If our visible universe is 144 billion light years in diameter, how many kilometres is this?

In this question, i is a unit vector due east and j is a unit vector due north. A cyclist rides at a speed of 4 m/s on a bearing

d1i1m1o1n [39]

Answer:

a) 1.04i + 3.86j

b) magnitude = 8; bearing = 302.7°

Step-by-step explanation:

a)

The first diagram represents the velocity vector of the cyclist.

To express this vector in the form xi + yj, we have to find the components of the vector in the horizontal (i) and vertical (j) directions.

If we consider the horizontal component of the vector to be $x$ , and the vertical component to be $y$ , then:

• horizontal component ⇒ $sin (15^{\circ}) = \frac{x}{4}$

⇒ $x = 4\space\ sin(15^{\circ})$

⇒ $x \approx \bf 1.04$

• vertical component ⇒ $cos(15^{\circ}) = \frac{y}{4}$

⇒ $y = 4 \space\ cos(15^{\circ})$

⇒ $y \approx \bf 3.86$

Now that we have the values of both the horizontal and vertical component, we can write the vector in the form of xi + yj:

vector ⇒ 1.04i + 3.86j

b)

The second diagram shows the first vector (red), the second vector (blue), and the resultant vector v (black). The dashed lines represent the components of the respective vectors.

To add two vectors given their magnitudes and direction, we have to add their components.

In order to find the horizontal and vertical components of the given vectors, we can use a method similar to that used above, so that:

○ For the first vector (magnitude 6):

• horizontal component ⇒ $x = 6 \space\ sin (60^{\circ})$

⇒ $\bf 5.2$

• vertical component ⇒ $y = 6 \space\ cos(60^{\circ})$

⇒ $y = \bf 3$

○ For the second vector (magnitude 2):

• horizontal component ⇒ $x = 2 \space\ cos (40^{\circ})$

⇒ $\bf 1.5$

• vertical component ⇒ $y = 2 \space\ sin(40^{\circ})$

⇒ $\bf 1.3$

Now we can add the respective components together:

v = 5.2i + 3j + 1.5i + 1.3j

⇒ (5.2 + 1.5)i + (3 + 1.3)j

⇒ 6.7i + 4.3j

∴ Magnitude of v ⇒ $|v| = \sqrt{(6.7)^2 + (4.3)^2}$

⇒ $|v| \approx \bf 8$

To find the bearing of v, we have to first calculate the angle marked $\alpha$ :

$tan \alpha = \frac{4.3}{6.7}$

⇒ $\alpha = tan^{-1}(\frac{4.3}{6.7})$

⇒ $\alpha = \bf 32.7^{\circ}$

∴ Bearing = 270° + 32.7°

= 302.7°

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The point where a function (for our problem its the cubic function shown)

cuts/crosses the x-axis

and/or

touches the x-axis

is a zero of the function.

From the graph shown, we can clearly see that it cuts the x-axis at -1 and touches the x-axis at 2.

So the zeros are at -1 and 2.

ANSWER: {-1,2}

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Step-by-step explanation:

-5 i think hope that helps

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