[1 6 9 3] [2 5 8 1] [0 2 2 4] what are the dimensions of the matrix
1 answer:
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You can solve this problem by using the pythagorean theorem.
Where the a- and b-values represent the sides of the triangle. Since the park is a sqaure, each side is equal. This means that both the a-value and the b-value are the same. Now, you must find the c-value.
145^2 + 145^2 = c^2
21,025 + 21,025 = c^2
42,050 = c^2 *now square root each side of the equation.*
205.06 = c
The length of the diagonal line is 205 meters.
Where the a- and b-values represent the sides of the triangle. Since the park is a sqaure, each side is equal. This means that both the a-value and the b-value are the same. Now, you must find the c-value.
145^2 + 145^2 = c^2
21,025 + 21,025 = c^2
42,050 = c^2 *now square root each side of the equation.*
205.06 = c
The length of the diagonal line is 205 meters.
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
- The area of segment ADB.
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
