Answer:
C: y+7=2/5(x+4)
Step-by-step explanation:
The point-slope form of the equation of a line with slope m through point (h, k) is ...
y -k = m(x -h)
You are given m=2/5 and (h, k) = (-4, -7). Put these numbers into the form and simplify the signs:
y -(-7) = 2/5(x -(-4)) . . . . . numbers put into the form
y +7 = 2/5(x +4) . . . . . . . signs simplified . . . . matches choice C
Okay so the problem starts you off with the fraction (1/4). Sharing equally among 3 people would mean that you'd want to divide it by three so your equation would initially look like (1/4) / 3. Having two dividing signs is confusing so what you do is change the dividing into multiplying by the reciprocal of 3 = 1/3... This would make your expression (1/4) * (1/3) and when you simplify this you should get 1/12 as the answer.
Answer:
No bcuz its not on the line
Step-by-step explanation:
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:
There's 3 questions which one do I answer?
Step-by-step explanation:
