1) P varies directly as the square of Q and inversely as the square root of R.If Q is reduced by 12% while R is increased by 21%,find the percentage change in P
2)The distance s metres of an object varies partly with time t seconds and partly with the square root of time.Given that s=16 when t=4 and s=48 when t =16, write an equation connecting s and t.
3)A quantity P varies jointly as the square root of Q and as the inverse of the square of R.Determine the Percentage change in P when Q is increased by 44% and R decreased by 28%
4)The mass of wire m grams(g)is partly constant and partly varies as the square of its thickness t mm When t=2 mm,m=40g and when t=4mm m=100g.Determine the value of. When t=7mm
5)The quantities P,Q and R are such that P varies directly as the square of Q and inversely as the square of R.P=8 when Q =2 and R=9.Determine the equation connecting P,Q and R
6)Two variables A and B are such that A varies partly as B and partly as the square root of B.Given that A=8.58 when B=1.69 and A =9.52 when B =1.968 find the law connecting A and B hence determine A when B =2.89
7)A quantity Q is partly constant and partly varies as the square of E.When E=2,Q=560 and when E=3,Q=510.Find an equation connecting Q and E hence determine E given Q=537.5
8)A quantity P is constant and partly varies as the inverse of Q^2.When Q=2 ,P =49 and a 1/2 and when Q=3,P =49 and 2/3.Determine the value of P when Q=5
9)A quantity y varies inversely as the square of x .The difference between the value of y when x =6 and when x=10 is 16.find the law connecting x And y
1) Let P = kQ^2 / sqrt(R), where k is the constant of variation.
If Q is reduced by 12%, then the new value of Q is 0.88Q.
If R is increased by 21%, then the new value of R is 1.21R.
Substituting these values into the equation for P, we get:
P' = k(0.88Q)^2 / sqrt(1.21R)
P' = (0.7744kQ^2) / (1.1sqrt(R))
The percentage change in P is given by the formula:
((P' - P) / P) * 100
=((0.7744kQ^2) / (1.1sqrt(R)) - kQ^2 / sqrt(R)) / (kQ^2 / sqrt(R)) * 100
=((0.7744 - 1) / 1) * 100
= -22.56%
Therefore, there is a decrease of 22.56% in P when Q is reduced by 12% and R is increased by 21%.
2) We are given two sets of values:
s=16 when t=4
s=48 when t=16
We can write the equation as follows:
s = at + b√t
Substituting the given values, we get:
16 = 4a + 2b
48 = 16a + 4b
Solving the system of equations, we find:
a = 4
b = -4
Therefore, the equation connecting s and t is:
s = 4t - 4√t
3) Let P = k√Q / (1 / R^2), where k is the constant of variation.
If Q is increased by 44%, then the new value of Q is 1.44Q.
If R is decreased by 28%, then the new value of R is 0.72R.
Substituting these values into the equation for P, we get:
P' = k√(1.44Q) / (1 / (0.72R)^2)
P' = (1.2k√Q) / (0.5184R^2)
The percentage change in P is given by the formula:
((P' - P) / P) * 100
=((1.2k√Q) / (0.5184R^2) - k√Q / (1 / R^2)) / (k√Q / (1 / R^2)) * 100
=((1.2 / 0.5184 - 1) / 1) * 100
= 37%
Therefore, there is an increase of 37% in P when Q is increased by 44% and R is decreased by 28%.
1) Let's start by setting up the equation for the direct variation and the inverse variation:
P = k * (Q^2) / √R
where k is the constant of variation.
Now, let's find the percentage change in P when Q is reduced by 12% and R is increased by 21%.
Let's assume the original values are Q₀ and R₀, and the new values are Q₁ and R₁.
The percentage change in Q is given by: %ΔQ = ((Q₁ - Q₀) / Q₀) * 100
Since Q is reduced by 12%, we have: %ΔQ = -12%
Therefore, Q₁ = Q₀ - 0.12 * Q₀ = 0.88 * Q₀
The percentage change in R is given by: %ΔR = ((R₁ - R₀) / R₀) * 100
Since R is increased by 21%, we have: %ΔR = 21%
Therefore, R₁ = R₀ + 0.21 * R₀ = 1.21 * R₀
Now, let's substitute these values into the equation for P:
P₁ = k * (Q₁^2) / √R₁
P₀ = k * (Q₀^2) / √R₀
To find the percentage change in P, we can use the formula:
%ΔP = ((P₁ - P₀) / P₀) * 100
Substituting the expressions for P₁ and P₀, we get:
%ΔP = ((k * (Q₁^2) / √R₁ - k * (Q₀^2) / √R₀) / (k * (Q₀^2) / √R₀)) * 100
Simplifying, we get:
%ΔP = (((Q₁^2) / √R₁ - (Q₀^2) / √R₀) / ((Q₀^2) / √R₀)) * 100
%ΔP = ((Q₁^2 / √R₁) - (Q₀^2 / √R₀)) / (Q₀^2 / √R₀) * 100
%ΔP = (Q₁^2 / √R₁) / (Q₀^2 / √R₀) - (Q₀^2 / √R₀) / (Q₀^2 / √R₀) * 100
%ΔP = (Q₁^2 / √R₁Q₀^2) - 1 * 100
%ΔP = (Q₁ / Q₀)^2 * (1 / √(R₁ / R₀)) - 1 * 100
Substituting the values for Q₁ and Q₀ and R₁ and R₀, we get:
%ΔP = (0.88^2 * 1 / √(1.21)) - 1 * 100
%ΔP = 0.7744 / √1.21 - 1 * 100
%ΔP ≈ -17.98%
Therefore, the percentage change in P is approximately -17.98%.
2) Let's assume the equation connecting s and t is of the form:
s = kt^a * √t^b
We can substitute the given values for s and t to find the values of k, a, and b.
When t = 4, s = 16:
16 = k * 4^a * √4^b
16 = k * 4^(a + b)
Similarly, when t = 16, s = 48:
48 = k * 16^a * √16^b
48 = k * 16^(a + b)
Dividing the second equation by the first equation, we get:
48 / 16 = 16^(a + b) / 4^(a + b)
3 = 4^(a + b)
Since 4 can be expressed as 2^2, we can rewrite the equation as:
3 = (2^2)^(a + b)
3 = 2^(2a + 2b)
Taking the logarithm of both sides, we get:
log(3) = (2a + 2b) * log(2)
Simplifying, we get:
2a + 2b = log(3) / log(2)
Now we have one equation with two variables. To solve for a and b, we need another equation. However, the problem does not provide any further information, so we cannot determine the exact values of a and b without additional information.