civonamzuk

Shortly after 10 'o clock, the hands of a clock made a 90-degree angle. What's the time?

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60 - x/12 + x = 90          x = 32⁸/₁₁ degrees       ( x = 360 / 11 )

Shortly after 10 'o clock, the hands of a clock made a 90-degree angle. What's the time?

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Here's another way to solve this problem.

Arc from 10 to 12 = 60º

Let x be the arc-path of minute hand in degrees.

60 + x - (x/12) = 90

x = 360/11 = 32.72727272º     (minute hand arc)

x/12 = 30/11 = 2.72727272º     (hour hand arc)

You're correct!

A = [4(5*6) + 2(5*5)] - 2(3*5) = 140 cm²

V = 6*5² - 3*5² = 75 cm³

Let PQRSTUVW be a rectangular prism, as shown. The area of face PQRS is 6. The area of face PQUT is 15. The area of face QUVR is 10. Find the volume of the rectangular prism.

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[PQRS] = 6           [PQUT] = 15            [QUVR] = 10

(xz) = 6         x = 6/z          z = 6/x

(yz) = 15       y = 15/z        z = 15/y

(xy) = 10       x = 10/y        y = 10/x

(yz) = (10/x)(6/x) = 15        x = 2          y = 10/2         z = 6/2

V = xyz

If [RSX] = 48, then [PQRS] = 129.967648

but...

If [RSX] = 45, then [PQRS] = 125

ΔABC is an isosceles triangle.

Angles B and C are congruent.       Their tangent = 10/3 = 3¹/₃

2(x / 3¹/₃ ) + x = 6

Square side   x = 3.75 cm

67 + 52 + x = 210

x = 91

Arc VX = x

[(80 - x) - x] / 2 = 30

x = 10

Arc VX = 10º             arc WY = 70º

"So there's this 45-45-90 triangle, and the hypotenuse is 5 sqrt 26. I know that the lengths of the other two sides are half of that."

The length of the legs is not 1/2 of the hypotenuse!!!

Hypotenuse = 5√26 = 25.49509757

Leg(s) = 5√13 = 18.02775638