* ANSWER PLS * Convert 55° to radians.
2 answers:
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Answer:
A. between 3.0 and 3.5 and between 4.0 and 4.5
Step-by-step explanation:
The idea here is that if a function is continuous and you have points on opposite sides of the x-axis, then there must be a zero-crossing between those points. As the table and graph show, ...
... (3.0, 4.0) and (3.5, -0.2)
are points on opposite sides of the x-axis. So, there must be a zero crossing between x=3.0 and x=3.5.
Likewise, ...
... (4.0, -0.8) and (4.5, 0.1)
are points on opposite sides of the x-axis. So, there must be a zero-crossing between x=4.0 and x=4.5
The appropriate answer choice lists both of these possible zero crossings.
Answer: Cylinder uses less material to make.
Box holds more soup then cylinder.
Step-by-step explanation:
Total surface area of cuboidal box =2(lw+wh+lh), l= length, w=width, h=height
Total surface area of cylinder = where r= radius and h is height.
Given , Dimension of box : 7.5 cm by 4.7 cm by 14.5 cm
Dimension of cylinder : radius of 3.3 cm and a height of 10 cm.
Total surface area of box =2((7.5)(4.7)+(4.7)(14.5)+(7.5)(147.5)) sq. cm
= 2(1209.65)
=2419.3 sq. cm
Total surface area of cylinder =
<em>[π=3.14]</em>
(Total surface area of cylinder)<(Total surface area of box )
So, cylinder uses less material to make.
Volume of box = l x w x h
= (7.5)(4.7)(14.5) = 511.13 cubic cm
Volume of cylinder =
= 341.946 cubic cm
As (Volume of box) > (Volume of cylinder )
So, Box holds more soup then cylinder.
Answer:
(a) 1/2
(b) 1/2
(c) 1/8
Step-by-step explanation:
Since, when a fair coin is tossed three times,
The the total number of possible outcomes
n(S) = 2 × 2 × 2
= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },
Here, B : { At least two heads are observed } ,
⇒ B = {HHH, HHT, HTH, THH},
⇒ n(B) = 4,
Since,
(a) So, the probability of B,
(b) A : { At least one head is observed },
⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},
∵ A ∩ B = {HHH, HHT, HTH, THH},
n(A∩ B) = 4,
(c) C: { The number of heads observed is odd },
⇒ C = { HHH, HTT, THT, TTH},
∵ A ∩ B ∩ C = {HHH},
⇒ n(A ∩ B ∩ C) = 1,