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blsea [12.9K]
3 years ago
14

BRAINLIESTTTT PLEASE ASAP!!!

Mathematics
1 answer:
podryga [215]3 years ago
8 0
I think that its d seems like it might hope this is right
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ANSWER: x = 130°

EXPLANATION: You must add given angles and then subtract from 180°.
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The number of subsets that can be created from the set {1, 2, 3} is:
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Answer:3

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The expression is 12n -4 = -7 solve for n
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What is the length of QR?
Elena L [17]

Answer:

C

Step-by-step explanation:

Since the triangle is right with hypotenuse QR

Use Pythagoras' identity to solve for QR

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3 years ago
Match each series with the equivalent series written in sigma notation
PIT_PIT [208]

Answer:

3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n

4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n

2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n

3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n

Step-by-step explanation:

Given

See attachment for complete question

Required

Match equivalent expressions

Solving (a):

3 + 12 + 48 + 192 + 768

The expression can be written as:

3 \to 3*4^{0 --- 0

12 \to 3 * 4^{1 ---- 1

48 \to 3 * 4^{2 --- 2

192 \to 3 * 4^{3 ---- 3

768 \to 3 * 4^{4 ---- 4

For the nth term, the expression is:

Term = 3 * 4^{n ---- n

So, the summation is:

3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n

Solving (b):

4 + 32 + 256 + 2048 + 16384

The expression can be written as:

4 \to 4 * 8^0 --- 0

32 \to 4 * 8^1 ---- 1

256 \to 4 * 8^2 --- 2

2048 \to 4 * 8^3 ---- 3

16384 \to 4 * 8^4 ---- 4

For the nth term, the expression is:

Term \to 4 * 8^n ---- n

So, the summation is:

4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n

Solving (c):

2 + 6 + 18 + 54 + 162

The expression can be written as:

2 \to 2 * 3^0 --- 0

6 \to 2 * 3^1 ---- 1

18 \to 2 * 3^2 --- 2

54 \to 2 * 3^3 ---- 3

162 \to 2 * 3^4 ---- 4

For the nth term, the expression is:

Term \to 2 * 3^n ---- n

So, the summation is:

2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n

Solving (d):

3 + 15 + 75 + 375 + 1875

The expression can be written as:

3 \to 3 * 5^0 --- 0

15 \to 3 * 5^1 ---- 1

75 \to 3 * 5^2 --- 2

375 \to 3 * 5^3 ---- 3

1875 \to 3 * 5^4 ---- 4

For the nth term, the expression is:

Term \to 3 * 5^n ---- n

So, the summation is:

3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n

5 0
2 years ago
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