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3 years ago

What is the solution to 6x^2 + 2x + 4 = 0?

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

8 0

Step 1) Divide 6 and 4 (the last 2 numbers) and bring everything else down

6x^24x+2x+4

Step 2) Place in parentheses

(6x^24x)+(2x+4)

Step 3) Figure out what you can take out of each parentheses.

**In the first one, you can take out 6x because both numbers have x and 6 is the largest number 6 and 24 can go into.**

**In the second one, you can only take out 2 because 2 is the largest number that goes into 2 and 4.**

Step 4) You will know you did it correctly when the parentheses match up

6x(x+4) + 2(x+4) <<< this is how it should look

Answer: (6x+2)(x+4)

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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

MY LAST QUESTION PLEASE HELP

liraira [26]

Answer:

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

cos theta = adj / hyp

cos 60 = 6/hyp

hyp = 6 / cos 60

hyp = 6 / (1/2)

htp = 12

3 0

3 years ago

What is the surface area of this 3-D solid?

Dvinal [7]

Answer:

I think its flat sorry if it's wrong

5 0

3 years ago

A student of the author earned grades of 92, 83, 77, 84, and 82 on herfive regular tests. She earned grades of 88 on the final e

goblinko [34]

The class is weighted as follows:

60% Regular Tests

10% Final Exam

15% Project

15% Homework

The total weighted points possible for the class are as follows:

(100+100+100+100+100)*.6 + 100*.1 + 100*.15 + 100*.15 = 340

To calculate her individual final weighted grade we plug in her scores for each category and complete the following computation:

(92+83+77+84+82)*.6 + 88*.1 + 95*.15 + 77*.15 = 285.4

So her weighted grade percent would be 285.4/340 = 83.9% which is a B.

8 0

3 years ago

What is the GCF of 12 and 6? Please let Dakotamcclea answer...

Hatshy [7]

The GCF of 12 and 6 is 6!

6 0

3 years ago

Read 2 more answers