C The triangles in this picture are similar. Find the height of the tree.

Me. Sanchez bought some books at a book sale, including 15 paperback books. 75% of all the books he bought were paperbacks

swat32

Answer:

Step-by-step explanation:

6 0

3 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

PLEASE HELP WILL GIVE BRAINLIEST!!!

geniusboy [140]

Answer:

b

please correct if wrong

Step-by-step explanation:

7 0

2 years ago

If three tangents to a circle form an equilateral triangle, prove that the tangent points form an equilateral triangle inscribed

weqwewe [10]

I added a figure so you can guide yourself throughout the proof I'm about to write, so I recommend that you download the picture beforehand and have this window and the picture's window open. Alright, let's get started!
Assuming that $FED$ is an equilateral triangle according to the wording of the problem, we have that the angles $\widehat{AFB}=\widehat{BEC}=\widehat{CDA}=60$ .
We also know that the circle in green is Inscribed in $FED$ .

The following applies to every inscribed circle inside a triangle:
The center of the inscribed circle of a triangle is the intercept of all three angle bisectors of the triangle.

The above theorem implies that the line $(FO)$ is an angle bisector because it goes through the vertex F and the center of the inscribed circle.

The previous statement implies that the angle $\widehat{OFB}=30$ .

Now let's work on the $OFB$ triangle.
Knowing that $\widehat{OBF}=90$ (because $(EF)$ is tangent to the circle at B and $OB$ is a radius of the circle. If you're lost here, remember that a tangent to a circle is always perpendicular to the radius of the circle.) we can then derive that $\widehat{FOB}=180-90-30=60$ (because the sum of the measures of all angles in a triangle is always equal to 180 degrees).

In the same way, we can prove also that:
$\widehat{BOE}=\widehat{EOC}=\widehat{COD}=\widehat{DOA}=\widehat{AOF}=60 degrees$ .
Knowing the above we notice that $\widehat{BOC}=\widehat{COA}=\widehat{AOB}=120degrees$ .

We're at the last part of our proof here:
Now notice that $\widehat{BOA}$ subtends the same arc on the circle that $\widehat{BCA}$ .
According to the inscribed angle theorem, an angle $\theta$ inscribed in a circle is half of the central angle $2\theta$ that subtends the same arc on the circle.
Therefore $\widehat{BCA}=\frac{\widehat{BOA}}{2}=60degrees$

We can prove in a similar fashion that: $\widehat{CAB}=\widehat{ABC}=60degrees$
Therefore all the angles of the $ABC$ triangle have a measure of 60 degrees, we conclude then that $ABC$ is equilateral.

5 0

3 years ago

Heyyy, please answer :]

Yuliya22 [10]

The solution is “no real solution”
Because the slope is the same, so they are parallel.

Hope that help!! :-)

6 0

2 years ago