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GalinKa [24]
3 years ago
6

Owners of a recreation area are filling a small pond with water. They are adding water at a rate of

Mathematics
1 answer:
Klio2033 [76]3 years ago
8 0
W divided by T
or 600/25 correct me if im wrong
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correct answer is "NONE" NOT $40. hOPE THIS HELPS!
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3 years ago
How do you model the square root of 16
kotegsom [21]
\sqrt{16} =\sqrt{4*4}=\sqrt{4}*\sqrt{4}=2*2=4


4 0
3 years ago
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Real quick please. At a carnival it costs $6.54 for 3 tickets. Write an equation that can be used to express the relationship be
BlackZzzverrR [31]

Answer:

t=2.18n

Step-by-step explanation:

Total cost is 6.54 for 3 tickets so divide 6.54 by 3 and you get price per ticket which is 2.18. Multiply that by how many tickets you want and you get your final price.

3 0
3 years ago
The sea ice area around the North Pole fluctuates between about 6 million square kilometers in September to 14 million square ki
Aleksandr [31]

Answer:

there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

Step-by-step explanation:

Given the data in the question;

Let S(t) represent the amount sea ice around the North Pole in millions of square meters at a given time t,

t is the number of months since January.

Now, we use a cosine curve to model this scenario

Vertical shift will be;

D = ( 6 + 14 ) / 2 = 20 / 2

D = 10

Next is the Amplitude;

|A| = ( 6 - 14 ) / 2

|A| = 4

Now, the horizontal stretch factor will be;

B = 2π / 12

B = π/6

Hence;

S(t) = 4cos( π/6 × t ) + 10 ----------- let this be equation 1

Now we find when there will be less than 9 million square meters of sea ice;

S(t) = 9

so we have

9 = 4cos( π/6 × (t-2) ) + 10

9 - 10 = 4cos( π/6 × (t-2) )

-1 = 4cos( π/6 × (t-2) )

-1/4 = cos( π/6 × (t-2) )

so we have;

cos⁻¹( -1/4 ) = π/6 × (t₁-2)  -------- let this be equation 2

2π - cos⁻¹( -1/4 ) = π/6 × (t₂-2)  -------- let this be equation 3

so we solve equation 2 and 3

we have'

t₁ - t₂ = 6/π × ( 2π - cos⁻¹( -1/4 ) - cos⁻¹( -1/4 ) )

t₁ - t₂ = 6/π × ( 2π - 2cos⁻¹( -1/4 )  

t₁ - t₂ = 6/π × ( π - cos⁻¹( -1/4 )  

t₁ - t₂ = 6/π × ( π - 104.4775 )

t₁ - t₂ = 6/π × ( π - 104.4775 )  

t₁ - t₂ = 5.035

therefore, there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

6 0
3 years ago
Please help asap 65 pts
Lady bird [3.3K]
B. It would be 5/2 and -3/4.
7 0
3 years ago
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