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Virty [35]
3 years ago
11

Which values are solutions to the inequality below ×<9​

Mathematics
1 answer:
iragen [17]3 years ago
5 0

Answer: theres not a picture soo idk

Step-by-step explanation:

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The sum of three consecutive numbers is equal to 123. What are the numbers?
Zina [86]

Answer:

40, 41, 42

Step-by-step explanation:

x +x+1+x+2=123

3x+3=123

3x=123-3

3x=120

x=120:3

x=40(the first number

40+1=41 (the second number

40+2=42 (the third number

4 0
3 years ago
How to change from expo to radical form
Natalka [10]

Answer:

See below.

Step-by-step explanation:

If the exponent is a/b  then a is the power and b is the value of the radical.

Example:   x^2/3  =   ∛x².    2 is the power and 3 is the radical.

Answers:

1)  7^2/3 =  ∛7²

5)  √3 = 3^1/2

9)  (∛27)⁴ = 27^4/3 =  3^4 = 81.

5 0
3 years ago
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Whats a synonym (another word) for the word median? *
borishaifa [10]

Another word for median is middle. The median is the middle number of a list of numbers.

7 0
3 years ago
What is the rate of change of the amount earned with respect to hours worked for this function?
Nutka1998 [239]

Answer:

72

Step-by-step explanation:

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5 0
1 year ago
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A growth medium is inoculated with 1,000 bacteria, which grow at a rate of 15% each day. What is the population of the culture 6
ddd [48]

Answer:

The population of bacteria after 6 days is 2,313.06

Step-by-step explanation:

Given as :

The initial population of bacteria = i = 1,000 bacteria

The growth rate of bacteria per day = 15%

Let The population of bacteria after 6 days = f

The time period of growth = 6 days

<u>Now, According to question</u>

The population of bacteria after 6 days = initial population × (1+\dfrac{\textrm rate}{100})^{\textrm time}

Or, f = i × (1+\dfrac{\textrm r}{100})^{\textrm t}

Or, f = 1000 × (1+\dfrac{\textrm 15}{100})^{\textrm 6}

Or, f = 1000 × (1.15)^{6}

Or, f = 1000 × 2.31306

∴  f = 2,313.06

So,The population of bacteria after 6 days = f = 2,313.06

Hence,The population of bacteria after 6 days is 2,313.06  Answer

3 0
3 years ago
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