Answer:
The matching is shown below.
Step-by-step explanation:
The given sets are defined as
A = {x | x < 1}
It means all the values of x which are less than 1.
B = {x | x ≥ 5}
It means all the values of x which are greater than or equal to 5.
C = {x | x = 5}
It means all the value of x is 5.
Union of two sets contains all the elements of both sets.
Intersection of two sets contains only common elements of both sets.
The matching is shown below.
Sets Correct value
1. A ∪ B {x | x < 1 or x ≥ 5}
2. A ∪ C {x | x < 1 or x = 5}
3. B ∪ C {x | x ≥ 5}
4. A ∩ B Ø
5. B ∩ C {x | x = 5}
A domain is an input that you put in an equation(x). A range is an output (y) that is the result of a certain input.
Hope this helps and please mark as brainliest!
Answer:
the baser to their question would be 2 since its lined up and thrn it would be 4
Answer:
Explanation:
<u>1) Write the changes in the temperature of the city over the 4 days in an easy way to read.</u>
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<u>2) Write the expression to show the average daily change in temperature.</u>
The average of a set of data is calculated with the expression:
- Average = (sum of the data) / amount of data
Hence, you need to add the four change of temperature data and divide by 4.
The expression is:
- Average = [ 1.34 °C + (- 5 / 7 °C) + ( - 0.75 °C) + 4/9 ° C ] / 4
<u>3) Write the steps to solve the expression:</u>
You can choose between adding fractions or adding the decimal forms of the numbers.
If you choose adding decimals, keep the complete decimals in your calculator, to avoid the accumulation of errors due to rounding. Round only the final result.
These are the steps.
<u>a. Find the decimal forms of the fractions:</u>
<u>b. Add the four data:</u>
- 1.34°C + (- 0.71428 °C) + (-0.75 °C) + 0.44444 °C ≈ 0.32019 °C
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<u>c. Divide the sum by the number of data (4)</u>
- 0.32019 °C / 4 ≈ 0.080004 °C
<u>4. Round to the nearest hundredth</u>
The place of the hundreth is the second after the decimal point, which is 8 in this case:
Answer:
up 2, right 3
Step-by-step explanation:
up 2, to the right 3.
