Asolid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. the radius of hemisphere is 3.5cm and the total wood used in the making of toy is 166 5/6cm³. find the height of the toy. also, find the cost of painting the hemi-spherical part of the toy at the rate of ₹ 10 per cm².(use π = 22/7 ).

Explanation below.

Step-by-step explanation:

Given: Radius of hemisphere = radius of cone = r = 3.5 cm

Volume of  total wood used in making a toy = 166 ⅚ = 1001/6 cm³

Let h be height of a cone.

Volume of  total wood used in making a toy = volume of hemisphere + volume of cone

(⅔) × πr³ + (⅓)π²h

= (⅓)πr²(2r + h)

1001/6 = (⅓) ×( 22/7) × (3.5)² × (h +2×3.5)

1001/6 = (⅓) ×( 22/7) × 3.5 ×3.5 ×( h+7)

1001/6 = (⅓) ×  22 × .5 ×3.5 × (h +7)

1001 × 3  = 6 × 22 × .5 ×3.5 × (h+7)

h +7  = 1001 × 3 / (6 ×  22 × .5 ×3.5)

h +7 = 1001 × 3 / 132 × 1.75

h +7 = 1001 × 3 × 100 / 132 × 175

h +7 = 91 × 3 × 4 / 12 × 7

h +7 = 91 × 12 / 12×7

h +7 = 13

h + 7 = 13 -7 = 6

Height of a cone(h) = 6

Height of the toy = Height of a cone + Height of a hemisphere

Height of the toy = 6 +3.5 = 9.5 cm

Curved surface area of hemisphere = 2πr²

CSA of hemisphere = 2 ×( 22/7) × 3.5 × 3.5

= 2 × 22 × .5 × 3.5 = 44 × 1.75 = 77 cm²

Rate of painting the hemispherical part of the toy = ₹ 10 per m².

Cost of painting the hemispherical part of the toy = 77 × 10 =  ₹ 770 .

Hence, the Height of the toy is 9.5 m & Cost of painting the hemispherical part of the toy is  ₹ 770.

walter needs to make at least 5 more cookies but no more than 40. 5 ≤ x ≤ 40

step-by-step explanation:

20 ≤ x + 15 ≤ 55

subtract 15 from all parts

20-15 ≤ x + 15-15 ≤ 55-15

5 ≤ x ≤ 40

this means how many more cookies he has to make

walter needs to make at least 5 more cookies, but no more than 40 cookies; 5 ≤ x ≤ 40

heyyy

step-by-step explanation: