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3 years ago

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Mathematics

1 answer:

Delvig [45]3 years ago

5 0

D is the right answer

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Find the length of LM if L is the midpoint of NM. NL = 3x + 1, LM = 8x - 24.

SashulF [63]

Answer:

LM = 16

Step-by-step explanation:

From the question given:

L is the midpoint of NM

NL = 3x + 1

LM = 8x - 24

LM =?

Next, we shall determine the value of x.

This can be obtained as follow:

Since L is the midpoint of NM, it means that NL and LM are equal i.e

NL = LM

Thus, we can obtain the value of x as follow:

NL = 3x + 1

LM = 8x - 24

NL = LM

3x + 1 = 8x - 24

Collect like terms

3x - 8x = - 24 - 1

-5x = - 25

Divide both side by - 5

x = -25/-5

x = 5

Finally, we shall determine the length of LM as follow:

LM = 8x - 24

x = 5

LM = 8(5) - 24

LM = 40 - 24

LM = 16

Therefore, the length of LM is 16

8 0

3 years ago

What sum will amount to rupees 4000 in 3 years at 6% p.a. compound interest

natali 33 [55]

Answer:

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Given:}}}}}}}\end{gathered}$

⇢ Principle = Rs.4000
⇢ Rate = 6%
⇢ Time = 3 year

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{To Find:}}}}}}}\end{gathered}$

⇢ Amount

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Using Formula:}}}}}}}\end{gathered}$

${\dag{\underline{\boxed{\sf{Amount ={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}$

$\dag{\underline{\boxed{\sf{Compound \: Interest = Amount- Principle }}}}$

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Solution:}}}}}}}\end{gathered}$

${\bigstar \:{\underline{\pmb{\frak{\red{Firstly,Finding \: the \: Amount }}}}}}$

$\quad {:\implies{\sf{Amount = \bf{P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}$

Substituting the values

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg(1 + \dfrac{6}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg(1 \times 100 + \dfrac{6}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{100 + 6}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{106}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg({\cancel{\dfrac{106}{100}}{\bigg)}}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{53}{50}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{53}{50} \times \dfrac{53}{50} \times \dfrac{53}{50}{\bigg)}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{148877}{125000}{\bigg)}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000 \times \dfrac{148877}{125000}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4{\cancel{000}} \times \dfrac{148877}{125{\cancel{000}}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\dfrac{148877 \times 4}{125}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\dfrac{595508}{125}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\cancel{\dfrac{595508}{125}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4764.064}}}}$

$\begin{gathered} \dag{\boxed{\textsf{\textbf{\underline{\color{green}{Amount = {Rs.4764.064}}}}}}}\end{gathered}$

Hence, The Amount is Rs.4764.064

$\begin{gathered}\end{gathered}$

${\bigstar \:{\underline{\pmb{\frak{\red{ Now,Finding \: The \: Compound \: Interest }}}}}}$

$\quad{: \implies{\sf{Compound \: Interest = \bf{Amount- Principle }}}}$

Substituting the values

$\quad{: \implies{\sf{Compound \: Interest = \bf{4764.064- 4000 }}}}$

$\quad{: \implies{\sf{Compound \: Interest =\bf{764.064}}}}$

$\begin{gathered} \dag{\boxed{\textsf{\textbf{\underline{\color{green}{Compound Interest = Rs.764.064}}}}}}\end{gathered}$

Henceforth,The Compound Interest is Rs.764064

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Learn More:}}}}}}}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered} \dag \: \underline{\bf{More \: Useful \: Formula}}\\ {\boxed{\begin{array}{cc}\dashrightarrow {\sf{Amount = Principle + Interest}} \\ \\ \dashrightarrow \sf{ P=Amount - Interest }\\ \\ \dashrightarrow \sf{ S.I = \dfrac{P \times R \times T}{100}} \\ \\ \dashrightarrow \sf{P = \dfrac{Interest \times 100 }{Time \times Rate}} \\ \\ \dashrightarrow \sf{P = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}} \\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}$