Answer:
<h2> I want to prove that every real polynomial of odd degree has at least one real root, using the intermediate value theorem.</h2><h2 /><h2>Let
</h2>
Let, one leg be x units.
since the other leg is 9.0 units shorter than the other,
the measure of the other leg =x-9
hypotenuse =13.0 units
According to Pythagorean theorem, the square of the hypotenuse is equal to the sum of squares of the other two sides.
x^2 +(x-9)^2 =13^2
x^2 + x^2 -18x +81 = 169
2x^2-18x-88=0
Divide the whole equation by 2
x^2 - 9x -44 =0
Let's use the quadratic formula to find the roots.
formula: x= [-b ± sqrt(b^2-4*a*c)]/2*a
a=1 b=-9 c=-44
x= [9±sqrt(81+176)]/2
=[9±sqrt(257)]/2
=[9±16.03]/2
=25.03/2 or -7.03/2
length of a triangle cannot be negative,
Therefore, x=25.03/2 =12.52
one leg =12.5 units
other leg =12.52-9 =3.5 units
Answer:
24
Step-by-step explanation:
To evaluate this, we just need to substitute 2 for whenever we see x.
40 - 8 (2)
From here, we can just use order of operations to simplify it down to a single number.
40 - 16
24
So, when x = 2, we get 24.
its upsidedown what are you doing
