Okay
we have the rate of change of AC = d(AC)/dt = -2
the rate of change od BC = d(BC)/dt
area = (1/2) *AC) (BC)
taking differential on both sides we ge
d(A)/dt = 1/2){ (BC) d(AC)/dt + (AC) d(BC)/dt)}....(1)
again
when AC= 3
applying pythagorous thm
we get
(5)^2 =(3)^2 +(BC)^2
hence we get BC = 4
now we need to find d(BC)/dt
we have
(5)^2 = (AC)^2 +(BC)^2
taking differenial
0=2(AC) d(AC/dt) +2BC d(BC)/dt
that is
d(BC)/dt = -(3) *(-2)/4 ..(at AC =3)
hence
d(BC)/dt = 3/2
substituting these values in equation (1)
d(A)/dt = (1/2) {4 * -2 + 3 *3/2}
which gives
<span>d(A)/dt = -7/4
</span>The rate, in square feet per second, at which <span>the area is changing when AC = 3 is -7/4 ft/sec.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
3(k-x)= -3x-9
3k - 3x = -3x - 9 ← equation without parentheses
3k - 3x + 3x = - 9
3k = -9
k = -9/3
k = -3 ← simplest form
Do 18 x 9 then u will get get the right answer
Found this. Hope it helps.
https://www.algebra.com/algebra/homework/word/coins/Word_Problems_With_Coins.faq.question.1059726.html
Answer:
t is less than or equal to 4
Step-by-step explanation:
