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irinina [24]
4 years ago
11

Help! Simplify (see photo) pls explain If you don’t know pls don’t answer thanks

Mathematics
1 answer:
Temka [501]4 years ago
7 0

Answer:

-8 \sqrt{6}

Step-by-step explanation:

4i\sqrt{-24} \\4i\sqrt{-1} *\sqrt{24} \\4i*i*\sqrt{4}*\sqrt{6}\\-4 * 2*\sqrt{6} \\-8 \sqrt{6}

.

\sqrt{-1} =i \\i*i = -1

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How much ribbon is required to wrap once around a basketball present? Basketballs are 78 cm in diameter.
Brut [27]

Answer:

C = 78 pi cm

Letting pi = 3.14

C = 244.92 cm

Letting pi =  pi button

C = 245.044227 cm

Step-by-step explanation:

We want to find the circumference

C = pi *d

C = pi *78

C = 78 pi cm

Letting pi = 3.14

C = 244.92 cm

Letting pi =  pi button

C = 245.044227 cm

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c

Answer:

Step-by-step explanation:

8 0
3 years ago
PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. the length TP in cm is:
kolezko [41]
<h3><u>Given</u><u> </u><u>:</u><u>-</u></h3>
  • PQ = 8cm
  • Radius = 5cm
  • Two Tangents = P & Q.
<h3><u>Construction</u><u> </u><u>:</u><u>-</u></h3>
  • Join OT.
<h3><u>⟼</u><u> </u><u>Solution</u><u> </u><u>:</u><u>-</u></h3>

Here, ΔTPQ is isosceles and TO is the angle bisector of ∠PTO.

[∵ TP=TQ = Tangents from T upon the circle]

⠀⠀⠀⠀⠀⠀⠀⠀∴ OT⊥PQ

  • So, PR = RQ = 4cm.

⠀⠀⠀

___________________________________________

\longmapsto{}By Applying Pythagoras Theorem in ∆OPR :

OR = √OP² - PR²

OR = √5² - 4²

OR = 3cm

__________________________________________

Now,

\leadsto∠TPR + ∠RPO = 90° (∵TPO=90°)

\leadsto∠TPR + ∠PTR (∵TRP=90°)

<u>\leadsto</u><u>∴ ∠RPO = ∠PTR</u>

⠀⠀

<u>∴ Right triangle TRP is similar to the right </u><u>triangle</u><u> </u><u>PR</u><u>O</u><u>.</u> [By A-A Rule of similar triangles]

⟼\frac{TP}{PO}  =  \frac{RP}{RO}

⟼\frac{TP}{5}  =  \frac{4}{3}

⟼TP=  \frac{20}{3}

<h3>Hence you got your answer here. </h3>

⠀⠀⠀⠀⠀

<h2>-MissAbhi</h2>

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2 years ago
Find the quotient:<br> 7/8
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Answer:

c.) 2 1/3

Step-by-step explanation:

i know

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