What is a “plasticity poem?

# What is a “plasticity poem?

• MATH

sec2(x)+0.5tan(x)−1=0sec2(x)+0.5tan(x)-1=0 Use a graphing utility to approximate the solutions (to…

see the attached graph. Solutions in the given interval are, x ≈≈ 0 , 2.700 , 3.100(pi) , 5.800 , 6.200(2pi)

• MATH

xcos(x)−1=0xcos(x)-1=0 Use a graphing utility to approximate the solutions (to three decimal…

see attached graph Solution in the given interval, x ≈≈ 4.900

• MATH

xtan(x)−1=0xtan(x)-1=0 Use a graphing utility to approximate the solutions (to three decimal…

see attached graph solutions in the given interval are, x ≈≈ 0.875 , 3.375

• MATH

cos(x)cot(x)1−sin(x)=3cos(x)cot(x)1-sin(x)=3 Use a graphing utility to approximate the solutions (to…

See the attached graph. Since the equation is undefined for x=pi/2 x=pi/6 , 5pi/6 x ≈≈ 0.500 , 2.600

• MATH

1+sin(x)cos(x)+cos(x)1+sin(x)=41+sin(x)cos(x)+cos(x)1+sin(x)=4 Use a graphing utility to approximate…

See attached graph x=π3,5π3π3,5π3 x ≈≈ 1.000 , 5.200

• MATH

4sin3(x)+2sin2(x)−2sin(x)−1=04sin3(x)+2sin2(x)-2sin(x)-1=0 Use a graphing utility to approximate the…

See the attached graph Solutions in the given interval are, x=π4,3π4,5π4,7π4,7π6,11π6x=π4,3π4,5π4,7π4,7π6,11π6 x≈0.800,2.300,3.600,3.900,5.500,5.800x≈0.800,2.300,3.600,3.900,5.500,5.800

• MATH

2sin(x)+cos(x)=02sin(x)+cos(x)=0 Use a graphing utility to approximate the solutions (to three decimal…

Solutions of 2sin(x)+cos(x)=02sin(x)+cos(x)=0 in the interval (0,2pi) See the attached graph, x ≈≈ 2.700 , 5.800

• MATH

y=sec4(πx8)−4y=sec4(πx8)-4 Find the x-intercepts of the graph.

x-intercepts are the points where y=0. So we have to solve equation sec4(πx8)−4=0,sec4(πx8)-4=0, or sec4(πx8)=4.sec4(πx8)=4. Then sec2(πx8)=2,sec2(πx8)=2, or sec(πx8)=±2–√.sec(πx8)=±2. sec(y) = 1/cos(y), therefore…

• MATH

y=tan2(πx6)−3y=tan2(πx6)-3 Find the x-intercepts of the graph.

x-intercepts are points where y=0, so we have to solve the equation tan2(πx6)−3=0.tan2(πx6)-3=0. This implies that tan(πx6)=3–√tan(πx6)=3 or tan(πx6)=−3–√,tan(πx6)=-3, and this in turn implies…

• MATH

y=sin(πx)+cos(πx)y=sin(πx)+cos(πx) Find the x-intercepts of the graph.

You need to evaluate the x intercepts of the graph, hence, you need to remember that the graph intercepts x axis at y = 0. Hence, you need to solve for x the equation y=f(x)=0y=f(x)=0 .

• MATH

y=sin(πx2)+1y=sin(πx2)+1 Find the x-intercepts of the graph.

y=sin(πx2)+1y=sin(πx2)+1 Before we solve for the x-intercepts, let’s determine the period of this function. Take note that if a trigonometric function has a form y= Asin(Bx + C) + D, its period…

• MATH

2sin(x2)+3–√=02sin(x2)+3=0 Solve the multiple-angle equation.

sin(x2)=−3–√2.sin(x2)=-32. The general solution is x2=(−1)k⋅arcsin(−3–√2)+kπ,k∈Z.x2=(-1)k⋅arcsin(-32)+kπ,k∈ℤ. So x=(−1)k+1⋅(2π3)+2kπ,k∈Z,x=(-1)k+1⋅(2π3)+2kπ,k∈ℤ, because arcsin(−3–√2)=−π3.arcsin(-32)=-π3.

• MATH

2cos(x2)−2–√=02cos(x2)-2=0 Solve the multiple-angle equation.

