Given that in a trianlgle the sides AB, BC, CA are in the ratio 3:4:6.
Let AB = 3k, BC = 4k and CA = 6k.
Then perimeter =3k+4k+6k = 13k
M, N, K are mid points of the sides.
By mid point theorem MN = 3k/2, NK = 4k/2 and KM = 6k/2
Hence perimeter of MNK = 13k/2 =5.2 (given)
Solve for k
k=2(5.2)/13 = 0.8
Hence sides are
AB = 3k = 3(0.8) = 2.4 in
BC = 4k= 3.2 in
CA = 4.8 in
Answer:
$31 per ticket
$0.33 per volleyball
2 strikeouts per inning
Step-by-step explanation:
Price per ticket:
279 = 9x
Divide both sides by 9
31 = x
Price per volleyball:.
6 = 18x
Divide both sides by 18
6/18 = x
Both the numerator and denominator have a common factor of 6
6 ÷ 6 / 18 ÷ 6 = x
1/3 = x
Convert to decimal for money
0.33 = x
Number of strikeouts per inning:
54 = 27x
Divide both sides by 27
2 = x
It’s changed by the perimeter by 2/5
Answer:
9.99493
Step-by-step explanation:
You divide the circumference by pi to get the diameter. Then if you want the radius, you just simply cut the diameter in half!
By converting into parametric equations,
<span>{<span><span>x<span>(θ)</span>=r<span>(θ)</span><span>cosθ</span>=<span>cos2</span>θ<span>cosθ</span></span><span>y<span>(θ)</span>=r<span>(θ)</span><span>sinθ</span>=<span>cos2</span>θ<span>sinθ</span></span></span></span>
By Product Rule,
<span>x'<span>(θ)</span>=−<span>sin2</span>θ<span>cosθ</span>−<span>cos2</span>θ<span>sinθ</span></span>
<span>x'<span>(<span>π2</span>)</span>=−<span>sin<span>(π)</span></span><span>cos<span>(<span>π2</span>)</span></span>−<span>cos<span>(π)</span></span><span>sin<span>(<span>π2</span>)</span></span>=1</span>
<span>y'<span>(θ)</span>=−<span>sin2</span>θ<span>sinθ</span>+<span>cos2</span>θ<span>cosθ</span></span>
<span>y'<span>(<span>π2</span>)</span>=−<span>sin<span>(π)</span></span><span>sin<span>(<span>π2</span>)</span></span>+<span>cos<span>(π)</span></span><span>cos<span>(<span>π2</span>)</span></span>=0</span>
So, the slope m of the curve can be found by
<span>m=<span><span>dy</span><span>dx</span></span><span>∣<span>θ=<span>π2</span></span></span>=<span><span>y'<span>(<span>π2</span>)</span></span><span>x'<span>(<span>π2</span>)</span></span></span>=<span>01</span>=0</span>
I hope that this was helpful.
