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3 years ago

A piecewise linear function is graphed below. (see image)

Mathematics

2 answers:

belka [17]3 years ago

7 0

Answer:

Step-by-step explanation:

Given is a piecewise graph

discontinuous at x=0

For all negative x, we have y = constant = 2

Hence I part is

y =2, for x<0

Next part is from x=0 to x=3

Passes through points (0,0) and (3,3)

Hence equation is y =3x for $0\leq x\leq 3$

For all values of x >3, we have a horizontal line as y = 3

Consolidated together

$y=x, for x$

ExtremeBDS [4]3 years ago

6 0

y = 2 for x < 0
= x for 0 ≤ x < 3
= 3 for x ≥ 3

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3 years ago

A mobile company charges a fixed rate of x cents per minute for the first 120 minutes of talk time and another rate of y cents p

patriot [66]

Answer:

$21.2

Step-by-step explanation:

From the problem given, we express them as algebra so that we can easily solve;

the total charge on a call greater than 120min is given as;

120x + y(n - 120) = total cost of call

where n is the number of minutes which is greater than 120;

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120x + y(175 - 120) = 26.80

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Now let us solve both equations for y and x;

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120x + 90y = 32.4 ----- (II)

subtract the two equations;

90y - 55y = 32.4 - 26.8

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y = $\frac{5.6}{35}$ = 0.16cents

then x;

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120x = 26.8 - 55(0.16)

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3 years ago

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Evaluate the definite integrals using properties of the definite integral and the fact that

anygoal [31]

I gather that the given functions satisfy the following definite integral relations:

$\displaystyle \int_{-1}^4 f(x) = -6$

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$\displaystyle \int_4^5 g(x) =5$

a) By linearity of the integral operator, we have

$\displaystyle \int_{-1}^4 10 f(x) \, dx = 10 \int_{-1}^4 f(x) \, dx = 10 \times (-6) = \boxed{-60}$

b) The integral over an interval is equal to the sum of integrals over a partition of that interval. In this case, the interval [-1, 5] can be written as the interval union [-1, 4] U [4, 5], so that

$\displaystyle \int_{-1}^5 f(x) \, dx = \int_{-1}^4 f(x) \, dx + \int_4^5 f(x) \, dx = -6 + 10 = \boxed{4}$

c) By linearity,

$\displaystyle \int_4^5 (f(x) - g(x)) \, dx = \int_4^5 f(x) \, dx - \int_4^5 g(x) \, dx = 10 - 5 = \boxed{5}$

d) By linearity,

$\displaystyle \int_4^5 (4f(x) + 5g(x)) \, dx = 4 \int_4^5 f(x) \, dx + 5 \int_4^5 g(x) \, dx = 4\times10 + 5\times5 = \boxed{65}$

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2 years ago