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Law Incorporation [45]
3 years ago
15

What is the slope of the graph of 14x + 7y = –9?

Mathematics
2 answers:
kondor19780726 [428]3 years ago
4 0
Y=mx+b
m=slope

or othe way

ax+by=c

slope=-a/b

14x+7y=-9
slope=-14/7=-2

slope=-2
Vera_Pavlovna [14]3 years ago
3 0
We will put ur equation in y = mx + b form, and the slope will be in the m position.

14x + 7y = -9
7y = -14x - 9
y = (-14/7)x - 9/7
y = -2x - 9/7

y = mx + b
y = -2x + (-9/7)....as u can see, the -2 is in the m position and is therefore the slope.
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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.

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\huge\textsf{Hey there!}

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\large\textsf{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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