Someone please solve this pleaase t
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2. The chord is bisected by the segment from the center, so you have a right triangle with legs 4 and 6. The hypotenuse is the radius of the circle, x.
x = √(4^2 +6^2) = √(16+36) = √52
x = 2√13
4. By the rules for secants, the product of the lengths of the parts of the chord is the same in each case.
3*7 = 2*x
x = 21/2
x = 10.5
6. The measure of angle "a" is the average of the two intercepted arcs.
a = (165° + 55°)/2
a = 110°
Of course, b is its supplement.
b = 180° - 110°
b = 70°
8. Secant rules again. The products of the near and far lengths are the same in each case. When the segment is a tangent, the near and far lengths are the same, so it is the square of the tangent length.
x² = 4(4+12) = 64
x = 8
10. This works the same way as problem 2. Here, it looks like you have a triangle with hypotenuse 6 and side length 3. The other side length is
half-chord = √(6² -3²) = √(36 -9) = √27 = 3√3
AB = 2*(half-chord) = 2*3√3
AB = 6√3
12. The equation of a circle of radius r and center (h, k) is
(x-h)² + (y-k)² = r²
Of course, the "passes through" point must satisfy the equation, so you have
(x-3)² + (y-0)² = r² = (-2-3)² + (-4-0)² = 25+16
(x-3)² + y² = 41
14. Same as 12. Radius is 4.
(x+5)² + (y-2)² = 16
16. Radius 2, center (0, -1).
x² + (y+1)² = 4
18. The angle bisectors for the axes are the lines
y = x
y = -x
Here's a plot.
x = √(4^2 +6^2) = √(16+36) = √52
x = 2√13
4. By the rules for secants, the product of the lengths of the parts of the chord is the same in each case.
3*7 = 2*x
x = 21/2
x = 10.5
6. The measure of angle "a" is the average of the two intercepted arcs.
a = (165° + 55°)/2
a = 110°
Of course, b is its supplement.
b = 180° - 110°
b = 70°
8. Secant rules again. The products of the near and far lengths are the same in each case. When the segment is a tangent, the near and far lengths are the same, so it is the square of the tangent length.
x² = 4(4+12) = 64
x = 8
10. This works the same way as problem 2. Here, it looks like you have a triangle with hypotenuse 6 and side length 3. The other side length is
half-chord = √(6² -3²) = √(36 -9) = √27 = 3√3
AB = 2*(half-chord) = 2*3√3
AB = 6√3
12. The equation of a circle of radius r and center (h, k) is
(x-h)² + (y-k)² = r²
Of course, the "passes through" point must satisfy the equation, so you have
(x-3)² + (y-0)² = r² = (-2-3)² + (-4-0)² = 25+16
(x-3)² + y² = 41
14. Same as 12. Radius is 4.
(x+5)² + (y-2)² = 16
16. Radius 2, center (0, -1).
x² + (y+1)² = 4
18. The angle bisectors for the axes are the lines
y = x
y = -x
Here's a plot.
The number of long distance calls to be made can be obtained by linear programming
- To minimize the phone bill, the number of long distance call to be made is none
Reason:
Maximum number of number of long distance calls = 15
Maximum number of local calls = 30
Number of minutes of calls to make each month = 240 minutes
Duration of each long distance call = 30 minutes
Duration of each local call = 10 minutes
Cost of long distance call = $0.08
Cost of local call = $0.03
Solution:
Let <em>X</em>, represent the number of long distance calls, and let <em>Y</em> represent the number local calls
The objective function is given as follows;
- P = 0.08·(30)·X + 0.03·(10)·Y
P = 2.4·X + 0.3·Y
The constraints are;
X ≥ 0
Y ≥ 0
X ≤ 15
Y ≤ 30
30·X + 10·Y ≥ 240
Solving we have;
- Y ≥ 24 - 3·X
The above inequality can be plotted with MS Excel
From the objective function, P = 2.4·X + 0.3·Y, the coefficient of the long
distance calls, <em>X</em> is larger than the coefficient of the local calls <em>Y</em>, therefore,
making the minimum possible number of long distance calls of 0, and 24
local calls will give a cost;
- P = 2.4 × 0 + 0.3 × 24 = 7.2
Therefore, to minimize the phone bill, the number of long distance calls to
be made is <u>zero long distance calls</u>
Learn more about linear optimization here:
