The sea ice area around the North Pole fluctuates between about 6 million square kilometers in September to 14 million square kilometers in March. Assuming sinusoidal fluctuation, during how many months are there less than 9 million square kilometers of sea ice?

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.

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29

jml angle lkj & angle mj = lm