<em>S</em><em>t</em><em>e</em><em>p</em><em>-</em><em>b</em><em>y</em><em>-</em><em>s</em><em>t</em><em>e</em><em>p</em><em> </em><em>e</em><em>xplanation</em><em>:</em>
<em>Since</em><em> </em><em>-3a</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>said</em><em> </em><em>to</em><em> </em><em>have</em><em> </em><em>a</em><em> </em><em>power</em><em> </em><em>of</em><em> </em><em>1</em><em>,</em><em> </em><em>we</em><em> </em><em>add</em><em> </em><em>the</em><em> </em><em>powers</em><em> </em><em>together</em><em> </em><em>and</em><em> </em><em>multiply</em><em> </em><em>the</em><em> </em><em>coefficients</em>
-9a³
Answer:
do you have a picture i would help then
Step-by-step explanation:
Answer:
what is the question
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, )
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