Answer:
$65.33
Step-by-step explanation:
Answer:
28
Step-by-step explanation:
bc = df x 2
14 x 2 = 28
I hope this helps
Use the rules of logarithms and the rules of exponents.
... ln(ab) = ln(a) + ln(b)
... e^ln(a) = a
... (a^b)·(a^c) = a^(b+c)
_____
1) Use the second rule and take the antilog.
... e^ln(x) = x = e^(5.6 + ln(7.5))
... x = (e^5.6)·(e^ln(7.5)) . . . . . . use the rule of exponents
... x = 7.5·e^5.6 . . . . . . . . . . . . use the second rule of logarithms
... x ≈ 2028.2 . . . . . . . . . . . . . use your calculator (could do this after the 1st step)
2) Similar to the previous problem, except base-10 logs are involved.
... x = 10^(5.6 -log(7.5)) . . . . . take the antilog. Could evaluate now.
... = (1/7.5)·10^5.6 . . . . . . . . . . of course, 10^(-log(7.5)) = 7.5^-1 = 1/7.5
... x ≈ 53,080.96
Answer:
* Elimination; a coefficient in Equation I is an integer multiple of a coefficient in Equation II.
* Elimination; a coefficient in Equation II is an integer multiple of a coefficient in Equation I.
Step-by-step explanation:
Equation I: 4x − 5y = 4
Equation II: 2x + 3y = 2
These equation can only be solved by Elimination method
Where to Eliminate x :
We Multiply Equation I by a coefficient of x in Equation II and Equation II by the coefficient of x in Equation I
Hence:
Equation I: 4x − 5y = 4 × 2
Equation II: 2x + 3y = 2 × 4
8x - 10y = 20
8x +12y = 6
Therefore, the valid reason using the given solution method to solve the system of equations shown is:
* Elimination; a coefficient in Equation I is an integer multiple of a coefficient in Equation II.
* Elimination; a coefficient in Equation II is an integer multiple of a coefficient in Equation I.
G=0 would be your answer as it is greater than the number
