Answer:
The length of the hypotenuse is 2 square root of 13 ⇒ c
Step-by-step explanation:
The rule of the area of the right triangle is A = 
× leg1 × leg2, where
leg1 and leg2 are the sides of the right angle
∵ The area of a right triangle is 12 in²
∵ The ratio of the length of its legs is 2: 3
→ Let leg1 = 2x and leg2 = 3x
∵ leg1 = 2x and leg2 = 3x
→ Substitute them in the rule of the area above
∴ 12 = × 2x × 3x
∵ 2x × 3x = 6x²
∴ 12 = × 6x²
∴ 12 = 3x²
→ Divide both sides by 3 to find x²
∴ 4 = x²
→ Take √ for both sides
∴ x = 2
→ Substitute x in the expressions of leg1 and leg2 to find them
∴ leg1 = 2(2) = 4 inches
∴ leg2 = 3(2) = 6 inches
∵ hypotenuse = 
∴ hypotenuse = 
∵ The simplest form of 
= 2
∴ The length of the hypotenuse = 2 inches
Sin2(x) +cos(x)=1
from the relation: (sin2(x) +cos2(x) =1 )
so , sin2(x)=1-cos2(x)
by subs. in the main eqn.
1-cos2(x) + cos(x) =1
by simplify the eqn.
cos(x) -cos2(x)=0
take cos(x) as a common factor
cos(x)* (1-cos(x))=0
then cos(x)=0 && cos(x)=1
cos(x)=0 if x= pi/2
& cos(x) = 1 if x = 0 , 2*pi
so the solution is x= {0,pi/2 , 2*pi}
The answer would be 6 (This is just to get 20 caracters NOT PART OF ANSWER)
The range is the ouutput from inputing the input
basically
25=k²+2k+1 and 64=k²+2k+1
the values that satisfy both equations (not at the same tim) are the valuess that are the domain
solve each
25=k²+2k+1
minus 25 both sides (or recognize the perfect square trinomial, but anyway)
0=k²+2k-24
factor
0=(k+6)(k-4)
set to zero
k+6=0
k=-6
k-4=0
k=4
k=-6 or 4
64=k²+2k+1
minus 64 both sides
0=k²+2k-63
facor
0=(k-7)(k+9)
set to zer
k-7=0
k=7
k+9=0
k=-9
k=-9 or 7
so the domain has the numbers
-9,-6,4,7
it seems we only want the positive square roots so
answer is {4,7} is the domain
Assuming that b>0 and c>0:
log(a)=3×log(b)-2×log(c)=
log(b³)-log(c²)=
log(b³/c²)
Hence a=b³/c²
