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lbvjy [14]
4 years ago
6

If , /, S is the center of the circle, and OM = 18, find NT. Round the answer to the hundredths place.

Mathematics
2 answers:
adell [148]4 years ago
8 0

Answer:

You are requested to kindly upload the image.

Step-by-step explanation:

Kruka [31]4 years ago
3 0
If my attached image is correct, then based on my computation NT is equal to C. 12.05

S is the center of the circle. OM = 18.

Based on the attached image, I concluded the following facts:
1) RM is the diameter
2) OM = NM ; NM = 18
3) ST and SU are radii so ST = SU
4) Q is the midpoint of SU ; QU = 8
5) P is the midpoint of ST. PT = QU thus PT = 8
6) P is the midpoint of NM thus NP = 9

Points NPT creates a right triangle. NT is the hypotenuse. We will use the Pythagorean theorem to solve for NT.

a² + b² = c²
9² + 8² = c²
81 + 64 = c²
145 = c²
√145 = √c²
12.0415 = c

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3 years ago
plz help i need helpplz help i need helpplz help i need helpplz help i need helpplz help i need helpplz help i need helpplz help
Volgvan

Answer:

the answer should be 9

Step-by-step explanation:

(256 * 3^{-5} * 6^{0} )^{-2} * (\frac{3^{-2}}{2^{3}} )^{4} * 2^{28}

(256*\frac{1}{3^{5}} *6^{0})^{-2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}

(256*\frac{1}{243} *6^{0})^{-2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}

(\frac{256}{243} *6^{0})^{-2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}

(\frac{256}{243})^{-2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}

(\frac{243}{256})^{2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}

\frac{59049}{65536}*(\frac{1}{2^{3}*3^{2}})^{4}*2^{28}

\frac{59049}{65536}*(\frac{1}{8*9})^{4}*2^{28}

\frac{59049}{65536}*(\frac{1}{72})^{4}*2^{28}

\frac{59049}{65536}*\frac{1}{26873856}*2^{28}

\frac{9}{65536}*\frac{1}{4096}*2^{28}

\frac{9}{65536*4096}*2^{28}

\frac{9}{268435456}*268435456

9

6 0
3 years ago
Read 2 more answers
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