Answer:
The slope would be 2
Step-by-step explanation:
Your slope-intercept form is always y=mx + b
The <em>b </em>is always your <em>constant</em>
The <em>mx </em>is your <em>slope</em>
The<em> y </em>is your<em> output</em>
Answer:
slope is -2/3
Step-by-step explanation:
<span><span>y = 2 + 2sec(2x)
The upper part of the range will be when the secant has the smallest
positive value up to infinity.
The smallest positive value of the secant is 1
So the minimum of the upper part of the range of
y = 2 + 2sec(2x) is 2 + 2(1) = 2 + 2 = 4
So the upper part of the range is [4, )
The lower part of the range will be from negative infinity
up to when the secant has the largest negative value.
The largest negative value of the secant is -1
So the maximum of the lower part of the range of
y = 2 + 2sec(2x) is 2 + 2(-1) = 2 - 2 = 0
So the lower part of the range is (, 0].
Therefore the range is (, 0] U [4, )
</span>
</span>
Answer:
x = 23
Step-by-step explanation:
This entire angle makes up 90° b/c of the square box
65 + x + 2 = 90
x + 67 = 90
x = 23
Answer:
a. The arm spans for Class A have more variability than the arm spans for Class B.
Step-by-step explanation:
The stemplot for the Arm span is given for Class A and Class B. The variability of Class A are higher than variability of Class B. The arm spans of Class A have values greater than Class B. The mean of Class A is greater than Class B.
