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sweet [91]
3 years ago
14

For the linear function f(x) = -2x+1 , find the following

Mathematics
1 answer:
Leya [2.2K]3 years ago
4 0

Given:

The function is:

f(x)=-2x+1

To find:

a. Slope and Y-intercept.

b. Is the function increasing, decreasing, or constant, justify your answer.

c. Graph the function, label x, and y-intercepts.

Solution:

a. We have,

f(x)=-2x+1                  ...(i)

The slope intercept form of a line is:

y=mx+b                    ...(ii)

Where, m is the slope and b is the y-intercept.

On comparing (i) and (ii), we get

m=-2,b=1

Therefore, the slope of the line is -2 and the y-intercept is 1.

b. If the slope of a linear function is negative, then the function is decreasing.

If the slope of a linear function is positive, then the function is increasing.

If the slope of a linear function is 0, then the function is constant.

The slope of the linear function is negative.

Therefore, the function is decreasing.

c. We have,

f(x)=-2x+1

At x=0, we get

f(0)=-2(0)+1

f(0)=1

So, the y-intercept is at (0,1).

At f(x)=0, we get

0=-2x+1

2x=1

x=\dfrac{1}{2}

x=0.5

So, the x-intercept is at \left(0.5,0\right).

Plot the points (0,1), \left(0.5,0\right) and connect them by a straight line as shown below:

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Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).​
Reil [10]

Answer:

Part A) sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}

Part B) tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}

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Step-by-step explanation:

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Part B) Find tan(\alpha)\ and\ cot(\beta)

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

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so

tan(\alpha)=cot(\beta)

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132

x=\sqrt{132}\ units

simplify

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Find the value of tan(\alpha) in the right triangle of the figure

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simplify

tan(\alpha)=\frac{4}{\sqrt{33}}

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tan(\beta)=\frac{4}{\sqrt{33}}

Part C) Find sec(\alpha)\ and\ csc(\beta)

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

\alpha+\beta=90^o ---> by complementary angles

so

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Find the value of sec(\alpha) in the right triangle of the figure

sec(\alpha)=\frac{1}{cos(\alpha)}

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cos(\alpha)=\frac{2\sqrt{33}}{14} ---> adjacent side divided by the hypotenuse

simplify

cos(\alpha)=\frac{\sqrt{33}}{7}

therefore

sec(\alpha)=\frac{7}{\sqrt{33}}

csc(\beta)=\frac{7}{\sqrt{33}}

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