If y varies directly as x, find the constant of variation if y = 12 when x = 1/3

Evaluate<br> —2y – 3x for x = 1 and y= -2.<br><br> The value of the expression is

bulgar [2K]

Answer:

1

Step-by-step explanation:

Given

- 2y - 3x ← substitute x = 1 and y = - 2 into the expression

= - 2(- 2) - 3(1)

= 4 - 3

= 1

7 0

3 years ago

Solve and get brainliest!!<br>please be sure of your answer!!!!

almond37 [142]

Answer:

I think it is supplementary

3 0

2 years ago

What does ^ this stand for

Talja [164]

Carets normally are used for an exponent so for example
2^3=8
Hope it helps feel free to ask question

3 0

3 years ago

Read 2 more answers

How much cardboard is needed to make the tissue box above? Square inches 8 3 3

Katen [24]

The amount of cardboard needed to make a cuboid box of dimensions 8 inch, 3 inches and 3 inches is 114 sq. inches.

<h3>What is the surface area of cuboid?</h3>

Let the three dimensions(height, length, width) be x, y,z units respectively.

The surface area of the cuboid is given by

$S = 2(a\times b + b\times c + c\times a)$

For this case, tissue box is almost cuboid shaped.

Also, its dimensions are given being 8 inches, 3 inches and 3 inches.

Suppose we measure the amount of cardboard needed in terms of area, then, the amount of cardboard needed to make that box(without any whole, full cuboid) is equal to the area of its surface(either outer or inner if we assume 0 inches thickness of cardboard),

Thus, we get:

Amount of cardboard needed = surface area of cuboid box with dimensions 8 by 3 by 3 (in inches)

= $2(8 \times 3 + 3 \times 3 + 3 \times 8) = 114 \: \rm in^2$

Thus, the amount of cardboard needed to make a cuboid box of dimensions 8 inch, 3 inches and 3 inches is 114 sq. inches.

Learn more about surface area of cuboid here:

brainly.com/question/13522634

#SPJ1

5 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