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Fantom [35]
3 years ago
9

Find the horizontal asymptote of of the graph of y=(-3x^6+5x+3)/9x^6+6x+4

Mathematics
1 answer:
NNADVOKAT [17]3 years ago
5 0

Answer:

  y = -1/3

Step-by-step explanation:

As x gets large, the value of the expression approaches the ratio of the highest-degree terms in numerator and denominator:

  y = -3x^6/(9x^6) = -3/9

  y = -1/3

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Answer:

f ( 7 ) = 342  

Step-by-step explanation:

Step 1:

f ( x ) = x³  - 1          Function

Step 2:

f ( 7 ) = 7³ - 1        Input x value

Step 3:

f ( 7 ) = 343 - 1        Exponents

Answer:

f ( 7 ) = 342          Subtract

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4 0
3 years ago
A point H is 20m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the po
astra-53 [7]

Answer:

a. See Attachment 1

b. PT = 12.3\ m

c. HT = 31.1\ m

d. OH = 28.4\ m

Step-by-step explanation:

Calculating PT

To calculate PT, we need to get distance OT and OP

Calculating OT;

We have to consider angle 50, distance OH and distance OT

The relationship between these parameters is;

tan50 = \frac{OT}{20}

Multiply both sides by 20

20 * tan50 = \frac{OT}{20} * 20

20 * tan50 = OT

20 * 1.1918 = OT

23.836  = OT

OT = 23.836

Calculating OP;

We have to consider angle 30, distance OH and distance OP

The relationship between these parameters is;

tan30 = \frac{OP}{20}

Multiply both sides by 20

20 * tan30 = \frac{OP}{20} * 20

20 * tan30 = OP

20 * 0.5774= OP

11.548 = OP

OP = 11.548

PT = OT - OP

PT = 23.836 - 11.548

PT = 12.288

PT = 12.3\ m (Approximated)

--------------------------------------------------------

Calculating the distance between H and the top of the tower

This is represented by HT

HT can be calculated using Pythagoras theorem

HT^2 = OT^2 + OH^2

Substitute 20 for OH and OT = 23.836

HT^2 = 20^2 + 23.836^2

HT^2 = 400 + 568.154896

HT^2 = 968.154896

Take Square Root of both sides

HT = \sqrt{968.154896}

HT = 31.1\ m <em>(Approximated)</em>

--------------------------------------------------------

Calculating the position of H

This is represented by OH

See Attachment 2

We have to consider angle 50, distance OH and distance OT

The relationship between these parameters is;

tan50 = \frac{OH}{OT}

Multiply both sides by OT

OT * tan50 = \frac{OH}{OT} * OT

OT * tan50 = {OH

OT * 1.1918 = OH

Substitute OT = 23.836

23.836 * 1.1918 = OH

28.4= OH

OH = 28.4\ m<em> (Approximated)</em>

5 0
2 years ago
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