Find the area A of JKL with vertices j(-5,5) k(0,9), and L(3,5)

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm

4 and 5)16.97 or 12√2

6)15.59 or 9√3

7)y = 15.59 or 9√3     x =27

8)17.80

so a parallelogram is just like a square base times height. multiply 40 and 39 to get 1560 inches.

a triangle is base times height divided by two. so multiply 7.9 and 5 and then divide it by 2 to get 19.75inches.

the kite can be split into two separate triangles and added together to get the total area. 30 x 4 /2 which equals 60. make sure to double it for the two triangles making 120.

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