2cos(x2)−2–√=0,2cos(x2)-2=0, 2cos(x2)=2–√,2cos(x2)=2, cos(x2)=2–√2.cos(x2)=22. The general solution is: x2=±arccos(2–√2)+2kπ,x2=±arccos(22)+2kπ, k∈Z.k∈ℤ. arccos(2–√2)=π4,arccos(22)=π4, so the final answer is:

• MATH

sec(4x)−2=0sec(4x)-2=0 Solve the multiple-angle equation.

sec(4x)−2=0,sec(4x)-2=0, sec(4x)=2.sec(4x)=2. Because sec(x)=1cos(x),sec(x)=1cos(x), obtain cos(4x)=12.cos(4x)=12. The general solution is 4x=±arccos(12)+2kπ,4x=±arccos(12)+2kπ,k∈Z.k∈ℤ. arccos(12)=π3,arccos(12)=π3, so the final answer is…

• MATH

tan(3x)−1=0tan(3x)-1=0 Solve the multiple-angle equation.

tan(3x)=1. tan(y) has period ππ and reaches any value once on each period. We know that tan(π4)=1,tan(π4)=1, so 3x=π4+k⋅π,k∈Z.3x=π4+k⋅π,k∈ℤ. The answer is x=π12+k⋅(π3),k∈Z.x=π12+k⋅(π3),k∈ℤ.

• MATH

2sin(2x)+3–√=02sin(2x)+3=0 Solve the multiple-angle equation.

2sin(2x)+3–√=0,2sin(2x)+3=0, sin(2x)=−3–√2.sin(2x)=-32. The general solution is 2x=(−1)k⋅arcsin(−3–√2)+kπ,2x=(-1)k⋅arcsin(-32)+kπ, k∈Z.k∈ℤ. Because arcsin(−3–√2)=−π3,arcsin(-32)=-π3, the final answer is…

• MATH

2cos(2x)−1=02cos(2x)-1=0 Solve the multiple-angle equation.

2cos(2x)−1=0,2cos(2x)-1=0, cos(2x)=12.cos(2x)=12. The general solution is 2x=±arccos(12)+2kπ,2x=±arccos(12)+2kπ, k∈Z.k∈ℤ. Because arccos(12)=π3,arccos(12)=π3,the final answer is x=±π6+kπ,x=±π6+kπ, k∈Z.k∈ℤ.

• MATH

sec(x)+tan(x)=1sec(x)+tan(x)=1 Find all the solutions of the equation in the interval 0,2π)0,2π).

sec(x)+tan(x)=1sec(x)+tan(x)=1 1cos(x)+sin(x)cos(x)=11cos(x)+sin(x)cos(x)=1 1+sin(x)=cos(x)1+sin(x)=cos(x)1+sin(x)−cos(x)=01+sin(x)-cos(x)=0 1+cos(π2−x)−cos(x)=01+cos(π2-x)-cos(x)=0 1+(2sin(π2−x+x2)sin(x−π2+x2)=01+(2sin(π2-x+x2)sin(x-π2+x2)=0 1+2sin(π4)sin(x−π4)=01+2sin(π4)sin(x-π4)=0…

• MATH

csc(x)+cot(x)=1csc(x)+cot(x)=1 Find all the solutions of the equation in the interval 0,2π)0,2π).

csc(x)+cot(x)=1csc(x)+cot(x)=1 1sin(x)+cos(x)sin(x)=11sin(x)+cos(x)sin(x)=1 1+cos(x)=sin(x)1+cos(x)=sin(x) 1+cos(x)−sin(x)=01+cos(x)-sin(x)=0 1+cos(x)−cos(π2−x)=01+cos(x)-cos(π2-x)=0  1+2sin(x+π2−x2)sin(π2−x−x2)=01+2sin(x+π2-x2)sin(π2-x-x2)=0 1+2sin(π4)sin(π4−x)=01+2sin(π4)sin(π4-x)=0…

• MATH

cos(x)+sin(x)tan(x)=2cos(x)+sin(x)tan(x)=2 Find all the solutions of the equation in the interval 0,2π)0,2π).

cos(x)+sin(x)tan(x)=2cos(x)+sin(x)tan(x)=2 cos(x)+sin(x)(sin(x)cos(x))=2cos(x)+sin(x)(sin(x)cos(x))=2 cos2(x)+sin2(x)cos(x)=2cos2(x)+sin2(x)cos(x)=2 1cos(x)=21cos(x)=2 cos(x)=12cos(x)=12 General solutions for cos(x)=1/2 are, x=π3+2πn,5π3+2πnx=π3+2πn,5π3+2πn Solutions for…

• MATH

2sec2(x)+tan2(x)−3=02sec2(x)+tan2(x)-3=0 Find all the solutions of the equation in the interval…

Find all solutions to the equation 2sec2(x)+tan2(x)−3=02sec2(x)+tan2(x)-3=0 in the interval [0,2π).[0,2π). 2sec2(x)+tan2(x)−3=02sec2(x)+tan2(x)-3=0 Use the pythagorean identity sec2(x)=1+tan2(x)sec2(x)=1+tan2(x) to substitute in for…

• MATH

2sin2(x)+3sin(x)+1=02sin2(x)+3sin(x)+1=0 Find all the solutions of the equation in the interval [0,2π)[0,2π).

