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Solnce55 [7]
3 years ago
6

A student graphs a line that passes through (3,15) and origin.

Mathematics
1 answer:
bogdanovich [222]3 years ago
7 0

Answer:

6,30 9,45 12,60

Step-by-step explanation:

This is a question that go on and on and on.

3,15

6,30

9,45

12,60

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Describe how the discriminant of a quadratic equation is related to the number of real solutions the equation has.
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Read 2 more answers
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>

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