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Thepotemich [5.8K]
3 years ago
11

Geometry math question please help

Mathematics
1 answer:
Nikolay [14]3 years ago
4 0

AD is perpendicular to BC so slope Ad = -1/slope of BC

Slope of BC = (2 - (-1)) / (- 6- 3) = 3 / -9 = -1/3

So slope of AD = -1 / (-1/3) = -3/-1 = 3 Answer

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5 days a week
9 hours a day
there are 52 weeks which means he works 260 days and 2340 hours a year
divide 29000 by 2340 and u get 12.4 per hour
7 0
2 years ago
What is the area of the figure?<br><br> Enter your answer in the box.<br><br> ______<br> ______ in²
Yuliya22 [10]
The area of the figure should be 112 in²
4 0
2 years ago
Janine has job offer sat two companies. One company offers a starting salary of 28,000 with a raise of 3,000 each year. The othe
DochEvi [55]
28,000 +3,000= 31,000 +3,000 = 34,000 +3,000 = 37,000 +3,000 = 40,000 +3,000 = 43,000 + 3,000 = 46,000

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I hope i did that right


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
2 years ago
1. What is the rate of change for the<br> equation 3x - 6y = 30 ?
vodka [1.7K]

Answer:

average rate of change is 1/2

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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