9514 1404 393
Answer:
see attached
Step-by-step explanation:
I find a graphing calculator to be the quickest way to create a graph of a system of equations. That result is attached.
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If you want to graph the equations by hand, you need to know a couple of points on each line. When the equations are in slope-intercept form, the y-intercept is often a good place to start. Another point is usually easy to find based on the slope of the line, starting at the y-intercept.
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Here, the equations are not in that form, but are in the form ax+by=c. In this form, it is often easy to find both the x- and y-intercepts and use those points to plot the line. Each intercept is found by setting the other variable to zero.
x-intercept: c/a
y-intercept: c/b
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For the given lines, the first equation has intercepts (2, 0) and (0, 2). The line has a slope of -1 and makes an isosceles triangle with the axes in the first quadrant.
The second equation has intercepts (-1, 0) and (0, 2). This line has a slope of +2 and makes a triangle with the axes in the second quadrant.
Answer: 8/4
Step-by-step explanation: if I’m wrong just watch the video for a hint it helps
Answer:
1 / 8², B
Step-by-step explanation:
8 / 8³
8³ = 8*8*8
So, the equation can also be written as:
8 / 8*8*8
divide the numerator and denominator by 8.
1 / 8*8
This is equal to:
1 / 8²
The answer is B.
we know that
the rate of a linear equation is equal to the slope
Step 1
Find the bear’s average heart rate when it’s active
Let
A(1.5,90) B ( 2,120)
the slope is equal to
m=(y2-y1)/(x2-x1)
m=(120-90)/(2-1.5)=30/0.5
m=60 heartbeats/min
therefore
the answer part a) is
the bear’s average heart rate when it’s active is 60 heartbeats/min
Step 2
Find the bear’s average heart rate when it’s hibernating
Let
A(1.5,18) B ( 2,24)
the slope is equal to
m=(y2-y1)/(x2-x1)
m=(24-18)/(2-1.5)=6/0.5
m=12 heartbeats/min
therefore
the answer part b) is
the bear’s average heart rate when it’s hibernating is 12 heartbeats/min
Answer:
(See explanation for further details).
Step-by-step explanation:
The sign does not affect the domain, as second-order polynomials are continuous functions. Nonetheless, the sign affect the range of function in the sense that elements can be lesser and equal to zero (a < 0) or greater and equal to zero (a > 0).
