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3 years ago

Find the difference 3 d and 7 d

Mathematics

2 answers:

Thepotemich [5.8K]3 years ago

8 0

Answer:

-4d

Step-by-step explanation:

3d - 7d

=> 3 x D - 7 x D

=> (3 - 7)(D)

=> -4 x D

=> -4d

hoa [83]3 years ago

5 0

Answer:

$-4d$

Step-by-step explanation:

We can subtract a term from another if they have the same variable. Both $3d$ and $7d$ have the variable $d$ , so we can subtract them.

When subtracting these terms, we subtract the coefficients and keep the variable.

So:

$3d - 7d = 3-7(d) = -4d$

Hope this helped!

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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

What is the vertex of the graph y = x^2 + 6x - 5 ?

Dafna11 [192]

Given: y = x^2 + 6x - 5. Then a = 1, b = 6 and c = -5.

The x-coord. of the vertex is given by x = -b / (2a), which here is x = -6 / (2*1) = -3.

Use the given formula y = x^2 + 6x - 5 to find the value of y when x = -3:

y = (-3)^2 + 6(-3) - 5 = 9 - 18 - 5 = -14

Then the vertex is (-3, -14).

6 0

4 years ago

Write algebraic expression 17 less than the product of 3 and m

motikmotik

"17 less than" implies subtraction of the product of 3 and m, so the equation would be, if solving for x:
x=3m-17

4 0

4 years ago

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tangare [24]

Answer:

340