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Bezzdna [24]
2 years ago
12

What is the value of x? How can you find the value of y if you know the value of x?

Mathematics
1 answer:
Y_Kistochka [10]2 years ago
6 0

Answer:

ummm it depends on the question for example if they give u a number for a certain variable if u could solve it like y-x=4  what is the value of x if y= 5 ?

5-x=4 so 5-4=1  ans: x= 1

or what is the value of y in terms of x    


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Show work please thank you
ira [324]
1) 12 * 9 = 108 cm² - part 1
2) 12 - 3 - 3 = 6 cm
3) 6 * 4 = 24 cm² - part 2
4) 108 + 24 = 132 cm ² - the area.

3 0
3 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
Determine the measures of the angles requested using the picture below.<br>​
lapo4ka [179]
SHDJJEHEH shdhshshshshduvuvuvububuyb
5 0
3 years ago
14. Is x + 2 a factor of the polynomial f(x)=2x-3x²-4x+1?
defon
Thanks for the free 10x+ points .
8 0
2 years ago
Please help! x^3 +y^3=9
Crank

Step-by-step explanation:

Taking the derivative of both sides with respect to x, we get

\dfrac{d}{dx}(x^3 + y^3) = \dfrac{d}{dx}(3)

\Rightarrow 3x^2 + 3y^2\dfrac{dy}{dx} = 0

Solving for \frac{dy}{dx}, we get

\dfrac{dy}{dx} = -\dfrac{3x^2}{3y^2} = -\dfrac{x^2}{y^2}

7 0
2 years ago
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