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4 years ago

In the diagram below, is the perpendicular bisector of JK. Which of the following statements must be true?

1 answer:

3 0

I need some details... If i can get a little more details with that "check all that apply" and i could get you an answer.

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Sam is 10 years younger than one-half the age of his aunt. Let a represent his aunt’s age. Which shows his aunt’s age written as

wariber [46]

Sam is 10 years old because 20 is half of 40 and 20 subtract 10 is 10

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3 years ago

Read 2 more answers

Jenny is making tie-dyed shirts for her class with her school colors,blue and yellow.she used a total of 8 ounces of fabric pain

Elodia [21]

Question:

Jenny is making tie-dyed shirts for her class with her school colors, blue and yellow. She used a total of 8 ounces of fabric paint. The amount of blue paint is 3 times the amount of yellow paint. Make a diagram to show the ratio of blue paint to yellow paint

Answer:

The ratio of blue paint to yellow paint is 3 : 1

Solution:

Let "x" be the amount of blue paint in ounces

Let "y" be the amount of yellow paint in ounces

She used a total of 8 ounces of fabric paint

Therefore,

x + y = 8 -------- eqn 1

The amount of blue paint is 3 times the amount of yellow paint

Amount of blue paint = 3 times the amount of yellow paint

x = 3y ------- eqn 2

Substitute eqn 2 in eqn 1

3y + y = 8

4y = 8

Divide both sides by 4

y = 2

Substitute y = 2 in eqn 2

x = 3(2)

x = 6

Therefore , the amount of blue paint is 6 ounces and the amount of yellow paint is 2 ounces

Find the ratio

To find out the ratio of blue paint to yellow paint, divide the amount of blue paint by the amount of yellow paint

$\frac{x}{y} = \frac{6}{2} = \frac{3}{1}$

Which means, of every 4 parts of paint, three parts are blue and one part is yellow

The diagram is attached below

4 0

4 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

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

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