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3 years ago

If the length of a rectangle is doubled, and its width is multiplied by 4, the area

Mathematics

1 answer:

agasfer [191]3 years ago

5 0

Answer:

The area would be multipled by 8

Step-by-step explanation:

Say you have a rectangle with the length of 5 and the width of 3, the area would be 15. Double the length, 10, and quadruple the width, 12. The area would be 120. 15 × 8 = 120.

Sorry if I'm wrong.

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A guy stole $100 from a stores cash register, let's call it Dave's. He bought $70 in groceries, and paid with the $100 bill. The

sergij07 [2.7K]

Well by calculations i would say 100

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3 years ago

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Suppose you could climb to the top of the Great Pyramid of Giza in Egypt. Which path would be shorter, climbing a lateral edge o

valentinak56 [21]

Check the picture below.

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3 years ago

SOLVE, GRAPH AND CHECK:<br> 6(4- x) < 12

jeka57 [31]

Answer:

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3 years ago

The assets (in billions of dollars) for a financial firm can be approximated by the function A(x)=318e^0.27x, where x=7 correspo

Mashutka [201]

The value of the asset in 2012, 2014, and 2017 are 8119.72, 13933.55 and, 31321.23, respectively. A function assigns the values.

<h3>What is a Function?</h3>

A function assigns the value of each element of one set to the other specific element of another set.

Given the assets (in billions of dollars) for a financial firm can be approximated by the function $A(x)=318e^{0.27x}$ , where x=7 corresponds to the year 2007. Therefore, the assets value in the following years will be,

A.) 2012 - $A(12)=318e^{0.27\times 12} = 8,119.72$

B.) 2014 - $A(14)=318e^{0.27\times 14} = 13,933.5$

C.) 2012 - $A(17)=318e^{0.27\times 17} = 31,321.23$

Hence, the value of the asset in 2012, 2014, and 2017 are 8119.72, 13933.55 and, 31321.23, respectively.

Learn more about Function:

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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