Answer:
x = 12
m∠A = 104º
m∠B = 32º
m∠C = 44º
shortest side = B
middle side = C
longest side = A
Step-by-step explanation:
Recall that all angles in a triangle add up to 180º. Also recall that the side opposite largest angle is the longest side, and vice versa for the side opposite the smallest angle.
Answer:
One other zero is 2+3i
Step-by-step explanation:
If 2-3i is a zero and all the coefficients of the polynomial function is real, then the conjugate of 2-3i is also a zero.
The conjugate of (a+b) is (a-b).
The conjugate of (a-b) is (a+b).
The conjugate of (2-3i) is (2+3i) so 2+3i is also a zero.
Ok so we have two zeros 2-3i and 2+3i.
This means that (x-(2-3i)) and (x-(2+3i)) are factors of the given polynomial.
I'm going to find the product of these factors (x-(2-3i)) and (x-(2+3i)).
(x-(2-3i))(x-(2+3i))
Foil!
First: x(x)=x^2
Outer: x*-(2+3i)=-x(2+3i)
Inner: -(2-3i)(x)=-x(2-3i)
Last: (2-3i)(2+3i)=4-9i^2 (You can just do first and last when multiplying conjugates)
---------------------------------Add together:
x^2 + -x(2+3i) + -x(2-3i) + (4-9i^2)
Simplifying:
x^2-2x-3ix-2x+3ix+4+9 (since i^2=-1)
x^2-4x+13 (since -3ix+3ix=0)
So x^2-4x+13 is a factor of the given polynomial.
I'm going to do long division to find another factor.
Hopefully we get a remainder of 0 because we are saying it is a factor of the given polynomial.
x^2+1
---------------------------------------
x^2-4x+13| x^4-4x^3+14x^2-4x+13
-( x^4-4x^3+ 13x^2)
------------------------------------------
x^2-4x+13
-(x^2-4x+13)
-----------------
0
So the other factor is x^2+1.
To find the zeros of x^2+1, you set x^2+1 to 0 and solve for x.
So the zeros are i, -i , 2-3i , 2+3i
Answer:
Zeros of a function: x = 1 and x = -10
Step-by-step explanation:
A zero of a function is the value of x when f(x) = 0. In other words it is the intersect point between x and y=f(x), when y =0.
First we need to make f(x)=0, so we would have: .
If we factorize this quadratic function, we would have:
So we get x = -10 and x = 1.
Hope it helps!