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crimeas [40]
3 years ago
11

What is the hcf of 45 and 100

Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
6 0

Answer:

45 100

/ \ / \

3 × 15 2×50

/ \ / \

3× 5 2×25

/ \

5×5

45 : 3×3×5

100: 2×2×5×5

common : 3×5

15

HCF: 15

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Please help me ! Reduced to lowest terms
Mandarinka [93]

Answer:

Step-by-step explanation:

7) 2ab + 4ac = 2*a*b + 2 *2 *a*c

                     = 2a(b + 2c)

\dfrac{2ab^{2}c}{2ab+4ac}=\dfrac{2*a* b^{2}*c}{2a(b+2c)}\\\\\\\text{2a in the denominator and numerator get cancelled}\\\\ = \dfrac{b^{2}c}{b+2c}

8) ac - a²c² = c( a - a²c)

bc - abc     = c(b - ab)

\dfrac{ac - a^{2}c^{2}}{bc-abc^{2}}=\dfrac{c(a-a^{2}c)}{c(b-ab)}\\\\

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9) b²+ 4b - 5 = b² + 5b - 1b - 5

                    = b(b + 5) - 1(b + 5)

                    = (b +5)(b - 1)

b² + 8b  + 15 = b² + 5b + 3b + 15

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\dfrac{b^{2}+4b-5}{b^{2}+8b+15}=\dfrac{(b+5)(b-1)}{(b+5)(b+3)}\\\\\\

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2 years ago
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Arte-miy333 [17]

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3 years ago
A spherical hot air balloon has a diameter of 55 feet when the balloon is inflated the radius increases at a rate of 1.5 feet pe
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Answer: 46.90mins

Step-by-step explanation:

The given data:

The diameter of the balloon = 55 feet

The rate of increase of the radius of the balloon when inflated = 1.5 feet/min.

Solution:

dr/dt = 1.5 feet per minute = 1.5 ft/min

V = 4/3·π·r³

The maximum volume of the balloon

= 4/3 × 3.14 × 55³

= 696556.67 ft³

When the volume 2/3 the maximum volume

= 2/3 × 696556.67 ft³

= 464371.11 ft³

The radius, r₂ at the point is

= 4/3·π·r₂³

= 464371.11 ft³

r₂³ = 464371.11 ft³ × 3/4

= 348278.33 ft³

348278.333333

r₂ = ∛(348278.33 ft³) ≈ 70.36 ft

The time for the radius to increase to the above length = Length/(Rate of increase of length of the radius)

The time for the radius to increase to the

above length

Time taken for the radius to increase the length.

= is 70.369 ft/(1.5 ft/min)

= 46.90 minutes

46.90mins is the time taken to inflate the balloon.

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What fraction of a given sample of radioactive nuclide remains after four half lives?
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