It's 2/11 because all the others round to 1/2.
Step-by-step explanation:
235•0.1=23.5
235•0.01=2.35
876•0.01=87.6
876•0.001=8.76
Answer:
Is here a pic that goes with it?
Step-by-step explanation:
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
Answer
Find out the length of OP .
To prove
As given
In △JKL, JO=44 in.
Now as shown in the diagram.
JP , MK, NL be the median of the △JKL and intresection of the JP , MK, NL be O .
Thus O be the centroid of the △JKL .
The centroid divides each median in a ratio of 2:1 .
Let us assume x be the scalar multiple of the OP and JO .
As given
JO = 44 in
2x = 44
x = 22 in
Thus the length of the OP IS 22 in .