Answer:
150.833... or 150 and 5/6 crates
Step-by-step explanation:
We can create an equation to solve this.
Let x represent the number of 120 kilogram crates that can be loaded into the chipping container.
120x (weight of the crates) + 5900 (weight already in the container) = 24000 (max weight in the container)
We'll start by subtracting 5900 from both sides to try to isolate x, the number of containers we still need to meet 24000 kilograms.
120x = 18100
Now we just have to isolate x by dividing by 120.
x = 150.83333333333333333333333333333
If you want to use a mixed fraction form for this solution since the 3's keep repeating, it would be 150 and 5/6 crates.
Answer:
<em>5(4) - 6 + 5(4) - 6</em>
Step-by-step explanation:
k(x) = 5x - 6
(k+k)(x) = 5x - 6 + 5x - 6
(k+k)(4) = 5(4) - 6 + 5(4) - 6
=> Option D is correct
Hope this helps!
Answer:
3493.70
Step-by-step explanation:
if you multipy all these in a calculator you get a long number, look at the third digit behind the decimal, if its 5+ then make the second digit behind the decimal one number higher. If its 4 or under than keep the number the same. To round, cut off all the digits behind the second digit thats past the secimal
Answer:
The radius of the circle P = 2√10 = 6.325
Step-by-step explanation:
∵ AB is a tangent to circle P at A
∴ (AB)² = BC × BE
∵ BC = 8 , AB = 12 , ED = 6
∵ BE = ED + DC + CB
∴ BE = 6 + CD + 8 = 14 + CD
∴ (12)² = 8 × (14 + DC) ⇒ (12)²/8 = 14 + CD ⇒ CD = (12)²/8 - 14
∴ CD = 4
Join PC and PE (radii)
In ΔBDC and ΔPDE ⇒ ∵ ∠PDC = Ф , ∴ ∠PDE = 180 - Ф
Use cos Rule:
∵ r² = (PD)² + (DC)² - 2(PD)(DC)cosФ
∴ r² = 16 + 16 - 32cosФ = 32 - 32cosФ ⇒ (1)
∵ r² = (PD)² + (DE)² - 2(PD)(DE)cos(180 - Ф) ⇒ cos(180 - Ф) = -cosФ
∴ r² = 16 + 36 + 48cosФ = 52 + 48cosФ ⇒ (2)
∵ (1) = (2)
∴ 32 - 32 cosФ = 52 + 48cosФ
∴ 32 - 52 = 48cosФ + 32cosФ
∴ -20 = 80cosФ
∴ cosФ = -20/80 = -1/4
∴ r² = 32 - 32(-1/4) = 32 + 8 = 40
∴ r = √40 = 2√10 = 6.325
Answer:
heres your answer
Step-by-step explanation:
here you go miss.
