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Veseljchak [2.6K]
3 years ago
8

HELP NEEDED QUICK what is pqr to the nearest degree?

Mathematics
1 answer:
AleksAgata [21]3 years ago
5 0

Answer:

your mom lol sooooooooooooooo

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Subtract 4 from 1/6 in parentheses
Rom4ik [11]
I am not sure if what I am about to explain is correct but I hope it is somewhere along the correct path

To solve the problem use the following steps below....

Step 1: Take your numbers and place them in parentheses

(4-1/6)

Step 2: Turn your whole number,4, into a fraction

6/6=1
6/6x4=24/6

Step 3: Place your new fraction back into the problem

(24/6-1/6)

Step 4: Use the operation,subtraction, and minus 1/6 from 24/6

(23/6)
24/6 minus 1/6=23/6

Step 5(is optional): Turn 23/6 into a mixed fraction by dividing 23 by 6

23<span>÷6= 3 5/6

Step 6(is optional): put your answer into a final statement

(4-1/6)=23/6=3 5/6

Answer: 3 5/6

I hope this helps and I apologize in advance if any false information was given.
</span>
3 0
3 years ago
Ted walked 5/8 mile to his soccer game then he walked 3/8 mile to his friends house which distance is shorter
tatiyna
3/8 because he walked longer when he went to his soccer game
5 0
3 years ago
What is the slope-intercept form of the equation of this line? y=−13x+4 y−5=−13(x+3) y=−3x+4 y−4=−13x The figure shows the graph
lisov135 [29]

Answer:

isfoijsofjsofdj

Step-by-step explanation:

3 0
2 years ago
what are the minimum first quartile median third quartile and the maximum of the data set? 9,20,4,18,4,18,20,9
den301095 [7]
<u>Answers</u>
1. Minimum = 4
2. First quartile = 6.5
3. Median = 13.5
4. Third quartile = 19
5. Maximum = 20

<u>Explanation</u>
To calculate the measure of central tendency, you first arrange the set of the data in ascending order.  
The set of data given will be;
4, 4, 9, 9, 18, 18, 20, 20.

Part 1:
The minimum value of the data is 4.

Part 2:
The first quatile is the median of the lower half which is comprised by:
4, 4, 9, 9

1st quartile = (4+9)÷2
                  = 13÷2
                  = 6.5

Part 3:
Median of the data is;

Median = (9+18)÷2
              =27÷2
            = 13.5

Part 4:
3rd quartile is the median of the upper half which comprises of; 
18, 18, 20, 20.

3rd quartile = (18+20)÷2
                  = 48÷2
                   = 19

Part 5
The maximum of the set of data is 20.
7 0
3 years ago
Solve for x: Log4(x+8)=3
viktelen [127]
If log_{4}(x+8)=3
Then 4^3=x+8
64=x+8
64-8=x
56=x
3 0
3 years ago
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