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seraphim [82]
3 years ago
5

ramesh took a loan of rs 50000 from urmila at the rate of 10%p.a. if he paid half of the principal and all the interest at the e

nd of 3 years in how many years should he pay the remaining amount with total interest of 20000 from the beginning?​
Mathematics
1 answer:
cricket20 [7]3 years ago
7 0

Answer:

Its not ramesh its rameesha

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3. You have a square plot of land that has an area of 144 square feet. You
ElenaW [278]

Answer:

113.04square feet

Step-by-step explanation:

square root of 144 is 12

use it as the dimensions of the plot.

the flower gardens largest possible diameter is 12

so divide 12 by a half

6=radius

3.14 times 6squared

you get 113.04

3 0
3 years ago
Can I please get some help with this?
Romashka-Z-Leto [24]

Answer: if this is any help look up slander on the internet and type in your math work book and but in your unit and your answer should be there.

3 0
3 years ago
Use the arc length formula and the given information to find θ. s = 4 m, r = 13 m; θ = ?
solong [7]
C=2πr and the circumference is measured at 360°

So if we were to set up a proportion we could say:

s/(2πr)=α/360

α=180s/(πr)

We are given that s=4m and r=13m so:

α=180(4)/(13π)

α=720/(13π) m  which is approximately:

α≈17.63°

Which is why radians are often used, because in radians...

α=[720/(13π)](π/180)

α=4/13 rads

Or the arc length in radians is simply expressed as:

s=rα

α=s/r  (which is much neater :P) then it is plain to see that:

α=4/13  rads


7 0
3 years ago
Read 2 more answers
Cos4theta+cos2theta/ cos4theta-cos2theta= _____
vovangra [49]

\bf \textit{Sum to Product Identities} \\\\ cos(\alpha)+cos(\beta)=2cos\left(\cfrac{\alpha+\beta}{2}\right)cos\left(\cfrac{\alpha-\beta}{2}\right) \\\\\\ cos(\alpha)-cos(\beta)=-2sin\left(\cfrac{\alpha+\beta}{2}\right)sin\left(\cfrac{\alpha-\beta}{2}\right) \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \cfrac{cos(4\theta )+cos(2\theta )}{cos(4\theta )-cos(2\theta )}\implies \cfrac{2cos\left( \frac{4\theta +2\theta }{2} \right)cos\left( \frac{4\theta -2\theta }{2} \right)}{-2sin\left( \frac{4\theta +2\theta }{2} \right)sin\left( \frac{4\theta -2\theta }{2} \right)} \implies \cfrac{cos\left( \frac{6\theta }{2} \right)cos\left( \frac{2\theta }{2} \right)}{-sin\left( \frac{6\theta }{2} \right)sin\left( \frac{2\theta }{2} \right)}

\bf \cfrac{cos(3\theta )cos(\theta )}{-sin(3\theta )sin(\theta )}\implies -\cfrac{cos(3\theta )}{sin(3\theta )}\cdot \cfrac{cos(\theta )}{sin(\theta )}\implies -cot(3\theta )cot(\theta )

8 0
3 years ago
Please answer quickly<br>expand and simplify: 4(5x-3y) (x-4y) -(3x-4y)(2x+3y)​
Gekata [30.6K]

Answer: 14x^2-93xy+60y^2 Hope that helps!

Step-by-step explanation:

1. Expand by distributing terms

(20x-12y)(x-4y)-(3x-4y)(2x+3y)

2. Use the Foil method:(a+b)(c+d)= ac+ad+bc+bd

20x^2-80xy-12yx+48y^2-(3x-4y)(2x+3y)

3. Use the Foil method : (a+b)(c+d)= ac+ad+bc+bd

20x^2-80xy-12yx+48y^2-(6x^2+9xy-8yx-12y^2)

4. Remove parentheses 20x^2-80xy-12yx+48y^2-6x^2-9xy+ 8yx+12y^2

5. Collect like terms (20x^2-6x^2)+(-80xy-12xy-9xy+8xy)+(48y^2+12y^2)

6. Simplify.

And your answer would be 14x^2-93xy+60y^2

5 0
3 years ago
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