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pantera1 [17]
3 years ago
12

I need help please, i don’t not understand

Mathematics
1 answer:
Leviafan [203]3 years ago
7 0
Write in 0 and 0 at the top:  this indicates that if there are zero pennies, the height of the tower is 0.  This is the point (0,0).  Another point is (4, 28):  a tower of 28 pennies is 4 cm tall.

Find the slope of the line connecting (0,0) and (4,28).
            4cm - 0 cm
    m = ----------------- = (1/7) (cm/penny)
            28 - 0

The equation we need, relating # of pennies to height of tower, is 

       t = (1/7) (cm/penny)x, where x is the # of pennies.

Thus, if x = 1, then t=1/7 cm

And, if x =15, then t = 15/7 cm

And so on.
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<em><u>ANSWER</u></em>

My answer is in the photo above

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2 years ago
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Which of these equations are in slope-intercept form?
maxonik [38]

Answer:

y = mx + b

Step-by-step explanation:

6 0
2 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
2 years ago
Find the degree of w
lorasvet [3.4K]

Sum of angles of a linear pair = 180. So angle with measurement 142 and the adjacent angle to it have a sum = 180.

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y = 38

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87+w=180

Subtracting 80 from both sides,

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4 0
3 years ago
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How could you find the area of the pink title? Choose all that apply. <br><br>​
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