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olasank [31]
3 years ago
6

Amy has a 55” television with a width of 72 cm and a height of 52 cm. Write two equivalent expressions for the area of her telev

ision screen.
Mathematics
1 answer:
Evgesh-ka [11]3 years ago
3 0

Area = Length x width.

1st equation: Area = 72 x 52

2nd equation: Using distributive property by rewriting 52 as a sum of two numbers. 30 + 22 = 52

So now we get the second equation of:  72(30+22)

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A landscaper buys a tree that grows at a rate of 18 inches every 3 months.
ipn [44]

1 it grows 18 inches every 3 months

 there are 12 months in 1 year

12 dived by 3 = 4 so there are 4 times it will grow 18 inches

18 * 4 = 72 inches, amount it grows in 1 year

 Answer is D


2. it grows 9 inches every 2 months

 12/2 = 6 times a year it will grow

 9 *6 = 54 inches total for 1 year

 Answer is C

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3 years ago
What is 3/4 of 76kg?
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Answer:

57,000 grams

Step-by-step explanation:

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3 years ago
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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
How do you graph y=-3x-3
netineya [11]

Answer:you go to negative (0,-3) on the y axis then u go down by 3 and then over by 1 and keep doing that to form ur line

Step-by-step explanation:

5 0
3 years ago
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When you translate, (x-4,y) will move ____
sesenic [268]

Answer:

The point will move to the left 4 place values.

Step-by-step explanation:

When you add or subtract from the x value, you are moving the point parallel to the x-axis, the amount you need to go. If subtracting, you move left, if adding, you move right.

When you add or subtract from the y value, you are moving the point parallel to the y-axis, the amount you need to go. If subtracting, you move down, if adding, you move up.

~

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3 years ago
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