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Aleksandr [31]
3 years ago
8

Which of the following shows a relationship of 4,000 as ten times greater than 400?

Mathematics
1 answer:
stira [4]3 years ago
8 0
400 times ten because when you multiply you get 40,000.
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What is the slope of the line represented by the equation y = = 4x+12<br> 2<br> N/P
Elena L [17]

Answer:

4/1

Step-by-step explanation:

hope that helps :))))

5 0
3 years ago
Anyone know how to solve this?
yuradex [85]

Answer:

<h2>\frac{16}{25}</h2>

Step-by-step explanation:

<h3>( -  \frac{4}{5})^{2}</h3><h3>{( - 1)}^{2}  \frac{ {4}^{2} }{ {5}^{2} }</h3><h3>1 \frac{ {4}^{2} }{ {5}^{2} }</h3><h3>\frac{ {4}^{2} }{ {5}^{2} }</h3><h3>\frac{16}{25}</h3><h3>Hope it is helpful....</h3>
5 0
3 years ago
I came up with 4 let me know if that's right
Anastasy [175]
Are you doing FLVS? IF so I need hep
7 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
Determine the equation of the tangent to x− 3x +1 at the point (3,1)
wariber [46]

9514 1404 393

Answer:

  y = 3x -8

Step-by-step explanation:

We assume you want the tangent to the parabola y = x² -3x +1 at the given point. The slope is found using the derivative of the function at that point.

  y' = 2x -3

At x=3, the slope is ...

  y' = 2(3) -3 = 3

The equation of the line through point (3, 1) with a slope of 3 is ...

  y -1 = 3(x -3) . . . .  use the point-slope form of the equation for a line

  y = 3x -9 +1 . . . . . eliminate parentheses, add 1

  y = 3x -8

4 0
3 years ago
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