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2 years ago

A 33 gram sample of a substance that's

Mathematics

1 answer:

Mazyrski [523]2 years ago

6 0

9514 1404 393

Answer:

6.2

Step-by-step explanation:

We presume your "k-value" is the k in the exponential decay term ...

e^(-kt) . . . where t is the number of time units

This is 1/2 when ...

ln(1/2) = -kt

t = ln(1/2)/(-k) = ln(2)/k

t = 0.69315/0.1124 ≈ 6.2

The half life is about 6.2 time units.

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A sprinter travels a distance of 100 m in a time of 11.21 seconds.

sergiy2304 [10]

Answer:

distance travelled by sprinter = 100 m

Time taken = 11.21sec

∵speed = distance/time taken

= 100/9.83

= 10.17 m/s

Now we know, 1 km/h = 1000/3600 m/s

so, 1 m/s = 3600/1000 km/h = 18/5 km/h

∴ 10.17 m/s = 10.17 × 18/5 = 10.17 × 18/5 = 36.612 km/h

Hence, average speed = 36.612 km/h

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Please help me with this sum

barxatty [35]

Answers:

781 × 8 = 6248

231 × 7 = 1617

421 × 6 = 2526

531 × 5 = 2655

681 × 2 = 1362

951 × 4 = 3804

741 × 3 = 2223

361 × 3 = 1083

931 × 7 = 6517

821 × 2 = 1642

8 0

2 years ago

Root 3 cosec140° - sec140°=4 prove that

Lorico [155]

Answer:

Step-by-step explanation:

We are to show that $\sqrt{3} cosec140^{0} - sec140^{0} = 4\\$

Proof:

From trigonometry identity;

$cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}$

$\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\$

From trigonometry, 2sinAcosA = Sin2A

$= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\= \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2 \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}$