

x  =  12

J = 14 + P

J+2 = 3 (P+2)           Sub in J = 14 + P

14 + P      + 2    =   3P  + 6

16+P   = 3P + 6

10       = 2P

P = 5                  then J = 14+P = 19 

If the odd numbers of the first die are painted red, the numbers 1, 3, and 5 are painted red.

On the second die, the only number that is painted red is the number 3.

To get a red numer to be rolled on the first die has a probability of 3/6.

To get a red number to be rolled on the second die has a probability of 1/6.

To get a red number to be rolled on both dice has a probability of (3/6) x (1/6)  =  3/36  =  1/12

-2u³y              25x³

-------     -     ---------      =     

15xz⁵          14u²y²

To subtract fractions, you need a common denominator; in this case, the common denominator is found by multiplying both denominators together:

     Denominator of each fraction =  15xz⁵ · 14u²y²  =  210xz⁵u²y²  

So, both the numerator and denominator of the first fraction must be multiplied by 14u²y²

     Numerator of the first fraction =  -2u³y · 14u²y²  =   -28u⁵y³

And, both the numerator and denominator of the second fraction must be multiplied by 15xz⁵

     Numerator of the second fraction  =  25x³ · 15xz⁵  =  375x⁴z⁵

The problem is now:     [ -28u⁵y³ / 210xz⁵u²y²  ]  -  [ 375x⁴z⁵ / 210xz⁵u²y²  ]

Combining the two numerators:  -28u⁵y³ - 375x⁴z⁵  over the common denominator:  210xz⁵u²y²  gives you your answer"

     =     ( -28u⁵y³ - 375x⁴z⁵ ) / 210xz⁵u²y² 

Probability that it will rain:

-- on both Saturday and Sunday  :  0.46 x 0.54  =  0.2484

-- on Saturday but not on Sunday:  0.46 x 0.46  =  0.2116

-- on Sunday but not on Saturday:  0.54 x 0.54  = 0.2916

On Saturday or Sunday (inclusive or  --  possibly both)  =  0.2484 + 0.2116 + 0.2916  =  0.7516

On Saturday or Sunday (exclusive or  -- one or the other, but not both)  =  0.2116 + 0.2916  =  0.5032

This may be somewhat confusing because "not Sunday"  =  1.00 - 0.54  =  0.46 (the same as "on Saturday") and

"not Saturday"  =  1.00 - 0.46  =  0.54 (the same as "on Sunday").

(75 + 97 + 60 + x) / 4   = 80

and 

(75 + 97 + 60 + x) /4  = 89           Solve for'x' in both equations to find the range of scores....