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anzhelika [568]
2 years ago
7

What is the answer to this? x + 8 >= 14

Mathematics
1 answer:
Anon25 [30]2 years ago
7 0

Answer:

x>=6

Step-by-step explanation:

subtract 8 on both sides

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There are 17 girls and 13 boys in a school club. What is the percent of the members are girls?
tresset_1 [31]

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131%

Step-by-step explanation:

17:13

17 ÷ 13 = 1.3076923076923 = 130.76923076923%

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greta made a desing with squares she colored 8 out of the 12 squares blue what fraction of the squares did she color blue
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2/3 because 8/12 divided down is 4/6 which reduced again is 2/3
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In the figure to the right, what is the measure of angle DAC ?
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Step-by-step explanation:

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3 years ago
Prove that for a triple element x, y, z the distributive law is always satisfied if one of the elements equals 0 or 1, or if two
solmaris [256]

The distributive law is satisfied if the numbers 1,2, and 3 are used as multyplying factors.

<h3>What is the distributive law?</h3>

This is basic mathematical law that states:

  • A multiplied by (b+c) equals (a x b) + (a x c)

<h3>Is this law satisfied if there are three numbers involved?</h3>

The law is always satisfied if one of the numbers is 1. Here is an example

  • 2 x (1+2+3) = 2 x 6 = 12

or

  • (2x1) + (2x2) + (2x3)
  • 2+4+6 = 12

Learn more about mathematical laws in: brainly.com/question/15190681

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3 0
2 years ago
Calculate the pH of a buffer solution made by mixing 300 mL of 0.2 M acetic acid, CH3COOH, and 200 mL of 0.3 M of its salt sodiu
Lesechka [4]

Answer:

Approximately 4.75.

Step-by-step explanation:

Remark: this approach make use of the fact that in the original solution, the concentration of  \rm CH_3COOH and \rm CH_3COO^{-} are equal.

{\rm CH_3COOH} \rightleftharpoons {\rm CH_3COO^{-}} + {\rm H^{+}}

Since \rm CH_3COONa is a salt soluble in water. Once in water, it would readily ionize to give \rm CH_3COO^{-} and \rm Na^{+} ions.

Assume that the \rm CH_3COOH and \rm CH_3COO^{-} ions in this solution did not disintegrate at all. The solution would contain:

0.3\; \rm L \times 0.2\; \rm mol \cdot L^{-1} = 0.06\; \rm mol of \rm CH_3COOH, and

0.06\; \rm mol of \rm CH_3COO^{-} from 0.2\; \rm L \times 0.3\; \rm mol \cdot L^{-1} = 0.06\; \rm mol of \rm CH_3COONa.

Accordingly, the concentration of \rm CH_3COOH and \rm CH_3COO^{-} would be:

\begin{aligned} & c({\rm CH_3COOH}) \\ &= \frac{n({\rm CH_3COOH})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}.

\begin{aligned} & c({\rm CH_3COO^{-}}) \\ &= \frac{n({\rm CH_3COO^{-}})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}.

In other words, in this buffer solution, the initial concentration of the weak acid \rm CH_3COOH is the same as that of its conjugate base, \rm CH_3COO^{-}.

Hence, once in equilibrium, the \rm pH of this buffer solution would be the same as the {\rm pK}_{a} of \rm CH_3COOH.

Calculate the {\rm pK}_{a} of \rm CH_3COOH from its {\rm K}_{a}:

\begin{aligned} & {\rm pH}(\text{solution}) \\ &= {\rm pK}_{a} \\ &= -\log_{10}({\rm K}_{a}) \\ &= -\log_{10} (1.76 \times 10^{-5}) \\ &\approx 4.75\end{aligned}.

7 0
3 years ago
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