<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:
≈ 1833
Step-by-step explanation:
To find ratio of proton mass to electron mass, we have to divide.
The numbers are given in <em>scientific notation</em>.
Let a number be 
and another be 
, when we divide, we will follow the rule shown below:
Now, we use the information given to find the ratio:
This means we can find the number by taking 4 decimal places to the right, so that would becomes:
The approximate ratio is 1833 [mass of proton is around 1833 times heavier than mass of electron]
Answer:
804 in^2
Step-by-step explanation:
The surface area of a sphere is given by A = 4πr^2, where r is the radius of the sphere.
Here, A = 4π*8^2 in^2 (must use units of measurement)
=256π in^2, or approximately 804 in^2
Answer:
21.2 yd
Step-by-step explanation:
8 + 6 + 7.2 = 14 + 7.2 = 21.2
Answer:
Step-by-step explanation:
Use SOH CAH TOA to recall how the trig functions fit on a triangle
SOH: Sin(Ф)= Opp / Hyp
CAH: Cos(Ф)= Adj / Hyp
TOA: Tan(Ф) = Opp / Adj
5)
Adj = 14
Hyp = 26
∠X
so use
CAH
Cos(X) = 14/26
X = arcCos(14/26)
X = 57.421°
X = 57.4 ° ( rounded to nearest 10th )
6)
∠X
Hyp = 46
Opp = 12
use SOH
Sin(x) = 12/46
X = arcSin(12/46)
X = 15.121°
X = 15.1 ° ( rounded to nearest 10th )
7)
∠X
Adj = 29
Opp = 24
use TOA
Tan(x) = 29 / 24
X = arcTan( 29 /24)
X = 50.389
X = 50.4 ° ( rounded to nearest 10th )
8)
∠X
Adj = 22
Opp = 6
use TOA agian
Tan(x) = 6 / 22
X = arcTan(6/22)
X = 5.194
X = 5.2 ° ( rounded to the nearest 10th )
:)
