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4 years ago

3 (2x-1 ) +7 = - 44 ( solve for x)

Mathematics

1 answer:

s344n2d4d5 [400]4 years ago

8 0

Answer:

x = - 8

Step-by-step explanation:

To solve for x, isolate it by "moving" everything else over to the other side. This is done in reverse BEDMAS order using the reverse operations of number you want to move.

3 (2x-1 ) +7 = - 44 Simplify first using distributive property over the brackets

6x - 3 + 7 = - 44 Combine like terms, terms that have the same variables

6x + 4 = - 44

6x + 4 - 4 = - 44 - 4 Subtract 4 from both sides to cancel it out on the left side

6x = - 48

6x/6 = - 48/6 Divide by sides by 6 to cancel out the 6 from 6x

x = - 8 Final answer

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Find an explicit formula for the arithmetic sequence -31,-27,-23,-19,...

Yuri [45]

First term (a) = -31

Common Difference (d) = -27 - (-31) = -27 + 31 = 4

nth term = a + (n - 1)d = -31 + (n - 1)4 = -31 + 4n - 4 = 4n - 35

Hence, the formula is 4n - 35.

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3 years ago

Kristin

Deffense [45]

I’ve attached my work...
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3 years ago

(39-3)÷44−7² any help would be Appreciated

goldfiish [28.3K]

Answer:

-48.1818

Step-by-step explanation: then u round up to 3 singnificant figures

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3 years ago

Question

krok68 [10]

Answer:

Acute angle between the two planes: approximately $43^\circ$ .

Step-by-step explanation:

Find the normal vector of each plane:

The normal vector of the plane $x - 2\, y + 5\, z = 3$ is $\displaystyle \begin{bmatrix}1 \\ -2 \\ 5\end{bmatrix}$ .
The normal vector of the plane $2\, x + y - 3\, z = 15$ is $\displaystyle \begin{bmatrix}2 \\ 1\\ -3\end{bmatrix}$ .

As the name suggests, there is a $90^\circ$ angle between a plane and its normal vector. The following four angles will correspond to the vertices of a quadrilateral:

The $90^\circ$ angle between the first plane and its normal vector.
The angle between the normal vector of each plane.
The $90^\circ$ angle between the second plane and its normal vector.
The smallest angle between these two planes.

The sum of these four angles should be $360^\circ$ . Two of these four angles were known to be $90^\circ$ . Once the third angle (the angle between the two normal vectors) is found, subtractions would give the measure of the other angle (the smallest angle between these two planes.)

Make use of the dot product to find the angle between these two normal vectors. Let $\theta$ denote the angle between these two vectors.

$\displaystyle \cos \theta = \frac{\begin{bmatrix}1 \\ -2 \\ 5\end{bmatrix} \cdot \begin{bmatrix}2 \\ 1 \\ -3\end{bmatrix}}{\sqrt{1^2 + (-2)^2 + 5^2} \cdot \sqrt{2^2 + 1^2 + (-3)^2}} \approx -0.73193$ .

Before continuing, notice that the smallest angle between the two planes would be $360^\circ - 90^\circ - 90^\circ - \theta = 180^\circ - \theta$ .

Consider the identity: $\cos\left(180^\circ - \theta\right) = -\cos \theta$ .

In other words, $\cos\left(180^\circ - \theta\right)$ , the cosine of the smallest angle between the two planes (which the question is asking for) will be the opposite of $\cos \theta$ , the cosine of the angle between the two normal vectors.