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2 years ago

10

Solve the inequality. Graph the solution. Describe the graph in the area provided. Be sure to indicate where there is an open or

closed circle and what direction the number
line is shaded in. You can upload a picture of your work and explanation instead of
typing if you prefer,
-27
12 (67– 12) or 315 – ) < 24

1 answer:

natka813 [3]2 years ago

7 0

Step-by-step explanation:

The given inequality is :

$-27\ge \dfrac{1}{2}(6x-12)$

Solving RHS of the inequality:

$-27\ge \dfrac{1}{2}\times 2(3x-6)\\\\-27\ge (3x-6)$

Adding 6 both sides of the inequality

$-27+6\ge (3x-6)+6\\\\-21\ge 3x\\\\x\le -7$

The attached figure shows the graph for the given inequality.

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Answer:

6

Step-by-step explanation:

The slope is: $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$ . In this case, for the numerator you get 30. For the denominator you get 5. Divide 30/5 to get 6 as your final slope.

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2 years ago

Read 2 more answers

3. A line goes through the points (3,4) and (-3,6).

Galina-37 [17]

Answer:

Part a) The slope is $m=-\frac{1}{3}$

Part b) The equation in point slope form is $y-4=-\frac{1}{3}(x-3)$

Part c) The equation in slope-intercept form is $y=-\frac{1}{3}x+5$

Step-by-step explanation:

we have the points (3,4) and (-3,6)

Part a) What is the slope of the line?

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the given points

$m=\frac{6-4}{-3-3}$

$m=\frac{2}{-6}$

$m=-\frac{1}{3}$

Part b) Write the equation of the line in point-slope form

$y-y1=m(x-x1)$

we have

$m=-\frac{1}{3}$

$point\ (3,4)$

substitute

$y-4=-\frac{1}{3}(x-3)$ ---> equation in point slope form

Part c) Write the equation of the line in slope-intercept form

$y=mx+b$

we have

$y-4=-\frac{1}{3}(x-3)$

Isolate the variable y

distribute right side

$y-4=-\frac{1}{3}x+1$

Adds 4 both sides

$y=-\frac{1}{3}x+1+4$

$y=-\frac{1}{3}x+5$ ---> equation in slope intercept form

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3 years ago

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Step-by-step explanation:

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3 years ago

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katovenus [111]

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2 years ago

give the standard form, degree, leading coefficient and the constant term of the function h(x) = (2x+1)² (2x-3)

Softa [21]

Answer:

The standard form is: $h(x) = 8x^3 - 4x^2 - 10x - 3$

The function is of the 3rd degree.

The leading coefficient is 8.

The constant term is -3.

Step-by-step explanation:

Placing in standard form:

To place in standard form, we solve the operations. So

$h(x) = (2x+1)^2(2x-3)$

$h(x) = (4x^2 + 4x + 1)(2x - 3)$

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Degree:

The degree is given by the highest power of x, which, in this exercise, is 3.

Leading coefficient:

The leading coefficient is the term that multiplies the highest power of x. In this exercise, the higher power of x is $x^3$ , which is multiplied by 8. So the leading coefficient is 8.

Constant term:

The constant term is the one which does not multiply a power of x. So in this exercise, it is -3.

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2 years ago