What is the answer to the problem 2/3x-40=5/6(x+36)

Evaluation the expression using a model 2/3 dived by 1/2.

m_a_m_a [10]

Answer:la respuesta es 8/0

Step-by-step explanation:

8 0

3 years ago

(multiple choice) How is e-mail different from other types of communication?

Dmitry_Shevchenko [17]

Answer: You can't hear the tone of voice or get clues from body language.

Step-by-step explanation:

When sending emails, the sender and the recipient communicate through a screen, thus each person is unable to see the other when conversing or hear the inflections or emphasis of voice in the message, although we can always attempt to interpret tone. Email allows you to store emails in the inbox or archive and access sent messages, but email lacks the ability to intimately understand specifics such as tone of voice or body language.

7 0

3 years ago

Read 2 more answers

Use the Pythagorean theorem to obtain the function D(x) for the length of the diagonal of a double square of width x.

Yuri [45]

Answer:

$D(x)= x\sqrt{2}$

Step-by-step explanation:

Given

$Side = x$

Required

Represent the diagonal as a function

Using Pythagoras theorem, the diagonal is:

$D^2= Side^2 + Side^2$

Substitute x for Side

$D^2= x^2 + x^2$

$D^2= 2x^2$

Take the square root of both sides

$D= \sqrt{2x^2$

Split

$D= \sqrt{2} * \sqrt{x^2$

$D= \sqrt{2} * x$

$D= x\sqrt{2}$

Express as a function.

$D(x)= x\sqrt{2}$

4 0

3 years ago

A company used 7.4 x 105 sheets of paper a month, the accounting department used 8.9 x 103 sheets. How many were used by other d

seraphim [82]

Answer:

731,100

7.311 × 10^5

Step-by-step explanation:

The computation of the number of sheet used by other department is given below:

Given that

The company used 7.4 × 10^5 sheets of paper So it would be 740,000

And, The accounting department used 8.9 × 10^3 sheets so it would be 8,900

Now the number of sheet used by other department is

= 740,000 - 8,900

= 731,100

= 7.311 × 10^5

4 0

2 years ago

Which one is bigger, 6 feet or 2 meters?

Vlad1618 [11]

CONVERSIONS, PHYSICAL CONSTANTS, AND UNITS

Length:
1 inch (in) = 2.54 centimeters (cm) exactly
1 foot (ft) = 0.305 meters (m)
1 yard (yd) = 0.914 meters (m)
1 mile (mi) = 5280 feet (ft) = 1.61 kilometers (km)
1 centimeter (cm) = 0.394 inches (in)
1 meter (m) = 100 centimeters (cm) = 39.4 inches (in) = 3.28 feet (ft) = 1.09 yards (yd)
1 kilometer (km) = 1000 meters (m) = 0.621 miles (mi)
1 rod = 16.5 feet (ft) = 5.5 yards (yd) = 5.03 meters (m)
1 astronomical unit (AU) = 1.50 x 108 kilometers (km) = 9.29 x 107 miles (mi)
1 light year (LY) = 9.461 x 1012 kilometers ( km) = 5.88 x 1012 miles (mi) = 0.307 parsec
1 parsec = approximately 3.26 light years (LY)

Area = length x width
1 square inch (in2) = 0.00694 square feet (ft2 = 6.45 square centimeters, cm2
1 square foot, (ft2) = 144 square inches (in2 = 0.0929 square meters, m2
1 square yard (yd2 = 1296 square inches (in2 = 9 square feet (ft2 = 0.836 square meters (m2
1 square mile (mi2 = 640 acres = 2.59 square kilometers (km2
1 acre = 160 square rods = 43,560 square feet (ft2) = 0.405 hectares
1 hectare = 10,000 square meters (m2 = 100 ares = 2.47 acres
1 square rod = 30,25 square yards (yd2 = 25.29 square meters (m2
1 are = 100 square meters (m2 = 0.01 hetares = 119.6 square yards (yd2

Volume = length x width x height
1 cubic inch (in3) = 0.000579 cubic feet (ft3) = 16.4 cubic centimeters (cm3)
1 cubic foot (ft3) = 1728 cubic inch (in3) = 0.0283 cubic meters (m3)
1 cubic yard (yd3) = 27 cubic feet (ft3) = 4.65 x 104 cubic inch (in3) = 0.765 cubic meters (m3)
1 cubic meters (m3) = 106 cubic centimeters (cm3) = 1,000 liters (L) = 35.3 cubic feet (ft3)
1 quart (qt) = 2 pints (pt) = 946 milliliters (mL) = 0.946 liters (L)
1 gallon (gal) = 4 quarts = 231 cubic inch (in3) = 3.79 liters (L)
1 liter (L) = 1000 cubic centimeters (cm3) = 1.06 quarts (qt) = 0.265 gallons (gal)

