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Can someone please help? Thanks in advance :)

To solve this, you can use quadratics. x + y = 177 and 4x + 11y = 1380 are the equations to solve for x and y. 177 is the total amount of people who entered the fair, and 1380 is the total money earned. x is the amount of children entered, while y is the amount of adults entered. If we were to do 4(x + y = 177), the equation would become 4x + 4y = 708. Then, we minus the 2 equations together, eliminating the x so we can solve for y. So, the final equation is 7y = 672, meaning y = 96. If we bring this into the first equation, x + y = 177, we can find that x = 81. So this means the answer is 81 children and 96 adults.

The answer would be 50, because adding a positive number and a negative number, is just basically the positive number minus the negative (-33 + 83 becomes 83 - 33).

what is -33+83= 

It's 50.  Would it be more clear if the numbers were swapped places with the numbers, thus:  83 – 33 = 

_.

I think the answer is way bigger than 6

base 7 doesnt work either...

Solve the following system:
{y = 5 x + 16 | (equation 1)
4 y - 2 x = -10 | (equation 2)

Express the system in standard form:
{-(5 x) + y = 16 | (equation 1)
-(2 x) + 4 y = -10 | (equation 2)

Subtract 2/5 × (equation 1) from equation 2:
{-(5 x) + y = 16 | (equation 1)
0 x+(18 y)/5 = (-82)/5 | (equation 2)

Multiply equation 2 by 5/2:
{-(5 x) + y = 16 | (equation 1)
0 x+9 y = -41 | (equation 2)

Divide equation 2 by 9:
{-(5 x) + y = 16 | (equation 1)
0 x+y = (-41)/9 | (equation 2)

Subtract equation 2 from equation 1:
{-(5 x)+0 y = 185/9 | (equation 1)
0 x+y = -41/9 | (equation 2)

Divide equation 1 by -5:
{x+0 y = (-37)/9 | (equation 1)
0 x+y = -41/9 | (equation 2)

x = -37/9         and           y = -41/9

Why base 6???

Try base 6.

a=1; b=1;d=b / (a+186); if(d==0.8, goto4, goto5);printa, b; a++;if(a<1000, goto2, 0);a=1;b++;if(b<1000, goto2, discard=0; 

OUTPUT: x = 4    and    y = 152

a 2^r = 1      and   a 64^r = 4

2^r = 1/a                   (2^6)^r = 4/a

                                 (2^r)^6 = 4/a

                                   (1/a)^6 = 4/a

                                    1 = a^6 * 4/a  = 4a⁵ 

                                            a = (1/4)^1/5 = .757858283

then    a 64^r = 4

.757858283 *  64^r = 4   

     r = .4   

AC/2 = 20      BC = 25          ∠ACB = arccos ((AC/2) / BC)

Angle ACB = arccos(20 / 25) = 36.86989765 degrees

Thanks!

30 : 40% = x : 60%     solve for x

pls respond!!!

Is it 12 ?   Hmmmm.....

80 83    87     94 96     currently 87 is the median   

                                    if the next two scores are >87

80  83 87 x x  94 96     one of the x's becomes the median  (x could be 87 -100)    (14 ways)

                                    if one is higher than 87 and one is lower

80 83 x 87 x 94 96      87 is the median  (1)

                                    if both are lower than 87

80 83 x x 87 94 96    one of the x's becomes the median  x could be any score 1 - 86  (86 ways)  ('0' score not possible)

So I THINK it is 14 + 1 + 86 different median scores

Basically the same question with different numbers....see if you can figure this one out:

https://web2.0calc.com/questions/algebra_21582

Find slope    =   m   = -2

point slope form

y-11 = -2 (x-1)     re-arrange

y -11 = -2x+ 2

y + 2x = 13          divide by 13

 x / (13/2)  + y/13 = 1                               a = 13/2

'Guest'   directions were correct..... try them to find your answer....

t = 1/2 sqrt (u/g)      multiply both sides of the equation by 2

2t = sqrt (u/g)         square both sides

4 t^2 = u/g              multiply both sides by g, then divide both sides by 4 t^2

g = u / ( 4t²)

Please check your work that is incorrect

its 12

a = 10/3.