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3 years ago

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A beaker was filled with 10 ounces of water before starting a science experiment. After the experiment, 6 ounces of water remain

ed in the beaker. How does the amount of water in the beaker before the experiment compare to the amount of water in the beaker after the experiment?
There were 16 more ounces of water before the experiment.
There were 4 fewer ounces of water before the experiment.
There were 4 fewer ounces of water after the experiment.
There were 16 more ounces of water after the experiment

2 answers:

KiRa [710]3 years ago

5 0

Answer: There were 4 fewer ounces of water after the experiment.

Step-by-step explanation:

Hi, to answer this question we have to analyze the information given:

Amount of water before the experiment: 10 ounces

Amount of water after the experiment: 6 ounces

10 > 6: there were fewer ounces of water after the experiment.

So, if we subtract the amount of water after the experiment (6) to the amount of water before the experiment (10), we obtain the water used for the experiment:

Mathematically speaking:

10-6 = 4

In conclusion, there were 4 fewer ounces of water after the experiment.

Allisa [31]3 years ago

3 0

There were 4 fewer ounces of water after the experiment

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Answer:

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Step-by-step explanation:

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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

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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

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Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

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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.

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And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

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$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

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

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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

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

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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)]

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