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Inga [223]
3 years ago
13

Which of the following inequalities matches the graph?

Mathematics
2 answers:
Neko [114]3 years ago
7 0

Answer:

c

Step-by-step explanation:

tino4ka555 [31]3 years ago
4 0

Answer:

  • C. 6x - y < -3

Step-by-step explanation:

<u>The graphed inequality shows that:</u>

  • The function is increasing, so it has a positive slope, shaded region is to the left of the dashed line, so it should have > sign after y.

<u>Above information lets us build the inequality:</u>

  • y > ax + b, where a> 0

<u>Now lets verify the options:</u>

A. 6x + y < 3 ⇒ y < -6x + 3

  • No

B. -6x + y < 3 ⇒ y < 6x + 3

  • No

C. 6x - y < - 3 ⇒ y > 6x + 3

  • Yes

D. Not applicable

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103times127 equal 13081.
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Explain how to graph a line when given the slope and one of the points on the line.
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Step-by-step explanation:

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Read 2 more answers
You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.
postnew [5]

<em><u>This problem seems to be wrong because no minimum point was found and no point of landing exists</u></em>

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

<u>Derivatives</u>

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

1)

The function provided in the question is

h(x)=2x^2-12x+20

Let's find the first derivative

h'(x)=4x-12

solving h'=0:

4x-12=0

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

2)

To find when will the ball land, we set h=0

2x^2-12x+20=0

Simplifying by 2

x^2-6x+10=0

Completing squares

x^2-6x+9+10-9=0

Factoring and rearranging

(x-3)^2=-1

There is no real value of x to solve the above equation, so the ball will never land.

This problem seems to be wrong because no minimum point was found and no point of landing exists

3 0
3 years ago
Help please with #3
Alenkasestr [34]

Answer: x = 2a over b + c

Step-by-step explanation:

1. Multiply both sides by two. This cancels out 2 and turns a into 2a. It should look like 2a = bx + cx

2) Since both b and c have x’s, we turn it into (b + c)x

3. Then divide both sides by b+c to get 2a over b + c = x

6 0
3 years ago
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