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quester [9]
3 years ago
15

7 Solve the equation x^2 - 8x + 15 = 0.

Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
6 0

Answer:

x = -5

Evidence to support reasoning:

Oksanka [162]3 years ago
3 0

Answer:

x=5   x=3

Step-by-step explanation:

x^2 - 8x + 15 = 0

Factor

What 2 numbers multiply to 15 and add to -8

-5*-3 = 15

-5-3 = -8

(x-5)(x-3) =0

Using the zero product property

x-5 =0    x-3 =0

x=5   x=3

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What is the equation for the plane illustrated below?
TiliK225 [7]

Answer:

Hence, none of the options presented are valid. The plane is represented by 3 \cdot x + 3\cdot y + 2\cdot z = 6.

Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

a\cdot x + b\cdot y + c\cdot z = d

Where:

x, y, z - Orthogonal inputs.

a, b, c, d - Plane constants.

The plane presented in the figure contains the following three points: (2, 0, 0),  (0, 2, 0), (0, 0, 3)

For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

xy-plane (2, 0, 0) and (0, 2, 0)

y = m\cdot x + b

m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Where:

m - Slope, dimensionless.

x_{1}, x_{2} - Initial and final values for the independent variable, dimensionless.

y_{1}, y_{2} - Initial and final values for the dependent variable, dimensionless.

b - x-Intercept, dimensionless.

If x_{1} = 2, y_{1} = 0, x_{2} = 0 and y_{2} = 2, then:

Slope

m = \frac{2-0}{0-2}

m = -1

x-Intercept

b = y_{1} - m\cdot x_{1}

b = 0 -(-1)\cdot (2)

b = 2

The equation of the line in the xy-plane is y = -x+2 or x + y = 2, which is equivalent to 3\cdot x + 3\cdot y = 6.

yz-plane (0, 2, 0) and (0, 0, 3)

z = m\cdot y + b

m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}

Where:

m - Slope, dimensionless.

y_{1}, y_{2} - Initial and final values for the independent variable, dimensionless.

z_{1}, z_{2} - Initial and final values for the dependent variable, dimensionless.

b - y-Intercept, dimensionless.

If y_{1} = 2, z_{1} = 0, y_{2} = 0 and z_{2} = 3, then:

Slope

m = \frac{3-0}{0-2}

m = -\frac{3}{2}

y-Intercept

b = z_{1} - m\cdot y_{1}

b = 0 -\left(-\frac{3}{2} \right)\cdot (2)

b = 3

The equation of the line in the yz-plane is z = -\frac{3}{2}\cdot y+3 or 3\cdot y + 2\cdot z = 6.

xz-plane (2, 0, 0) and (0, 0, 3)

z = m\cdot x + b

m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}

Where:

m - Slope, dimensionless.

x_{1}, x_{2} - Initial and final values for the independent variable, dimensionless.

z_{1}, z_{2} - Initial and final values for the dependent variable, dimensionless.

b - z-Intercept, dimensionless.

If x_{1} = 2, z_{1} = 0, x_{2} = 0 and z_{2} = 3, then:

Slope

m = \frac{3-0}{0-2}

m = -\frac{3}{2}

x-Intercept

b = z_{1} - m\cdot x_{1}

b = 0 -\left(-\frac{3}{2} \right)\cdot (2)

b = 3

The equation of the line in the xz-plane is z = -\frac{3}{2}\cdot x+3 or 3\cdot x + 2\cdot z = 6

After comparing each equation of the line to the definition of the equation of the plane, the following coefficients are obtained:

a = 3, b = 3, c = 2, d = 6

Hence, none of the options presented are valid. The plane is represented by 3 \cdot x + 3\cdot y + 2\cdot z = 6.

8 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)]

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Approximately equal to symbol
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BC² = 10² + 30²

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