The first one is B. I'm sorry, but I don't know the other 2.
To solve this problem you must follow the proccedure below:
a<span>. Find the perimeter and area of the cracker remaining:
The perimeter of a quarter circle is:
P=(</span>πr/2)+2r
P=(πx3 cm/2)+2(3 cm)
P=10.71 cm
The perimeter of <span>the cracker remaining is:
</span><span>
Pt=3 cm+6 cm+3 cm+10.71 cm
Pt=22.71 cm
The area of a quarter circle is:
A=</span>πr²/4
A=π(3 cm)²/4
A=7.06 cm²
<span>
The area of</span><span>of the cracker remaining is:
</span><span>
At=Area of a square-Area of quarter circle
At=L</span>²-(πr²/4)
At=(6 cm)²-(π(3 cm)²/4)
At=28.93 cm²
<span>
b. About how many bites can you get from the entire cracker?</span>
Number of bites=L²/(π(3 cm)²/4)
Number of bites=5
Answer:
Yes this statement is true because 0/3 is 0
Answer:
A is right
B is acute
C is obtuse
D is right
Step-by-step explanation:
90 is right
anything under 90 is acute
over 90 obtuse
The circumference of the circle is 53.41.
