Since in the above case, the beaker has two sections each with different radius and height, we will divide this problem into two parts.
We will calculate the volume of both the beakers separately and then add them up together to get the volume of the beaker.
Given, π = 3.14
Beaker 1:
Radius (r₁) = 2 cm
Height (h₁) = 3 cm
Volume (V₁) = π r₁² h₁ = 3.14 x 2² x 3 = 37.68 cm³
Beaker 2:
Radius (r₂) = 6 cm
Height (h₂) = 4 cm
Volume (V₂) = π r₂² h₂ = 3.14 x 6² x 4 = 452.16 cm³
Volume of beaker = V₁ + V₂ = 37.68 + 452.16 = 489.84 cm³
Answer:
It’s y2-y1 over x2-x1
Y2 is 5 and y1 is 5
5-5
X2 is 7 and x1 is -2
7- negative 2
5-5 is 0
7- negative 2 is 9
Slope is 0
Step-by-step explanation:
Hope this helps:)
Answer:
See below ~
Step-by-step explanation:
Finding t₁₂ :
⇒ t₁₂ = t₁ + 11d
⇒ t₁₂ = 6 + 11(2)
⇒ t₁₂ = 6 + 22
⇒ t₁₂ = 28
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Finding S₁₂ :
⇒ S₁₂ = 12/2 × 2t₁ + (12 - 1)d
⇒ S₁₂ = 6 × 2(6) + 11(2)
⇒ S₁₂ = 6 × 12 + 22
⇒ S₁₂ = 6 × 34
⇒ S₁₂ = 204
(a) If
is the mass (in mg) remaining after
years, then
(b)
(c)
Answer:
h=2
a=4
k= -21
Step-by-step explanation:
your h and k are vertex . h= x coordinate of vertex and k= y
so formula for vertex coordinates is x= -b/2a and y= (4ac-b²)/4a
your a b c are 4 -16 and -5 respectively according to the main equation
your a is concavity which is the same number as the 'a' of the equation
