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4 years ago

Using elimination, find the ordered pair. 6x + 15y = 15 -6x +7y = -37

Mathematics

1 answer:

sveta [45]4 years ago

3 0

Answer:

Step-by-step explanation:

$\underline{+\left\{\begin{array}{ccc}6x+15y=15\\-6x+7y=-37\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad22x=-22\qquad\text{divide both sides by 22}\\.\qquad\qquad x=-1\\\\\text{Put the value of x to the first equation:}\\6(-1)+15y=15\\-6+15y=15\qquad\text{add 6 to both sides}\\15y=21\qquad\text{divide both sides by 15}\\y=\dfrac{21}{15}\to y=\dfrac{7}{5}$

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PLEASE HELP! I WILL GIVE BRAINLIEST <3

Helga [31]

Answer:

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Step-by-step explanation:

6 0

3 years ago

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At a lunch buffet, the price for an adult plate is $19, and the price for a child plate is $8. On Tuesday, a total of 32 plates

mestny [16]

Answer:

Adult plate = 21 , child plate = 11

Step-by-step explanation:

number of adult plate = a , no of child plate = c

$487 = $19a + $8c (equation 1)

32 = a + c (equation 2)

Multiply equation 2 by 8 gets you:

256 = 8a + 8c (equation 3)

Solve equation 1 and equation 3 simultaneously:

487 = 19a + 8c (equation 1) minus

256 = 8a + 8c (equation 3) :

231 = 11a

a = 231/11

a= 21

Substitute a = 21 into equation 2 :

32 = 21 + c

c = 32 - 21

c = 11

So 21 adult plates and 11 child plates were sold.

6 0

3 years ago

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Which set of absolute values is compared correctly?

stepan [7]

I would say D because, it makes sense the most, absolute values make the negatives positive, so the order that D goes in is correct.

8 0

3 years ago

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Which of these strategies would eliminate a variable in the system of equations?

mariarad [96]

Answer:

Multiply the bottom equation by -3/2 then add the equations.

Multiply the top equation by-3. then add the equations

Step-by-step explanation:

Given the simultaneous equation

2x- 6y=6 ... 1

6x - 4y = 2 ... 2

To eliminate a variable, we have to make the coefficient of one of the variable to be the same.

Multiply equastion 1 by -3

-6x+18y= -18

6x - 4y = 2

Add the result:

-6x + 6x + 18y-4y = -18+2

18y-4y = -18+2

14y = -18

y = -9/7

Another way is to Multiply the bottom equation by -3/2 then add the equations.

Multiplying equation 2 by -3/2 will give;

6x(-3/2) - 4y(-3/2) = 2(-3/2)

-9x + 6y = -3

Add to equation 1;

2x- 6y=6

-9x + 2x + 0 = -3+6

-7x = 3

x = -3/7

Hence the correct two options are;

Multiply the bottom equation by -3/2 then add the equations.

Multiply the top equation by-3. then add the equations

4 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