X cannot be solved.
This is not a solvable equation
no matter what, it will not equal
so your answer is unsolvable
hope this helps
Answer:
See Explanation
Step-by-step explanation:
![log(x + y) = log3 + \frac{1}{2} logx+ \frac{1}{2} logy \\ \\ log(x + y) = log3 + logx ^{\frac{1}{2}} + logy ^{\frac{1}{2}}\\ \\ log(x + y) = log3 + log(xy) ^{\frac{1}{2}} \\ \\ log(x + y) = log[3(xy) ^{\frac{1}{2}}] \\ \\ x + y = 3(xy) ^{\frac{1}{2}} \\ \\ squaring \: both \: sides \\ {(x + y)}^{2} = \bigg(3(xy) ^{\frac{1}{2}} \bigg)^{2} \\ \\ {x}^{2} + {y}^{2} + 2xy = 9xy \\ \\ {x}^{2} + {y}^{2} = 9xy - 2xy \\ \\ \purple{ \bold{{x}^{2} + {y}^{2} = 7xy}} \\ thus \: proved](https://tex.z-dn.net/?f=log%28x%20%2B%20y%29%20%3D%20log3%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20logx%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20logy%20%5C%5C%20%20%5C%5C%20log%28x%20%2B%20y%29%20%3D%20log3%20%2B%20%20%20%20logx%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%2B%20%20%20logy%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%20%20%5C%5C%20%20log%28x%20%2B%20y%29%20%3D%20log3%20%2B%20%20%20%20log%28xy%29%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5C%5C%20%20%5C%5C%20log%28x%20%2B%20y%29%20%3D%20%20log%5B3%28xy%29%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5D%20%5C%5C%20%20%5C%5C%20x%20%2B%20y%20%3D%203%28xy%29%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5C%5C%20%20%5C%5C%20squaring%20%5C%3A%20both%20%5C%3A%20sides%20%5C%5C%20%20%7B%28x%20%2B%20y%29%7D%5E%7B2%7D%20%20%3D%20%20%5Cbigg%283%28xy%29%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cbigg%29%5E%7B2%7D%20%20%5C%5C%20%20%5C%5C%20%20%7Bx%7D%5E%7B2%7D%20%20%2B%20%20%7By%7D%5E%7B2%7D%20%20%2B%202xy%20%3D%209xy%20%5C%5C%20%20%5C%5C%20%20%7Bx%7D%5E%7B2%7D%20%20%2B%20%20%7By%7D%5E%7B2%7D%20%20%3D%209xy%20-%202xy%20%5C%5C%20%20%5C%5C%20%20%20%5Cpurple%7B%20%5Cbold%7B%7Bx%7D%5E%7B2%7D%20%20%2B%20%20%7By%7D%5E%7B2%7D%20%20%3D%207xy%7D%7D%20%5C%5C%20thus%20%5C%3A%20proved)
A. The point estimate of μ1 − μ2 is calculated using the value of x1 - x2, therefore:
μ1 − μ2 = x1 – x2 =
7.82 – 5.99
μ1 − μ2 = 1.83
B. The formula for
confidence interval is given as:
Confidence interval
= (x1 –x2) ± z σ
where z is a value
taken from the standard distribution tables at 99% confidence interval, z =
2.58
and σ is calculated
using the formula:
σ = sqrt [(σ1^2 /
n1) + (σ2^2 / n2)]
σ = sqrt [(2.35^2 /
18) + (3.17^2 / 15)]
σ = 0.988297
Going back to the
confidence interval:
Confidence interval
= 1.83 ± (2.58) (0.988297)
Confidence interval
= 1.83 ± 2.55
Confidence interval
= -0.72, 4.38
Answer:
3
Step-by-step explanation:
Use the Pythagorean Theorem.
=
+
1369=+100
-100 -100
1269=
=
3=x
For this case we have the following polynomials:
3x2
x2y + 3xy2 + 1
We have then:
For 3x2:
Classification: polynomial of one variable:
Degree: 2
For x2y + 3xy2 + 1:
Classification: polynomial of two variables
Degree: 2 + 1 = 3
Answer:
The polynomial 3x2 is of one variable with a degree of 2.
The polynomial x2y + 3xy2 + 1 is of two variables a with a degree of 3.
