Answer:
2i: 169.71
2ii: 0.17L
3a: 4×10⁻⁵
3b: 110011
Step-by-step explanation:
2i. The surface of the top and bottom of the tin is two times (top and bottom) π·r² = 2·π·3² = 18π cm².
The circumference of the circle is 2·π·r = 6π cm².
The area of the material connecting top and bottom is a rectangle of the tin height times the circumference: 6·6π = 36π cm².
This gives a total of 18π + 36π = 54π cm².
With π approximated by 22/7 the total surface area is 54*22/7 ≈ 169.71.
Notice how the calculation is simple by waiting until the very last moment to substitute π.
2ii. The volume is the area π·r² of the circle times the height of the tin: 9π*6 = 54π cm³ ≈ 169.71 cm³.
Since 1L = 1000 cm³ the volume is 0.16971 litres, which should be rounded to 0.17 L.
3a: If we rewrite P as 36 x 10⁻⁴ and realize that 36/2.25 = 16, then the fraction can be written as
16 x 10⁻⁴⁻⁶ = 16 x 10⁻¹⁰.
The square root of that is taking it to the power of 1/2, so (16x10⁻¹⁰)^0.5 = 4x10⁻⁵ = 0.00004
3b: 1111 1111 is 255 in decimal. 101 is 5 in decimal. 255/5 is 51 in decimal. 51 in binary is 110011.
F(x) = √(5x + 7)
g(x) = √(5x - 7)
(f + g)(x) = f(x) + g(x)
(f + g)(x) = √(5x + 7) + √(5x - 7)
Answer: C)
Distance between two points, d, is given by

where: (x1, y1) = (2, 2) and (x2, y2) = (5, 5)
Answer:
3.897 
Step-by-step explanation:
equilateral triangles are also equiangular, meaning the have equal angles.
Triangle sum theory says that angles of a triangle add up to 180.
That means each angle is 60.
A = bh/2
You need the (h). The base of is 3. Perimeter = 9, so each side is 3
Draw a perpendicular line for the height. The line cuts the base in half (1.5)
Using trigonometry you can find the height.
tan 60° = h/1.5
h = height, 1.5 is half of 3, 60° is the base angle.
multiply each side by 1.5
1.5(tan 60°) = h
h=2.598
then substitute h into formula
A= <u>(2.598)(3) </u>
2
A = 3.897 rounded
well, we know the ceiling is 6+2/3 high, and Eduardo has 4+1/2 yards only, how much more does he need, well, is simply their difference, let's firstly convert the mixed fractions to improper fractions and then subtract.
![\stackrel{mixed}{6\frac{2}{3}}\implies \cfrac{6\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{20}{3}} ~\hfill \stackrel{mixed}{4\frac{1}{2}}\implies \cfrac{4\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{9}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{20}{3}-\cfrac{9}{2}\implies \stackrel{using ~~\stackrel{LCD}{6}}{\cfrac{(2\cdot 20)-(3\cdot 9)}{6}}\implies \cfrac{40-27}{6}\implies \cfrac{13}{6}\implies\blacktriangleright 2\frac{1}{6} \blacktriangleleft](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B6%5Cfrac%7B2%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B6%5Ccdot%203%2B2%7D%7B3%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B20%7D%7B3%7D%7D%20~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B4%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B9%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B20%7D%7B3%7D-%5Ccfrac%7B9%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Busing%20~~%5Cstackrel%7BLCD%7D%7B6%7D%7D%7B%5Ccfrac%7B%282%5Ccdot%2020%29-%283%5Ccdot%209%29%7D%7B6%7D%7D%5Cimplies%20%5Ccfrac%7B40-27%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B13%7D%7B6%7D%5Cimplies%5Cblacktriangleright%202%5Cfrac%7B1%7D%7B6%7D%20%5Cblacktriangleleft)