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2 years ago

Determine the quadrant(s) in which (x, y) could be located. (Select all that apply.)

Mathematics

1 answer:

djyliett [7]2 years ago

7 0

Answer: D. Quadrant IV

Step-by-step explanation: Numbers with a positive or negative sign are referred to be directed numbers. These numerals are easily found on the number line. 2, 5, -1, -3, -7, 9, and so on are some examples.

As a result: x > 0 and y 0 are both true.

Consider the case of specific directed numbers for x and y. Assume x = 5 and y = -2, and the following is true: (5, -2). In a Cartesian plane, this point can be found in the fourth quadrant.

As a result, in a Cartesian plane, the quadrant in which (x, y) might be found is Quadrant IV.

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The value of a stock increases at a rate of 1/2% per year. If the initial value of the stock $40 a share, when will the value of

natka813 [3]

Answer:

After 50 years the stock value will be $50 per share.

Step-by-step explanation:

Simple Interest Equation (Principal + Interest)

A = P(1 + rt)

Where:

A = Future amont = $50

P = Principal Amount = $40

r = Rate of Interest per year in decimal; r = R/100 = 0.5/100 = 0.005

t = Time Period involved in months or years

Plug in the values

50 = 40(1 + 0.005t)

50 / 40 = (1 + 0.005t)

5/4 = 1 + 0.005t

5/4 - 1 = 0.005t

0.25 = 0.005t

t = 0.25 / 0.005

t = 50 years

4 0

3 years ago

Read 2 more answers

I need help quick!! Plz I’m timed and it’s almost up!!

Pepsi [2]

Answer:

Step-by-step explanation:

4 0

2 years ago

Determine whether the given vectors are orthogonal, parallel or neither. (a) u=[-3,9,6], v=[4,-12,-8,], (b) u=[1,-1,2] v=[2,-1,1

nevsk [136]

Answer:

a) $u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

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

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

b) $u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

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

c) $u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

Step-by-step explanation:

For each 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.

Part a

u=[-3,9,6], v=[4,-12,-8,]

The dot product on this case is:

$u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

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:

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

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

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

Part b

u=[1,-1,2] v=[2,-1,1]

The dot product on this case is:

$u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

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:

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

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

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

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

Part c

u=[a,b,c] v=[-b,a,0]

The dot product on this case is:

$u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

5 0

3 years ago

Read 2 more answers

Can someone help me please?

Molodets [167]

-2 should be correct. It is an input for that given graph.

5 0

3 years ago

Help me plz!!!!!!! test tomorrow

Alex

Answer:

The number of movies purchased is 5

Step-by-step explanation:

40.00 + 3.50 = 43.50

43.50 + 3.50 = 47.00

47.00 + 3.50 = 50.50

50.50 + 3.50 = 54.00

54.00 + 3.50 = 57.50

so if you add 3.50 to 40, 5 times you'll get 57.50

Hope it helped

7 0

3 years ago