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vagabundo [1.1K]
3 years ago
11

What is the slope of the line?

Mathematics
1 answer:
katrin [286]3 years ago
5 0

Answer:

3

Step-by-step explanation:

down 3 over 1 3/1 or 3.

Hope this helps plz hit the crown ;D

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1.00 is 5% of what number
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20

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If you work this out you get 1:

5/100*20

5 divided by 100 is 0.05 then 0.05 times 20 is 1

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adele Govern saves $150 per month and deposits the money in an account earning 3% annual interest compounded monthly. how much w
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Given:
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Pls. see attachment for my answer.


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Write the equation of the line that passes through the given points. Include your work in your final answer. Type your answer in
faust18 [17]
First, find the slope of this line, if possible.  Recall that slope = m = rise / run.

We see that x does not change:

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4 0
3 years ago
Explain why the graphs of reciprocals of linear functions (except horizontal ones) always have vertical asymptotes,
Naily [24]

Answer:

The reason is because linear functions always have real solutions while some quadratic functions have only imaginary solutions

Step-by-step explanation:

An asymptote of a curve (function) is the line to which the curve is converging or to which the curve to line distance decreases progressively towards zero as the x and y coordinates of points on the line approaches infinity such that the line and its asymptote do not meet.

The reciprocals of linear function f(x) are the number 1 divided by function that is 1/f(x) such that there always exist a value of x for which the function f(x) which is the denominator of the reciprocal equals zero (f(x) = 0) and the value of the reciprocal of the function at that point (y' = 1/(f(x)=0) = 1/0 = ∞) is infinity.

Therefore, because a linear function always has a real solution there always exist a value of x for which the reciprocal of a linear function  approaches infinity that is have a vertical asymptote.

However a quadratic function does not always have a real solution as from the general formula of solving quadratic equations, which are put in the form, a·x² + b·x + c = 0 is  \dfrac{-b \pm \sqrt{b^{2} - 4\cdot a\cdot c}}{2\cdot a}, and when 4·a·c > b² we have;

b² - 4·a·c < 0 = -ve value hence;

√(-ve value) = Imaginary number

Hence the reciprocal of the quadratic function f(x) = a·x² + b·x + c = 0, where 4·a·c > b² does not have a real solution when the function is equal to zero hence the reciprocal of the quadratic function which is 1/(a·x² + b·x + c = 0) has imaginary values, and therefore does not have vertical asymptotes.

6 0
3 years ago
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