See the attached image for the drawings of the problems. The figures are not to scale. The decimal values in each figure are approximate to one decimal place.
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Focus on Figure 1
sin(angle) = opposite/hypotenuse
sin(A) = BC/AC
sin(36) = 13.2/x
x*sin(36) = 13.2
x = 13.2/sin(36)
x = 13.2/0.58778525229248 <<-- make sure calc is in degree mode
x = 22.4571813404936
x = 22.5
This value is approximate (rounded to one decimal place)
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Move onto Figure 2
cos(angle) = adjacent/hypotenuse
cos(D) = DE/FD
cos(50) = y/57.4
57.4*cos(50) = y
y = 57.4*cos(50)
y = 57.4*0.64278760968653
y = 36.8960087960069
y = 36.9
This value is approximate (rounded to one decimal place)
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Finally, move onto Figure 3
tan(angle) = opposite/adjacent
tan(G) = JH/GH
tan(18) = z/10
10*tan(18) = z
z = 10*tan(18)
z = 10*0.3249196962329
z = 3.249196962329
z = 3.2
This value is approximate (rounded to one decimal place)
Answer:
The end answer is 669.4
Step-by-step explanation:
So you take the volume of the fish tank: 51*39*28=55692, but the tank is only 95% full, so you take 55692 and divide it by 100 and multiply it by 95, you get 52907.4. Then you calculate the volume of the log: 3.14*5^2*44=3454.
Now that we know these values, we take the volume of the water which is 52907.4 and add the volume of the log 3454 which gets you 56361.4.
Now we know the combined volume of the water and the log, we can remove the volume of the tank, which will still be holding the water that hasn't exceeded the capacity of the tank: 56361.4-55692=669.4
So then we know what has exceeded the capacity of the tank, which is how much water has spilled
So the water spilled equals 669.4 cubic centimetres of water.
Answer: ok so meep is gunna meep to the meep power of meep and 6 more meeps
Step-by-step explanation:
i got it right when i looked it up lol
Answer:

Step-by-step explanation:



Answer:
x₍ₙ₎ = x₍₁₎ * (-5)ⁿ⁻¹
Step-by-step explanation:
x₍ₙ₎ = x₍₁₎ * (-5)ⁿ⁻¹
x₍₁₎ = -2
x₍₂₎ = (-2) x (-5)²⁻¹ = (-2) x (-5) = 10
x₍₃₎ = (-2) x (-5)³⁻¹ = (-2) x (-5)² = -50