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3 years ago

2x < - x + 20 + 7 I don't get how to answer this question

Mathematics

1 answer:

dimulka [17.4K]3 years ago

6 0

ANSWER

X < 9

Solve

2x < - x + 20 + 7

2x < - x + 27 change the signs

2x + x < 27 add the like terms

3x < 27 divide by 3

x < 9

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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 $\times$ 3^-2)^3 (5 $\times$ 3^2)^2 / (3^-2)(5 $\times$ 2)^2 = 2

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

3) (3^7 $\times$ 4^7) (2 $\times$ 5)^-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 $\times$ 3^-2)^3 (5 $\times$ 3^2)^2 / (3^-2)(5 $\times$ 2)^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)

⇒ (2)(10^2) (3^-2) / (3^-2) (10^2)

⇒ 2

Step 2 :

(3^3) (4^0)^2 (3 $\times$ 2)^-3 (2^2)

Any number with power 'zero' is 1.

⇒ (3^3) (1^2) (3^-3) (2^-3) (2^2)

⇒ (2^(-3+2))

⇒ 2^-1 = 1/2

Step 3 :

(3^7 $\times$ 4^7) (2 $\times$ 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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2 years ago

The difference of any two negative integers is a negative integer

Vinvika [58]

Answer: look at the bottom

Step-by-step explanation:

Difference of two negative numbers may or may not be negative. It depends on the numbers taken as well as on the sequence in which they are taken. Then, a-b=b-a=0, (which is neither negative nor positive..!!!) So difference of two negative numbers may be positive, negative or even zero.

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2 years ago

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Gala2k [10]

Answer:

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