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4 years ago

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In Exercises 29 and 30, describe the possible echelon forms of the standard matrix for a linear transformation T . Use the notat

ion of Example 1 in Section 1.2. 29. T W R3 ! R4 is one-to-one. 30. T W R4 ! R3 is onto.

1 answer:

TiliK225 [7]4 years ago

4 0

Answer:

<u>note:</u>

<u><em>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</em></u>

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(X^2+x-6)(2x^2+4x) factor completely

MissTica

Answer:

$2x\left(x-2\right)\left(x+3\right)\left(x+2\right)$

Step-by-step explanation:

Lets go ahead and take a step by step approach to solving your problem.

First we must start by factoring $x^2+x-6$ .

To do this, we must break the expression into groups! Although I will also provide a definition to help you understand! For $ax^2+bx+c$ we need to find u, v > $u\cdot \:v=a\cdot \:c$ and $u+v=b$ which we will group into $\left(ax^2+ux\right)+\left(vx+c\right)$ .

The values we have are... $a=1,\:b=1,\:c=-6$ and $u\cdot v=-6,\:u+v=1$ .

Now we need the factors of 6 which are 1, 2, 3 and 6. We also need the negative factors which you get simply by multiplying the positives by -1 or just reversing them.

Now we need to check for every two factors if u + v = 1.

$\mathrm{Check}\:u=1,\:v=-6: \:u\cdot v=-6,\:u+v=-5 \Rightarrow \mathrm{False}$

$\mathrm{Check}\:u=2,\:v=-3: \:u\cdot v=-6,\:u+v=-1 \Rightarrow \mathrm{False}$

$\mathrm{Check}\:u=3,\:v=-2: \:u\cdot v=-6,\:u+v=1 \Rightarrow \mathrm{True}$

$\mathrm{Check}\:u=6,\:v=-1: \:u\cdot v=-6,\:u+v=5 \Rightarrow \mathrm{False}$

Therefore $u=3,\:v=-2$ .

Now we want to $\mathrm{Group\:into\:}\left(ax^2+ux\right)+\left(vx+c\right)$ . Which is $\left(x^2-2x\right)+\left(3x-6\right)$ .

Now we must factor x from $x^2-2x$ .

Lets apply the exponent rule of $a^{b+c}=a^ba^c$ which means $x^2=xx$ .

Therefore we now have xx - 2x. Lets factor out the common term of x to get x(x - 2).

Now factor 3 out of $3x-6$ .

Rewrite 6 as 3 * 2. $3x-3\cdot \:2$ .

Now factor out the common term of 3 > $3\left(x-2\right)$ .

$x\left(x-2\right)+3\left(x-2\right)$ > Factor out the common term of x - 2 > $\left(x-2\right)\left(x+3\right)$ .

Now lets factor $2x^2+4x$ .

Apply the previous exponent rule of $a^{b+c}=a^ba^c$ > $x^2=xx$ > $2xx+4x$ .

Now rewrite 4 as 2 * 2 > $2xx+2\cdot \:2x$ .

Now factor out the common term of 2x to get 2x(x + 2).

Combine it all and we get $2x\left(x-2\right)\left(x+3\right)\left(x+2\right)$ .

Hope this helps!

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3 years ago

En un triangulo rectangulo ABC (B=90) REDUCE:

Dimas [21]

Answer:

puc provar, però no sé lemme see

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3 years ago

Need help on this question thank you

Ivahew [28]

Answer:

B.

Step-by-step explanation:

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4 years ago

A markdown that is greater than 99% but less than 100%?

Eva8 [605]

Magic is the answer

please revise your incomplete question next time before you post it

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deff fn [24]

Hello there

the answer is 3 x 3 x 5 x 7 x 7 = 2,205

it can also be written in exponential form

hope it help

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3 years ago

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