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prohojiy [21]
3 years ago
8

Which construction is the Perpendicular Bisector Construction?

Mathematics
1 answer:
Leni [432]3 years ago
6 0

Answer:

Step-by-step explanation:

This construction shows how to draw the perpendicular bisector of a given line segment with compass and straightedge or ruler. This both bisects the segment (divides it into two equal parts, and is perpendicular to it. It finds the midpoint of the given line segment.

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Mrs.Ramirez bought 2 qt of orange juice. If the juice is served in 6-oz glasses,how many glasses can be completely filled?
alekssr [168]
Hey there!

<span>There is 32 oz per qt. 2qts=64oz. Divide 64 by 6. </span><span> 6 goes evenly into 64 10 times with a remainder. So the answer is 10 glasses. 
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Hope this helps!
4 0
3 years ago
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3 years ago
Solve for x<br> 6/x^2+2x-15 +7/x+5 =2/x-3
timama [110]

Answer:

x = ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3))/(15 (2140 - 9 sqrt(56235))^(1/3)) - 1/3 or x = 1/15 (17 5^(2/3) (-1/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) - 1/3 or x = -1/3 - 17/(3 (10700 - 45 sqrt(56235))^(1/3)) - (2140 - 9 sqrt(56235))^(1/3)/(3 5^(2/3))

Step-by-step explanation:

Solve for x:

6/x^2 + (2 x - 8)/(x + 5) = 2/x - 3

Bring 6/x^2 + (2 x - 8)/(x + 5) together using the common denominator x^2 (x + 5). Bring 2/x - 3 together using the common denominator x:

(2 (x^3 - 4 x^2 + 3 x + 15))/(x^2 (x + 5)) = (2 - 3 x)/x

Cross multiply:

2 x (x^3 - 4 x^2 + 3 x + 15) = x^2 (2 - 3 x) (x + 5)

Expand out terms of the left hand side:

2 x^4 - 8 x^3 + 6 x^2 + 30 x = x^2 (2 - 3 x) (x + 5)

Expand out terms of the right hand side:

2 x^4 - 8 x^3 + 6 x^2 + 30 x = -3 x^4 - 13 x^3 + 10 x^2

Subtract -3 x^4 - 13 x^3 + 10 x^2 from both sides:

5 x^4 + 5 x^3 - 4 x^2 + 30 x = 0

Factor x from the left hand side:

x (5 x^3 + 5 x^2 - 4 x + 30) = 0

Split into two equations:

x = 0 or 5 x^3 + 5 x^2 - 4 x + 30 = 0

Eliminate the quadratic term by substituting y = x + 1/3:

x = 0 or 30 - 4 (y - 1/3) + 5 (y - 1/3)^2 + 5 (y - 1/3)^3 = 0

Expand out terms of the left hand side:

x = 0 or 5 y^3 - (17 y)/3 + 856/27 = 0

Divide both sides by 5:

x = 0 or y^3 - (17 y)/15 + 856/135 = 0

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:

x = 0 or 856/135 - 17/15 (z + λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:

x = 0 or z^6 + z^4 (3 λ - 17/15) + (856 z^3)/135 + z^2 (3 λ^2 - (17 λ)/15) + λ^3 = 0

Substitute λ = 17/45 and then u = z^3, yielding a quadratic equation in the variable u:

x = 0 or u^2 + (856 u)/135 + 4913/91125 = 0

Find the positive solution to the quadratic equation:

x = 0 or u = 1/675 (9 sqrt(56235) - 2140)

Substitute back for u = z^3:

x = 0 or z^3 = 1/675 (9 sqrt(56235) - 2140)

Taking cube roots gives (9 sqrt(56235) - 2140)^(1/3)/(3 5^(2/3)) times the third roots of unity:

x = 0 or z = (9 sqrt(56235) - 2140)^(1/3)/(3 5^(2/3)) or z = -((-1)^(1/3) (9 sqrt(56235) - 2140)^(1/3))/(3 5^(2/3)) or z = ((-1)^(2/3) (9 sqrt(56235) - 2140)^(1/3))/(3 5^(2/3))

Substitute each value of z into y = z + 17/(45 z):

x = 0 or y = (9 sqrt(56235) - 2140)^(1/3)/(3 5^(2/3)) - (17 (-1)^(2/3))/(3 (5 (2140 - 9 sqrt(56235)))^(1/3)) or y = 17/3 ((-1)/(5 (2140 - 9 sqrt(56235))))^(1/3) - ((-1)^(1/3) (9 sqrt(56235) - 2140)^(1/3))/(3 5^(2/3)) or y = ((-1)^(2/3) (9 sqrt(56235) - 2140)^(1/3))/(3 5^(2/3)) - 17/(3 (5 (2140 - 9 sqrt(56235)))^(1/3))

Bring each solution to a common denominator and simplify:

x = 0 or y = ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3))/(15 (2140 - 9 sqrt(56235))^(1/3)) or y = 1/15 (17 5^(2/3) ((-1)/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) or y = -(2140 - 9 sqrt(56235))^(1/3)/(3 5^(2/3)) - 17/(3 (5 (2140 - 9 sqrt(56235)))^(1/3))

Substitute back for x = y - 1/3:

x = 0 or x = 1/15 (2140 - 9 sqrt(56235))^(-1/3) ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3)) - 1/3 or x = 1/15 (17 5^(2/3) (-1/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) - 1/3 or x = -1/3 - 1/3 5^(-2/3) (2140 - 9 sqrt(56235))^(1/3) - 17/3 (5 (2140 - 9 sqrt(56235)))^(-1/3)

5 (2140 - 9 sqrt(56235)) = 10700 - 45 sqrt(56235):

x = 0 or x = ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3))/(15 (2140 - 9 sqrt(56235))^(1/3)) - 1/3 or x = 1/15 (17 5^(2/3) (-1/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) - 1/3 or x = -1/3 - (2140 - 9 sqrt(56235))^(1/3)/(3 5^(2/3)) - 17/(3 (10700 - 45 sqrt(56235))^(1/3))

6/x^2 + (2 x - 8)/(x + 5) ⇒ 6/0^2 + (2 0 - 8)/(5 + 0) = ∞^~

2/x - 3 ⇒ 2/0 - 3 = ∞^~:

So this solution is incorrect

6/x^2 + (2 x - 8)/(x + 5) ≈ -3.83766

2/x - 3 ≈ -3.83766:

So this solution is correct

6/x^2 + (2 x - 8)/(x + 5) ≈ -2.44783 + 1.13439 i

2/x - 3 ≈ -2.44783 + 1.13439 i:

So this solution is correct

6/x^2 + (2 x - 8)/(x + 5) ≈ -2.44783 - 1.13439 i

2/x - 3 ≈ -2.44783 - 1.13439 i:

So this solution is correct

The solutions are:

Answer:  x = ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3))/(15 (2140 - 9 sqrt(56235))^(1/3)) - 1/3 or x = 1/15 (17 5^(2/3) (-1/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) - 1/3 or x = -1/3 - 17/(3 (10700 - 45 sqrt(56235))^(1/3)) - (2140 - 9 sqrt(56235))^(1/3)/(3 5^(2/3))

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What two rays that intersect at a common endpoint?
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We have a parabola.

As it is concave down, we have one maximum and no minimum.

We can identify the maximum at point (-2, 4).

Answer: (-2,4)

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