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2 years ago

Expand and simplify 3(x-2) + 2(4x-1)

Mathematics

1 answer:

Zanzabum2 years ago

3 0

Answer:

11x-8

Step-by-step explanation:

$3(x - 2) + 2(4x - 1) \\ multiply \: inside \: the \: bracket \\ 3x - 6 + 8x - 2 \\ 11x - 8$

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The vector (a) is a multiple of the vector (2i +3j) and (b) is a multiple of (2i+5j) The sum (a+b) is a multiple of the vector (

kow [346]

Answer:

$\|a\| = 5\sqrt{13}$ .

$\|b\| = 3\sqrt{29}$ .

Step-by-step explanation:

Let $m$ , $n$ , and $k$ be scalars such that:

$\displaystyle a = m\, (2\, \vec{i} + 3\, \vec{j}) = m\, \begin{bmatrix}2 \\ 3\end{bmatrix}$ .

$\displaystyle b = n\, (2\, \vec{i} + 5\, \vec{j}) = n\, \begin{bmatrix}2 \\ 5\end{bmatrix}$ .

$\displaystyle (a + b) = k\, (8\, \vec{i} + 15\, \vec{j}) = k\, \begin{bmatrix}8 \\ 15\end{bmatrix}$ .

The question states that $\| a + b \| = 34$ . In other words:

$k\, \sqrt{8^{2} + 15^{2}} = 34$ .

$k^{2} \, (8^{2} + 15^{2}) = 34^{2}$ .

$289\, k^{2} = 34^{2}$ .

Make use of the fact that $289 = 17^{2}$ whereas $34 = 2 \times 17$ .

$\begin{aligned}17^{2}\, k^{2} &= 34^{2}\\ &= (2 \times 17)^{2} \\ &= 2^{2} \times 17^{2} \end{aligned}$ .

$k^{2} = 2^{2}$ .

The question also states that the scalar multiple here is positive. Hence, $k = 2$ .

Therefore:

$\begin{aligned} (a + b) &= k\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 2\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 16\, \vec{i} + 30\, \vec{j}\\ &= \begin{bmatrix}16 \\ 30 \end{bmatrix}\end{aligned}$ .

$(a + b)$ could also be expressed in terms of $m$ and $n$ :

$\begin{aligned} a + b &= m\, (2\, \vec{i} + 3\, \vec{j}) + n\, (2\, \vec{i} + 5\, \vec{j}) \\ &= (2\, m + 2\, n) \, \vec{i} + (3\, m + 5\, n) \, \vec{j} \end{aligned}$ .

$\begin{aligned} a + b &= m\, \begin{bmatrix}2\\ 3 \end{bmatrix} + n\, \begin{bmatrix} 2\\ 5 \end{bmatrix} \\ &= \begin{bmatrix}2\, m + 2\, n \\ 3\, m + 5\, n\end{bmatrix}\end{aligned}$ .

Equate the two expressions and solve for $m$ and $n$ :

$\begin{cases}2\, m + 2\, n = 16 \\ 3\, m + 5\, n = 30\end{cases}$ .

$\begin{cases}m = 5 \\ n = 3\end{cases}$ .

Hence:

$\begin{aligned} \| a \| &= \| m\, (2\, \vec{i} + 3\, \vec{j})\| \\ &= m\, \| (2\, \vec{i} + 3\, \vec{j}) \| \\ &= 5\, \sqrt{2^{2} + 3^{2}} = 5 \sqrt{13}\end{aligned}$ .

$\begin{aligned} \| b \| &= \| n\, (2\, \vec{i} + 5\, \vec{j})\| \\ &= n\, \| (2\, \vec{i} + 5\, \vec{j}) \| \\ &= 3\, \sqrt{2^{2} + 5^{2}} = 3 \sqrt{29}\end{aligned}$ .

6 0

3 years ago

Which of the following functions represents a direct variation?

AnnyKZ [126]

It's B. Since the slope is -8/1 x. Slope is rise/run. The other answers doesn't make sense since the x is put in a different place.

7 0

3 years ago

Find the slope of the line that passes through the points 2,4,6 and 12

kolezko [41]

Answer:

The slope is 2.

Step-by-step explanation:

If you use the formula that is shown in the picture, you can calculate that 12-4/6-2 is the slope. If you simplify, you'll get that 8/4 is the slope which simplifies to 2.

3 0

3 years ago

I need help with number 3 please!

WITCHER [35]

The answer to the question

8 0

3 years ago

The following two vectors satisfy the same system of linear equations. u=⎡⎣⎢⎢6−5−5⎤⎦⎥⎥, v=⎡⎣⎢⎢746⎤⎦⎥⎥ Find x and y that make the

kondaur [170]

Answer:

x=-30/71, y=0

Step-by-step explanation:

if a vector satisfies an equation of the form ax+by+cz=0 then the vector is

parallel to the plane ax+by+cz=0, and, the cross product of two vectors results in an orthogonal vector to both.

So, <a,b,c> the normal vector of the plane ax+by+cz=0, can be found as the cross product of the two parallel vetors to the plane:

<a,b,c> = <6,-5,-5>×<7,4,6>= <-10,-71,59>

so, the homogeneous system is:

-10x-71y+59z=0

by replacing the vector <3,x,y> in the system

-30-71x+59y=0

So, there are 2 unknown variables and 1 equation, it means that 1 variable is free

so, y=0 is a random defition and x can be obtain with the equation

-30-71x=0 -> x=-30/71

3 0

3 years ago