What is the length of EF in the right triangle below?
1 answer:
Answer:
A
Step-by-step explanation:
Since the triangle is right use Pythagoras' identity to find EF
The square on the hypotenuse is equal to the sum of the squares on the other two sides, that is
EF² + DF² = DE² ← substitute values
EF² + 12² = 18², that is
EF² + 144 = 324 ( subtract 144 from both sides )
EF² = 180 ( take the square root of both sides )
EF = → A
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Ooh, fun
geometric sequences can be represented as
so the first 3 terms are
the sum is -7/10
and their product is -1/125
from the 2nd equation we can take the cube root of both sides to get
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
subsituting -1/5 for ar
which simplifies to
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for
so
for 2r²-5r+2=0
a=2
b=-5
c=2
so
or 
use them to solve for the value of a
try for r=2 and 1/2
or 
test each
for a=-1/10 and r=2
a+ar+ar²=
it works
for a=-2/5 and r=1/2
a+ar+ar²=
it works
both have the same terms but one is simplified
the 3 numbers are
, 
, and 
geometric sequences can be represented as
so the first 3 terms are
the sum is -7/10
and their product is -1/125
from the 2nd equation we can take the cube root of both sides to get
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
subsituting -1/5 for ar
which simplifies to
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for
so
for 2r²-5r+2=0
a=2
b=-5
c=2
so
use them to solve for the value of a
try for r=2 and 1/2
test each
for a=-1/10 and r=2
a+ar+ar²=
it works
for a=-2/5 and r=1/2
a+ar+ar²=
it works
both have the same terms but one is simplified
the 3 numbers are