Quadratics Help please?
2 answers:
Remember that
for a right triangle
with legs a and b and hptonuse c
a²+b²=c²
given
a=x+5
b=x-3
c=9
(x+5)²+(x-3)²=9²
x²+10x+25+x²-6x+9=81
2x²+4x+34=81
2x²+4x-47=0
use quadratic formula
for 0=ax²+bx+c
x=
so
a=2
b=4
c=-47
x=
x=
x=
x=
x=
x can't be negative because lengths can't be negative
so
x=-1+3.5√2
if we evaluate
x=3.94975
round
x=3.950 cm
for a right triangle
with legs a and b and hptonuse c
a²+b²=c²
given
a=x+5
b=x-3
c=9
(x+5)²+(x-3)²=9²
x²+10x+25+x²-6x+9=81
2x²+4x+34=81
2x²+4x-47=0
use quadratic formula
for 0=ax²+bx+c
x=
so
a=2
b=4
c=-47
x=
x=
x=
x=
x=
x can't be negative because lengths can't be negative
so
x=-1+3.5√2
if we evaluate
x=3.94975
round
x=3.950 cm
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Answer:
The expression used to find the change in temperature per hour is Algebraic expression
Thus per hour; the temperature falls at the rate of
Step-by-step explanation:
A temperature falls from 0 to in
Which expression finds the change in temperature per hour.
From the above given information;
The initial temperature is 0
The final temperature is
The change in temperature is
Thus;
-12.25 ° = 3.5 hours
To find the change in x° per hour; we have;
x° = 1 hour
The expression used to find the change in temperature per hour is Algebraic expression
From above if we cross multiply ; we have;
(- 12.25° × 1 hour) = (x° × 3.5 hour)
Divide both sides by 3.5 hours; we have:
x° = - 3.5
x° =
Thus per hour; the temperature falls at the rate of