Find the period of cos (3x+5)+7.

Step-by-step explanation:

Given function is: f(x) = cos(3x + 5) +7 Let the function be periodic with period A, then according to the

property of periodicity,

» f(x+A) = f(x)

→ cos(3(x + A) + 5) + 7 = cos(3x+5)+7

→ cos (3(x+A)+5) = cos(3x + 5)

→ cos(3x + 5+ 3A) = cos(3x + 5)

: We know that, cos(0° (threata) +2π) = cos(0° (threata) )

→ 3A = 2π

» A = 2π / 3

Thus, the period of given function cos(3x+5) = 2π/3

Given function is: f(x) = cos(3x + 5) +7 Let

the function be periodic with period A,

then according to the

property of periodicity,

> f(x+A) = f(x)

→ Cos(3(x + A) + 5) + 7 = cos(3x+5)+7

Step-by-step explanation:

→ COS (3(x+A)+5) = cos(3x + 5)

COS(3x + 5+ 3A) = cos(3x + 5)

: We know that, cos(Oº (threata) +21) =

cos(Oº (threata))

→ 3A = 21

» A = 2/3

Thus, the period of given function

cos(3x+5) = 21/3

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.