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velikii [3]
3 years ago
8

Please give answer ofHCF of 15 and 18​

Mathematics
1 answer:
lana66690 [7]3 years ago
6 0

Answer: The prime factorization of 15 is: 3 x 5. The prime factorization of 18 is: 2 x 3 x 3. The prime factors and multiplicities 15 and 18 have in common are: 3. 3 is the gcf of 15 and 18.

Step-by-step explanation:

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No they are not… not the same size
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Y + 1= 4/5(x+4) in standard form is
aleksklad [387]

Answer:

4x - 5y = -11

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Can someone please explain this
S_A_V [24]
Oh. It's rounding. So, for example, depending what you were told to round to, the second problem would be 760-330=430
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11. Kamala borrowed 26400 from a bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at
oee [108]
<h2>Answer </h2>

Amount (A) = P[1 + (r/100)]n

Principal (P) = ₹ 26400

Time period (n) = 2 years 4 months

Rate % (R) = 15% compounded annually

<h3>Steps </h3>

First, we will calculate Compound Interest (C.I) for the period of 2 years

A = P[1 + (r/100)]n

= 26400[1 + (15/100)]²

= 26400[(100/100) + (15/100)]²

= 26400 × 115/100 × 115/100

= 26400 × 23/20 × 23/20

= 26400 × 1.3225

= 34914

C.I. = A - P

= 34914 - 26400

= 8514

Now, we will find Simple Interest (S.I) for the period of 4 months

Principal for 4 months after C.I. for 2 years = ₹ 34,914

<h3>We know that ,</h3>

S.I = PRT/100

Here T = 4 months = 4/12 years = 1/3 years

S.I. for 4 months = (1/3) × 34914 × (15/100)

= (1/3) × 34914 × (3/20)

= 34914/20

= 1745.70

Total interest for 2 years 4 months = 8514 + 1745.70

= 10259.70

Total amount for 2 years 4 months = 26400 + 10259.70

= ₹ 36659.70

<h3>So , the correct answer is ₹ 36659.70 . </h3>

6 0
2 years ago
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Prove that
Dovator [93]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

secx = \frac{1}{cosx} , cosecx = \frac{1}{sinx}

cotx = \frac{cosx}{sinx} , tanx = \frac{sinx}{cosx}

Consider the left side

secA cosecA - cotA

= \frac{1}{cosA} × \frac{1}{sinA} - \frac{cosA}{sinA}

= \frac{1}{cosAsinA} - \frac{cosA}{sinA}

= \frac{1-cos^2A}{cosAsinA}

= \frac{sin^2A}{cosAsinA} ( cancel sinA on numerator/ denominator )

= \frac{sinA}{cosA}

= tanA = right side ⇒ proven

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3 years ago
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