Yes, because you can simplify it to .2 which can be written as a fraction. Irrational numbers are those ones that keep going with an infinite string of numbers past the decimal.
Answer:
x = -1 and y = 18
Step-by-step explanation:
Given the system of equations
y + 8x = 10... 1
2y - 4x = 40 ... 2
From 1; y = 10-8x
Substitute into 2;
2(10-8x) - 4x = 40
20 - 16x - 4x = 40
20 - 20x = 40
-20x = 40 - 20
-20x = 20
x = -1
Recall that y = 10-8x
y = 10 - 8(-1)
y = 10 + 8
y = 18
therefore your answer is x = -1 and y = 18
5.78778778 ….. is an irrational number because the decimal can't be represented as a ratio or fraction.
<h3>What are irrational numbers?</h3>
Irrational numbers are the set of real numbers that cannot be expressed in the form of a fraction, x/y where x and y are integers.
In other words, an Irrational Number is a real number that cannot be written as a simple fraction.
For example π = 3.1459....
π cannot be written as a fraction or ratio.
The decimals of irrational number goes on forever without repeating.
The decimals are continuous.
Therefore, 5.78778778 ….. is an irrational number because the decimal can't be represented as a ratio or fraction.
learn more on irrational number here: brainly.com/question/3386568
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Answer:
0
Step-by-step explanation:
8c+2c-10c
First, you add 8c to 2c an get 10c.
10c-10c
Then you subtract 10c and 10c which both cancel out and you will get 0
0
Hope it helps have a great day!!!
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
, 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.