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3 years ago

Which graph shows a negative rate of change for the interval 0 to 2 on the x-axis?

2 answers:

5 0

It seems that the four graphs are the same and they do not have a negative change rate in the interval 0 to 2 in the x-axis.

A negative change rate means that when x increases the value of the function (y) decreases; this is, the function is decreasing in the interval being estudied, which is the same that going downward.

So, you must look for in your graphs where the equation is going downward.

For example, in the graph attached, that happens in any interval from negative infitity to 1.5.

The vertex will help you to identify it.

Given that the graph goes downward from negative infinity to the vertex, any interval that includes that range will have negative change.

You must look for a parabola that opens upward and whose vertex is in x = 2.

Read more on Brainly.com - brainly.com/question/3774202#readmore

vitfil [10]3 years ago

5 0

Answer:

Step-by-step explanation:

Check for the slope/trend.

The graph with decreasing values of y (negative slope) as x goes from 0 to 2 is your answer.

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Find the value of Y<br> .....

vekshin1

Answer:

$y = 6 \sqrt{3} \: units$

Step-by-step explanation:

In order to find the value of y, first we need to find the length of the perpendicular dropped from one of the vertices of the triangle to its opposite side.

By geometric mean theorem:

Length of the perpendicular

$=\sqrt{9\times 3}$

$=\sqrt{27}\: units$

Next, by Pythagoras theorem:

${y}^{2} = {9}^{2} + {( \sqrt{27} )}^{2} \\ \\ {y}^{2} = 81 + 27 \\ \\ {y}^{2} = 108 \\ \\ y = \sqrt{108} \\ \\ y = \sqrt{36 \times 3} \\ \\ y = \sqrt{ {6}^{2} \times 3} \\ \\ y = 6 \sqrt{3} \: units$

4 0

2 years ago

can some one also please help my one that i should do after i finds the area of the square aka how to find the area of the trian

Andre45 [30]

Answer:

6 square units

Step-by-step explanation:

Given shape is of a trapezoid:

$area \: of \: trapezoid \\ = \frac{1}{2} \times (4 + 2) \times 2 \\ \\ = 6 \: {units}^{2}$

Area of shape = area of square + area of triangle

$= 2^2 + \frac{1}{2} \times 2\times 2\\= 4 + 2\\= 6\: units^2 \\$

4 0

3 years ago

Determine if the following relation is a function or not

Darya [45]

1 is not. 2 is. 3 is not. 4 is. 5 is. 6 is. 7 is not. 8 is.

7 0

2 years ago

Find the limit. (If an answer does not exist, enter DNE.)<br> lim ( (1/x)-(1/lxl)) <br> x→0^+

jek_recluse [69]

Answer:

Step-by-step explanation:

The + on the zero means the limit as x approaches 0 from the right side

$\lim_{x \to 0^{+} } (\frac{1}{x} - \frac{1}{|x|})\\ \lim_{x \to 0^{+} } (\frac{1}{x} - \frac{1}{x})\\ \lim_{x \to 0^{+} } (0)\\0$

4 0

2 years ago

A health psychologist knew that corporate executives in general have an average score of 2.55 with a standard deviation of 0.5 o

Mumz [18]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Population mean (μ) = 2.55

Population standard deviation (σ) = 0.5

Sample size (n) = 30

Sample mean (x) = 2.76

α = 0.05

STEP 1:

Stress score in general executive (s1)

Stress score in exercising executive (s2)

Null : s1 = s2

Alternative : s1 < s2

STEP 2:

Shape of distribution = normal

Population mean (μ) = 2.55

Population standard deviation (σ) = 0.5

Sample size (n) = 30

Sample mean (x) = 2.76

α = 0.05

Decision rule :

α = 0.05 which corresponds to a t score (t0) ;

df = n - 1 = 30 - 1= 30 at 0.05 = 1.699

If :

(Test statistic (t) > t0) ; reject the Null

(right tailed test)

Test statistic (t) :

(x - μ) / (σ/√n)

(2.76 - 2.55) / (0.5/√30)

0.21 / 0.0913

= 2.30

t > t0

2.30 > 1.699

t is more extreme than t0

Hence, reject the null at α = 0.05

7 0

3 years ago