If I varies inversely as d-2 and I - 20- when d - 3- what is I when d - 10-666dfrac-2-3-181-812

Question

Toppr Admin · Accepted Answer

Given that I varies inversely with d2 -xA0-Let the proportional constant be k-We get I-kd2At I-20- d is 3- so we get k-20-xD7-9-180So- the equation becomes I-180d2When d-10- we get I-180100-1-8

If $I$ varies inversely as $d^{2}$ and $I = 20$ , when $d = 3$ , what is $I$ when $d = 10$ ?
$6$
$66 \frac{2}{3}$
$18$
$1.8$
$12$

Given that $I$ varies inversely with $d^{2}$
Let the proportional constant be $k$ .
We get $I = \frac{k}{d^{2}}$
At $I = 20$ , $d$ is $3$ , so we get $k = 20 \times 9 = 180$
So, the equation becomes $I = \frac{180}{d^{2}}$
When $d = 10$ , we get $I = \frac{180}{100} = 1.8$

If I varies inversely as d2 and I=20, when d=3, what is I when d=10?66623181.812

Given that I varies inversely with d2 Let the proportional constant be k.We get I=kd2At I=20, d is 3, so we get k=20×9=180So, the equation becomes I=180d2When d=10, we get I=180100=1.8

If $I$ varies inversely as $d^{2}$ and $I = 20$ , when $d = 3$ , what is $I$ when $d = 10$ ?
$6$
$66 \frac{2}{3}$
$18$
$1.8$
$12$

Given that $I$ varies inversely with $d^{2}$
Let the proportional constant be $k$ .
We get $I = \frac{k}{d^{2}}$
At $I = 20$ , $d$ is $3$ , so we get $k = 20 \times 9 = 180$
So, the equation becomes $I = \frac{180}{d^{2}}$
When $d = 10$ , we get $I = \frac{180}{100} = 1.8$