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2 years ago

Does sine inverse have an exponent? If not, why is there a negative one there? If so, why can't you just divide?

Mathematics

1 answer:

Fofino [41]2 years ago

6 0

Answer:

see below

Step-by-step explanation:

the actual way of writing sin inverse is arcsine, not sin⁻¹

since the output of sin is from -1 to 1, the domain of arsince is from -1 to 1. This is different than 1/sin(x), whose domain is all real numbers. we now write sin⁻¹ to fit it on the tiny buttons on our calculators

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K(n)=n^3-3n; Find k(-5)

kozerog [31]

Answer:

The answer to your question is: k(-5) = -115

Step-by-step explanation:

Function k(n) = n³ - 3n find k(-5)

k(-5) = (-5)³ - 3(-5) substitution

k(-5) = -125 + 15 simplify

k(-5) = -115

5 0

3 years ago

Thank u for all the amazing answers everyone :)

Dennis_Churaev [7]

I'm sorry if I wasn't there to help but thank the others, and thank you :)

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3 years ago

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Gina can run 21 laps in 7 minutes.what is her average running rate in laps per minute

ivann1987 [24]

Average running time = $\frac{laps run}{time taken}$
= $\frac{21 laps}{7 mins}$
= 3 laps / min

5 0

3 years ago

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Please and Thank you!

babymother [125]

Answer:

A. 871.97 m

Step-by-step explanation:

Model this is a right triangle.

Use the sine trigonometric ratio to find the height of the weather balloon from the ground.

Sine trigonometric ratio

$\sf \sin(\theta)=\dfrac{O}{H}$

where:

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

Given:

$\theta$ = 33°
O = height
H = 1601 m

Substitute the given values into the formula and solve for O:

$\implies \sf \sin(33^{\circ})=\dfrac{height}{1601}$

$\implies \sf height=1601\sin(33^{\circ})$

$\implies \sf height=871.97\:m\:\:(nearest\:hundredth)$

3 0

1 year ago

Read 2 more answers

Convert the inequalities into equations, and then use the substitution method to find four possible vertices in the form (x, y,

alina1380 [7]

Answer:

For 2x + 5y < 100, x + y < 30. Vertices : (50,0) , (0,20) , (35,0) , (0,35) , (25,0)

Step-by-step explanation:

Considering inequality equations

i) 2x + 5y < 100 ;

ii) x + y < 35

Concerting them to equality conditions

i) 2x + 5y =100.

Intercepts : If y = 0, x = 50. So, (50,0). If x = 0, y = 20. So, (0,20)

ii) x + y = 35. Intercepts :

Intercepts : If y = 0, x = 35. So, (35,0). If x = 0, y = 35 So, (0,35)

Solving them as per equality (with substitution)

From ii), x = 35 - y . Putting the value in i)

2 (35 - y) + 5y = 100 → 70 - 2y + 5y = 100 → 70 + 3y = 100

3y = 100 - 70 → 3y = 30 → y = 30 / 3 = 10

Putting y = 10 in ii) : x = 35 - 10 = 25

As per the 'less than' inequalities : the inner 'towards origin' area from both lines is in common shaded area.

Henceworth, vertices are : (50,0) , (0,20) , (35,0) , (0,35) & intersection (25,0)

8 0

3 years ago