Help me please.. Explain also :(
1. E is inversely proportional to Z and Z = 4 when E = 6.
2. P varies inversely as Q and Q = 2/3 when P = 1/2.
3.R is inversely proportional to the square of I and I = 25 when R = 100.
4. F varies inversely as the cube root of S and S = 64 when F = 4.
5. Z is inversely proportional to the sum of x and y and the sum of x and y is 10 when Z is 4
6. A varies inversely as the cube of b and b is 1/3 when a is 27
1) E=K/Z
6=K/4
k=24
2) P=K/Q
1/2=K/2/3
K=2/6 or K= 1/3
y varies inversely as x. When y = 18, x = 6. Find the constant of variation and determine the value for y when x = 27
If p is inversely proportional to q and p=5 when q=4 .what is their relationship.what is the value of p when q=25
H varies directly as J and inversely with M. If the constant of variation is 3/4, find H when J=16 and M=24.
Pôtáng îna to mensi haha nga saldi wenno addang
Of course, I'd be happy to help explain these problems to you!
1. E is inversely proportional to Z and Z = 4 when E = 6:
To find the relationship between E and Z, we can write it as E = k/Z, where k is a constant of proportionality. Since we know that Z = 4 when E = 6, we can substitute these values into the equation: 6 = k/4. To solve for k, we can multiply both sides by 4, resulting in 24 = k. So the equation relating E and Z is E = 24/Z.
2. P varies inversely as Q and Q = 2/3 when P = 1/2:
Similar to the previous problem, we can write P = k/Q. We can substitute Q = 2/3 and P = 1/2 into the equation: 1/2 = k/(2/3). To simplify, we multiply both sides by 2/3, giving us 1/2 * (2/3) = k. The result is 1/3 = k. Therefore, the equation relating P and Q is P = 1/3Q.
3. R is inversely proportional to the square of I and I = 25 when R = 100:
Using the same logic, we can write R = k/I^2. Substituting I = 25 and R = 100, we have 100 = k/(25^2). Simplifying further, we get 100 = k/625. We can then multiply both sides by 625 to find k: 62500 = k. Thus, the equation relating R and I is R = 62500/I^2.
4. F varies inversely as the cube root of S and S = 64 when F = 4:
The inverse relationship can be presented as F = k/S^(1/3). Substituting S = 64 and F = 4, we can rewrite the equation: 4 = k/(64^(1/3)). Simplifying, we have 4 = k/4. By multiplying both sides by 4, we find that k = 16. Therefore, the equation relating F and S is F = 16/S^(1/3).
5. Z is inversely proportional to the sum of x and y and the sum of x and y is 10 when Z is 4:
In this scenario, we can write Z = k/(x + y). Substituting Z = 4 and (x + y) = 10, we have 4 = k/10. To solve for k, we multiply both sides by 10, resulting in 40 = k. Therefore, the equation relating Z, x, and y is Z = 40/(x + y).
6. A varies inversely as the cube of b and b is 1/3 when a is 27:
Again, we can use the formula A = k/b^3. Substituting A = 27 and b = 1/3, we can rewrite the equation as 27 = k/(1/3)^3. Simplifying, we have 27 = k/(1/27). To solve for k, we multiply both sides by 1/27, giving us k = 729. Thus, the equation relating A and b is A = 729/b^3.
I hope these explanations help! Let me know if you have any other questions.
What are you supposed to do? Writethe equation that relates the two variables?
1. E*Z = 24, so E = 24/Z
2. P*Q = constant = (2/3)*(1/2) = 1/3, so P = 1/(3 Q)
6. a*b^3 = constant = 27*(1/3)^3 = 1
so
a = 1/b^3
You should be able to do the others.