Answer:
55- 3m is the awnser for this question
<u>We'll assume the quadratic equation has real coefficients</u>
Answer:
<em>The other solution is x=1-8</em><em>i</em><em>.</em>
Step-by-step explanation:
<u>The Complex Conjugate Root Theorem</u>
if P(x) is a polynomial in x with <em>real coefficients</em>, and a + bi is a root of P(x) with a and b real numbers, then its complex conjugate a − bi is also a root of P(x).
The question does not specify if the quadratic equation has real coefficients, but we will assume that.
Given x=1+8i is one solution of the equation, the complex conjugate root theorem guarantees that the other solution must be x=1-8i.
Answer:
23
Step-by-step explanation:
we are given that angle NRQ is 78 degrees
we can see from the figure that the sum of the given angles is angle NRQ
so
(8x + 7) + (4x - 1) = 78
12x + 6 = 78
12x = 72
x = 6
Now, we have to find angle PRQ
replacing x with 6 in the equation of angle PRQ
PRQ = 4(6) - 1
PRQ = 23
No.
Since w = 8, you must replace the variable with the number associated to it.
So, it would be 5 + 8 = 58
5 + 8 is not equal to 58, it is equal to 13.
If you wanted to find out what w was, simply subtract 5 on both sides.
w = 58 - 5
Answer:
A. I can't quite see the question, but I'm pretty sure it's A
Step-by-step explanation:
Sin(A) = 1/3
Sin^2(A) + Cos^2(A) = 1
(1/3)^2 + cos^2(A) = 1
1/9 + cos^2(A) = 1
cos^2(A) = 1 - 1/9
cos^2(A) = 8/9
cos(A) = √(8/9)
√8 = √(2 * 2 * 2) = 2√2
√9 = 3
cos(A) = 2√2/3
