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3 years ago

8(3n + 5) = -32 what does n =

Mathematics

1 answer:

olga_2 [115]3 years ago

8 0

Answer: n = -3

Step-by-step explanation:

8(3n + 5) = -32

Divide both sides by 8.

(3n + 5) = -4

Subtract 5 by both sides.

3n = -4 - 5

3n = -9

Divide both sides by 3.

n = -3

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3 years ago

1. Simplify the expression: 4w3z4 + 3w3 + 4z4 + 2w3z4 + z

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Step-by-step explanation:

You multiply them together them add to simplify them.

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3 years ago

F(x)=x2+5x-1 is shifted 3 units left the result is g(x) what is g(x)?

Len [333]

Replace the x in x^2 + 5x - 1 by x+3:-

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3 years ago

Base: z(x)=cosx Period:180 Maximum:5 Minimum: -4 What are the transformation? Domain and Range? Graph?

garik1379 [7]

Answer:

The transformations needed to obtain the new function are horizontal scaling, vertical scaling and vertical translation. The resultant function is $z'(x) = \frac{1}{2} + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)$ .

The domain of the function is all real numbers and its range is between -4 and 5.

The graph is enclosed below as attachment.

Step-by-step explanation:

Let be $z (x) = \cos x$ the base formula, where $x$ is measured in sexagesimal degrees. This expression must be transformed by using the following data:

$T = 180^{\circ}$ (Period)

$z_{min} = -4$ (Minimum)

$z_{max} = 5$ (Maximum)

The cosine function is a periodic bounded function that lies between -1 and 1, that is, twice the unit amplitude, and periodicity of $2\pi$ radians. In addition, the following considerations must be taken into account for transformations:

1) $x$ must be replaced by $\frac{2\pi\cdot x}{180^{\circ}}$ . (Horizontal scaling)

2) The cosine function must be multiplied by a new amplitude (Vertical scaling), which is:

$\Delta z = \frac{z_{max}-z_{min}}{2}$

$\Delta z = \frac{5+4}{2}$

$\Delta z = \frac{9}{2}$

3) Midpoint value must be changed from zero to the midpoint between new minimum and maximum. (Vertical translation)

$z_{m} = \frac{z_{min}+z_{max}}{2}$

$z_{m} = \frac{1}{2}$

The new function is:

$z'(x) = z_{m} + \Delta z\cdot \cos \left(\frac{2\pi\cdot x}{T} \right)$

Given that $z_{m} = \frac{1}{2}$ , $\Delta z = \frac{9}{2}$ and $T = 180^{\circ}$ , the outcome is:

$z'(x) = \frac{1}{2} + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)$

The domain of the function is all real numbers and its range is between -4 and 5. The graph is enclosed below as attachment.

8 0

3 years ago

Which of the following is the graph of the quadratic function? y=x^2+4-12

tiny-mole [99]

Answer:

Graph B

Step-by-step explanation:

The solutions to a quadratic equation are the points on which we have the graph of the curve touching the x-axis

Now the first thing we will do here is to solve the quadratic equation graph;

x^2 + 4x -12 = 0

x^2 +6x - 2x -12 = 0

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(x-2)(x + 6) = 0

x = 2 or -6

So the graph that touches the x-axis at the points x = 2 and x = -6 is the solution to the quadratic equation

Graph B is the closest to what we have as answer

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3 years ago