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Assuming that QED denotes Q*E*D:

(5+2i)(5-2i)(i) = (5^2 - (2i)^2)(i) = (25 - (-4))(i) = 29i.

QED=29i

ANSWER= 29i

Slope intercept form    y = mx + b    m is slope  b is y- axis intercept (the value when x = 0)

slope =  (y1-y2) / (x1-x2)    = (3- -3) / (3 - -6)  = 6/9  = 2/3

y = 2/3 x + b    now put in one of the points to find the value of 'b'

3 = 2/3 (3) + b

3 = 2+b      b= 1

y = 2/3 x + 1

If  f((0)  =  47.....then  the  constant term  is  47

If f(1)  =   32......then the  sum of the coefficients +  the  constant term    = 32

This  implies that

sum of coefficients  + constant term =   32

sum of coefficients  + 47  =  32         subtract  47  from both sides

sum of the coefficients    =    -15

thank you guys now I get it I'm really missing being in school stupid coronavirus 

Can you please explain your work.

It means to make him be proud of you.

Thank you! You too~

That's impressive, Chris!!!!

Hm... try messaging CPhill or Melody.

Last one

Area =  (1/2)  (XZ)(YZ)  * sin ( Z)   =  (1/2) (9)(4) sin (58°)  =  18 * sin (58°)  ≈  15.3  units^2

Second one

Area of sector   =  (1/2) r^2  ( theta in rads)   =  (1/2) (5^2) (5pi/12)      =

125 pi

_____    in^2

  24

As before  using Heron's formula

s =   [7 + 7 + 6 ]  /2  =  20 /2  =10

Area  =   √[10 * (10 - 7) (10 - 7) (10- 6)  ]  = √[10 * 3 * 3 * 4  ]  = √ 360  = √ [36 * 10 ] = 

6√10  units^2

That's the spirit! LOL!

A is one leg of the triangle while B is the other leg(order does not matter). C is the hypotenuse also known as the diagonal. A squared + B squared is equal to C squared.

Nice, Chris!

That's amazing, Guest!

For example, in this case,

10^2+24^2=c^2, and if you square root c^2, you will get the third side length.

Hope this helped!

We  have a right triangle  with legs of   10  and 24

To find the remaining side  ( the side opposite the right angle = the hypotenuse )  we  have that

hypotenuse  =  √  [ ( leg 1) ^2   + ( leg 2) ^2 ]  = √ [10^2 + 24^2 ] = √ [100 + 576 ]  =

√ [ 676 ]  = 

26

It is a^2+b^2+c^2, where a, b, and c are side lengths.

Second one

One  equation  is

a / 3  = 5 / 8

Another one  is just a variation of this

a / 5 =  3 / 8

First one ....not possible

All the   sides of the smaller figure  would  have to be  1/2  the length of the corresponding sides of the larger figure

But  note  that    AE / VZ  =  10 /12  =  5 /6

So.....we cannot dilate  the first figure  to get the second

a=1; b=1;c=lcm(a, b); d=gcd(a, b);if(c==108 and d==2, goto5, goto7);print"LCM of",a,"and", b,"=",c;print"GCD of",a,"and", b,"=",d; a++;if(a<200, goto2, 0);a=1;b++;if(b<200, goto2, discard=0;

OUTPUT:

LCM of 108 and 2 = 108
GCD of 108 and 2 = 2
LCM of 54 and 4 = 108
GCD of 54 and 4 = 2
LCM of 4 and 54 = 108
GCD of 4 and 54 = 2
LCM of 2 and 108 = 108
GCD of 2 and 108 = 2

OK, man....glad to help where I can....

i=0;j=0;m=0;t=0;a=(4, 5, 7);r= (3, 2, 4);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

  /* CHINESE REMAINDER THEOREM + MODULAR MULTIPLICATIVE INVERSE */

OUTPUT = 140 m +  67, where m =0, 1, 2, 3.......etc.

The minimum number of cookies =67

Since (x - 7) is a factor   then 7 is a zero

Using a little  synthetic  division to find the  remaining polynomial we have that

7  [ 1    -14       61      -84   ]

               7      -49       84

     ______________________

       1      -7      12        0

The remaining polynomial  is

x^2   -7x  +  12

Set to  0  and factor and we have that

(x - 3) (x - 4)  =  0

Setting each factor to 0  and solving for x  gives the other two roots  (zeroes)  of  x = 3  and  x = 4