Please refer to the attachment

Mathematics

1 answer:

MariettaO [177]3 years ago

5 0

We have:-

P=₹26,400
R=15% p.a compounded yearly
n=2 years and 4 months

$\tt \: So,amount \: for \: 2 \: years = p \large(1 + \frac{r}{100}) {}^{n}$

$\tt=₹26,400 \large(1 + \frac{15}{100}) {}^{2}$

$\tt=₹26,400 \: \large( \frac{23}{20}) {}^{2}$

$\tt =₹26,400 \times \frac{23}{20} \times \frac{23}{20} \\$

$\tt = 66 \times 529=₹34,914$

$\tt \: So,SI \: for \: 4 \: months = \frac{P \times R \times n }{100 \times 12}$

$\tt \: = \frac{ 34,914 \times 15 \times 4 }{100 \times 12} \\$

$\tt\boxed{ = 1745.70}$

Thus,the amount after 2years and 4months =₹34,914+₹1745.70=₹36,659.70

$\bold \red{I \: hope \: this \: may \: help \: you}$

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How can you check to see if a value of x is a zero of the polynomial

Mila [183]

Answer:

substitute that value for x in the polynomial and see if it evaluates to zero

Step-by-step explanation:

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3 -3 = 0

___

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In the attachment, Horner Form is shown at the bottom.