1angle = 90
4 angles =360
the answer is false
Answer:
y-6=-7(x-3)
Step-by-step explanation:
m=y2-y1/x2-x1
m=-8-6/5-3
m=-14/2
m=-7
y-y1=m(x-x1) point slope
<span>(4x-2)×6(2x+7) =
</span>
(4x-2) <span>× 12x + 42 =
48x^2 -24x + 168x -84 = 0
</span>
<span>48x^2 +144x -84 = 0
a = 48 b = 144 and c = -84
and the solution is
x1=.5 and x2=-3.5
</span>
Here are your measures of variability. The range is found by subtracting the highest and the lowest (29-5=24). To find the interquartile range, you will find the median of the lower half of the data and the median of the higher half of sta and subtract these 2 numbers. Here is your list. I have PUT PARENTHESES around the upper and lower quartiles: 5, 17, (18), 20, 20, 21, 23, (26), 28, 29. It is like finding the middle of the entire set of data and then finding the middle of each half. Subtract 26 and 18 to find the interquartile range of 8 touchdowns.
As we can see, the sin and cos of 45 degrees are 1/sqrt (2). Thus 1×1 is a number such that 1×1 > 0 and 1xi2 = x2, so it is the square root. This section is concerned with rewriting an expression such as. Rationalizing a denominator involving a square root
Solving the above equations
24+ Sqrt (Cos(X)) *Cos(300X) +Sqrt (Abs(X))-0.7) *(4-X*X) ^0.01 Sqrt(6-X^2)
Integrate [Sqrt [(-2 x^2)/ (-6 + x^2) + (-(x/(x^2) ^ (3/4)) - (0.028 x)/ (4 - x^2) ^0.99 + (Cos [300 x] Sin[x])/Sqrt [Cos[x]] + 600 Sqrt [Cos[x]] Sin [300 x]) ^2/4], {x, -4.5, 4.5}]
Plot [{Sqrt [Cos[x]] Cos [300 x] + Sqrt [Abs[x]] - 0.7 (4 - x x) ^0.01, Sqrt [6 - x^2], -Sqrt [6 - x^2]}, {x, -4.5, 4.5}]
More graph-related problems are given below,
brainly.com/question/16304194?referrer=searchResults
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