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.
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It could be 180 but then idk if it needs a picture because I don’t see it
Answer:
C) Neither
Step-by-step explanation:
These two equations are not parallel, because they don't have the same slope. And they're not perpendicular to each other either, because perpendicular lines have negative reciprocals of the slopes. Therefore, the answer is C) Neither.
Answer:
so 14 can go into 28, 2 times and 19 can go into 38 2 times so the techaer can make 2 group with 14 girls and 19 boys in each group.
Step-by-step explanation:
<span>Total = Principal × ( 1 + Rate )<span>^years
</span></span><span>Total = 900 * (1.03)^5
</span>Total = 900 *
<span>
<span>
<span>
1.1592740743
</span>
</span>
</span>
Total =
<span>
<span>
<span>
1,043.35
</span></span></span>
