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2 years ago

Consider the points Q(23, 48) and R(7, 62). What is the component form of Vector Q R?

Mathematics

2 answers:

aev [14]2 years ago

8 0

Answer:

The correct option is;

LeftAngleBracket negative 16, 14 RightAngleBracket which is $\left \langle -16, 14 \right \rangle$

Step-by-step explanation:

The given coordinates of the points are;

The starting point, Q(23, 48) and the final point, R(7, 62)

Therefore, we have;

The x component = The x-value of R - The x-value of R = 7 - 23 = -16

The y component = The y-value of R - The y-value of R = 62 - 48 = 14

Therefore, the component form of the vector QR = $\left \langle -16, 14 \right \rangle$

Alika [10]2 years ago

4 0

Answer:

B on Edg.

Step-by-step explanation:

Yay

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Find the exact value of the trigonometric function given that sinU=-7/25 and cosV=-4/5 (Both U and V are in Quadrant III)

Contact [7]

cosU:-

$\\ \rm\longmapsto \sqrt{1-sin^2U}=\sqrt{1-49/625}=24/25$

As U lies in Q3

cosU=-24/25

sinV

$\\ \rm\longmapsto \sqrt{1-cos^2V}=\sqrt{1-16/25}=3/4$

As V lies in Q3

sinV=-3/5

sin(V-U)=sinVcosU-cosVsinU=(-3/5)(-24/25)-(-4/5)(-7/25)=72/125-28/125=72-28/125=44/125
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1 year ago

Find the greatest common factor of 8m 3 and 6m 4

irina [24]

ANSWER

$2{m}^{3}$

EXPLANATION

The expressions are

$8 {m}^{3} = {2}^{3} \times {m}^{3}$

and

$6 {m}^{4} = 2 \times 3 {m}^{4}$

The greatest common factor is the product the least powers of the common factors.

$= 2 {m}^{3}$

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2 years ago

3 moon + 1 star = 29 1 moon +1 moon = 15 Find value of 1 moon

Montano1993 [528]

Hi!

I can help you with joy!

3 moons+1 star equals 29.

All right.

Actually, there are a few numbers that can be hidden behind the moon:

3*8=24 + 1 star (5) = 29

But 1 moon + 1 moon equals 15.

So one moon equals 7.5.

$*GentleGirlie*$

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2 years ago

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Lelechka [254]

Answer:

Step-by-step explanation:

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For any circle with Cartesian equation
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So in the case that
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we essentially have that
$a = 5, b = 0, r^2 = 38$ .

So the centre of the circle is $(5,0)$ , and the radius is $\sqrt{38}$ .

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3 years ago