Simplify this equation 3x^2×4x^5y^4

Please help and thank you

worty [1.4K]

I believe 2 dollars and 50 cents

6 0

3 years ago

Read 2 more answers

Which statement describes the function below f(x)=2x^3-3x+1

Ilia_Sergeevich [38]

The given function is f(x)=2x^3+2x^2-x. In this case, the highest degree among the terms is 3. According to the Fundamental Theorem of Algebra, the number of zeros or roots of the equation is equal to the highest degree among the terms. Hence for every value of y, there are 3 equivalent x's. The answer thus is B. It is a many-to-one function.

3 0

3 years ago

a store received 12 boxes of doughnuts. Each box contained 12 doughnuts. Which expression represents how many doughnuts there ar

Pachacha [2.7K]

Answer:

12x12=144 doughnuts

Step-by-step explanation:

3 0

2 years ago

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)]

8 0

3 years ago

Note that WXYZ has vertices W(-1, 2), X(-5, 7), y(-1, -2), and Z (3, -7).

svp [43]

a. Slope of WZ = -2.25; Slope of WX = 5

b. WZ = √97; WX = √41

c. WXYZ is not a <em>rectangle, rhombus, nor a square</em>. We can conclude that: <em>D. WXYZ is none of these</em>.

<h3>Slope of a Segment</h3>

Slope = change in y/change in x

Given:

W(-1, 2), X(-5, 7), Y(-1, -2), and Z (3, -7)

a. Slope of WZ and slope of WX:

Slope of WZ = (-7 - 2)/(3 -(-1)) = -2.25

Slope of WX = (7 - 2)/(-1 -(-1)) = 5

b. Use distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , to find WZ and WX:

$WZ = \sqrt{(3 - (-1))^2 + (-7 - 2)^2}\\\\\mathbf{WZ = \sqrt{97} }$

$WX = \sqrt{(-5 -(-1))^2 + (7 - 2)^2}\\\\\mathbf{WX = \sqrt{41} }$

c. The quadrilateral WXYZ have adjacent sides that are not perpendicular to each other and have different slopes and different lengths, so therefore, WXYZ is not a rectangle, rhombus, nor a square. We can conclude that: <em>D. WXYZ is none of these</em>.

Learn more about slopes on:

brainly.com/question/3493733

5 0

2 years ago