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3 years ago

What is the domain of the function shown in the mapping?

Mathematics

1 answer:

EastWind [94]3 years ago

4 0

Answer:

it's A

Step-by-step explanation:

The domain is all the input values which is x|x=-5,-3,1,2,6

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What is the quotient 2 1/5 over negative 1/10

valina [46]

Answer:

Step-by-step explanation:

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3 years ago

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Which statement is true for an equilateral triangle?

bezimeni [28]

It has all angles measuring 60°

8 0

3 years ago

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Prove the trigonometric identity

Annette [7]

Answer:

Proved See below

Step-by-step explanation:

Man this one is a world of its own :D Just a quick question are you a fellow Add Math student in O levels i remember this question from back in the day :D Anyhow Lets get started

For this question we need to know the following identities:

$1+tan^{2}x=sec^2x\\\\1+cot^2x=cosec^2x\\\\sin^2x+cos^2x=1$

Lets solve the bottom most part first:

$1-\frac{1}{1-sec^2x} \\\\$

Take LCM

$1-\frac{1}{1-sec^2x} \\\\\frac{1-sec^2x-1}{1-sec^2x} \\\\\frac{-sec^2x}{1-sec^2x} \\\\\frac{-(1+tan^2x)}{-tan^2x}$

now break the LCM

$\frac{-1}{-tan^2x}+\frac{-tan^2x}{-tan^2x}\\\\\frac{1}{tan^2x}+1\\\\cot^2x+1$

because 1/tan = cot x

and furthermore,

$cot^2x+1\\cosec^2x$

now we solve the above part and replace the bottom most part that we solved with $cosec^2x$

$\frac{1}{1-\frac{1}{cosec^2x} } \\\\\frac{1}{1-sin^2x} \\\\\frac{1}{cos^2x}\\\\sec^2x$

Hence proved! :D

4 0

3 years ago

What would the correct answer be

Karo-lina-s [1.5K]

Think it would be true although I don’t know what the transistive property is.

Hope this helps!

4 0

3 years ago

Read 2 more answers

Assume a plane is flying directly north at 200 mph, but there is a wind blowing west at 23 mph. Part I: Express both the velocit

yuradex [85]

Define i as a unit vector in the eastern direction.
Define j as a unit vector in the northern direction.

Part I
Because the wind is blowing west, its velocity vector is
-23i mph or as (-23, 0) mph
Because the plane is traveling north, its velocity vector is
200j mph or as (0, 200) mph

Part II
The actual velocity of the plane is the vector sum of the plane and wind velocities.
That is,
200j - 23i or (-23, 200) mph

Part III
The ground speed of the plane is the magnitude of its vector.
The ground speed is
√[200² + (-23)²] = 201.32 mph

The ground speed of the plane is 201.3 mph (nearest tenth)

Not:
The direction of the plane is
tan⁻¹ 23/200 = 6.56° west of north.

7 0

3 years ago