at first, theta is in basic terms a place holder, you ought to place any image there to symbolize a variable, consisting of x, or z. there are a number of strategies, attempt multiplying the two facets by potential of a million sinx LS = cosx RS = (a million-sinx)(secx tanx) RS = secx tanx sinxsecx -sinxtanx on account that secx = a million/cosx and tanx = sinx/cosx weve: RS = secx tanx tanx sin^2x secx RS = secx (a million-sin^2x) RS = secx(cos^2x) (by potential of the pythagorean identification: sin^2x cos^2x = a million) RS = (a million/cosx)(cos^2x) RS = cosx for this reason RS = LS, so the unique identification is authentic.

first of all, theta is largely a place holder, you could desire to place any image there to signify a variable, including x, or z. there are various procedures, attempt multiplying the two sides via a million sinx LS = cosx RS = (a million-sinx)(secx tanx) RS = secx tanx sinxsecx -sinxtanx on the grounds that secx = a million/cosx and tanx = sinx/cosx weve: RS = secx tanx tanx sin^2x secx RS = secx (a million-sin^2x) RS = secx(cos^2x) (via the pythagorean id: sin^2x cos^2x = a million) RS = (a million/cosx)(cos^2x) RS = cosx for this reason RS = LS, so the unique id is actual.

first of all, theta is largely a place holder, you could desire to place any image there to signify a variable, including x, or z. there are various procedures, attempt multiplying the two sides via a million sinx LS = cosx RS = (a million-sinx)(secx tanx) RS = secx tanx sinxsecx -sinxtanx on the grounds that secx = a million/cosx and tanx = sinx/cosx weve: RS = secx tanx tanx sin^2x secx RS = secx (a million-sin^2x) RS = secx(cos^2x) (via the pythagorean id: sin^2x cos^2x = a million) RS = (a million/cosx)(cos^2x) RS = cosx for this reason RS = LS, so the unique id is actual.

B. A(r) = pi* r^2

(the area of the circular base is pi times radius squared)

A. r(x) = x/2

(the radius is half the side of the square base)

A. r(x) = x/2

(the radius is half the side of the square base)

r(x) = x

A. Write the radius r of the tank as a function of the length x of the sides of the square.

r(x) = x/2

B. Write the area A of the circular base of the tank as a function of the radius r

A(r) = r

A. Write the radius r of the tank as a function of the length x of the sides of the square.

r(x) = x/2

C. Find and interpret (A o r)(x)

A(r(x)) = (x/2) = x/4

A. r(x) = x/2

(the radius is half the side of the square base)

B. A(r) = pi* r^2

(the area of the circular base is pi times radius squared)

C. (A o r)(x) = pi*(x/2)^2 = (pi*x^2)/4

(plug your answer to part A in for r in your answer to part B.)

A. r(x) = x/2

(the radius is half the side of the square base)

B. A(r) = pi* r^2

(the area of the circular base is pi times radius squared)

This post is last updated on hrtanswers.com at Date : 1st of September – 2022