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2 years ago

Please help which of the following conditions does the graph meet?

Mathematics

2 answers:

aliina [53]2 years ago

4 0

<h3>Answer: Choice A</h3>

Explanation:

The left shaded portion describes all (x,y) points such that x < -1. Basically, it's any point to the left of that dashed vertical line through x = -1. Points on the dashed line are not part of the solution set.

The shaded portion on the right side visually describes any point (x,y) such that $x \ge 2$ . Here we have a solid boundary line, and any point on this line is included in the solution set.

These two regions are connected with an "or" because you can be in either region, but not both at the same time.

Notice how y is never mentioned other than when we talk about (x,y). This is because we're just dealing with the x axis number line.

ANEK [815]2 years ago

3 0

Answer:

{(x,y):x≥-1) or {(x,y):x≤2)

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Please help it’s a pic

Evgesh-ka [11]

Answer:

58°

Step-by-step explanation:

<MQR = <XQL (vertically opposite angles)

28 + 5b = 70 - 2b

28 - 70 = -2b - 5b

-42 = -7b

-42/-7 = b

6 = b

<MQR = 28 + 5b

= 28 + 5(6)

= 28 + 30

= 58°

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2 years ago

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Which numbers belong to the solution set of the inequality? 9x>117

evablogger [386]

(13,infinity] hope this helps

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2 years ago

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Verdich [7]

Answer:

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Step-by-step explanation:

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3 years ago

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Varvara68 [4.7K]

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2 years ago

What would y=x^2 +x+ 2 be in vertex form

balu736 [363]

Answer:

$y = (x + \frac{1}{2} )^{2} + \frac{7}{4}$

Step-by-step explanation:

$y = {x}^{2} + x + 2$

We can covert the standard form into the vertex form by either using the formula, completing the square or with calculus.

$y = a(x - h)^{2} + k$

The following equation above is the vertex form of Quadratic Function.

Vertex — Formula

$h = - \frac{b}{2a} \\ k = \frac{4ac - {b}^{2} }{4a}$

We substitute the value of these terms from the standard form.

$y = a {x}^{2} + bx + c$

$h = - \frac{1}{2(1)} \\ h = - \frac{ 1}{2}$

Our h is - 1/2

$k = \frac{4(1)(2) - ( {1})^{2} }{4(1)} \\ k = \frac{8 - 1}{4} \\ k = \frac{7}{4}$

Our k is 7/4.

Vertex — Calculus

We can use differential or derivative to find the vertex as well.

$f(x) = a {x}^{n}$

Therefore our derivative of f(x) —

$f'(x) = n \times a {x}^{n - 1}$

From the standard form of the given equation.

$y = {x}^{2} + x + 2$

Differentiate the following equation. We can use the dy/dx symbol instead of f'(x) or y'

$f'(x) = (2 \times 1 {x}^{2 - 1} ) + (1 \times {x}^{1 - 1} ) + 0$

Any constants that are differentiated will automatically become 0.

$f'(x) = 2 {x}+ 1$

Then we substitute f'(x) = 0

$0 =2x + 1 \\ 2x + 1 = 0 \\ 2x = - 1 \\x = - \frac{1}{2}$

Because x = h. Therefore, h = - 1/2

Then substitute x = -1/2 in the function (not differentiated function)

$y = {x}^{2} + x + 2$

$y = ( - \frac{1}{2} )^{2} + ( - \frac{1}{2} ) + 2 \\ y = \frac{1}{4} - \frac{1}{2} + 2 \\ y = \frac{1}{4} - \frac{2}{4} + \frac{8}{4} \\ y = \frac{7}{4}$

Because y = k. Our k is 7/4.

From the vertex form, our vertex is at (h,k)

Therefore, substitute h = -1/2 and k = 7/4 in the equation.

$y = a {(x - h)}^{2} + k \\ y = (x - ( - \frac{1}{2} ))^{2} + \frac{7}{4} \\ y = (x + \frac{1}{2} )^{2} + \frac{7}{4}$

7 0

2 years ago