I have 5 problems I seem to not be able to comprehend out of 150 that are due Tuesday. I have tried every which way to solve them on my own..please help.
Find the variation constant in which y varies inversely as x, and the following condition exists.
Y=30 when x = 1/6
k will = ____.
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The strength S of a rectangle beam varies jointly as its width w and the square of its thickness t.
If a wood beam 6 inches wide and 5 inches thick supports 500 pounds, how much can a similar beam 6 inches wide and 3 inches thick support? Round to the nearest pound if needed.
Thank you ahead of time for any consideration .
since y varies inversely as x, xy=k, a constant
1/6 * 30 = k
5 = k
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S(w,t) = kwt^2
500 = k*6*25
500 = 150k
k=4
So, now we know k, and we can figure S from any w and t:
S(6,3) = 4*6*9 = 216
To find the variation constant in the first problem, where y varies inversely as x, we can use the formula:
y = k/x
We are given the condition y = 30 when x = 1/6. Plugging these values into the formula, we get:
30 = k/(1/6)
To solve for k, we can multiply both sides of the equation by (1/6):
30 * (1/6) = k
The left side simplifies to:
5 = k
Therefore, the variation constant (k) is equal to 5.
Moving on to the second problem, where the strength S of a rectangle beam varies jointly as its width w and the square of its thickness t:
S = kw * t^2
We are given that a wood beam with a width of 6 inches and a thickness of 5 inches supports 500 pounds. Let's plug these values into the formula to solve for k:
500 = k * 6 * 5^2
First, we calculate 5^2 = 25:
500 = k * 6 * 25
Divide both sides by 6 * 25 to isolate k:
k = 500 / (6 * 25)
Simplify:
k = 500 / 150
k ≈ 3.3333
Rounding to the nearest pound, the variation constant (k) is approximately 3.