There is a problem with your question: If pq = 10 cm, qr = 8 cm and pr = 5.6 cm then if qx is perpendicular to pr through q it does NOT equal 7.2 cm; it is approx 8.0 cm: Let X be the distance px, then xr = 5.6 - X Using Pythagoras: In pxq: 10² = qx² + X² → qx² = 10² - X² In rxq: 8² = qx² + (5.6 - X)² → qx² = 8² - (5.6 - X)² → 10² - X² = 8² - (5.6 - X)² → 10² - X² = 8² - 5.6² + 2×5.6×X - X² → 2×5.6×X = 10² - 8² + 5.6² → X = (10² - 8² + 5.6²)/(2×5.6) → qx² = 10² - ( (10² - 8² + 5.6²)/(2×5.6) )² → qx = √(10² - ( (10² - 8² + 5.6²)/(2×5.6) )²) ≈ 7.989265 cm ≈ 8.0 cm Similarly, for py: py = √(10² - ( (10² - 5.6² + 8²)/(2×8) )²) ≈ 5.59248 cm ≈ 5.6 cm The obtuse triangle has angles: qpr ≈ 53°, prq ≈ 93°, rqp ≈ 34°; the perpendiculars qx and py lie outside the triangle; angle prq ≈ 93° which is not far off a right angle making sides pr and qr approximately perpendicular, and shows that the perpendiculars to the sides next to it (ie the perpendiculars to pr and qr) will be approximately equal to the lengths of the other side (next to it, ie length of perpendicular to pr will be approx qr, and the length of the perpendicular to qr will be approx pr).
4R size is 4" × 6"; 10 cm x 15 cm Other photographic print sizes are 3R (3½" × 5"; 9 cm x 13 cm), 4D (4½" × 6"), the 5R (5" × 7"; 13 cm x 18 cm) and 6R (6" × 8"; double the 4R). The print photo size 8R (8" × 10"; 20 cm x 25 cm) is also well known.
Since, 10 cm = 1 dm , 10 cm x 10 cm x 10 cm = 1 dm x 1 dm x 1 dm , Then, 1000 cm^3 = 1 dm^3
Vol = 10 cm * 5 cm * 40 cm = 2,000 cm3
The circumference of a circle is given by 2 X pi X radius Thus, the circumference of an 8 cm circle is 2 X 3.14 X 8 = 50.24 cm
8 in = 20.32 cm 10 in = 25.4 cm
A = (Pi x D) x H A = (3.1416 x 8 cm) x 10 cm A = 251.3274 cm2
8 cm x 10 = 80 cm.
The volume of a carton (or rectangular prism) is width x length x height, so in this case, the volume is 10x8x8 = 640 cubic centimeters.
80mm (cm x 10 = mm)
There is a problem with your question: If pq = 10 cm, qr = 8 cm and pr = 5.6 cm then if qx is perpendicular to pr through q it does NOT equal 7.2 cm; it is approx 8.0 cm: Let X be the distance px, then xr = 5.6 - X Using Pythagoras: In pxq: 10² = qx² + X² → qx² = 10² - X² In rxq: 8² = qx² + (5.6 - X)² → qx² = 8² - (5.6 - X)² → 10² - X² = 8² - (5.6 - X)² → 10² - X² = 8² - 5.6² + 2×5.6×X - X² → 2×5.6×X = 10² - 8² + 5.6² → X = (10² - 8² + 5.6²)/(2×5.6) → qx² = 10² - ( (10² - 8² + 5.6²)/(2×5.6) )² → qx = √(10² - ( (10² - 8² + 5.6²)/(2×5.6) )²) ≈ 7.989265 cm ≈ 8.0 cm Similarly, for py: py = √(10² - ( (10² - 5.6² + 8²)/(2×8) )²) ≈ 5.59248 cm ≈ 5.6 cm The obtuse triangle has angles: qpr ≈ 53°, prq ≈ 93°, rqp ≈ 34°; the perpendiculars qx and py lie outside the triangle; angle prq ≈ 93° which is not far off a right angle making sides pr and qr approximately perpendicular, and shows that the perpendiculars to the sides next to it (ie the perpendiculars to pr and qr) will be approximately equal to the lengths of the other side (next to it, ie length of perpendicular to pr will be approx qr, and the length of the perpendicular to qr will be approx pr).
Well honey, 8 x 10 inches is 20.32 x 25.4 cm. Just remember, when it comes to measurements, it's all about the conversion game. Hope that helps, darling!
10 cm = 0.1 meterThe volume of the slab is (6 x 8 x 0.1) = 4.8 cubic meters
1 ft = 30.48 cm (exactly) ⇒ 8 ft x 16 ft = (8 x 30.48) cm x (16 x 30.48) cm = 243.84 cm x 487.68 cm
1 litre.
4R size is 4" × 6"; 10 cm x 15 cm Other photographic print sizes are 3R (3½" × 5"; 9 cm x 13 cm), 4D (4½" × 6"), the 5R (5" × 7"; 13 cm x 18 cm) and 6R (6" × 8"; double the 4R). The print photo size 8R (8" × 10"; 20 cm x 25 cm) is also well known.
The mass of the paraffin can be calculated using the formula: mass = density x volume. Plugging in the values, mass = 0.8 g/cm³ x 10 cm³ = 8 grams. Therefore, the mass of the paraffin is 8 grams.