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Leya [2.2K]
3 years ago
12

In the diagram below, which distance represents the distance from point D to

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
4 0
Answer
A.CD
Cause theCD is the mid point of triangle ADB which divides triangle in two half if you look carefully you can understand the diagram ☺️☺️
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5. Original Price: $40 Sale Price: $30
telo118 [61]
It’s %25 off think about it 40-10=30 and that’s %25 of it hope I help you have a good day
5 0
3 years ago
Can you help me plz
Gelneren [198K]

i think the answer would be C

4 0
3 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
The guitar that you always wanted is on sale in the music store. It costs 2/3 the original price. The guitar now costs $1996.67,
AysviL [449]

Answer:

3,327.9

Step-by-step explanation:

now: 1996.67

original: 3,327.9

1996.67/.6 = 3,327.9

4 0
3 years ago
Write an equation that is parallel to y=-5/2x-5
NeX [460]

       A parallel equation (when graphed) will have the same slope, but a different y-intercept.

       As long as you keep y = -\frac{5}{2}x + b, you can input anything for b to solve this question.

       Given:

y = -\frac{5}{2}x - 5

       Equation of a parallel line:

y = -\frac{5}{2}x + 6, y = -\frac{5}{2}x + 1,356, y = -\frac{5}{2}x - 8, etc

     Example answer you can use:

             y = -\frac{5}{2}x - 8

6 0
2 years ago
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