x = business man's salary
y = employee's salary
x + 10y = 64000
x = 6y
6y + 10y = 64000
16y = 64000
y = 64000/16 = 4000
x = 6*4000
x = 24000
business man's salary was $24,000
Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
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You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
Answer:
Value of House = $98000
Step-by-step explanation:
The sum of all of the assets and liabilities + value of the house = $101,800
Value of House + $900 - $3400 + $16900 - $16300 + $4500 + $1200 = $101,800
=> Value of House + $3800 = $101,800
=> Value of House = $101,800 - $3800
=> Value of House = $98000
Answer:
More students like PE
Step-by-step explanation:
100÷20=5
1 student = 5%
4×5=20
20% students favourite is social studies
75%>20%
Answer:
Step-by-step explanation:
Domain: the set of all permissible input values. In this particular case, x cannot be -3, because that would cause division by zero. Thus, the domain is (-∞, -3) ∪ (-3,∞)
Range: the range consists of the set of all values of y that are possible. Looking at the parent function, y = 1/x, we see that the range of that function is (-∞, 0) ∪ (0,∞); in other words, y gets close to but never actually reaches zero. This is also true of y = -1 / (x + 3).
Thus, the desired range is (-∞, 0) ∪ (0,∞).
