All categories

3 years ago

what is the answer to Graph the function. f(x)=cos(x)−3 Use 3.14 for π i need it now im struggling on this test

Mathematics

1 answer:

kvv77 [185]3 years ago

8 0

Answer:

we can draw the graph of cosine with amplitude 2.

Step-by-step explanation:

In the given function g(x) = 2cosx we know the following parameters which are helpful to draw the graph.

Its a cosine graph given in the form of y = acos(Bx + c) + d

Here a is amplitude, period = , c = phase shift, d = vertical shift

We know a = 2, B = , c = 0, d = 0.

You might be interested in

A store finds that its sales revenue changes at a rate given by S'(t) = −30t2 + 420t dollars per day where t is the number of da

allochka39001 [22]

Answer:

Step-by-step explanation:

Give the rate of change of sales revenue of a store modeled by the equation $S'(t)= -30t^{2} + 420t$ . The Total sales revenue function S(t) can be gotten by integrating the function given as shown;

$\int\limits {S'(t)} \, dt = \int\limits ({-30t^{2}+420t }) \, dt \\S(t) = \frac{-30t^{3} }{3}+\frac{420t^{2} }{2}\\ S(t)= -10t^{3} +210t^{2} \\$

a) The total sales for the first week after the campaign ends (t = 0 to t = 7) is expressed as shown;

$Given\ S(t) = -10t^{3} + 210t^{2}$

$S(0) = -10(0)^{3} + 210(0)^{2}\\S(0) = 0\\S(7) = -10(7)^{3} + 210(7)^{2}\\S(7) = -3430+10,290\\S(7) = 6,860$

Total sales = S(7) - S(0)

= 6,860 - 0

Total sales for the first week = $6,860

b) The total sales for the secondweek after the campaign ends (t = 7 to t = 14) is expressed as shown;

Total sales for the second week = S(14)-S(7)

Given S(7) = 6,860

To get S(14);

$S(14) = -10(14)^{3} + 210(14)^{2}\\S(14) = -27,440+41,160\\S(14) = 13,720$

The total sales for the second week after campaign ends = 13,720 - 6,860

= $6,860

8 0

3 years ago

Which equation best represents the relationship between x and y ?

Klio2033 [76]

Answer:

ummmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

Step-by-step explanation:

7 0

3 years ago

(6-3(cube root of 6)/(cube root of 9)

MatroZZZ [7]

For this case we must simplify the following expression:

$\frac {6-3 \sqrt [3] {6}} {\sqrt [3] {9}}$

Multiplying the numerator and denominator by $(\sqrt [3] {9}) ^ 2$

$\frac {6-3 \sqrt [3] {6}} {\sqrt [3] {9}} * \frac {(\sqrt [3] {9}) ^ 2} {(\sqrt [3] { 9}) ^ 2} =$

We rewrite:

$\frac {\frac {6-3 \sqrt [3] {6}} * (\sqrt [3] {9}) ^ 2} {\sqrt [3] {9} * (\sqrt [3] {9 }) ^ 2} =$

By properties of powers we have that:

$a ^ m * a ^ n = a ^ {m + n}\\\frac {(6-3 \sqrt [3] {6}) * (\sqrt [3] {9}) ^ 2} {(\sqrt [3] {9}) ^ 3} =\\\frac {(6-3 \sqrt [3] {6}) * (\sqrt [3] {9}) ^ 2} {9} =$