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.
Answer:
18 gallons
Step-by-step explanation:
We'll find out how much of the total container was lost when pouring out by subtracting 2/3 from 1/2.
2/3 = 4/6
1/2 = 3/6
4/6 - 3/6 = 1/6
1/6 of the total amount the container could hold was lost when 3 gallons was poured out. This means that 1/6 of the container equals to 3 gallons. We can then simply multiply 3 by 6 to find the number of gallons of liquid the container can hold.
3 * 6 = 18 gallons
Checking my answer:
18 * 2/3 = 6 * 2 = 12
12 - 3 = 9
1/2 of 18 = 9
This means that our answer is correct.
Answer:
x=1/2,-2/3
Step-by-step explanation:
Answer:
f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)
Step-by-step f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)explanation:
f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(xf(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(xf(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)
