Which of the following ordered pairs could be placed in the table below and still have the relation qualify as a linear function? Input | Outputs

(x) (y)

−5, 7

−3, 15

−1, 23

? ,?

(0, 31)

(−7, 30)

(1, 31)

(0, −1)

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.

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

isocolese triangle

we know that the line connect a and b is the hyptonuse

so first find the hypotonuse length using the distance formula, call the length c

since it is an isocolese triangle, the lengths of the legs are the same

using pythagorean theorem,

in this case, a=b since isocolese triangle so we get

or

we can find c so we can find the distances

we know that if we draw 2 circles, each has center at a and b, and find where they intersect, we can the 2 possible locations for c

if a has center at (xa,ya) and b has center at (xb,yb), then we can solve where

and

intersect to find where they intersect

i'm not going to put an example because i have to go to class. i hope this explanation is enough