Solve the equation 2sin2(x)+3sin(x)+1=02sin2(x)+3sin(x)+1=0 in the interval [0,2pi). 2sin2(x)+3sin(x)+1=02sin2(x)+3sin(x)+1=0 (2sin(x)+1)(sin(x)+1)=0(2sin(x)+1)(sin(x)+1)=0 Set each factor equal to zero and solve for the x value(s). 2sin(x)+1=02sin(x)+1=0…

• MATH

2cos2(x)+cos(x)−1=02cos2(x)+cos(x)-1=0 Find all the solutions of the equation in the interval 0,2π)0,2π).

Find all solutions to the equation 2cos2(x)+cos(x)−1=02cos2(x)+cos(x)-1=0 in the interval [0,2π).[0,2π). 2cos2(x)+cos(x)−1=02cos2(x)+cos(x)-1=0 (2cos(x)−1)(cos(x)+1)=0(2cos(x)-1)(cos(x)+1)=0 Set each factor equal to zero and solve for the x value(s)….

• MATH

sin(x)−2=cos(x)−2sin(x)-2=cos(x)-2 Find all the solutions of the equation in the interval 0,2π)0,2π).

sin(x)−2=cos(x)−2sin(x)-2=cos(x)-2 sin(x)−2−cos(x)+2=0sin(x)-2-cos(x)+2=0 sin(x)−cos(x)=0sin(x)-cos(x)=0 sin(x)−cos(x)cos(x)=0sin(x)-cos(x)cos(x)=0 tan(x)−1=0tan(x)-1=0 tan(x)=1tan(x)=1 General solutions for tan(x)=1 are x=π4+πnx=π4+πn Solutions for the range…

• MATH

2sin(x)+csc(x)=02sin(x)+csc(x)=0 Find all the solutions of the equation in the interval 0,2π)0,2π).

2sin(x)+csc(x)=02sin(x)+csc(x)=0 2sin(x)+1sin(x)=02sin(x)+1sin(x)=0 2sin2(x)+1sin(x)=02sin2(x)+1sin(x)=0solving above , 2sin2(x)+1=02sin2(x)+1=0 2sin2(x)=−12sin2(x)=-1 sin2x=−12sin2x=-12 sin(x)=±i2–√sin(x)=±i2 Solutions for the range 0≤x≤2π0≤x≤2π No…

• MATH

sec(x)csc(x)=2csc(x)sec(x)csc(x)=2csc(x) Find all the solutions of the equation in the interval 0,2π)0,2π).

sec(x)csc(x)=2csc(x)sec(x)csc(x)=2csc(x) sec(x)csc(x)−2csc(x)=0sec(x)csc(x)-2csc(x)=0 csc(x)(sec(x)−2)=0csc(x)(sec(x)-2)=0 csc(x)=0,sec(x)=2csc(x)=0,sec(x)=2 No solutions for x for csc(x) in the range 0≤x≤2π0≤x≤2π General solutions for sec(x)=2 are…

• MATH

sec2(x)−sec(x)=2sec2(x)-sec(x)=2 Find all the solutions of the equation in the interval [0,2π)[0,2π).

Solve the equation sec2(x)−sec(x)=2,[0,2π).sec2(x)-sec(x)=2,[0,2π). sec2(x)−sec(x)−2=0sec2(x)-sec(x)-2=0 (sec(x)−2)(sec(x)+1)=0(sec(x)-2)(sec(x)+1)=0 Set each factor equal to zero and solve for the x values. sec(x)−2=0sec(x)-2=0 sec(x)=2sec(x)=2 cos(x)=12cos(x)=12…

• MATH

2sin2(x)=2+cos(x)2sin2(x)=2+cos(x) Find all the solutions of the equation in the interval [0,2π)[0,2π).

Find all solutions for 2sin2(x)=2+cos(x)2sin2(x)=2+cos(x) in the interval [0, 2π).2π).Use the pythagorean identity sin2(x)+cos2(x)=1.sin2(x)+cos2(x)=1. Solve for sin2(x)sin2(x) and the pythagorean identity would be…

• MATH

3tan3(x)=tan(x)3tan3(x)=tan(x) Find all the solutions of the equation in the interval 0,2π)0,2π).

