Resolver los siguientes polinomios con procedimiento a. 10 × 4 + 25 ÷ 5 - 12
b. 200 × 7 + 4 × 3 + 140 ÷ 7- 32
c. 6 × 4 + 28 ÷ 4 + 36 ÷ 9 - 11
d. 38^4 × 2³ + 20² ÷ 10² - 5³
e. √49 × log4 256 + 6^5 ÷ 3^4 - log 10000

answer: 60

step-by-step explanation:

1/2a + (2/3×30)   = 50

1/2a + 20 = 50

1/2a = 30

a = 60

probably .5

i can on others if necessary.

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: