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Ipatiy [6.2K]
3 years ago
9

Help? Please. I really need help

Mathematics
1 answer:
tia_tia [17]3 years ago
3 0

Answer:

D

Step-by-step explanation:

Any other product in Quadrant IV will be negative also, so Quadrant IV is part of my answer. The points (x, y), with xy < 0, lie in Quadrants II and IV.

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What does k equal?<br> As in y=kx<br> Pls help
Studentka2010 [4]

Answer:

Direct Variation Use y=kx. Means “y varies directly with x.” k is called the constant of variation. “y varies inversely with x.” k is the constant of variation.

8 0
3 years ago
Help me i need this homework for today pls my peeps
Doss [256]
Ok sorry but there is no picture :(
3 0
3 years ago
A house was purchased for $89,000. After 6 years the value of the house was $101,000. Find a linear equation that models the val
mamaluj [8]

Answer:  y = 2000x + 89000

Step-by-step explanation:

Given that;

initial purchase amount = $89,000

price after 6 years = $ 101,000

years = 6

now

slope = ( 101000 - 89000) / 6

slope = 12000 / 6

slope = 2000

therefore the linear equation that models the value of the house after x years will be;

y = 2000x + 89000

7 0
3 years ago
Find each key feature of the function shown in the graph. Write the range and domain in interval notation, and assume all values
gayaneshka [121]

The domain of the graph is  -8 while the range of the graph is -4\leq x.

  • The domain of the graph is input values of "x" for which the function exists while the range is the output values "y" for which the function exists.

For the graph, the domain will be the values of the graph along the x-axis.

  • Domain = -8 (Note that -8 is not included)

  • For the range, the interval is given as: -4\leq x. Note that the value of 4 is not included since it is opened.

Learn more on domain and range here: brainly.com/question/1942755

3 0
2 years ago
find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[1,2,-2]. B
SashulF [63]

Answer:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

Step-by-step explanation:

For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

a=[1,2,-2], b=[4,0,-3,]

The dot product on this case is:

a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

|a|= \sqrt{(1)^2 +(2)^2 +(-2)^2}=\sqrt{9} =3

|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5

And finally we can calculate the angle between the vectors like this:

cos \theta = \frac{ab}{|a| |b|}

And the angle is given by:

\theta = cos^{-1} (\frac{ab}{|a| |b|})

If we replace we got:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

3 0
3 years ago
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