3tan3(x)=tan(x)3tan3(x)=tan(x) 3tan3(x)−tan(x)=03tan3(x)-tan(x)=0 tan(x)(3tan2(x)−1)=0⇒tan(x)(3tan2(x)-1)=0⇒ tan(x)=0,(3tan2(x)−1)=0tan(x)=0,(3tan2(x)-1)=0General solutions for tan(x)=0 are, x=0+πnx=0+πn Solutions in the range 0≤x≤2π0≤x≤2π are,

• MATH

sec2(x)−1=0sec2(x)-1=0 Find all the solutions of the equation in the interval 0,2π)0,2π).

Solve: sec2(x)−1=0,[0,2π)sec2(x)-1=0,[0,2π) tan2(x)=1tan2(x)=1  tan(x)=±1–√tan(x)=±1 tan(x)=±1tan(x)=±1 x=π4,x=3π4,x=5π4,x=7π4x=π4,x=3π4,x=5π4,x=7π4

• MATH

cos3(x)=cos(x)cos3(x)=cos(x) Find all the solutions of the equation in the interval 0,2π)0,2π).

cos3(x)=cos(x)cos3(x)=cos(x) Let cos(x)=ycos(x)=y y3=yy3=y y3−y=0y3-y=0 y(y+1)(y−1)=0y(y+1)(y-1)=0 solve for y, y=0 , -1 , 1 Therefore cos(x)=0 , cos(x)=-1 and cos(x)=1 General solutions for cos(x)=0 are,

• MATH

(2sin2(x)−1)(tan2(x)−3)=0(2sin2(x)-1)(tan2(x)-3)=0 Solve the equation.

Solve the equation (2sin2(x)−1)(tan2(x)−3)=0(2sin2(x)-1)(tan2(x)-3)=0 Set each factor equal to zero and solve for the x values. 2sin2(x)−1=02sin2(x)-1=0 2sin2(x)=12sin2(x)=1 sin2(x)=12sin2(x)=12 sin(x)=±12−−√sin(x)=±12 sin(x)=±2–√2sin(x)=±22…

• MATH

sin(x)(sin(x)+1)=0sin(x)(sin(x)+1)=0 Solve the equation.

Solve sin(x)(sin(x)+1)=0sin(x)(sin(x)+1)=0 sin(x)=0sin(x)=0 x=0+πnx=0+πn sin(x)+1=0sin(x)+1=0 sin(x)=−1sin(x)=-1 x=3π2+2πnx=3π2+2πn

• MATH

cos(2x)(2cos(x)+1)=0cos(2x)(2cos(x)+1)=0 Solve the equation.

cos(2x)(2cos(x)+1)=0cos(2x)(2cos(x)+1)=0 solving each part, cos(2x)=0 General solutions for cos(2x)=0 are, 2x=π2+2πn,3π2+2πn2x=π2+2πn,3π2+2πn x=π+4πn4,3π+4πn4x=π+4πn4,3π+4πn4 Solving 2cos(x)+1=0, 2cos(x)=−12cos(x)=-1…

• MATH

tan(3x)tan(x−1)=0tan(3x)tan(x-1)=0 Solve the equation.

tan(3x)tan(x−1)=0tan(3x)tan(x-1)=0 solving each part, tan(3x)=0 General solutions for tan(3x)=0 are, 3x=0+πn3x=0+πn x=πn3x=πn3 General solutions for tan(x-1)=0 are, x−1=0+πnx-1=0+πn x=1+πnx=1+πn so the solutions are,…

• MATH

tan2(3x)=3tan2(3x)=3 Solve the equation.

tan2(3x)=3tan2(3x)=3 tan(3x)=±3–√tan(3x)=±3 General solutions for tan(3x)=-3–√3are 3x=2π3+πn3x=2π3+πn x=3πn+2π9x=3πn+2π9 General solutions for tan(3x)=3–√3 are 3x=π3+πn3x=π3+πn x=3πn+π9x=3πn+π9 Therefore…

• MATH

2sin2(2x)=12sin2(2x)=1 Solve the equation.

