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1.letfbethefunctiongivenbyf(x)=2e^(4x^2).forwhatvalueofxistheslopeofthelinetangenttothegraphoffat(x,f(x))equaltoA.0.618B.0.276C.0.318D.0.342E.0.5512.arailroadtrackandaroadcrossatnightangles.anobserverstands
题目详情
1.let f be the function given by f(x)=2e^(4x^2).for what value of x is the slope of the line tangent to the graph of f at( x,f(x) ) equal to
A.0.618 B.0.276 C.0.318 D.0.342 E.0.551
2.a railroad track and a road cross at night angles.an observer stands on the road 70m south of the crossing and watches an eastbound train traveling at 60m/s.at how many meters per second is the train moving away from the observer 4 seconds after it passes through the intersection?
A.57.60 B.57.88 C.59.20 D.60.00 E.67.40
3.if the derivate of f is given by f ' (x)=e^x-3x^2,at which of the following values of x does f have a relative max value?
A.-0.46 B.0.20 C.0.91 D.0.95 E.3.73
4.At time t≥0,the acceleration of a particle moving on the x-axis is a a(t)=t+sint.at t=0,the velocity of the particle is -2.for what value of t will the velocity of the particle be zero?
A.1.02 B.1.48 C.1.85 D.2.81 E.3.14
A.0.618 B.0.276 C.0.318 D.0.342 E.0.551
2.a railroad track and a road cross at night angles.an observer stands on the road 70m south of the crossing and watches an eastbound train traveling at 60m/s.at how many meters per second is the train moving away from the observer 4 seconds after it passes through the intersection?
A.57.60 B.57.88 C.59.20 D.60.00 E.67.40
3.if the derivate of f is given by f ' (x)=e^x-3x^2,at which of the following values of x does f have a relative max value?
A.-0.46 B.0.20 C.0.91 D.0.95 E.3.73
4.At time t≥0,the acceleration of a particle moving on the x-axis is a a(t)=t+sint.at t=0,the velocity of the particle is -2.for what value of t will the velocity of the particle be zero?
A.1.02 B.1.48 C.1.85 D.2.81 E.3.14
▼优质解答
答案和解析
1.let f be the function given by f(x)=2e^(4x^2).for what value of x is the slope of the line tangent to the graph of f at( x,f(x) ) equal to
A.0.618 B.0.276 C.0.318 D.0.342 E.0.551
【Solution】:
df/dx = 16xe^(4x²)
Let df/dx = 3
We get e^(4x²) = 3/16x,
By GC,x = 0.17
Ans:None of the above.
2.a railroad track and a road cross at night angles.an observer stands on the road 70m south of the crossing and watches an eastbound train traveling at 60m/s.at how many meters per second is the train moving away from the observer 4 seconds after it passes through the intersection?
A.57.60 B.57.88 C.59.20 D.60.00 E.67.40
【Solution】:
Let x be the distance between the train and the junction(intersection).
Let r be the distance between the train and the observer.
So,r² = 70² + x²
2rdr/dt = 2xdx/dt
t = 4 sec,x = 4×60 = 240 (m/s),r = √(4900 + 57600) = 250 (m)
∴ dr/dt = 240×60÷250 = 57.60 (m/s)
Ans :A
3.if the derivate of f is given by f ' (x)=e^x-3x^2,at which of the following values of x does f have a relative max value?
A.-0.46 B.0.20 C.0.91 D.0.95 E.3.73
【Solution】:
∵df/dx = e^x -3x²,
∴d²f/dx² = e^x -6x
Let df/dx = 0,x₁= -0.46,x₂= 0.91 (By GC)
d²f/dx²|(x₁) = 3.39 > 0 (Rej)
d²f/dx²|(x₂) = -2.98 < 0
Ans:C
4.At time t≥0,the acceleration of a particle moving on the x-axis is a a(t)=t+sint.at t=0,the velocity of the particle is -2.for what value of t will the velocity of the particle be zero?
A.1.02 B.1.48 C.1.85 D.2.81 E.3.14
【Solution】:
a = dv/dt = t + sint
v = ½t² - cost + c
-2 = 0 - 1 + c,c = -1
v = ½t² - cost - 1
Let v = 0,
ie.½t² - cost - 1 = 0
t₁= - 1.48 (s) (Rej)
t₂= + 1.48 (s)
Ans:B
A.0.618 B.0.276 C.0.318 D.0.342 E.0.551
【Solution】:
df/dx = 16xe^(4x²)
Let df/dx = 3
We get e^(4x²) = 3/16x,
By GC,x = 0.17
Ans:None of the above.
2.a railroad track and a road cross at night angles.an observer stands on the road 70m south of the crossing and watches an eastbound train traveling at 60m/s.at how many meters per second is the train moving away from the observer 4 seconds after it passes through the intersection?
A.57.60 B.57.88 C.59.20 D.60.00 E.67.40
【Solution】:
Let x be the distance between the train and the junction(intersection).
Let r be the distance between the train and the observer.
So,r² = 70² + x²
2rdr/dt = 2xdx/dt
t = 4 sec,x = 4×60 = 240 (m/s),r = √(4900 + 57600) = 250 (m)
∴ dr/dt = 240×60÷250 = 57.60 (m/s)
Ans :A
3.if the derivate of f is given by f ' (x)=e^x-3x^2,at which of the following values of x does f have a relative max value?
A.-0.46 B.0.20 C.0.91 D.0.95 E.3.73
【Solution】:
∵df/dx = e^x -3x²,
∴d²f/dx² = e^x -6x
Let df/dx = 0,x₁= -0.46,x₂= 0.91 (By GC)
d²f/dx²|(x₁) = 3.39 > 0 (Rej)
d²f/dx²|(x₂) = -2.98 < 0
Ans:C
4.At time t≥0,the acceleration of a particle moving on the x-axis is a a(t)=t+sint.at t=0,the velocity of the particle is -2.for what value of t will the velocity of the particle be zero?
A.1.02 B.1.48 C.1.85 D.2.81 E.3.14
【Solution】:
a = dv/dt = t + sint
v = ½t² - cost + c
-2 = 0 - 1 + c,c = -1
v = ½t² - cost - 1
Let v = 0,
ie.½t² - cost - 1 = 0
t₁= - 1.48 (s) (Rej)
t₂= + 1.48 (s)
Ans:B
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