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mestny [16]
3 years ago
12

How would you graph this correctly onto a graph?

Mathematics
1 answer:
saw5 [17]3 years ago
8 0

Answer:

Here is how you graph your equations correctly :D

Step-by-step explanation:

Graph y > x − 2.

Use the slope-intercept form to find the slope and y-intercept.

Slope: 1

y-intercept: (0, -2)

Graph a dashed line, then shade the area above the boundary line since y is greater than x − 2 .

y > x - 2

Graphy < −4x + 2.

Use the slope-intercept form to find the slope and y-intercept.

Slope: -4

y-intercept: (0, 2)

Graph a dashed line, then shade the area below the boundary line since y is less than − 4 x + 2 .

y < -4x + 2

Plot each graph on the same coordinate system.

y > x - 2

Y < -4x + 2

I hope this helps.

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Use four rectangles to estimate the area between the graph of the function f(x) = V3x + 5 and the x-axis on the interval[0, 4] u
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Answer:

  12.123

Step-by-step explanation:

You want the area under the curve f(x) = √(3x+5) on the interval [0, 4] estimated using the left sum and four subintervals.

<h3>Riemann sum</h3>

When the interval [0, 4] is divided into four equal parts, each has unit width. That means the area of the rectangle defined by the curve and the interval width will be equal to the value of the curve at the left end of the interval.

The area we want is the sum ...

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As the attachment shows, that sum is ...

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__

<em>Additional comment</em>

The table values in the attachment are rounded to 7 decimal places. Trailing zeros are not shown. Actual values used have 12 significant digits, as the total shows.

Such a sum is called a Riemann sum, named for a German mathematician. Four such sums are commonly used, and further refinements are possible. Those are the left sum (as here), the right sum, the midpoint sum, and a sum using a trapezoidal approximation of the rectangle area.

For left, right, and midpoint sums, n function values are required for n subintervals. When the trapezoidal approximation is used, n+1 function values are required.

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