3)The original rational number was 17/12.
4)The age of Ruby and Reshma are 20 and 28 years respectively.
Explanation:
3)Let the numerator be x.
So denominator will be (x - 5)
If we add 5 to numberator then it will be (x + 5)
New number is (x + 5)/(x - 5)=11/6
Solve for X by cross multiplying
6(x + 5)=11(x - 5)
6x + 30=11x - 55
5x=85
x=17
So original rational number was 17/12.
4)The ages of Ruby and Reshma are in the ratio 5:7
So, let the present ages of Ruby and Reshma be 5x and 7x respectively.
Also, it is given that four years from now the ratio of their ages will be 3:4
So, the equation - 5x + 4 / 7x + 3 = 3/4
⇒4(5x+4)=3(7x+4)
⇒20x+16=21x+12
⇒21x−20x=16−12
⇒x=4
⇒5x=20 and 7x=28
The age of Ruby and Reshma are 20 and 28 years respectively.
The interquartile range is 4
Answer:
SAS
Step-by-step explanation:
JK is congruent to LM (given)
L is midpoint of JN (given)
JL is congruent to LN (definition of midpoint)
JK is parallel to LM (given)
Angle LJK is congruent to Angle NLM (corresponding angles)
20 times 10=200 fluid ounces
128 fluid ounces per gallon
1 gallon is too little
2 gallons is to much
but you can't buy 1 and 72/100 of a bottle so just buy 2
2 bottles
The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be 3.75 square units.
<h3>What is an area bounded by the curve?</h3>
When the two curves intersect then they bound the region is known as the area bounded by the curve.
The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be
Then the intersection point will be given as
Then by the integration, we have
![\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214}[ (5 \sin \theta)^2 - 4^2] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [25\sin ^2 \theta - 16] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [ \dfrac{25}{2}(1 - \cos 2\theta ) - 16] d\theta \\](https://tex.z-dn.net/?f=%5Crightarrow%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cint%20_%7B0.927%7D%5E%7B2.214%7D%5B%20%285%20%5Csin%20%5Ctheta%29%5E2%20-%204%5E2%5D%20d%5Ctheta%20%5C%5C%5C%5C%5C%5C%5Crightarrow%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cint%20_%7B0.927%7D%5E%7B2.214%7D%20%5B25%5Csin%20%5E2%20%5Ctheta%20-%2016%5D%20d%5Ctheta%20%5C%5C%5C%5C%5C%5C%5Crightarrow%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cint%20_%7B0.927%7D%5E%7B2.214%7D%20%5B%20%5Cdfrac%7B25%7D%7B2%7D%281%20-%20%5Ccos%202%5Ctheta%20%29%20-%2016%5D%20d%5Ctheta%20%5C%5C)
![\rightarrow \dfrac{1}{2} [\dfrac{25 \theta }{5} - \dfrac{25 \cos 2\theta }{2} - 16\theta]_{0.927}^{2.214} \\\\\\\rightarrow \dfrac{1}{2} [\dfrac{25(2.214 - 0.927) }{5} - \dfrac{25 (\cos 2\times 2.214 - \cos 2\times 0.927) }{2} - 16(2.214 - 0.927]\\](https://tex.z-dn.net/?f=%5Crightarrow%20%5Cdfrac%7B1%7D%7B2%7D%20%5B%5Cdfrac%7B25%20%5Ctheta%20%7D%7B5%7D%20-%20%5Cdfrac%7B25%20%5Ccos%202%5Ctheta%20%7D%7B2%7D%20-%2016%5Ctheta%5D_%7B0.927%7D%5E%7B2.214%7D%20%5C%5C%5C%5C%5C%5C%5Crightarrow%20%5Cdfrac%7B1%7D%7B2%7D%20%5B%5Cdfrac%7B25%282.214%20-%200.927%29%20%7D%7B5%7D%20-%20%5Cdfrac%7B25%20%28%5Ccos%202%5Ctimes%202.214%20-%20%5Ccos%202%5Ctimes%200.927%29%20%7D%7B2%7D%20-%2016%282.214%20-%200.927%5D%5C%5C)
On solving, we have
Thus, the area of the region is 3.75 square units.
More about the area bounded by the curve link is given below.
brainly.com/question/24563834
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