Answer:
Domain is your x max and Min, Range is your y max and Min
Step-by-step explanation:
Domain would be -5 to 8 or (-5,8)
range would be -3 to 2 or (-3,2)
Answer:
KL = 27
JK = 16
MK = 30
NL = 23
m∠JKL = 132°
m∠KLJ = 22°
m∠KMJ = 54°
m∠KJL = 26°
Step-by-step explanation:
The given parameters of the quadrilateral JKLM are;
JM = 27, ML = 16, JL = 46, NK = 15, KLM = 48, JKM = 78, MJL = 22
Taking the sides as parallel, we have that quadrilateral JKLM is a parallelogram
Therefore;
KL = JM = 27
JK = ML = 16
m∠KLJ = m∠MJL = 22°
MN = NK = 15
MK = MN + NK = 15 + 15 = 30
NL = JL/2 = 46/2 = 23
m∠KJM = m∠KLM = 48°
m∠KJL = m∠KLM - m∠MJL = 48° - 22° = 26°
m∠KML = m∠JKM = 78°
m∠MKL = 180° - m∠KML - m∠KLM = 180° - 78° - 48° = 54°
m∠MKL = 54°
m∠JKL = m∠JKM + m∠MKL = 78° + 54° = 132°
m∠KMJ = m∠MKL = 54°
Each person received one quarter, which can be written as 1/4 or 0.25.
In a 45 45 90 triangle, the hypotenuse equals the length of the leg multiplied by the square root of 2.
Therefore, x = 6 * square toot (2)
which equals
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8.48528</span></span></span>...
Source:
http://www.1728.org/trig2.htm
<u>ANSWER: </u>
The slope of the given equation is -2 and a point on the given line is (-1, 3)
<u>SOLUTION:</u>
Given, linear equation in two variables is y – 3 = -2(x + 1) ----- eqn (1)
We need to find the slope of the given equation and a point through which the given line passes.
Given equation is in the form of point slope form .i.e 
--- eqn 2
where,"m" is the slope of the line
is point on that line.
Now, by comparing (1) and (2)
m = -2
so, the slope of the given equation is -2 and a point on the given line is (-1, 3)
