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maks197457 [2]
3 years ago
12

Tell whether each graph is a function and justify your answer. Which graph is not a good representation of a​ real-world situati

on? Explain.

Mathematics
1 answer:
mina [271]3 years ago
5 0

Using the function concept, it is found that:

  • Graph B is a function because each value of x corresponds to exactly one y-value.

In a function, <u>one value of the input can be related to only one value of the output</u>.

  • In a graph, it means that for each value of x(horizontal axis), there can be only one respective value of y(vertical axis).

In this problem, at Graph A, when x = 5, for example a vertical line crosses the function 3 times, hence there are 3 respective values of y for x = 5, and the same is valid for other values of x, hence it is not a function.

At Graph B, <u>for each value of x, there is only one value of y</u>, hence it is a function.

Hence:

Graph B is a function because each value of x corresponds to exactly one y-value.

To learn more about the function concept, you can take a look at brainly.com/question/12463448

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4 0
3 years ago
F(x) = 3x + 3; f(33)
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f(33) = 3(33) +  3
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f(33) = 102</span>
3 0
4 years ago
Awnsers please<br>solve it
sukhopar [10]

1)30

2)700

3)7

4)9 000

5)30 000

6)500

7)800 000

8)6 000

5 0
3 years ago
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andrew-mc [135]

For the reaction,

CO(g) + H_2 O(g) = CO_2(g) + H_2(g)

Initial concentration:

<u>0.1M</u> , <u>0.1M</u> , <u>0</u> , <u>0</u>

Let 'x' mole per litre of each of theproduct be formed.

At equilibrium:

<u>(0.1 – x)M</u> , <u>(0.1 – x)M</u> , <u>xM</u> , <u>xM</u>

where x is the amount of Carbon dioxide and Hydrogen, at equilibrium.

Hence, equilibrium constant can be written as,

K_c = \frac{x²}{(0.1 – x)²} = 4.24

→ x² = 4.24 (0.01 + x² – 0.2x)

→ x² = 0.0424 + 4.24 x² - 0.848x

→ 3.24x² - 0.848x + 0.0424 = 0

<em>a = 3.24, b = -0.848, c = 0.0424</em>

(for quadratic equation ax² + bx+c=0)

x =  \frac{( - b \: ± \: \sqrt{ {b}^{2} - 4ac) } }{2a}

=  > x =  \frac{ - ( - 0.848 \: ± \:  \sqrt{( - 0.848)^{2} - 4(3.24)(0.0424) } }{2 \times 3.24}

=  > x =  \frac{ - 0.848±0.4118}{6.48}

x_1 =  \frac{0.848 - 0.4118}{6.48} = 0.067

x_2 =  \frac{0.848 + 0.4118}{6.48} = 0.194

Here, the value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration.

∴ The equilibrium concentrations are :-

[CO_2] [H_2] = x = 0.067M

[CO] [H_2 O] = 0.1 - 0.067 = 0.033M

7 0
3 years ago
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