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3 years ago

Please help me with this quesion

Mathematics

2 answers:

Mama L [17]3 years ago

8 0

Answer:

i think it is c

Step-by-step explanation:

MrMuchimi3 years ago

7 0

Answer:

IM GUESSING THE SECOND ONE

Step-by-step explanation:

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The hyperbola below is a graph of the equation x^2/3^2 - y^2/2^2=1. Which of the points satisfy the inequality x^2/3^2 - y^2/2^2

kirill [66]

$( \frac{x}{3})^{2} - ( \frac{y}{2}) ^{2} \ \textgreater \ 1,
 x^{2} \ \textgreater \ 3^2(1+ ( \frac{y}{2} )^{2} ),
x^2\ \textgreater \ \frac{9}{4} (4+y^2)
if y=0,x^2\ \textgreater \ 9
$
x<-3,or x>3
only A and B satisfy the given condition.

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3 years ago

PLease Help Me [99 points]

xxTIMURxx [149]

A graph is a function if there is only one X value for any Y value.

Since the graph is in the shape of a U, there can only be one X for every Y value, so this graph is a function.
The answer is C.

6 0

3 years ago

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The 18 faculty members in a college math department range in age from 32 to 68. A stem plot follows: The first and third quartil

valentina_108 [34]

Answer:

$Q_{1}=39, \:and\: Q_{3}=58$

Step-by-step explanation:

1) This Stem and Leaf plot works like a Histogram. Completing the question, check below the Stem plot. Look the data out of the Stem plot and within (in the graph below)

32, 34 , 38 , 39 , 39 , 40 , 43 , 45 , 46 , 49 , 53 , 54 , 57 , 58 , 59 , 59 , 63 , 68

2) Notice how the first digit is on the Stem column (check below)

3) To find the first and Third Quartiles

Let's use the formulas to find the position, then the value. As it follows:

$Q_3=\frac{3}{4}(n+1) \Rightarrow Q_3=\frac{3}{4}(18+1) Q_3=14.25 \approx14th =58$

$Q_1=\frac{(n+1)}{4} \Rightarrow Q_1=4.75 th \approx Q_1 =5th =39$

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

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Explain how to convert 6 feet into meters.

SCORPION-xisa [38]

Its 1.8288 meters use the conversion its equals this plz mark brainlest

7 0

3 years ago

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