Ya, calculus and related rates, such fun!
everything is changing with respect to t
altitude rate will be dh/dt and that is 1cm/min
dh/dt=1cm/min
area will be da/dt which is increasing at 2cm²/min
da/dt=2cm²/min
base=db/dt
alright
area=1/2bh
take dervitivie of both sides
da/dt=1/2((db/dt)(h)+(dh/dt)(b))
solve for db/dt
distribute
da/dt=1/2(db/dt)(h)+1/2(dh/dt)(b)
move
da/dt-1/2(dh/dt)(b)=1/2(db/dt)(h)
times 2 both sides
2da/dt-(dh/dt)(b)=(db/dt)(h)
divide by h
(2da/dt-(dh/dt)(b))/h=db/dt
ok
we know
height=10
area=100
so
a=1/2bh
100=1/2b10
100=5b
20=b
so
h=10
b=20
da/dt=2cm²/min
dh/dt=1cm/min
therefor
(2(2cm²/min)-(1cm/min)(20cm))/10cm=db/dt
(4cm²/min-20cm²/min)/10cm=db/dt
(-16cm²/min)/10cm=db/dt
-1.6cm/min=db/dt
the base is decreasing at 1.6cm/min
Answer:
12+33 and 33 - (-12)
Step-by-step explanation:
12 - (-33) = 45
12 + 33 gives us 45
33 - (-12) gives us 45
Answer:
CODE: 1977.98
Step-by-step explanation:
A.
(To get the closest answer, round the circumference to the nearest ten thousandth.)
C = 2(3.14)r Circumference formula: C = 2πr
C = 2(3.14)(3)
C = 18.84
B.
A = (3.14)r²
A = (3.14)(3)²
A = (3.14)(9)
A ≈ 28.26
C. (It's asking for the circumference.)
C = 2(3.14)r
C = 2(3.14)(58)
C ≈ 364.24
D. (It's a linear pair, which is 180 degrees.)
4x + 2x = 180
6x = 180
x = 30
m∠ABD = 4x
m∠ABD = 4(30)
m∠ABD = 120°
E. (∠GHI & ∠JHK are vertical angles, so they are congruent.)
x + 7 = 3x - 21
28 = 2x
14 = x
F. (x = 14)
m∠JHK = 3x - 21
m∠JHK = 3(14) - 21
m∠JHK = 42 - 21
m∠JHK = 21°
G. (Supplementary - two angles that add up to 180 degrees.)
180 - 84
= 96°
CODE: E(C - D) - F(G - B) - A
CODE: 14(364.24 - 120) - 21(96 - 28.26) - 18.84
CODE: 14(244.24) - 21(67.74) - 18.84
CODE: 3419.36 - 1422.54 - 18.84
CODE: 1977.98
It’s left 2 up 3
Thanks buddy thanks again
Answer:
i think 4x+3y = -24 is answer
