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3 years ago

Some helppp me plzzz

Mathematics

2 answers:

Eddi Din [679]3 years ago

5 0

Answer:

-2, 4

-1, -2

0, 0

1, 2

2, 4

Black_prince [1.1K]3 years ago

3 0

Step-by-step explanation:

X. Y

-2 = 4

-1 = 2

0 = 0

1 = -2

2 = -4

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A plant cell is 0.00001267 meter wide.How is this number written in scientific notation?

MA_775_DIABLO [31]

<span>The answer is 0.00001267 m = 1.267 * 10^(-5) m. In scientific notation, the number is written as two factors - a number m from 0-10 and 10 raised to the n power: m x 10^n. 0.00001267 m = 1.267 x 0.00001 m. 0.00001 m = (0.1)^5 = 10^(-5). So: 0.00001267 m = 1.267 x 0.00001 m = 1.267 x (0.1)^5 m = 1.267 x 10^(-5) m.</span>

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3 years ago

Determine all real values of p such that the set of all linear combination of u = (3, p) and v = (1, 2) is all of R2. Justify yo

Rama09 [41]

Answer:

p ∈ IR - {6}

Step-by-step explanation:

The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2

is all R2 ⇔

$u\neq 0_{R2}$

$v\neq 0_{R2}$

And also u and v must be linearly independent.

In order to achieve the final condition, we can make a matrix that belongs to $R^{2x2}$ using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.

Let's make the matrix :

$A=\left[\begin{array}{cc}3&1&p&2\end{array}\right]$

We used the first vector ''u'' as the first column of the matrix A

We used the second vector ''v'' as the second column of the matrix A

The determinant of the matrix ''A'' is

$Det(A)=6-p$

We need this determinant to be different to zero

$6-p\neq 0$

$p\neq 6$

The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that $p\neq 6$

We can write : p ∈ IR - {6}

Notice that is $p=6$ ⇒

$u=(3,6)$

$v=(1,2)$

If we write $3v=3(1,2)=(3,6)=u$ , the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.