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DedPeter [7]
3 years ago
5

What is the length of ST¯¯¯¯¯? Enter your answer as a decimal in the box. Round your final answer to the nearest hundredth. in.

Circle A with a tangent and a secant segment. Point S at 11 o clock, point R at 2 o clock, and point Q at 6 o clock are located on the circle. Point T at 2 o clock is located outside of the circle. Tangent S T and secant Q R T is drawn that intersect each other at point T. Segment R T is labeled as 7 inches. Segment R Q is labeled as 23 inches.

Mathematics
2 answers:
allochka39001 [22]3 years ago
8 0

Answer:

_ST = TQ-TR (23-7)

Yanka [14]3 years ago
6 0

Answer:

16 in.

Step-by-step explanation:

23-7=16 hence you cane see that the angles are 16 inches.

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The two values of roots of the polynomial x^{2}-11 x+15 are \frac{11+\sqrt{61}}{2} \text { or } \frac{11-\sqrt{61}}{2}

<u>Solution:</u>

Given, polynomial expression is x^{2}-11 x+15

We have to find the roots of the given expression.

In order to find roots, now let us use quadratic formula.

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Given that x^{2}-11 x+15

Here a = 1, b = -11 and c = 15

On substituting the values we get,

x=\frac{-(-11) \pm \sqrt{(-11)^{2}-4 \times 1 \times 15}}{2 \times 1}

\begin{array}{l}{x=\frac{11 \pm \sqrt{121-60}}{2}} \\\\ {x=\frac{11 \pm \sqrt{61}}{2}} \\\\ {x=\frac{11+\sqrt{61}}{2} \text { or } \frac{11-\sqrt{61}}{2}}\end{array}

Hence, the roots of given polynomial are \frac{11+\sqrt{61}}{2} \text { or } \frac{11-\sqrt{61}}{2}

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