Which ordered pair makes both inequalities true?

y > –2x + 3

y < x – 2

On a coordinate plane, 2 straight lines are shown. The first solid line has a positive I I NEED HELP ASAP, QUICK slope and goes through (0, negative 2) and (2, 0). Everything to the right of the line is shaded. The second dashed line has a negative slope and goes through (0, 3) and (1, 1). Everything to the right of the line is shaded.
(0,0)
(0,–1)
(1,1)
(3,0)

(3,0)

Step-by-step explanation:

y > –2x + 3

y < x – 2

Graph the inequalities

We graph above the dotted line for y > –2x + 3   and below the dotted line for  y < x – 2

The solution is where the shaded areas overlap.

(3,0) is a solution

(3,0)

Step-by-step explanation:

y > –2x + 3

y < x – 2

(0,0)

0 > -2(0) + 3

0 > 3 --> true

0 < 0 - 2

0 < -2 --> false

(0,–1)

-1 > -2(0) + 3

-1 > 3 --> false

(1,1)

1 > -2(1) + 3

1 > 1 --> false

(3,0)

0 > -2(3) + 3

0 > -6 + 3

0 > -3 --> true

0 < 3 - 2

0 < 1 --> true

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752