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Effectus [21]
3 years ago
9

Given the following linear function, sketch the graph of the function and find the domain and range. f(x)=2/3-3

Mathematics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

f(x)= 2/3-3 is undefined.

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Solving a Work Problem
RUDIKE [14]

Answer:

3.4 hours

Step-by-step explanation:

6 0
3 years ago
From deltamath.com Help me please i really don't get it.
Brums [2.3K]

Answer:

cos = adjacent/hypotenuse

3/square root13

now you do square root 13 x3 divided by square root 13^2

= 3square root13/13

hope that answers your question

3 0
3 years ago
Ashley is riding her bicycle. She takes 2 hours to ride 12.8 kilometers. What is her speed?
Lemur [1.5K]

Answer:

6.4 kmph

Step-by-step explanation:

12.8/2=6.4 kmph

7 0
3 years ago
What is m∠G to the nearest tenth? A right angle triangle EFG. The length of EF is 28.2 and GF is 45.8. Angle GFE is given as 90
mixas84 [53]

Answer:

31.6°

Step-by-step explanation:

The triangle given us a right angled Triangle

We solve the above question using the trigonometric function of Tangent

The formula is given as:

tan θ = opp/adj

θ = m∠G = ?

Opposite = EF is 28.2

Adjacent = GF is 45.8

tan θ = 28.2/45.8

θ = arc tan (28.2/45.8)

θ = 31.621459805°

Approximately

m∠G = 31.6°

7 0
3 years ago
At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m
beks73 [17]
<h2>Answer:</h2>

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=4.9t^{2}

and,

v_{x}=1.39a_{x}+2.5

<h2>Step-by-step explanation:</h2>

In the question,

Taking the elevation of pool along the y-axis, and length of the board along the x-axis.

On drawing the illustration in the co-ordinate system we get,

lₓ = 2 m

uₓ = 2.5 m/s

and,

h_{y}=9.5\,m

So,

From the equations of the laws of motion we can state that,

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}

So,

On putting the values we can say that,

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=(0)t+\frac{1}{2}(9.8)t^{2}\\t^{2}=\frac{9.5}{4.9}\\t^{2}=1.93\\t=1.39\,s

Now,

The <u>equation of the motion in the horizontal</u> can be given by,

v_{x}=u_{x}+a_{x}t\\v_{x}=2.5+a_{x}(1.39)\\So,\\v_{x}=1.39a_{x}+2.5

<em><u>Therefore, the equations of the motions in the horizontal and verticals are,</u></em>

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=4.9t^{2}

and,

v_{x}=1.39a_{x}+2.5

6 0
3 years ago
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