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kobusy [5.1K]
3 years ago
10

Please help need this answer

Mathematics
1 answer:
I am Lyosha [343]3 years ago
8 0

A function is a relation for which each value from the set the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair.

(2, 0), (-3, 3), (9, 1), (-3, 5)  NOT

(9, 1), (-3, 4), (2, 1), (9, 2)     NOT

(2, 4), (-3, 2), (9, 1), (-7, 2)     YES

(2, 4), (-3, 6), (2, 3), (-7, 2)     NOT

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frozen [14]
Probably kJ/month if that’s what you want
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3 years ago
Evaluate each expression for x=3, y= -4 and z= 6.
Elena L [17]

Answer:

yes so basically where the negative is outside the parenthesis is a understood 1 or in your case a negative 1,is this what you needed help with?

8 0
2 years ago
What is the solution set of this inequality? -8x-1>3x+14​
Nezavi [6.7K]

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5 the guy above is right

Step-by-step explanation:

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3 years ago
Evaluate the square root
Alexandra [31]

\sqrt{a}=b\iff b^2=a\ for\ a\geq0\ and\ b\geq0\\\\\sqrt{\dfrac{1}{144}}=\dfrac{\sqrt1}{\sqrt{144}}=\dfrac{1}{12}\\\\\\\sqrt1=1\ because\ 1^2=1\\\\\sqrt{144}=12\ because\ 12^2=144

3 0
3 years ago
How to solve an expression with variables in the exponents?
gtnhenbr [62]

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.

__

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.

__

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.

8 0
2 years ago
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