A Number P Varies Directly As q And Partly Inversely As Q ^ 2, Given That P=11 When q=2 and p=25.16 when q=5. calculate the value of p when q=7
Since it varies directly as q and partly inversely as q^2 it is represented has this below
p = aq + b/q^2
The constants of proportionality here are a and b
11 = 2a + b/4 ----- i x 5
25.16 = 5a + b/25 ------ ii x 2
Multiply equation i by 5 and equation ii by 2
55 = 10a + 5b/4 ----- iii
50.32 = 10a + 2b/25 ----- iv
Subtract equation iv from equation iii
b = 4
Put the value of b in equation iv
50.32 = 10a + 8/25
Multiply through by 25
1258 = 250a + 8
a = 5
p =5q + 4/q^2
Therefore using this equation put into the equation the value of q which is 7
p = 5(7) + 4/(7)^2
p = 35 +4/49
p = 35.082
the description tells us that we have an equation of the form
p = mq + n/q^2
Now just plug in your values to find m and n.
11 = 2m+n/4
25.16 = 5m+n/25
That gives you a final equation of
p = 5q + 4/q^2
Now just plug in q=7 to find p
Looking for answers
Well, let's break down this hilarious math problem, shall we? We have a number P that varies directly as q and partly inversely as Q^2. That's quite a complex relationship. But fear not, my friend, we'll solve it together!
Firstly, since P varies directly with q, we can write an equation as P = kq, where k is the constant of proportionality. Easy peasy, lemon squeezy.
Now, we also know that P varies partly inversely as Q^2. So, we can add another term to our equation, making it P = kq/Q^2. Don't you just love math complications?
To find the value of k, we can use the given information. When q = 2, P = 11. Plugging these values into our equation, we get 11 = 2k/2^2. Simplifying this, we find k = 44.
So now we have our equation as P = 44q/Q^2. Let's plug in the next set of values, where q = 5 and P = 25.16. Solving for Q, we get Q = √(44 * (5/25.16)). Calculating this monstrosity, we find Q ≈ 2.827.
Finally, we substitute Q = 7 into our equation P = 44q/Q^2 and solve for P. P = 44(7)/(2.827)^2. Goodness gracious, my circuits are overheating! After all the calculations, we find that P is approximately equal to 39.15.
So, when q = 7, the value of P is approximately 39.15. Voila! Math problem solved, and we even managed to have a laugh along the way.
To solve this problem, we need to use the given relationship between the variables and use the given values to calculate the constant of variation.
Let's break down the information given:
1. "A Number P Varies Directly As q": This means that when q increases, P also increases, and when q decreases, P also decreases.
2. "And Partly Inversely As Q^2": This means that as Q^2 increases, P decreases, and as Q^2 decreases, P increases.
3. "P = 11 when q = 2": This gives us a specific example of the relationship between P and q.
4. "P = 25.16 when q = 5": This gives us another specific example of the relationship between P and q.
To determine the relationship between P and q, we can set up an equation combining the two factors: direct variation and partial inverse variation.
The equation can be written as: P = kq / Q^2, where k is the constant of variation.
To find the value of the constant k, we can substitute one of the given sets of values (P = 11 when q = 2) into the equation:
11 = k * 2 / (2^2)
11 = 2k / 4
44 = 2k
k = 22
Now that we have the value of k, we can use it to find the value of P when q = 7.
P = 22 * 7 / (7^2)
P = 22 * 7 / 49
P = 154 / 49
P ≈ 3.143
Therefore, when q = 7, the value of P is approximately 3.143.