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Leno4ka [110]
2 years ago
10

Find the area of the triangle below. Round your answer to the nearest tenth.

Mathematics
1 answer:
zaharov [31]2 years ago
7 0

Answer:

22in

Step-by-step explanation:

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Round 958,622 to the nearest hundred <br>​
HACTEHA [7]

Answer:

958,600

Step-by-step explanation:

3 0
3 years ago
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If two lines are parallel, corresponding angles are _________.
jolli1 [7]
B. congruent, im pretty sure
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Stephen rode his bike 23.4865 miles On monday and 38.243 miles on Tuesday. How many miles did he ride in all
professor190 [17]

Answer:

61.7295 (61.7) miles

If I'm reading this correctly, this is just a simple addition equation. You'd add up 23.4865 and 38.243 to get 61.7295 miles, or 61.7 miles.

Hope I helped! ☺

6 0
3 years ago
Find out net profit after tax from the following information:
WARRIOR [948]

Answer:

Net Profit after tax Rs 15,000

Step-by-step explanation:

The computation of the net profit after tax is shown below:

Gross profit Rs. 1,25,000

Less:

Selling and distribution expenses Rs. 21,000

General and administrative expenses Rs. 75,000

Interest on loan Rs. 5,000

Gain on sale of plant Rs. 4,000

Profit before tax Rs 20,000

Less: income tax expense at 25% of Rs 20,000 Rs 5000

Net Profit after tax Rs 15,000

3 0
3 years ago
Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction
Oliga [24]

The base case is the claim that

\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}

which reduces to

\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86

which is true.

Assume that the inequality holds for <em>n</em> = <em>k </em>; that

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}

By the induction hypothesis,

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}

Now compare this to the upper bound we seek:

\dfrac{2k+1}{k+1}  > \dfrac{2k+2}{k+2}

because

(2k+1)(k+2) > (2k+2)(k+1)

in turn because

2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0

6 0
2 years ago
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