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3 years ago

7+8 = 879+3 + 6 = 6394+ 5+ 4+ 5 = 54542 + 3 + 1 + 2 + 3 = ?mout answer here

Mathematics

2 answers:

Bess [88]3 years ago

6 0

Answer:

The answer is 32132.

(Answer - no. from right to left)

blagie [28]3 years ago

3 0

Answer:

32132 is the answer.

Step-by-step explanation:

it takes the back of the numbers. right to left

You might be interested in

If rectangle ABCD is rotated counter clock wise 90º about the origin, what will be the coordinates of the vertex D'?

7nadin3 [17]

So, the best way to do this is translate it to clockwise. 90 degrees counterclockwise is equal to 270 degrees clockwise. So, basically, to rotate, you would follow the following format for each point-
(X,Y) -> (-Y,X)
Now, you do it for each of the points.
A= (-5,5), so A' would be (-5,-5)
B= (-1,5), so B' would be (-5,-1)
C= (-5,4), so C' would be (-4,-5)
D= (-1,4) so D' would be (-4,-1)
Notice, how all the points end up in the square below it. Each quadrant has a specific number. The top right is quadrant 1, the top left is quadrant 2, the bottom left is quadrant 3, and the bottom right is quadrant 4. If you are rotating 270 degrees clockwise, you move to the right, like a clock. That puts the new rectangle in quadrant 3. That is a way to check your work.
Now, just so you know for future reference, the following are also different formats for different problems--
A 90 degree Clockwise rotation about the origin will be (X,Y) -> (Y, -X) *Note, -x just stands for the opposite. Say your original x is a negative number. Then the prime (new) x will be positive.
A 180 degree Clockwise rotation about the origin would be (X,Y) -> (-X,-Y) *Note, -y also stands for the opposite.
A 270 degree clockwise rotation about the origin would be (X,Y) -> (-Y,X).
For translating---
90 degrees Clockwise = 270 degrees Counter
270 degrees Clockwise = 90 degrees Counter

Hope this helped!

4 0

3 years ago

Which equation represents the statement: Three times the sum of a number and 5 is –30?

castortr0y [4]

Answer:

3(n+5)=-30

Step-by-step explanation:

Three times the sum of a number and 5 is - 30

Vocabulary:

times = multiplication, represented with *

sum = addition, represented by +

unknown number = n

Three * the sum of a number and 5

The sum of a number and 5 is the same as saying n + 5

So we are multiplying 3 by n + 5

Because n is an unknown variable we have to separate the sum of n and 5 from being multiplied by 3 with parenthesis. We do this because if we want to multiply 3 by the sum of n and 5. If we put 3 * n + 5 we are only multiplying the unknown variable by 3 not including the 5 that is added to it.

So we get 3( n + 5 ) is -30

3(n+5)=-30 is the answer.

3 0

2 years ago

Read 2 more answers

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

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

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