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zlopas [31]
3 years ago
12

I need help on (13.)

Mathematics
1 answer:
Trava [24]3 years ago
5 0
(-6,2),(-7,10)
slope = (10 - 2) / (-7 - (-6) = 8 / (-7 + 6) = 8/-1 = -8

y = mx + b
slope(m) = -8
use either of ur points...(-6,2)...x = -6 and y = 2
now we sub and find b, the y int
2 = -8(-6) + b
2 = 48 + b
2 - 48 = b
-46 = b

so ur equation is : y = -8x - 46....or 8x + y = -46
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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)]

8 0
3 years ago
Todd orders pictures from a photographer. Each picture costs $7.50. A one-time shipping fee of $3.25 is added to the cost of the
Vsevolod [243]

Answer: The equation is y = 3.25 + 7.5x and Todd bought 11 pictures

Step-by-step explanation:

4 0
3 years ago
What is the answer for Ab•2b
7nadin3 [17]

Answer:

2Ab^2

Step-by-step explanation:

I think its right

4 0
3 years ago
If a wheel with a radius of 80 inches spins at a rate of 50 revolutions per minute, find the approximate linear velocity in mile
Lubov Fominskaja [6]
\bf \textit{arc's length}\\\\
s=rw\qquad 
\begin{cases}
r=radius\\
w=\textit{angular speed}
\end{cases}\qquad w=\cfrac{50 rev}{min}\cdot \cfrac{2\pi }{rev}
\\\\\\
w=\cfrac{100\pi }{min}\qquad s=80in\cdot \cfrac{100\pi }{min}\implies s=\cfrac{8000\pi\  in}{min}\\\\
-------------------------------\\\\
\textit{now to convert it to miles/hr}
\\\\\\
\cfrac{8000\pi\ in}{min}\cdot \cfrac{ft}{12in}\cdot \cfrac{mile}{5280ft}\cdot \cfrac{60min}{hr}\implies \cfrac{8000\pi \cdot 60\ in}{12\cdot 5280\ hr}
3 0
3 years ago
Your are designing a rectangular birthday card for a friend. You want the card's length to be 1 inch more than twice the card's
klemol [59]

The width of the card to the nearest tenth of an inch is 6.15 inches

<h3>Area of rectangle</h3>

  • Width = x
  • Length = 2x + 1
  • Area = 88 square inches

Area of a rectangle = Length × Width

88 = (2x + 2) × x

88 = 2x² + 2x

2x² + 2x - 88 = 0

x = -b ± √b² - 4ac / 2a

= -2 ± √2² - 4(2)(-88) / 2(2)

= -2 ± √4 - (-704) / 4

= -2 ± √708 / 4

= -1/2 ± √177/2

x = 6.15 or -7.15 inches

The width of the rectangle can not be negative, so, 6.15 inches is the width.

Learn more about rectangle:

brainly.com/question/13048427

#SPJ1

5 0
2 years ago
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