B is the answerr okkkkkkk
The paper is 16 cm by 34 cm to start with, so we started with area
.. (16 cm)*(34 cm) = 544 cm^2
A circle of diameter 16 cm has been removed. The area of that circle is
.. π*(d/2)^2 = 3.14*(16 cm/2)^2 = 200.96 cm^2
Then the area of the paper that remains is
.. 544 cm^2 -200.96 cm^2
.. = 343.04 cm^2
You should google the worksheet that this is from, easier results.
What is the interquartile range of the sequence 5,5,8,8,13,14,16,16,19,22,23,27,31 ?
Romashka-Z-Leto [24]
Answer:
The Interquartile range is 10.
Step-by-step explanation:
First, we will need to find the mean, the mean of this sequence is 16, you will now need to find quartile 1 and quartile 3. Quartile 1 is 13, and quartile 3 is 23. Lastly, subtract Quartile 3 and Quartile 1 will be the answer.
So, 23-13=10
The Answer will be 10, the interquartile range is 10.
Hope this helps!
Answer:
6c³
Step-by-step explanation:
18c³ = 2 × 3 × 3 × c × c × c
24c³ = 2 × 2 × 2 × 3 × c × c × c
Now we show the common factors in bold:
18c³ = 2 × 3 × 3 × c × c × c
24c³ = 2 × 2 × 2 × 3 × c × c × c
The common factor are:
2, 3, c, c, c
GCF = 2 × 3 × c × c × c = 6c³
