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2 years ago

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Mathematics

1 answer:

juin [17]2 years ago

7 0

Your answer is 139 hope

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In matrix theory, many of the familiar properties of the real number system are not valid. If a and b are real numbers, then ab

kozerog [31]

Answer:

B= $\left[\begin{array}{ccc}0&0\\0&1\\\end{array}\right]$

Step-by-step explanation:

Let's do the multiplication AB.

If A= $\left[\begin{array}{ccc}1&0\\0&0\\\end{array}\right]$

then the first row of A is= (1 0) by the first column of B= (0 0) is equal to zero.

the first row of A is= (1 0) by the second column of B= (0 1) is equal to zero too because 1.0+0.1=0.

the second row of A is= (0 0) by any colum of B is equal to zero too.

So we have found an example that works!

7 0

3 years ago

Help me please I’m going INSANE

Ksivusya [100]

Answer:

24/4

Step-by-step explanation:

all you have to do is multiply 4x6=24

7 0

3 years ago

Read 2 more answers

Is (x+7) a factor of f(x) x3-3x2+ 2x-8?

Snowcat [4.5K]

Answer:

(x+7) is not a factor

Step-by-step explanation:

we know that

If (x+7) is a factor of f(x)

then

For $x=-7, f(-7)=0$

Verify

we have

$f(x)=x^{3}-3x^{2} +2x-8$

Substitute x=-7

$f(-7)=(-7)^{3}-3(-7)^{2} +2(-7)-8$

$f(-7)=-343-147 -14-8$

$f(-7)=--512$

$-512\neq 0$

therefore

(x+7) is not a factor

3 0

3 years ago

This is for middle school 8th grade please help me :)

Crazy boy [7]

Answer:

what do you need?

Step-by-step explanation:

I don't see a question.

8 0

3 years ago

The logarithmic expression ln(√e/y^3) can be rewritten as

kvv77 [185]

Answer:

$\dfrac{1 - 6 \ln y}{2}$

Step-by-step explanation:

$\ln \dfrac{\sqrt{e}}{y^3} =$

$= \ln \dfrac{e^\frac{1}{2}}{y^3}$

$= \ln e^\frac{1}{2} - \ln y^3$

$= \dfrac{1}{2} \ln e - 3 \ln y$

$= \dfrac{1}{2} - 3 \ln y$

$= \dfrac{1 - 6 \ln y}{2}$

6 0

2 years ago