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Fudgin [204]
2 years ago
3

A LIST OF 300 NUMBERS START AT 1.AFTER THAT EVERY NUMBER IS TRIPLED THE NUMBER WHICH IT HAD PRECEDES IT.THE 200TH NUMBER ON THE

LIST IS
A:600
B:900
C:3 to the 199 power
D:3 to the 200 power
Mathematics
1 answer:
katrin2010 [14]2 years ago
5 0
Since each term is a multiple of the previous term, this is a geometric sequence, because there is a common ratio.  (The constant found when dividing any term by the previous term.)  Any geometric sequence can be expressed as:

a(n)=ar^(n-1), where a=initial term, r=common ratio, n=term number:

We are told that a=1 and r=3 so

a(n)=3^(n-1), so the 200th term is:

a(200)=3^199

So your answer is C. <span />
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In triangle FGH, FG = 12.3 cm, GH = 10.6 cm, Measure of angle G = 80 degrees, and Measure of angle H = 55 degrees. If Triangle A
aalyn [17]

Question:

In triangle FGH, FG = 12.3 cm, GH = 10.6 cm, m angle G=80 degrees, and m angle H=55 degrees. If triangle ABC GHF, which statement is true?

(A) AB = 12.3 cm

(B) AB = 10.6 cm

(C) m∠C = 55°

(D) m∠C = 80°

Answer:

Option B:

AB = 10.6 cm

Solution:

The image of the triangle is attached below.

In ΔFGH,

FG = 12.3 cm, GH = 10.6 cm, m∠G = 80° and m∠H = 55°

Option A: AB = 12.3 cm

Given ΔGHF \sim ΔABC.

Corresponding parts of congruence triangles are congruent.

GH = AB = 10.6 cm

So, option A is not true.

Option B: AB = 10.6 cm

Already proved in option A that AB = 10.6 cm

So, option B is true.

Option C: m∠C = 55°

ΔGHF \sim ΔABC

Corresponding parts of congruence triangles are congruent.

m∠G = m∠A = 80°

m∠H = m∠B = 55°

Sum of the angles of a triangle = 180°

⇒ m∠A + m∠B + m∠C = 180°

⇒ 80° + 55° + m∠C = 180°

⇒ m∠C = 180° – 80° – 55°

⇒ m∠C = 45°

So, option C is not true.

Option D: m∠C = 80°

Already proved in option C that m∠C = 45°.

So, option D is not true.

Hence option B is the true statement that AB = 10.6 cm.

7 0
2 years ago
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__________ is the source error that can be avoided by locating questions sensitive to bias and changing or dropping them.
brilliants [131]
I think the correct answer from the choices listed above is option B. Deliberate bias <span> is the source error that can be avoided by locating questions sensitive to bias and changing or dropping them. Hope this answers the question. Have a nice day.</span>
8 0
2 years ago
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In 154 N38 what digit should be replaced to N to make the number divisible by 3,6 and 9?​
JulsSmile [24]

Answer:

N = 6 so that number becomes divisible by 3, 6 and 9.

Step-by-step explanation:

In Number Theory there is a rule of thumb which states that sum of digits of a multiple of 3 equal 3 or a multiple of three. If we know that n = 154N38, then its sum of digits is:

x = 1 + 5+4+N+3+8

x = 21+N (Eq. 1)

We have to determine which digits corresponds to multiples of three, there are four digits:

N = 0

x = 21+0

x = 21 (3\times 7 = 21)

N = 3

x = 21+3

x = 24 (3\times 8 = 24)

N = 6

x = 21+6

x = 27 (3\times 9 = 27)

N = 9

x = 21+9

x = 30 (3\times 10 = 30)

We get the following four distinct options: 154038, 154338, 154638, 154938. Now we find which number is divisible by 6 and 9 by factor decomposition:

154038 = 2\times 3\times 25673

154338 =2\times 3\times 29 \times 887

154638 = 2\times 3\times 3\times 11\times 11\times 71

154938 = 2\times 3\times 7\times 7\times 17\times 31

It is quite evident that N = 6 so that number becomes divisible by 3, 6 and 9.

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<span>The answer is She should have multiplied by 2 instead of dividing by 2.
First one.</span>
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2 years ago
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What is the common ratio of the geometric sequence whose second and fourth terms are 6 and 54, respectively?
Gnom [1K]
The common ratio is 3.
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2 years ago
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