The figure shows secants ps and pt and tangent qr intersecting at point p. triangle stp is an isosceles triangle with legs ps and pt find x and y

x°=70° and y°=40°

Step-by-step explanation:

We are given that  secants PS and PT and tangent QR intersecting at point P. Triangle STP is an isosceles triangle with legs PS and PT.

Now, we know that , thus

⇒2m∠PST=m(TP)

⇒2m∠PST=140°

⇒m∠PST=70°

Since, the ΔPST is an isosceles triangle with SP=TP, therefore

∠PST=∠PTS=70°

Now,in ΔPST, applying the angle sum property, we have

∠PST+∠PTS+∠SPT=180°

⇒70°+70°+∠SPT=180°

⇒140°+∠SPT=180°

⇒∠SPT=180°-140°

⇒∠SPT=40°

therefore, y°=40°

Now, using  the  theorem that angle made with the tangent is equal to the interior angle of the inscibed triangle, thus

x°=∠PTS

⇒x°=70°

Thus, the value of x and y is: 70 and 40 respectively.

answer:

step-by-step explanation:

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! : )

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