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poizon [28]
3 years ago
6

Find the equation of the line with slope m

Mathematics
1 answer:
sukhopar [10]3 years ago
4 0

Answer:

<em>y = - </em>\frac{1}{2}<em> x - 4 </em>

Step-by-step explanation:

( x_{1} , y_{1} )

y -  y_{1} = m ( x - x_{1} )

~~~~~~~~~~~~~~~~~

m = - \frac{1}{2}

( - 10, 1 )

y - 1 = - \frac{1}{2} [ x - ( - 10 )]

y - 1 =  - \frac{1}{2} x + ( - \frac{1}{2} )(10)

<em>y = - </em>\frac{1}{2}<em> x - 4</em>

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Find the 8th term of the geometric sequence 2, -10, 50, ...
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-156250

Step-by-step explanation:

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2, -10, 50, -250, 1250, -6250, 31250, -156250

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3 years ago
Who knows the answer to do this with full work
wolverine [178]

Answer:

Step-by-step explanation:

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3 years ago
What value of a makes the equation true?
zimovet [89]
The answer is A heheue

5 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
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