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sveta [45]
3 years ago
10

What's the answer to number two

Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
5 0
You have to show more of the paper so i can calculate it. Sorry
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This graph displays a linear function. What is the rate of change?
ivolga24 [154]

Answer:

5/1 or 5

Step-by-step explanation: You find two points that the line goes on. So in this it would be 1,1 for the first one, the second one is 2,4. Then you go from the first point up or down to get to the same level at the second point. Then you go over to the left or right to meet the second point, in this graph you will go up 5 times to get to the same level. Then over 1 time to get to the point. Then you divide the rise over the run the rise is 5 and the run is 1, so you would do 5 divided by 1 and the answer of that is 5.

Hope this helps you in the future

                                                            Smiles,  MysticCryptic (Aka) Jordan

8 0
2 years ago
87 less than the quotient of an unknown number and 43 is -75
choli [55]
The statement is expressed algebraically as
(x/43) - 87 = -75
8 0
3 years ago
there are 4800 plastic spoons in a case there 6 box3s of spoons in each case how many spoons are in each box
Slav-nsk [51]

Sorry if im wrong but i am pretty sure that is 266

3 0
2 years ago
Which expression represents the quotient of d and 11​
inna [77]

Answer:   :} where is the rest of the question

Step-by-step explanation:

3 0
2 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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