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bogdanovich [222]
3 years ago
15

Jarvis left out a statement from the proof shown.which statement is best explained by the corresponding angles postulate

Mathematics
2 answers:
Salsk061 [2.6K]3 years ago
7 0
The postulate of the corresponding angles establishes that when a transversal line cuts two parallel lines, the corresponding angles are congruent. These angles are on the same side of the parallel lines and on the same side of the transversal line.

  Then, if we based on this definition and analize the figure attached, we can notice that the angles ∠1 and ∠3 are corresponding angles, so they are congruent. In this case the angle ∠1 is internal and the angle ∠3 is external.

 The answer is: ∠1 and ∠3 are congruent (See the image attached).

mrs_skeptik [129]3 years ago
3 0

Answer:

∠1 ≅ ∠3

Step-by-step explanation:    

Jarvis left out a statement from the proof shown. Which statement is best explained by the corresponding angles postulate?

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Write in descending order.<br> 360t + 10t3 - 120t2
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30t - 120t² + 10t³

Step-by-step explanation:

30t + 10t³ - 120t²

In order of degrees of function, t³ > t² > t

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30t - 120t² + 10t³ likewise in terms if increasing degree of function,

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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
2 years ago
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