Given:
Volume of toy=231 cm³
Diameter of a hemisphere= 7 cm
cone and hemisphere have equal radius.
radius of hemisphere = radius of cone = 3.5 cm
Height of hemisphere= radius of hemisphere= 3.5 cm
Let H be the height of toy
H= height of cone+ height of hemisphere
H = h + r , ( h = height of cone)
H = h + 3.5
volume of toy = volume of cone + volume of hemisphere
Volume of toy= 1/3πr²h + 2/3πr³
Volume of toy=πr²/3(h+2r)
231 = (22/7)×(3.5)² ×(1/3)(h+2×3.5)
231×3 =( 22×3.5×3.5)/7 (h+7)
h+7 = (231×3×7)/(3.5×3.5×22)
h+7 = (3×3)/(.5×.5×2)
h+7= 900 /50= 90/5= 18
h+7= 18
h=18-7
h= 11 cm
Height of toy = h+r
Height of toy = 11+3.5= 14.5
Height of toy =14.5 cm
Hence, the height of the toy = 13.5 cm
Answer:
c
Step-by-step explanation:
i am a really good guesser and i think it is c
<h3>Answer: Approximately 6.4 units</h3>
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Explanation:
The origin is the point (0,0)
Use the distance formula to find the distance from (0, 0) to (4, -5)
Let
be our two points. Plug those values into the distance formula below and use a calculator to compute
The distance between the two points (0,0) and (4,-5) is approximately 6.4 units.
Answer:
10
Step-by-step explanation:
5 - 5 = 0 You would want to subtract five from zero giving you -5.
Answer:
there is no greatest load
Step-by-step explanation:
Let x and y represent the load capacities of my truck and my neighbor's truck, respectively. We are given two relations:
x ≥ y +600 . . . . . my truck can carry at least 600 pounds more
x ≤ (1/3)(4y) . . . . . my truck carries no more than all 4 of hers
Combining these two inequalities, we have ...
4/3y ≥ x ≥ y +600
1/3y ≥ 600 . . . . . . . subtract y
y ≥ 1800 . . . . . . . . multiply by 3
My truck's capacity is greater than 1800 +600 = 2400 pounds. This is a lower limit. The question asks for an <em>upper limit</em>. The given conditions do not place any upper limit on truck capacity.
