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2 years ago

First teo answer get these points

Mathematics

2 answers:

djverab [1.8K]2 years ago

5 0

BoopAnswer:

Step-by-step explanation:

Nope

zheka24 [161]2 years ago

3 0

Answer:

Hi!

Explanation:

How you day is going well!

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Please answer this correctly

OlgaM077 [116]

Answer:

$21.57

Step-by-step explanation:

4.5 (1.22) + 0.8 (1.70) + 4.6 (3.20) =

5.49 + 1.36 + 14.72

21.57

4 0

4 years ago

Solve for x.<br> (−19)2=x

hram777 [196]

Answer:

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Simone applied the distributive property using the greatest common factor to determine the expression that is equivalent to 24 +

adell [148]

Answer:

The correct answer is A

Step-by-step explanation:

The reasoning is that I just got it correct on Edgenuity :p

5 0

4 years ago

Is 41/50 closer to 9/11 or 10/11? Verify your answer

professor190 [17]

First,
41/50 = 41 divide by 50 = 0.82

9/11 = 9 divided by 11 = 0.81

10/11= 10 divided by 11= 0.90

Answer: 9/11 is closer.

6 0

3 years ago

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

Natalija [7]

Given info:

Compound Interest = Rs.1290

Rate of Interest = 15% p.a

Time = 2 years

<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>

C.I = Compound Interest

P = Principle

R = Rate of Interest

N = Time

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

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.

3 0

3 years ago