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First, you'd work out the centre value
It would be the opposite value to what is in the brackets
Thus, the centre value is (2, -4)
Plot this value on a graph
In order to find the radius, you must square root the 25
Thus, the radius would be 5
Plot some points with a radius of 5 from the centre and then draw the circle
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Answer:
285.6 square units
Step-by-step explanation:
The shaded area can be described a a whole circle less a 90° segment, or as a 3/4 circle with a right triangle added. The formula for the segment area is that of a sector less the triangle, so the computation is the same either way.
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<h3>3/4 circle area</h3>
The area of a circle is given by ...
A = πr²
so 3/4 of a circle is 3/4 this value:
3/4 circle area = (3/4)(π)(10²) = 75π . . . . square units
<h3>Triangle area</h3>
The area of the isosceles right triangle is given by ...
A = 1/2s²
A = 1/2(10²) = 50 . . . . square units
<h3>Shaded area</h3>
The shaded area is the sum of these two ...
shaded area = 3/4 circle area + right triangle area
shaded area = 75π + 50 ≈ 285.6 . . . . square units
P = 2L + 2W, keep that in mind (W=width, L=length)
Just fill it in. 82=2(5+2w) + 2w. First solve for w.
82 = 10+ 10w + 2w,
82 = 10 + 12w. Now subtract the 10 from both sides, and get:
72 = 12w, now divide both sides by 12, and get:
6 = w.
W is 6.
Now to solve for L, go back to the problem we had, L = 5 + 2w. Fill in w,
L = 5 + 2(6), and then it is L = 5 + 12.
It is a simple addition equation now.
L = 17, W = 6
If you multiple 13 18=234 then subtract that by 13 18 203 that would be the area so 13 is less than 18
To simplify this, use the FOIL method. First, Outer, Inner, Last.
First: Multiply the first numbers in each set of parentheses.
Outer: Multiply the two outermost numbers in the equation.
Inner: Multiply the two innermost numbers in the equation.
Last: Multiply the last numbers in each set of parentheses.
Then, we put them all together in an equation, from the highest exponent to the lowest, then highest prefix (number before the variable) to lowest.
And simplify.
