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vladimir2022 [97]
2 years ago
5

yuki says that if all lines are congruent, then all line segments must be congruent. Is yuki correct? explain

Mathematics
1 answer:
Lady_Fox [76]2 years ago
4 0

We want to see if we can extrapolate a property of lines to a property of segments.

We will see that Yuki is not correct.

We start with the statement:

"All lines are congruent, then all line segments must be congruent"

Where two things are congruent if these have the same shape and size.

The thing with lines, is that we know that all lines are infinite.

So we could say that all<u> lines have the same length</u>, so they are congruent.

But this is not the case for segments, segments have a <u>definite length </u>(because they have a beginning and an end).

Then we can find two segments with different lengths, and because these segments have different lengths, these segments can't be congruent.

Thus Yuki is not correct in her statement.

If you want to learn more, you can read:

brainly.com/question/4094374

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Rachna borrowed a certain sum at the rate of 15% per annum. if she paid at the end of the two years Rs.1290 as interest Compound
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Given info:

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<h3>Formula we have to know:-</h3>

\dag{\underline{\boxed{\sf{C.I = P  \bigg(1  + \dfrac{R}{100} \bigg)^{n}  - 1}}}}

<u>Where</u>

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R = Rate of Interest

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\textsf{ \underline{Solution-}}\\

Here

C.I = Rs.1290

R = 15%

N = 2 years

Principle = ?

Now,Calculating the sum (Principle) borrowed by Rachna

\quad{: \implies{\sf{C.I =  \Bigg[P  \bigg(1  + \dfrac{R}{100} \bigg)^{n}  - 1 \Bigg]}}}

Substituting the given values

\quad{: \implies{\sf{1290 =  \Bigg[P  \bigg(1  + \dfrac{15}{100} \bigg)^{2}  - 1 \Bigg]}}}

\quad{: \implies{\sf{1290 =  \Bigg[P  \bigg(\dfrac{(1 \times 100) + 15}{100} \bigg)^{2}  - 1 \Bigg]}}}

\quad{: \implies{\sf{1290 =  \Bigg[P  \bigg(\dfrac{115}{100} \bigg)^{2}  - 1 \Bigg]}}}

\quad{: \implies{\sf{1290 =  \Bigg[P  \bigg(\dfrac{115}{100} \times \dfrac{115}{100}  \bigg)  - 1 \Bigg]}}}

\quad{: \implies{\sf{1290 =  \Bigg[P  \bigg(\dfrac{13225}{10000} \bigg) - 1 \Bigg]}}}

\quad{: \implies{\sf{1290 =  \Bigg[P  \bigg( \cancel{\dfrac{13225}{10000}} \bigg) - 1 \Bigg]}}}

\quad{: \implies{\sf{1290 =  \Bigg[P  \bigg({1.3225 - 1} \bigg) \Bigg]}}}

\quad{: \implies{\sf{1290 =  P  \times  {0.3225}}}}

\quad{: \implies{\sf{\dfrac{1290}{0.3225}  =  P}}}

\quad{: \implies{\sf{\dfrac{1290 \times 1000}{0.3225 \times 1000}  =  P}}}

\quad{: \implies{\sf{\dfrac{12900000}{3225}  =  P}}}

\quad{: \implies{\sf{\cancel{\dfrac{12900000}{3225}}  =  P}}}

\quad{: \implies{\sf{Rs.4000  =  P}}}

\quad{\dag{\underline{\boxed{\tt{\blue{Principle} =  \purple{Rs.4000 }}}}}}

\begin{gathered}\end{gathered}

<u>Hence,</u>

The sum (Principle) is Rs.4000.

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