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Darya [45]
3 years ago
15

Pls help with this and show work on piece of paper

Mathematics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0
2 because 12x3=36 and you couldn’t ride that many bc u wouldn’t have enough $$$ so you have to ride two
Lapatulllka [165]3 years ago
5 0

Answer:

????

i dont know

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Solve: VII multiplied by IX. Show your answer in standard form. Standard Number 1 Roman Numeral 5 V 10 50 100 500 1,000 X-003 A)
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63

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Which is the best estimation of
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The answer to this would be:

B) between 2.6 and 2.7

Step-by-step explanation:

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Unit 2 (12 points)
klio [65]
1+3=4

A counterexample is an a example that proves the statement false.

1+3 are not even numbers but they equal an even one, so it just proved the statement wrong.
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What is 7 divided by 630
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90

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3 years ago
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David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below. GFC of 80, 32, and 48: 16 GCF o
BlackZzzverrR [31]

Answer: The correct statements are

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

David applied the distributive property.

Step-by-step explanation:

GCF = Greatest common factor

1) GCF of coefficients : (80,32,48)

80 = 2 × 2 × 2 × 2 × 5

32 = 2 × 2 × 2 × 2 × 2

48 = 2 × 2 × 2 × 2 × 3

GCF of coefficients : (80,32,48) is 16.

2) GCF of variables :(b^4,b^2,b^4)

b^4= b × b × b × b

b^2 = b × b

b^4 =b × b × b × b

GCF of variables :(b^4,b^2,b^4) is b^2

3) GCF of c^3 and c: c is not the GCF of the polynomial. The variable c is not common to all terms, so a power of c should not have been factored out.

4) 80b^4-32b^2c^3+48b^4c

=16b^2(5b^2-2c^3+3b^2c)

David applied the distributive property.

7 0
3 years ago
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