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beks73 [17]
3 years ago
5

Need help with this one

Mathematics
2 answers:
denis-greek [22]3 years ago
5 0
Here is your answer

\huge\frac {24m{n}^{2}p}{3mnp}

\huge=\frac{24×m×n×n×p}{3×m×n×p}

On solving, we get

8n

HOPE IT IS USEFUL
Gennadij [26K]3 years ago
3 0

I believe that the answer is 8n because you are dividing and most of problem cancels out. Once everything is cancelled out you are left with 8n. Let me know if that helped.

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Brilliant_brown [7]

Answer:

Scientist

Step-by-step explanation:

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3 years ago
Which shows how to solve the equation 3/4x=6 for x in one step?
a_sh-v [17]

Well going off what I know (you don't have the equations up)

You have to isolate x

3/4x = 6

(4) 3/4x = 6 *multiplying by 4 cancels the fraction.

3x = 24

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7 0
3 years ago
Read 2 more answers
−4x+35&gt;2<br> Whats the innequality
geniusboy [140]

-4x + 35 > 2

Subtract 35 from both sides:

-4x> -33

Divide both sides by -4:

X > -33 s -4

Because you are dividing the inequality. Y a negative value, reverse the inequality sign.

X < 33/4

7 0
4 years ago
Sally and Jim went berry picking. Sally picked 3 fewer pints than twice as many as Jim. If together the picked 21 pints of berri
Aloiza [94]
2x - 3 = sally's berries

x= jim's berries

2x - 3 + x = 21 } equation
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+3 +3
3x = 25
---- ----
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8 0
4 years ago
<img src="https://tex.z-dn.net/?f=%5Cboxed%7B%5Cblue%7B%5Cmathscr%7BHello%5C%3ABrainliacs%7D%7D%7D" id="TexFormula1" title="\box
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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