All categories

2 years ago

Help!!!! If you wanted to model movements with both of the robot's arms in the air, what function

Mathematics

1 answer:

mixas84 [53]2 years ago

3 0

Answer:

f(x)=x^2

Step-by-step explanation:

if you wanted have a function where both of the robot's arms are in the air the function could be x to the power of any even number such f(x)=x^2, f(x)=x^4, f(x)=x^6, or even f(x)=x^10. And you could still do the same transformations with these equations.

You might be interested in

Alaska has a land area of about 1,700,000 km² Florida has a land area 1/10 the size of Alaska what is the land area Florida

ElenaW [278]

1/10 of 1,700,000 km² is 170,000 km² .

5 0

2 years ago

What is an equation of the line that passes through the points (-4, -5) and (-8, -2)?

nikklg [1K]

Solution:

Step-1: Find the slope of the line.

Formula of slope: y₂ - y₁/x₂ - x₁

y₂ - y₁/x₂ - x₁ = Slope
=> -2 - (-5)/-8 - (-4) = Slope
=> -2 + 5/-8 + 4 = Slope
=> 3/-4 = Slope

Step-2: Use the point slope formula to determine the slope.

Point slope form formula: y - y₁ = m(x - x₁)

y - y₁ = m(x - x₁) = Equation of line
=> y - (-5) = -3/4{x - (-4)} = Equation of line
=> y + 5 = -3/4{x + 4} = Equation of line
=> y + 5 = -3x/4 - 3 = Equation of line
=> y = -3x/4 - 8 = Equation of line

The equation of the line is y = -3x/4 - 8.

7 0

2 years ago

Read 2 more answers

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

Hence, the following are identities

$\begin{gathered} B)sin\theta=\frac{1}{csc\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

The marked are the trigonometric identities

3 0

1 year ago

What multiplies to 24 but adds to 23

laiz [17]

1 and negative 24 is the answer

4 0

2 years ago

Read 2 more answers

Apply Crammer's Rule to find the solution to the following quations .2x + 3y = 1, 3x + y = 5

Bond [772]

Answer:

The solution to the equation system given is:

x = 2
y = -1

Step-by-step explanation:

First, we must know the equations given:

2x + 3y = 1
3x + y = 5

Following Crammer's Rule, we have the matrix form:

$\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] =\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}1\\5\end{array}\right]$

Now we solve using the determinants:

$x=\frac{\left[\begin{array}{ccc}1&3\\5&1\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(1*1)-(5*3)}{(2*1)-(3*3)} = \frac{1-15}{2-9} =\frac{-14}{-7} = 2$

$y=\frac{\left[\begin{array}{ccc}2&1\\3&5\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(2*5)-(3*1)}{(2*1)-(3*3)}=\frac{10-3}{2-9} =\frac{7}{-7}=-1$

Now, we can find the answer which is x= 2 and y= -1, we can replace these values in the equation to confirm the results are right, with the first equation:

2x + 3y = 1
2(2) + 3(-1)= 1
4 - 3 = 1
1 = 1

And, with the second equation:

3x + y = 5
3(2) + (-1) = 5
6 - 1 = 5
5 = 5

4 0

2 years ago