Mass
1 slug = 14.6 kilograms (kg)
1 kilogram (kg) = 1,000 grams = 0.0685 slugs
1 atomic mass unit (u) = 1.66 x 10-27 kilogram (kg) =1.66 x 10-24
1 electron mass = 9.11 x 10-31 kilogram (kg) = 9.11 x 10-28 grams (g) = 5.46 x 10-4 atomic mass unit (u)
1 proton mass = 1.00728 atomic mass unit (u)
1 neutron mass = 1.00866 atomic mass unit (u)

Conversion Between Weight and Mass on Earth
A slug is the mass unit in the British system and is equal to 32.2 pounds (lb)
A kilogram weighs 9.81 newtons (N) or 2.21 pounds (lb)
A mass of one (1) gram (g) weighs 981 dynes or 0.0353 ounces (oz)
A one (1) ounce weight has a mass 28.4 grams (g) or 0.0284 kilograms (kg)
A one (1) pound weight has a mass of 454 grams (g) or 0.454 kilograms (kg)
A metric ton has a mass of 1,000 kilograms (kg)
A one (1) ton (regular) weight has a mass of 2,000 pounds (lb)

Velocity and Speed = distance/time
1 ft/s (ft s-1) = 0.305 m/s (m s-1)
1 mi/hr (mi hr-1) = 1.47 ft/s (ft s-1) = 1.61 km/hr (km hr-1) = 0.447 m/s (m s-1)
1 m/s (m s-1) = 100 cm/s (cm s-1) = 3.28 ft/s (ft s-1)
1 km/hr (km hr-1) = 0.278 m/s (m s-1) = 0.621 mi/hr (mi hr-1) = 0.912 ft/s (ft s-1)
Speed of light in a vacuum = 3.00 x 108 m/s (m s-1) = 3.00 x 1010 cm/s (cm s-1) = 186,000 mi/s (mi s-1)
Speed of sound in air at 0 oC (32 oF) = 331 m/s (m s-1) = 1090 ft/s (ft s-1) = 41 mi/hr (mi hr-1)

Acceleration = velocity / time
1 ft/s2 = 0.305 m/s2 = 30.5 cm/s2
1 mi/hr/s = 1.47 ft/s2 = 1.61 km/hr/s = 0.447 m/s2
1 m/s2 = 100 cm/s2 = 3.28 ft/s2
Acceleration due to gravity on Earth = 9.81 m/s2 = 981 cm/s2 = 32.2 ft/s2

Force = mass x acceleration
1 lb = 16 oz = 4.45 newtons (N)
1 N (kg· m/s2) = 0.223 lb

Temperature
oF = 9/5 x oC + 32
oC = (oF - 32) x 5/9
Absolute Temperature, kelvin, K = oC + 273.15
0 K = -273.15 oC = -459.72 oF
Normal Body Temperature = 98.6 oF

Pressure = Force/Area
1 Pascal, Pa = 1 N/m2 = 1.45 x 10-4 lb/in2
1 bar = 105 N/m2 = 14.50 lb/in2
1 atmosphere, atm = 760 mm Hg = 76.0 cm Hg = 760 torr = 14.50 lb/in2 = 1.01325 x 105 N/m2 (or Pa)
Air pressure at sea level is approximately = 1 atm = 1 bar = 105 N/m2 (or Pa) = 14.50 lb/in2
Water pressure increases 9810 N/m2 for each meter of depth or 0.433 lb/in2 for each foot of depth

Work and Energy or Heat Work = Force x parallel distance
1 joule, J = 0.738 ft . lb = 107 ergs
1 calorie, cal = 4.18 J = 0.00442 British Thermal Units, Btu
1 foot pound, ft . lb = 1.36 J
1 Btu = 252 cal = 778 ft·lb = 1.054 x 103 J
1 electron Volt, eV = 1.602 x 10-19 J
931.5 million electron volts,MeV is equivalent to 1 atomic mass unit, u

Power = Work/time
1 horse power, hp = 550 ft·lb/s = 0.746 kilowatts, kW = 746 W
1 Watt, W = 1 J/s = 0.738 ft·lb/s = 0.00134 hp
1 Btu/hr = 0.293 W
1 kW = 1.34 hp

Radioactivity
1 curie = 3.70 x 1010 disintegrations/second = 3.70 x 1010 bequerels

SI BASE UNITSPhysical Quantity Name of UnitSymbolLengthmetermMass kilogramkgTimesecondsTemperaturekelvinKElectric currentampereALuminous intensitycandelacdAmount of substancemolemol

SI PREFIXESMACROMICROFactorPrefixSymbolFactorPrefixSymbol1018exaE10-1decid1015petaP10-2centic1012teraT10-3millim109gigaG10-6microμ106megaM10-9nanon103kilok10-12picop102hectoh10-15femtof101dekada10-18attoa

3 0

3 years ago

Read 2 more answers