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3 years ago

Find a number less than 100 that is increased by one-fifth of its value when its digits are reversed.

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

7 0

Lets say the number is ab. Its value is 10a + b

When it is reversed it is 10b + a = 12a+1.2b (from the condition its value should increase 0.2 times).

11 a = 8.8b

a/b = 8.8/11 = 0.8/1 = 8/10 = 4/5 ( we do this because a and b should be natural numbers less than 10).

answer is 45.

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Natasha_Volkova [10]

The correct answers are:
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2) tan(x) = $\sqrt{3}$

Explanation:
Given:
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Step 1:
Since, according to the Trigonometric identity:
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Step 2:
Plug in the value of cos(x) in equation (1):
$sin^2(x) + ( \frac{1}{2} )^2 = 1 \\ sin^2(x) + \frac{1}{4} = 1 \\ sin^2(x) = \frac{3}{4}$

Step 3:
Take square-root on both sides:
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Now to find the tan(x), we would use the following formula:

tan(x) = $\frac{sin(x)}{cos(x)}$ --- (2)

Plug in the values of sin(x) and cos(x) in equation (2):
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Hence tan(x) = $\sqrt{3}$

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3 years ago

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AysviL [449]

Answer:

Open circle. Points right

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2 years ago

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creativ13 [48]

Answer:

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Step-by-step explanation:

The first term of the sequence is a1 = -12.

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d = -13 -(-12) = -1

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an = a1 +d(n -1)

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