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1 year ago

A boat is heading towards a lighthouse, whose beacon-light is 139 feet above the

Mathematics

1 answer:

Novay_Z [31]1 year ago

5 0

The horizontal distance between light house and boat is 1588.78 feet approximately.

The figure is given by,

Here, AB = height of the lighthouse bacon light above the water = 139 feet

Now angle ACB = 5 degree

Let the horizontal distance of light house from the boat = BC = x feet

So by trigonometric function we get,

tan 5 = AB/BC

tan 5 = 139/x

x = 139/tan 5 = 1588.78 (approximately)

Hence the horizontal distance between light house and boat is 1588.78 feet approximately.

To know more about Trigonometric Function refer to:

brainly.com/question/1143565

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Round 1627187 to the nearest ten?

Eduardwww [97]

Answer:

1627190

Step-by-step explanation:

(see attached for reference)

Given the number 1627187, we can see that the number in the tens place is the number 8.

How we round this depends on the number immediately to the right of this number. (i.e the digit in the ones place)

Case 1: If the digit in the ones place is less less than 5, then the number in the tens place remains the same and replace all the digits to its right with zeros

Case 2: If the digit in the ones places is 5 or greater, then we increase the digit in the tens place and replace all the digits to its right with zeros.

In our case, the digit in the ones places is 7, this greater than 5, hence according to Case 2 above, we increase the digit in the tens place by one (from 8 to 9) and replace all the digits to its right by zeros giving us:

1627190

7 0

3 years ago

Read 2 more answers

Jaquon made 252 calls in 36 days. What is the average number of calls he made in a day?

Rainbow [258]

Do 252 divided by 36 and you should get 7. so the average number of calls would be 7

5 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=f%28x%29%20-%20%5Cfrac%7Bx%5E%7B2%7D-4%20%7D%7Bx%5E%7B4%7D%20%2Bx%5E%7B3%7D%20-4x%5E%7B2%7D-4%

Llana [10]

a) The given function is

$f(x)=\frac{x^2-4}{x^4+x^3-4x^2-4}$

The domain refers to all values of x for which the function is defined.

The function is defined for

$x^4+x^3-4x^2-4\ne0$

This implies that;

$x\ne -2.69,x\ne 1.83$

b) The vertical asymptotes are x-values that makes the function undefined.

To find the vertical asymptote, equate the denominator to zero and solve for x.

$x^4+x^3-4x^2-4=0$

This implies that;

$x= -2.69,x=1.83$

c) The roots are the x-intercepts of the graph.

To find the roots, we equate the function to zero and solve for x.

$\frac{x^2-4}{x^4+x^3-4x^2-4}=0$

$\Rightarrow x^2-4=0$

$x^2=4$

$x=\pm \sqrt{4}$

$x=\pm2$

The roots are $x=-2,x=2$

d) The y-intercept is where the graph touches the y-axis.

To find the y-inter, we substitute;

$x=0$ into the function

$f(0)=\frac{0^2-4}{0^4+0^3-4(0)^2-4}$

$f(0)=\frac{-4}{-4}=1$

e) to find the horizontal asypmtote, we take limit to infinity

$lim_{x\to \infty}\frac{x^2-4}{x^4+x^3-4x^2-4}=0$

The horizontal asymtote is $y=0$

f) The greatest common divisor of both the numerator and the denominator is 1.

There is no common factor of the numerator and the denominator which is at least a linear factor.

Therefore the function has no holes.

g) The given function is a proper rational function.

There is no oblique asymptote.

See attachment for graph.

6 0

3 years ago

Please help with any of this Im stuck and having trouble with pre calc is it basic triogmetric identities using quotient and rec

german

How I was taught all of these problems is in terms of r, x, and y. Where sin = y/r, cos = x/r, tan = y/x, csc = r/y, sec = r/x, cot = x/y. That is how I will designate all of the specific pieces in each problem.

Let's start with sin here. $\frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}}$ Therefore, because sin is y/r, r = $\sqrt{5}$ and y = +2. Moving over to cot, which is x/y, x = -1, and y = 2. We know y has to be positive because it is positive in our given value of sin. Now, to find cos, we have to do x/r.

cos = $\frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}$

Let's start with secant here. Secant is r/x, where r (the length value/hypotenuse) cannot be negative. So, r = 9 and x = -7. Moving over to tan, x must still equal -7, and y = $4\sqrt{2}$ . Now, to find csc, we have to do r/y.

csc = $\frac{9}{4\sqrt{2}} = \frac{9\sqrt{2}}{8}$

The pythagorean identities are

sin^2 + cos^2 = 1,

1 + cot^2 = csc^2,

tan^2 + 1 = sec^2.

Let's take a look at the information given here. We know that cos = -3/4, and sin (the y value), must be greater than 0. To find sin, we can use the first pythagorean identity.

sin^2 + (-3/4)^2 = 1

sin^2 + 9/16 = 1

sin^2 = 7/16

sin = $\sqrt{7/16}$ = $\frac{\sqrt{7}}{4}$

Now to find tan using a pythagorean identity, we'll first need to find sec. sec is the inverse/reciprocal of cos, so therefore sec = -4/3. Now, we can use the third trigonometric identity to find tan, just as we did for sin. And, since we know that our y value is positive, and our x value is negative, tan will be negative.

tan^2 + 1 = (-4/3)^2

tan^2 + 1 = 16/9

tan^2 = 7/9

tan = - $\sqrt{7/9} = \frac{-\sqrt{7}}{3}$

Let's take a look at the information given here. If we know that csc is negative, then our y value must also be negative (r will never be negative). So, if cot must be positive, then our x value must also be negative (a negative divided by a negative makes a positive). Let's use the second pythagorean identity to solve for cot.

1 + cot^2 = $(\frac{-\sqrt{6}}{2})^{2}$

1 + cot^2 = 6/4

cot^2 = 2/4

cot = $\frac{\sqrt{2}}{2}$

tan = $\sqrt{2}$

Next, we can use the third trigonometric identity to solve for sec. Remember that we can get tan from cot, and cos from sec. And, from what we determined in the beginning, sec/cos will be negative.

$(\frac{2}{\sqrt{2}})^2$ + 1 = sec^2