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Natali [406]
3 years ago
6

Determine the function which corresponds to the given graph. (3 points) a natural logarithmic function crossing the x axis at ne

gative two and y axis at one. The asymptote is x = -3

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Answer:

  y=\log_3{(x+3)}

Step-by-step explanation:

The parent log function has a vertical asymptote at x=0, so the asymptote at x=-3 indicates a left shift of 3 units.

The parent log function crosses the x-axis 1 unit to the right of the vertical asymptote, which this one does, indicating there is no vertical shift.

The parent log function has an x-value equal to its base when it has a y-value of 1. Here, the y-value of 1 corresponds to an x-value 3 units to the right of the vertical asymptote, so the base of this logarithm is 3.

The function is ...

  \boxed{y=\log_3{(x+3)}}

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Ships a and b or 1425 feet apart and detect a submarine below them
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Answer:

AX = 1084.20

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Step-by-step explanation:

See attachment for complete question

Let the position of the submarine be represented with X

Given

AB = 1425

\angle A = 59^{\circ}

\angle B = 47^{\circ}

First, we calculate angle at X.

\angle X + \angle A + \angle B = 180

\angle X + 59^{\circ} + 47^{\circ}= 180^{\circ}

\angle X = 180^{\circ} -59^{\circ} - 47^{\circ}

\angle X = 74^{\circ}

Solving (a): Distance AX: The distance between ship A and the submarine

To do this, we apply sine formula which states

\frac{a}{sin\ A} = \frac{b}{sin\ B} = \frac{c}{sin\ C}

In this case:

\frac{AB}{sin\ X} = \frac{AX}{sin\ B}

Substitute values for AB, \angle X and \angle B

\frac{1425}{sin(74^{\circ})} = \frac{AX}{sin(47^{\circ})}

Make AX the subject

AX = \frac{1425}{sin(74^{\circ})} * sin(47^{\circ})

AX = \frac{1425}{0.9613} * 0.7314

AX = \frac{1425 * 0.7314}{0.9613}

AX = \frac{1042.245}{0.9613}

AX = 1084.20

Solving (b): Distance BX: The distance between ship B and the submarine

To do this, we apply sine formula which states

In this case:

\frac{AB}{sin\ X} = \frac{BX}{sin\ A}

Substitute values for AB, \angle X and \angle A

\frac{1425}{sin(74^{\circ})} = \frac{BX}{sin(59^{\circ})}

Make BX the subject

BX = \frac{1425}{sin(74^{\circ})} * sin(59^{\circ})

BX = \frac{1425}{0.9613} * 0.8572

BX = \frac{1425* 0.8572}{0.9613}

BX = \frac{1221.51}{0.9613}

BX = 1270.69

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3 years ago
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