Answer: The given triangle LMN is an obtuse-angled triangle.
Step-by-step explanation: We are given to use Pythagorean identities to prove whether ΔLMN is a right, acute, or obtuse triangle.
From the figure, we note that
in ΔLMN, LM = 5 units, MN = 13 units and LN = 14 units.
We know that a triangle with sides a units, b units and c units (a > b, c) is said to be
(i) Right-angled triangle if 
(ii) Acute-angled triangle if 
(iii) Obtuse-angled triangle if 
For the given triangle LMN, we have
a = 14, b = 13 and c = 5.
So,
Therefore,
Thus, the given triangle LMN is an obtuse-angled triangle.
Answer:
(x, y) = (0, 1/2) or (1, 3)
Step-by-step explanation:
The first equation factors as ...
x(3x -y) = 0
This has solutions x=0 and y=3x.
__
<u>x = 0</u>
Using this in the second equation gives ...
2y -0 = 1
y = 1/2
(x, y) = (0, 1/2) is a solution
__
<u>y = 3x</u>
Using the expression for y in the second equation, we get ...
2(3x) -5x = 1
x = 1 . . . . . . . . . simplify
y = 3x = 3 . . . . using x=1 in the first equation
(x, y) = (1, 3) is a solution
_____
Interestingly, the (red line) graph of 3x^2 -xy = 0 produced by this graphing calculator has a "hole" at x=0, It says that point is (0, undefined). In a sense, y is undefined, in that it can be <em>anything</em>. A more appropriate graph would graph that equation as the two lines x=0 and y=3x.
First find the rate, so dollars/1 hour
To do this, divide both sides by 1.5 so u get the Pay of one hour because 1.5/1.5 is one and the result is 16/1 hour, now multiply 16 by the hour so 16(8) then 16(12) and so on, you can do it well! I don’t have a calculator lol
There could be extra tips or they could be taxed differently
Answer:
a) 5.5
b) None
Step-by-step explanation:
The given data set is {96,89,79,85,87,94,96,98}
First we must find the mean.
We now find the absolute value of the distance of each value from the mean.
This is called the absolute deviation
{
}
{
}
We now find the mean of the absolute deviations
The least absolute deviation is 1.5. This is not within one absolute deviation.
Therefore none of the data set is closer than one mean absolute deviation away from the mean.
