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luda_lava [24]
3 years ago
5

Find the distance between the points given. (-1, -1) and (1, 3)

Mathematics
1 answer:
ra1l [238]3 years ago
5 0

Answer:

Step-by-step explanation:

DISTANCE FORMULA

D=(SQRT(X2-X1)^2+(Y2-Y1)^2)

SQRT(1-(-1))^2+(3-(-1)^2)

SQRT((2)^2+(4)^2)

SQRT(4+16)

SQRT(20)

2SQRT(5)

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Simplify each exponential expression using the properties of exponents and match it to the correct answer.
saveliy_v [14]

1) (2\times3^-2)^3 (5\times3^2)^2 / (3^-2)(5\times2)^2 = 2

2) (3^3) (4^0)^2 (3\times2)^-3 (2^2) = 1/2

3) (3^7\times4^7) (2\times5)^-3 (5)^2 / (12^7) (5^-1) (2^-4) = 2

4) (2.3)^-1 (2^0) / (2.3)^-1 = 1

<u>Step-by-step explanation</u>:

Step 1 :

(2\times3^-2)^3 (5\times3^2)^2 / (3^-2)(5\times2)^2

⇒ (2^3) (3^-6) (5^2) (3^4) / (3^-2) (10^2)

⇒ (2.2^2.5^2) (3^-6.3^4) / (3^-2) (10^2)

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Step 2 :

(3^3) (4^0)^2 (3\times2)^-3 (2^2)

Any number with power 'zero' is 1.

⇒ (3^3) (1^2) (3^-3) (2^-3) (2^2)

⇒ (2^(-3+2))

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Step 3 :

(3^7\times4^7) (2\times5)^-3 (5)^2 / (12^7) (5^-1) (2^-4)

⇒ (12^7) (2^-3) (5^-3) (5^2) / (12^7) (5^-1) (2^-4)

⇒ (12^7) (2^-3) (5^-3) (5^2) / (12^7) (5^-1) (2^-4)

⇒ (2^-3) (5^(-3+2)) / (5^-1) (2^-4)

⇒ (2^-3) (5^-1) / (5^-1) (2^-4)

⇒ 1 / (2^-1)

⇒ 2

Step 4 :

(2.3)^-1 (2^0) / (2.3)^-1

⇒ (2^0)

⇒ Any number with power 'zero' is 1

⇒ 1

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Answer:

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Step-by-step explanation:

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