So in order for us to know the area of the square that is not covered by the circle, we need to find first both the areas of the square and the circle.
So for the area of the square it is A = sxs. And for the circle is A = pi*r^2.
Let us find the area of the square first given that the side is 3 inches.
So A = 3*3
A = 9 square inches.
Next is the area of the circle. Since the center of the circle is the same with the center of the square, the radius would be 1.5.
SO, A = (3.14)(1.5)^2
A = 2.25 (3.14)
A = 7.065 square inches.
Next, we deduct the area of circle from area of square and the result would be 1.935 <span>in². So the answer for this would be option B.
Hope this answer helps.</span>
Answer:

Step-by-step explanation:
<u>Step 1: Find the square root of -1</u>
<u />

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Answer: 
The given expression is :
2xx+3-5(x-2)
We know that, 

Aleena expands 2xx+3-5(x-2) as 2x²+3-5x-10. She is mistaken because the sign before 10 should be +10 instead of -10. Hence, the correct expanded form is 2x²+3-5x+10. Hence, this is the required solution.
Answer:sorry for the delate
Step-by-step explanation:
9514 1404 393
Answer:
x = 1 or 5
Step-by-step explanation:
The notion of "cross-multiplying" is the idea that the numerator on the left is multiplied by the denominator on the right, and the numerator on the right is multiplied by the denominator on the left. This looks like ...

Then the solution proceeds by eliminating parentheses, and solving the resulting quadratic equation.

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<em>Comment on "cross multiply"</em>
Like a lot of instructions in Algebra courses, the idea of "cross multiply" describes <em>what the result looks like</em>. It doesn't adequately describe how you get there. The <em>one and only rule</em> in solving Algebra problems is "<em>whatever is done to one side of the equation must also be done to the other side of the equation</em>." If you multiply one side by one thing and the other side by a different thing, you are violating this rule.
What looks like "cross multiply" is really "<em>multiply by the product of the denominators</em> and cancel like terms from numerator and denominator." Here's what that looks like with the intermediate steps added.
