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3 years ago

10

Nicolette was swinging as high as she could on the 20 foot tall swing set on the playground while her lab partner, Robin, collec

ted
data on the swings.
For each swing Robin calculated the time and the height of Nicolette's swing as she reached the highest point, h. If she decided to
plot this information on a scatter plot, what would be a reasonable range of her graph?
A.
150 <h < 625
OB. 6<n< 10
6<h < 25
OD. 0<h <1

1 answer:

djverab [1.8K]3 years ago

3 0

The answer is obvious a) i think lol

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Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

Applying the coordinates 2,6 and -2 to the basis E we obtain

$2+6t-2t^{2}$

That was the original result of T[e1(t)]

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3 years ago

How can a Line Plot help you find an average with data given in Fractions?

umka2103 [35]

Answer:

I did not understand its confusing

7 0

2 years ago

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Which of the following equations have complex roots?

mestny [16]

Answer:

A

Step-by-step explanation:

Complex roots of quadratic functions occur when the <u>discriminant is negative</u>.

<u>Discriminant</u>

$b^2-4ac\quad\textsf{when}\:\:ax^2+bx+c=0$

Evaluate the discriminant of each of the given equations.

$\textsf{A.} \quad 3x^2+2=0$

$\implies a=3, \quad b=0, \quad c=2$

$\implies b^2-4ac=0^2-4(3)(2)=-24$

As -24 < 0 the equation will have complex roots.

$\textsf{B.} \quad 2x^2+1=7x$

$\implies 2x^2-7x+1=0$

$\implies a=2, \quad b=-7, \quad c=1$

$\implies b^2-4ac= (-7)^2-4(2)(1)=41$

As 41 > 0 the equation does not have complex roots.

$\textsf{C.} \quad 3x^2-1=6x$

$\implies 3x^2-6x-1=0$

$\implies a=3, \quad b=-6, \quad c=-1$

$\implies b^2-4ac=(-6)^2-4(3)(-1)=48$

As 48 > 0 the equation does not have complex roots.

$\textsf{D.} \quad 2x^2-1=5x$

$\implies 2x^2-5x-1=0$

$\implies a=2, \quad b=-5, \quad c=-1$

$\implies b^2-4ac=(-5)^2-4(2)(-1)=33$

As 33 > 0 the equation does not have complex roots.

Learn more about discriminants here:

brainly.com/question/27444516

brainly.com/question/27869538

Learn more about complex roots here:

brainly.com/question/26344541

6 0

2 years ago

Choose a system of equations with the same solution as the following system:

Lina20 [59]

Answer:

Option A.

Step-by-step explanation:

We are given 6x+2y =-6 ......(1) and 3x-4y=-18 .....(2)

Solving equations (1) and (2) we get

12x+3x =-30, ⇒x= -2 and from equation (1), 2y= -6-6(-2) =6, ⇒ y=3

Therefore, the solution for the given set of equations is x= -2 and y=3

A) Solving the equations 8x+4y = -4 ....... (3)and 17x+2y =-28, we get

8x -34x = -4 -(-56) =52, ⇒-26x =52, ⇒x=-2 and from equation )(3), we get -16+4y =-4, ⇒ y=3

Therefore, the solution x=-2 and y =3 match with the original solution.

B)Solving the equations 12x + 4y = 12 ....... (4)and 21x + 2y = −36, we get

12x -42x = 12 -(-72) =84, ⇒-30x =84, ⇒x=-84/30.

Therefore, the solution x=-84/30 does not match with the original solution.

C) Solving the equations 6x + 8y = −36 ....... (4) and 15x + 6y = −60, we get 18x -60x = -108 -(-240) =132, ⇒-42x =132, ⇒x=-132/42

Therefore, the solution x=--3 do not match with the original solution.

D) Solving the equations 6x + y = 15 ....... (5) and 15x − y = −9, we get 6x +15x = 6, ⇒21x =6, ⇒x=2/7

Therefore, the solution x=2/7 do not match with the original solution.

Therefore, option A) will have the same solution. (Answer)

4 0

2 years ago

Math problem.. Transfrormations.. who's good at that

Oksanka [162]

Me i recently finished doing transformations. What's your problem?

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2 years ago

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