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alexandr402 [8]
3 years ago
5

The point A(5, -2) has been transformed to A'(-5, 2). The transformation is described as ______.

Mathematics
1 answer:
sesenic [268]3 years ago
3 0

Answer:The transformation is described as a rotation of 180 degrees clockwise around the origin.

Step-by-step explanation:

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a man retires at age 50 with $625,000 in savings.after 7 years of retirement, he has spent $259,000 of his savings and is consid
vaieri [72.5K]
Im not entirely sure what you are looking for, but if you are trying to find out how much he has left, you would take 625,000-259,000=366,000

He would have $366,000 left from his retirement.

Hope this helps! :)
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3 years ago
Gabe drove 124 miles in 150 minutes. find Gabe's speed in kilometers per hour. HELP PLEASE FOR MY TEST TMMRW
mario62 [17]

First, you need to convert the minutes into hours, to do this you divide the minutes by 60 (the amount of minutes in an hour). Once you've gotten the value (which is 2.5), you divide that by the amount of miles drove, then you get the unit rate, or in other words       the kilometers per hour, which equals 49.6 km/h

4 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 two numbers that have a sum of 4 and a product of -5
nekit [7.7K]

Answer:

-1 and 5

Step-by-step explanation:

7 0
3 years ago
Solve
NISA [10]

Answer:

r = 0.01

Step-by-step explanation:

In order to find r we have to square both sides of the equation

0.1 =  \sqrt{r} \\  {0.1}^{2}  =  { \sqrt{r} }^{2}

The *square root* sign cancels out the *squared* sign therefore:

{0.1}^{2}  = r \\ 0.01 = r

6 0
2 years ago
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