Principles and Applications of Mathematics for Communications-electronicsHeadquarters, Department of the Army, 1992 |
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Página 13
... feet of iron pipe is 8 pounds . What is the weight of 255 feet of the same pipe ? Let the un- known quantity be repre- sented by the letter y . Since ratios must express a relation between quantities of the same kind , one ratio must be ...
... feet of iron pipe is 8 pounds . What is the weight of 255 feet of the same pipe ? Let the un- known quantity be repre- sented by the letter y . Since ratios must express a relation between quantities of the same kind , one ratio must be ...
Página 13
... feet at 68 ° F is .248 ohm , what is the resistance of 1,500 feet ; of 1,200 feet ; of 1,850 feet ; of 3,600 feet ? f . If 21 - gage wire weighs 2.452 pounds per 1,000 feet , what is the weight of 1,150 feet ; 1,540 feet ; 1,680 feet ...
... feet at 68 ° F is .248 ohm , what is the resistance of 1,500 feet ; of 1,200 feet ; of 1,850 feet ; of 3,600 feet ? f . If 21 - gage wire weighs 2.452 pounds per 1,000 feet , what is the weight of 1,150 feet ; 1,540 feet ; 1,680 feet ...
Página 82
... feet long . Find the length of the hypotenuse . Dividing 10 and 24 by 2 gives 5 and 12 , the two sides of a 5-12-13 ... feet and the hypotenuse 25 feet . 135. Quadrilaterals A quadrilateral is a plane figure bounded by four straight ...
... feet long . Find the length of the hypotenuse . Dividing 10 and 24 by 2 gives 5 and 12 , the two sides of a 5-12-13 ... feet and the hypotenuse 25 feet . 135. Quadrilaterals A quadrilateral is a plane figure bounded by four straight ...
Página 84
... feet 6 inches . A = r2 5.5 \ 2 2 141. Area of Ring = 157.08 = 2 = 78.54 square inches A ring is the area between the circumfer- ences of two concentric circles . The area of a ring may be found by subtracting the area of the small ...
... feet 6 inches . A = r2 5.5 \ 2 2 141. Area of Ring = 157.08 = 2 = 78.54 square inches A ring is the area between the circumfer- ences of two concentric circles . The area of a ring may be found by subtracting the area of the small ...
Página 85
... feet , how far from its top is an object on the ground 60 feet from the base of the pole ? k . How many square feet of lumber are needed AGO 558A 85.
... feet , how far from its top is an object on the ground 60 feet from the base of the pole ? k . How many square feet of lumber are needed AGO 558A 85.
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20 ohms ac circuit ac series circuit amperes antilog applied voltage base bers binary numbers capacitive reactance Centimeters/second Change Colang Tong complex numbers Cosec Cubic meters current flowing decibels decimal number decimal point denominator digit Divide equal Example exponents factors feet Figure Find the area Find the voltage formula fraction graph hairline hypotenuse inductive reactance Kilometers leads the applied logarithm mantissa Meters/minute Meters/second Microhenrys Miles/hour Miles/minute monomial Multiply negative numbers oblique triangle ohms Ounces avoirdupois Paragraph parallel circuit percent phase angle Plot problem quadratic equation quantity R₁ radians radical radicand ratio rectangle resistor right triangle ABC scale scientific notation series circuit side opposite Simplify slide rule square root Statcoulombs Statvolts Step Subtract Tang Colang tion triangle ABC fig trigonometric functions unknown vector diagram voltage drop volts watts yards zero