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The  diagonals  bisect each other

One  side  length =  sqrt [ ( 12/2 )^2  + (16/2)^2 ]  =  sqrt [ 6^2  + 8^2 ]  = sqrt  [ 100] =  10

Perimeter  =   4 * 10  = 40

Tysm!

x^2 - 2x  - 3  =  0

( x - 3) ( x + 1)  = 0

Setting  each factor  to 0  and  solving for x  we get   x  = 3  and  x = -1

If  p   =  -1    then    ( -1 + 1) ( -1 + 1)  = (0) * (0)  =   0

If  p =  3   then ( 3 + 1) (3 + 1) =  (4) * (4)  =  16

NP  !!!!

Here's a graph  :  https://www.desmos.com/calculator/srqrtecj8e

tysm!!!!!

This region will form a triangle

To find  the height, let  x =  0

So  2(0) +y  = 6

y  = 6  = the height  of the triangle

To  find the  base  let  y   = 0

2x + 0  =  6

2x  = 6

x= 3

Area =  (1/2) base * height   =  (1/2) (3) ( 6)  = (1/2) (18) =    9  sq units

1 = 90º

2 = 53º

3 = 37º

4 = 37º

janine = x      Ingrid = x+ 2  

x/7   + ( x+2)/3   = 14      multiply through y 21

3x    + 7 (x+2) =  294

10x = 280

x = janine = 28 y/o     ingrid = 30 y/o

a^3  +  3ab^2  =  679

3a^2b  + b^3   =  -652

Note  that   ( a + b)^3  =  ( a^3  + 3a^2b) + (3ab^2  + b^2) =  (a^3 + 3ab^2 )  + ( 3a^2b + b^3)

So

( a + b)^3  =   679  - 652

(a + b)^3  =  27         take the cube root of both sides

(a + b)  =  3

R + G =   100      ⇒  G =  100  - R      (1)

R - 3  =   2G  + 2          (2)

Sub (1)  into (2)

R - 3  =  2(100 - R  + 2)    simplify

R - 3 =  200 - 2R  + 4

R - 3  =  204  - 2R

R + 2R  = 204  + 3

3R =  207

R =  207 / 3     =    69            

G =  100  - 69  =   31 

Thanks, Idyllic....here's  another  method

36  =   2^2 * 3^2

90  =   2 * 3^2  *  5

Look at the  prime factors  that are common to  each each  =  2  and 3

Take  each  with their  lowest powers    =  2 * 3^2  =  2 * 9  =  18  =  GCF

16 =  2^4

24 = 2^3 * 3

40  = 2^3 * 5

Only  common prime to each = 2

Lowest power  =  3

GCF  = 2^3  =  8

https://web2.0calc.com/questions/hyperbola-help_1

https://web2.0calc.com/questions/halp-5th-or-6th-repost-lost-track-lol#r2

AD / BD  =  BD /  DC

AD * DC =  BD^2

4  *  9  =   BD^2

36  =  BD^2

6  = BD

B.    As more time elapses, the candle grows shorter. 

_.

3rd place =  x

1st place =  2x + 500

2nd place  =  (2x + 500) - 400 =  2x   + 100

So

x  + ( 2x + 500)  + (2x + 100) =  1850

5x   +  600  =  1850

5x  =  1850  - 600

5x  = 1250

x = 1250  / 5  =   $ 250  =  3rd place

2(250) + 100  =  $ 600 =  2nd place

2(250) + 500 =  $1000 = 1st place

Let  the number of dimes =  x    and the number of quarters = 6x

So

Value of a dime * numbereof them  + Value of a quarter * number of them  = Total value in the  bank

10x   + (6x) (25)  =  2880    simplify

10x  +  150x  = 2880

160x  = 2880

x =  2880 /  160   =   18   dimes

6x=  6 (18) =  108  quarters

Stars and Bars ==> number of ways = 18!/(3!*7!*8!) = 5250960.

in trapezoids, opposite angles are always supplementary (they add up to 180). and because this trapezoid is isoceles, we can create these equations:

g = 4h

g + h = 180

now we can use substitution to solve.

5h = 180

h = 36.

and we know angle h is equal to angle e, so the answer is just 36 degrees.

hope this helped! please let me know if you are confused about anything i did :)

algebraic solver => a = 7.44856, b = -3.44856, so a + b = 4.

we can make a system of equations for this one. but first, let's define some variables for Marie, Brenda, and Deepa: m, b, and d, respectively. 

here is our system of equations:

m = 2b, or b = m/2

d = m - 3

d + m + b = 22

in the first two equations, you can see that they both can be written in terms of Marie's run distance. now we can use subtitution and substitute those expressions into the last equation.

m - 3 + m + m/2 = 22

2m + m/2 = 25

4m/2 + m/2 = 25

5m = 50

m = 10

now to solve for the others, we can simply plug back our value of m into the previous equations.

d = 10 - 3 = 7.

b = 10/2 = 5.

hope this helped! please let me know if you are confused about anything i did 

i'm not really sure how to explain this, but if you are new to solving the GCF, an easy way to do this is to just list out all the factors of the numbers.

this method works best for smaller numbers. sometimes, you can occasionally tell the GCF of some numbers just by first glance. if you are interested in other (more effective) methods, feel free to ask!

a)

factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

factors of 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

as you can see, they both share the common factor 18, so the GCF of 36 and 90 would just be 18.

b)

factors of 16: 1, 2, 4, 8, 16

factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

factors of 40: 1, 2, 4, 5, 8, 10, 20, 40

as you can see, they both share the common factor 8. they also share other factors, like 1, 2, and 4, but 8 is the largest one. so the GCF of 16, 24, and 40 is 8.

hope this helped! please let me know if you still have any further questions. :)

in a quadrilateral (shape with 4 sides), the sum of all the interior angles is 360.

so the measure of angle x would be

360 - 103 - 133 - 58 = 66 degrees.

hope this helped! please let me know if you have any questions 

I am so sorry for being this late, but I recently had to solve this problem too. 

If you look at the equation \((a^mx-a^n)(a^py-a^2)=a^4b^4\), you may notice that you can factor the previous equation to be in this form. 

First, you distribute \(a^4\)into the parentheses to get \(a^4b^4 - a^4\). If you look at the other equation, you may see that their answer is very similar to \(a^4b^4 - a^4\). So, you first add \(a^4\) to both sides to get \((a^3x-a^2)(a^4y-a^2)=a^4b^4\). Factoring the left side gives \((a^3x-a^2)(a^4y-a^2)=a^4b^4\). So, \((m,n,p)=(3,2,4)\), which means \(mnp=3\cdot2\cdot4=\boxed{24}\).