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Luba_88 [7]
3 years ago
15

U and v are position vectors with terminal points at (-1, 5) and (2, 7), respectively. Find the terminal point of -2u + v.

Mathematics
1 answer:
Papessa [141]3 years ago
4 0

Answer:

  (4, -3)

Step-by-step explanation:

-2u +v = -2(-1, 5) +(2, 7) = (-2(-1)+2, -2(5)+7)

  = (4, -3)

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When Famke charges her phone, it gains 2 percentage points each minute. When Frmke uses her phone, it loses 5 percentage points
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6 0
3 years ago
Find equations of the spheres with center(3, −4, 5) that touch the following planes.a. xy-plane b. yz- plane c. xz-plane
postnew [5]

Answer:

(a) (x - 3)² + (y + 4)² + (z - 5)² = 25

(b) (x - 3)² + (y + 4)² + (z - 5)² = 9

(c) (x - 3)² + (y + 4)² + (z - 5)² = 16

Step-by-step explanation:

The equation of a sphere is given by:

(x - x₀)² + (y - y₀)² + (z - z₀)² = r²            ---------------(i)

Where;

(x₀, y₀, z₀) is the center of the sphere

r is the radius of the sphere

Given:

Sphere centered at (3, -4, 5)

=> (x₀, y₀, z₀) = (3, -4, 5)

(a) To get the equation of the sphere when it touches the xy-plane, we do the following:

i.  Since the sphere touches the xy-plane, it means the z-component of its centre is 0.

Therefore, we have the sphere now centered at (3, -4, 0).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, -4, 0) as follows;

d = \sqrt{(3-3)^2+ (-4 - (-4))^2 + (0-5)^2}

d = \sqrt{(3-3)^2+ (-4 + 4)^2 + (0-5)^2}

d = \sqrt{(0)^2+ (0)^2 + (-5)^2}

d = \sqrt{(25)}

d = 5

This distance is the radius of the sphere at that point. i.e r = 5

Now substitute this value r = 5 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 5²  

(x - 3)² + (y + 4)² + (z - 5)² = 25  

Therefore, the equation of the sphere when it touches the xy plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 25  

(b) To get the equation of the sphere when it touches the yz-plane, we do the following:

i.  Since the sphere touches the yz-plane, it means the x-component of its centre is 0.

Therefore, we have the sphere now centered at (0, -4, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (0, -4, 5) as follows;

d = \sqrt{(0-3)^2+ (-4 - (-4))^2 + (5-5)^2}

d = \sqrt{(-3)^2+ (-4 + 4)^2 + (5-5)^2}

d = \sqrt{(-3)^2 + (0)^2+ (0)^2}

d = \sqrt{(9)}

d = 3

This distance is the radius of the sphere at that point. i.e r = 3

Now substitute this value r = 3 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 3²  

(x - 3)² + (y + 4)² + (z - 5)² = 9  

Therefore, the equation of the sphere when it touches the yz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 9  

(b) To get the equation of the sphere when it touches the xz-plane, we do the following:

i.  Since the sphere touches the xz-plane, it means the y-component of its centre is 0.

Therefore, we have the sphere now centered at (3, 0, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, 0, 5) as follows;

d = \sqrt{(3-3)^2+ (0 - (-4))^2 + (5-5)^2}

d = \sqrt{(3-3)^2+ (0+4)^2 + (5-5)^2}

d = \sqrt{(0)^2 + (4)^2+ (0)^2}

d = \sqrt{(16)}

d = 4

This distance is the radius of the sphere at that point. i.e r = 4

Now substitute this value r = 4 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 4²  

(x - 3)² + (y + 4)² + (z - 5)² = 16  

Therefore, the equation of the sphere when it touches the xz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 16

 

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What’s all the answers to lesson 9 algebra 1 a unit 5
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If you invest $823 at an interest rate of 3 percent,how much money will you have in total afer nine years?​
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Answer:

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Step-by-step explanation:

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