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goldenfox [79]
2 years ago
10

Mr. Gonzales is replacing a cylindrical air-conditioning duct. He estimates the radius of the duct by folding a ruler to form tw

o 6-in. tangents to the duct. The tangents form an angle. Mr. Gonzales measures the angle bisector from the vertex to the duct. It is about 2.75 inches long. What is the radius of the duct?
Mathematics
1 answer:
GalinKa [24]2 years ago
4 0

Answer:

  • <em><u>5.6875 in</u></em>

Explanation:

At the point of tangency, the <em>tangent </em>to a circle and the <em>radius</em> form a right triangle (the radius is perpendicular to the tangent).

Here you are given the length of the tangent (6in), and the distance from the bisected vertex to the circle (2.75 in)

I tried to upload the drawing but the tool is not allowing it now.

In the figure:

  • The length of the tangent (6 in) is one leg of the triangle
  • The distance from vertex and the circle (2.75in)  along with the radius forms the hypotenuse of the right triangle: 2.75 + r.
  • The other leg is the radius, r.

Then, you can use Pythagorean theorem:

  • (r)² + (6)² = (r +2.75)²

Solve:

  • r² + 36 =  r² + 5.5r + 7.5625
  • 5.5r = 36 - 7.5625
  • 5.5r = 28.4375
  • r = 5.6875

The solution is in inches: r = 5.6875 inches ← answer

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Answer:

  2 small; 3 large

Step-by-step explanation:

Using s and L for the number of small and large cars, respectively, we have ...

  5s +8L = 34 . . . . number of passengers

  s + L = 5 . . . . . . . number of cars

  5s +8(5 -s) = 34 . . . . substitute for L

  -3s +40 = 34

  -3s = -6 . . . . . . subtract 40

  s = 2 . . . . . . . . divide by -2

  L = 5-2 = 3

2 small cars and 3 large cars were rented.

7 0
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Geoff planted dahlias in his garden. Dahlias have bulbs that divide and reproduce underground. In the first year, Geoff’s garden
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3 years ago
The curves r1(t) = 4t, t2, t3 and r2(t) = sin(t), sin(5t), 2t intersect at the origin. find their angle of intersection, θ, corr
san4es73 [151]
First get the tangent vectors by differentiating r1 and r2 
r_1 '(t) = (4,2t,3t^2) \\  \\  r_2'(t) = (cos (t), 5 cos(5t), 2)
Evaluate at t=0
r_1' = (4,0,0) \\  \\ r_2' = (1,5,2)

Use identity for angle between 2 vectors:
u*v = |u| |v| cos \theta

Evaluate dot product and unit vectors:
u*v = (4,0,0)*(1,5,2) = 4 \\  \\ |u| = \sqrt{4^2 +0^2+0^2} = 4 \\  \\ |v| = \sqrt{1^2 + 5^2+2^2} = \sqrt{30}

Sub into identity and solve for theta:
4 = 4 \sqrt{30} cos \theta \\  \\ cos\theta = \frac{1}{\sqrt{30}} \\  \\ \theta = 79.48

Answer:
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3 years ago
37 points Which expressions are equivalent to 6a(5+d) ?
uysha [10]

Answer:

B AND C

Step-by-step explanation:

5 0
3 years ago
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