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ICE Princess25 [194]
3 years ago
12

Find the distance of a chord 8cm long from the centre of a circle radius 5cm​

Mathematics
1 answer:
KiRa [710]3 years ago
4 0

Answer:

3 cm

Step-by-step explanation:

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No spoliation skdnfnd
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17 Find the volume of the figure below. (1 Point) V = luh​
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Please help me with 5
Fantom [35]
Bluegill
3/5 × 1100
660

catfish
70% × (1100-660)
70% × 440
308

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1100 - (660 + 308) = 132
6 0
3 years ago
Suppose that the position of one particle at time is given by x1=3sin t, y1 = 2 cos t, 0 ≤ t ≤ 2π and the position of a second p
Mashcka [7]

Answer:

there is no collision between the particles

Step-by-step explanation:

for the first particle

x1=3sin t, y1 = 2 cos t, 0 ≤ t ≤ 2π

for the second particle

x2 = -3 + cos t, y2 = 1 + sin t, 0 ≤ t ≤ 2π

then for the collision

x1=x2 → 3*sin t = -3 + cos t → sin t= -1 + (cos t)/3→ 1+ sin t = (1/3)cos t  

y1=y2 → 1 + sin t = 2 cos t → (1/3)cos t  = 2 cos t →(1/3) = 2

since 1/3 ≠ 2 there is no collision between the particles

6 0
3 years ago
Prove the following statement.
jolli1 [7]
<h2>Step-by-step explanation:</h2>

As per the question,

Let a be any positive integer and b = 4.

According to Euclid division lemma , a = 4q + r

where 0 ≤ r < b.

Thus,

r = 0, 1, 2, 3

Since, a is an odd integer, and

The only valid value of r = 1 and 3

So a = 4q + 1 or 4q + 3

<u>Case 1 :-</u> When a = 4q + 1

On squaring both sides, we get

a² = (4q + 1)²

   = 16q² + 8q + 1

   = 8(2q² + q) + 1

   = 8m + 1 , where m = 2q² + q

<u>Case 2 :-</u> when a = 4q + 3

On squaring both sides, we get

a² = (4q + 3)²

   = 16q² + 24q + 9

   = 8 (2q² + 3q + 1) + 1

   = 8m +1, where m = 2q² + 3q +1

Now,

<u>We can see that at every odd values of r, square of a is in the form of 8m +1.</u>

Also we know, a = 4q +1 and 4q +3 are not divisible by 2 means these all numbers are odd numbers.

Hence , it is clear that square of an odd positive is in form of 8m +1

3 0
3 years ago
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