Given: δtsp, ts = sp = 10cm, tp = 12cm. find: three altitudes of δtsp.

The altitudes of ΔTSP is SC is 8 unit ,

PA is 5 unit ,

TB is 5 unit .

Step-by-step explanation:

Given as :

The Triangle Δ TSP having side TS , SP , TP

The measure of TS = 12 cm

The measure of SP = 10 cm

The measure of TP = 10 cm

Let The altitude from point T on side PS = TB

The altitude from point S on side PT = SC

The altitude from point P on side TS = PA

The altitude divide the sides

TS = TA + AS

PS = PB + PS

PT = PC + CT

So, From Pythagoras theorem

PA² = PS² - AS²

PA² = 10² - 5²

Or, PA² = 100 - 25

Or, PA =

∴  PA = 5 unit

Again

TB² = TS² - BS²

TB² = 10² - 5²

Or, TB² = 100 - 25

Or, TB =

∴ TB =  5 unit

Similarly

SC² = TS² - TC²

SC² = 10² - 6²

Or, SC² = 100 - 36

Or, SC =

∴ SC = 8 unit

Hence The altitudes of ΔTSP is SC is 8 unit , PA is 5 unit , TB is 5 unit . Answer

go hook jvc hgv cj xk jvc 6756

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source