Answer:
- The shaded region is 9.83 cm²
Step-by-step explanation:
<em>Refer to attached diagram with added details.</em>
<h2>Given </h2>
Circle O with:
- OA = OB = OD - radius
- OC = OD = 2 cm
<h2>To find</h2>
<h2>Solution</h2>
Since r = OC + CD, the radius is 4 cm.
Consider right triangles OAC or OBC:
- They have one leg of 2 cm and hypotenuse of 4 cm, so the hypotenuse is twice the short leg.
Recall the property of 30°x60°x90° triangle:
- a : b : c = 1 : √3 : 2, where a- short leg, b- long leg, c- hypotenuse.
It means OC: OA = 1 : 2, so angles AOC and BOC are both 60° as adjacent to short legs.
In order to find the shaded area we need to find the area of sector OADB and subtract the area of triangle OAB.
Area of <u>sector:</u>
- A = π(θ/360)r², where θ- central angle,
- A = π*((mAOC + mBOC)/360)*r²,
- A = π*((60 + 60)/360))(4²) = 16.76 cm².
Area of<u> triangle AO</u>B:
- A = (1/2)*OC*(AC + BC), AC = BC = OC√3 according to the property of 30x60x90 triangle.
- A = (1/2)(2*2√3)*2 = 4√3 = 6.93 cm²
The shaded area is:
- A = 16.76 - 6.93 = 9.83 cm²
<h3>
Answer: 14x - 8</h3>
=======================================================
Explanation:
I'll use the quadratic formula to find the roots or x intercepts. This slight detour allows us to factor without having to use guess-and-check methods.
The equation is of the form ax^2+bx+c = 0
This leads to...
Now use those roots to form these steps
Refer to the zero product property for more info.
Therefore, the original expression factors fully to (4x-5)(3x+1)
Use the FOIL rule to expand it out and you should get 12x^2-11x-5 again.
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We did that factoring so we could find the side lengths of the rectangle.
I'm using the fact that area = length*width
- L = length = 4x-5
- W = width = 3x+1
The order of length and width doesn't matter.
From here, we can then compute the perimeter of the rectangle
P = 2(L+W)
P = 2(4x-5+3x+1)
P = 2(7x-4)
P = 14x - 8
Answer:
Your answer is -5
Step-by-step explanation:
On paper;
It's simple arithmetic. The Answer is 604661760000000000000000000000000000000000000000000000000000000000000000000000000000 and in scientific notation
60466176 x 10⁸³
<span>I think you know by now that I strongly encourage everyone to shoot a proper round and whatever the score is, to submit it to our Records Officer, Giles Conn. Think of it as an annual competition (a) to wear him out, and (b) to see if we can altogether, beat last year's tally. Also, for the outdoor season rounds, you can have a go at achieving the St Wilfred trophy. I've won it 3 years running (last year jointly with Terry Skinner), but they wouldn't let me keep it this time, sadly.</span>