2sin2(2x)=12sin2(2x)=1 sin2(2x)=12sin2(2x)=12 sin(2x)=±12–√sin(2x)=±12 General solutions for sin(2x)=-1/2–√2 2x=5π4+2πn,7π4+2πn2x=5π4+2πn,7π4+2πn x=5π+8πn8,7π+8πn8x=5π+8πn8,7π+8πn8 General solutions for…

• MATH

sin2(x)=3cos2(x)sin2(x)=3cos2(x) Solve the equation.

sin2(x)=3cos2(x)sin2(x)=3cos2(x) sin2(x)cos2(x)=3sin2(x)cos2(x)=3 tan2(x)=3tan2(x)=3 tan(x)=±3–√tan(x)=±3General solutions for tan(x)=3–√3 are x=π3+πnx=π3+πn General solutions for tan(x)=-3–√3 are x=2π3+πnx=2π3+πn So…

• MATH

4cos2(x)−1=04cos2(x)-1=0 Solve the equation.

Solve the equation 4cos2(x)−1=04cos2(x)-1=0 4cos2(x)=14cos2(x)=1 cos2(x)=14cos2(x)=14 cos(x)=±14−−√cos(x)=±14 cos(x)=±12cos(x)=±12 x=π3+πnx=π3+πn x=2π3+πnx=2π3+πn

• MATH

3cot2(x)−1=03cot2(x)-1=0 Solve the equation.

Solve the equation 3cot2(x)−1=03cot2(x)-1=0 3cot2(x)=13cot2(x)=1 cot2(x)=13cot2(x)=13 cot(x)=±13−−√cot(x)=±13 cot(x)=±13–√cot(x)=±13 tan(x)=±3–√tan(x)=±3 x=π3+πnx=π3+πnx=2π3+πnx=2π3+πn

• MATH

3sec2(x)−4=03sec2(x)-4=0 Solve the equation.

Solve the equation 3sec2(x)−4=03sec2(x)-4=0 3sec2(x)=43sec2(x)=4 sec2(x)=43sec2(x)=43 sec(x)=±43−−√sec(x)=±43 sec(x)=±23–√±23 Because cosine and secant are reciprocal functions cos(x)=±3–√2cos(x)=±32 x=π6+πnx=π6+πn…

• MATH

3sin(x)+1=sin(x)3sin(x)+1=sin(x) Solve the equation.

Solve the equation 3sin(x)+1=sin(x)3sin(x)+1=sin(x) 3sin(x)+1−sin(x)=03sin(x)+1-sin(x)=0 2sin(x)+1=02sin(x)+1=0 sin(x)=−12sin(x)=-12 x=7π6+2πnx=7π6+2πn x=11π6+2πnx=11π6+2πn

• MATH

cos(x)+1=−cos(x)cos(x)+1=-cos(x) Solve the equation.

Solve the equation cos(x)+1=−cos(x)cos(x)+1=-cos(x) 2cos(x)+1=02cos(x)+1=0 2cos(x)=−12cos(x)=-1 cos(x)=−12cos(x)=-12 x=2π3+2πn,x=4π3+2πnx=2π3+2πn,x=4π3+2πn

• MATH

tan(x)+3–√=0tan(x)+3=0 Solve the equation.

tan(x)+3–√=0tan(x)+3=0  tan(x)=−3–√tan(x)=-3 x=2π3+πnx=2π3+πn General solution for x,

• MATH

3–√csc(x)−2=03csc(x)-2=0 Solve the equation.

Solve the equation: 3–√csc(x)−2=03csc(x)-2=0 csc(x)=23–√csc(x)=23 Since sin(x) and csc(x) are reciprocal functions sin(x)=3–√2sin(x)=32 x=π3+2πnx=π3+2πn x=2π3+2πnx=2π3+2πn

• MATH

tan(t)cot(t)=1tan(t)cot(t)=1 Verfiy the identity.

Verify the identity: tan(t)cot(t)=1tan(t)cot(t)=1 Simplify the left side of the equation by using the reciprocal identity cot(t)=1/tan(t). tan(t)⋅1tan(t)=1tan(t)⋅1tan(t)=1 tan(t)tan(t)=1tan(t)tan(t)=1 1=11=1

• SCIENCE

Where do Avalanches happen?

Avalanches are natural events (or hazards due to their destructive capacity) that are characterized by movement of very large snow mass over a slope. Extremely large quantity of snow moves very…

• SOCIAL SCIENCES

When a monopoly’s fixed costs increase, why does the price stay the same and the profit increase?

I believe that you must have made a mistake in typing in this question. When a monopoly’s fixed costs increase, it is true that the price the monopoly charges does not increase. However, it is…

• HISTORY

How did the U.S. economy end up suffering both from inflation and high unemployment?

In the 1970s, the United States experienced something economically that had never happened in our country. We had both high unemployment and high inflation at the same. We had strategies to deal…