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3 years ago

Two consecutive sides or a rhombus has lengths of 3cm and 5cm true or false?

Mathematics

1 answer:

Vesnalui [34]3 years ago

5 0

Answer:

False

A rhombus is a quadrilateral having ALL FOUR sides EQUAL.

Step-by-step explanation:

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The path of a race will be drawn on a coordinate grid like the one shown below. The starting point of the race will be at (−5.3,

m_a_m_a [10]

The Cartesian coordinate plane, specify locations by a pair of coordinates that indicate the magnitude in perpendicular, x, and y directions

The correct responses are;

Part A: The starting point is in the second quadrant and the finishing point is in the fourth quadrant.

Part B: The starting point and the check point are on different quadrant. Both points are on the same horizontal line.

Reasons:

The location of the start point of the race = (-5.3, 1)

Location of the finishing point = (1, -5.3)

Part A: The signs of the x and y-values in the quadrants are;

$\begin{tabular}{|cc|c|}Quad II, x -ve, y +ve&&Quad I, x, and y are +ve\\&&\\Quad III, x -ve, y -ve&&Quad IV, x +ve, y -ve\end{array}\right]$

The coordinates of the starting point is (-5.3, 1), therefore, given that the x-

coordinate is negative and the y-coordinate is positive, the quadrant of the

starting point is quadrant II (the second quadrant).

The quadrant of the starting point is the second quadrant.

The coordinates of the finishing point is (1, -5.3), the x-value is positive and

the y-value is negative, which indicates that the quadrant of the finishing

point is quadrant IV (the fourth quadrant).

The quadrant of the finishing point is the fourth quadrant.

Part B: The location of the checkpoint is; (5.3, 1)

The difference is that the x-value of the checkpoint is positive while the x-value of the starting point is negative, and they are therefore on different quadrants.

The similarity is that, the magnitude of the x, and y values of the coordinates of the check point and the starting point are the same. Therefore, the checkpoint and the starting point are on the same horizontal line.

Learn more about the Cartesian coordinate plane here:

brainly.com/question/17877300

brainly.com/question/16177011

8 0

3 years ago

A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now

olga_2 [115]

Answer:

1) 288.8 km due North

2) 144.9 km due East

3) 323.1 km

4) 207°

Step-by-step explanation:

Bearing: The angle (in degrees) measured clockwise from north.

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Cosine rule

$c^2=a^2+b^2-2ab \cos C$

where a, b and c are the sides and C is the angle opposite side c

-----------------------------------------------------------------------------------------------

Draw a diagram using the given information (see attached).

Create a right triangle (blue on attached diagram).

This right triangle can be used to calculate the additional vertical and horizontal distance the ship sailed after sailing north for 250 km.

Question 1

To find how far North the ship is now, find the measure of the short leg of the right triangle (labelled y on the attached diagram):

$\implies \sf \cos(75^{\circ})=\dfrac{y}{150}$

$\implies \sf y=150\cos(75^{\circ})$

$\implies \sf y=38.92285677$

Then add it to the first portion of the journey:

⇒ 250 + 38.92285677... = 288.8 km

Therefore, the ship is now 288.8 km due North.

Question 2

To find how far East the ship is now, find the measure of the long leg of the right triangle (labelled x on the attached diagram):

$\implies \sf \sin(75^{\circ})=\dfrac{x}{150}$

$\implies \sf x=150\sin(75^{\circ})$

$\implies \sf x=144.8888739$

Therefore, the ship is now 144.9 km due East.

Question 3

To find how far the ship is from its starting point (labelled in red as d on the attached diagram), use the cosine rule:

$\sf \implies d^2=250^2+150^2-2(250)(150) \cos (180-75)$

$\implies \sf d=\sqrt{250^2+150^2-2(250)(150) \cos (180-75)}$

$\implies \sf d=323.1275729$

Therefore, the ship is 323.1 km from its starting point.

Question 4

To find the bearing that the ship is now from its original position, find the angle labelled green on the attached diagram.

Use the answers from part 1 and 2 to find the angle that needs to be added to 180°:

$\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{Total\:Eastern\:distance}{Total\:Northern\:distance}\right)$

$\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{150\sin(75^{\circ})}{250+150\cos(75^{\circ})}\right)$

$\implies \sf Bearing=180^{\circ}+26.64077...^{\circ}$

$\implies \sf Bearing=207^{\circ}$

Therefore, as bearings are usually given as a three-figure bearings, the bearing of the ship from its original position is 207°

8 0

2 years ago

Read 2 more answers

I need the solution

Nataly [62]

The only thing you can do here is to try to simplify, so I'll assume that's what you want to do.

$\dfrac{b^2-a^2}{2a^2+ab-3b^2}$

The numerator is a difference of squares, so it's easy to factorize:

$b^2-a^2=(b-a)(b+a)$

In the denominator, we factorize by grouping:

$2a^2+ab-3b^2=2a^2-2ab+3ab-3b^2=2a(a-b)+3b(a-b)=(2a+3b)(a-b)$

So we have

$\dfrac{b^2-a^2}{2a^2+ab-3b^2}=\dfrac{(b-a)(b+a)}{(2a+3b)(a-b)}=-\dfrac{b+a}{2a+3b}$

5 0

4 years ago

Can someone help me please? Thanks!

bazaltina [42]

Answer:

(-2, 6), (-7, 4), (-2, 4)

Step-by-step explanation:

by reflecting in the line y = 0, change all x coordinates to the opposite and y coordinates remain

Topic: Graphs

If you like to venture further, please give my Instagram page a follow (learntionary). I'll be constantly posting math tips and notes! Thanks!

5 0

3 years ago

Read 2 more answers

2/3, 2/8, 2/6 how to do least to greatest?

Morgarella [4.7K]

Answer: 2/8, 2/6, 2/3

Step-by-step explanation:

First, find the LCM of 3, 6, and 8, in this case 24. Then, make the denominator of each fraction 24: 16/24, 6/24, 8/24. Then, simply sort the fractions from least to greatest by their numerators: 6/24, 8/24, 16/24

Hope it helps <3

5 0

3 years ago