Your aanswer is 13.9 because the 13 stays the same and in decimals .1 is a tenth so there for .9 is your tenth 13.9
Answer:
Step-by-step explanation:
Domain : Set of all possible input values (x-values) on a graph
Codomain : Set of all possible out values for the input values (y-values) on the graph
Range : Actual output values for the input values (x-values) given on the graph.
Therefore, for the given graph,
Domain : (-∞, ∞)
Codomain : (-∞, 2]
Range : (-∞, 2]
From the given graph every input value there is a image or output value.
Therefore, the given function is onto.
Answer:
2x2=4
Step-by-step explanation:
In order to accurately answer any fraction, it is best to have a common denominator (bottom of the fraction)
Since in
, 100 is the denominator, you need to make the other fraction with the same denominator
So Multiply
by <u />
So
or
You can do this another way which involves the opposite way of multiplying which is by dividing
Since you have to get a common denominator to measure accurately no matter what, you can divide
One thing to know is that when dividing fractions, both the denominator and numerator have to be divided by a factor to be divided and result in a whole number
÷
=
So just have to multiple
⇒
⇒
The results are the same
Again
is greater than
Answer:
x ≈ {-11.789, +0.501, 11.288}
Step-by-step explanation:
The cubic g(x) - f(x) = 0 has three real solutions. It can be rewritten as ...
Since the solutions are irrational, they are best found using a spreadsheet or graphing calculator. My favorite graphing calculator shows the approximate solutions below.
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<em>Comment on the problem statement</em>
Your expression for f(x) is ambiguous in that many Brainly questions have the exponentiation indicator replaced by a blank: 3 x -1 often means 3^x -1 and sometimes means 3^(x-1). We have taken the expression at face value and have assumed it is a linear expression. If otherwise, the problem is basically worked the same way: write a function h(x) = g(x) - f(x) and look for solutions to h(x) = 0. Graphing can be useful.