Applying the required rule or theorem, it can be concluded that the second biker is farther from the parking lot. The distance of the bikers to the parking lot are:

i. First biker = 17.0 km

ii. Second biker = 20.22 km

The path of travel of both bikers would form a triangle. Applying the Pythagoras theorem to the path of the first biker would give his distance from the starting point. While applying the cosine rule to the path of second rider would gives his distance to the starting point.

Thus,

a. To determine the distance of the first biker from the parking lot.

Let the required distance be represented by x. Applying the Pythagoras theorem, we have:

$hyp^{2}$ = $adj 1^{2}$ + $adj 2^{2}$

$x^{2}$ = $8^{2}$ + $15^{2}$

= 64 + 225

= 289

x = $\sqrt{289}$

= 17

x = 17 km

Thus, the first biker is 17.0 km from the starting point.

b. To determine the distance of the second biker from the parking lot.

Let the required distance be represented by x. So that applying the cosine rule, we have:

$c^{2}$ = $a^{2}$ + $b^{2}$ - 2ab Cos θ

$x^{2}$ = $8^{2}$ + $15^{2}$ - 2(15*8) Cos (180 - 20)

= 64 + 225 - 240 Cos 160

= 289 - 240 * -0.5

$x^{2}$ = 289 + 120

= 409

x = $\sqrt{409}$

= 20.2237

x = 20.22 km

Thus, the second biker is 20.22 km from the starting point.

Therefore, the second biker is farther from the parking lot.

A sketch of the path of travel for the two bikers is attached for more clarifications.

Visit: brainly.com/question/22699651

7 0

2 years ago

Graph the following equation: y= x-2 y= -1/5x - 8

never [62]

Answer:

Use the app desmos is really good

6 0

3 years ago

20 POINTS HELPPPPP

kiruha [24]

For this case we have the following expression:

$\frac {216 ^ {n-2}} {(\frac {1} {36}) ^ {3n}} = 216$

We multiply both sides by: $(\frac {1} {36}) ^ {3n}$

$216 ^ {n-2} = 216 * (\frac {1} {36}) ^ {3n}$

We divide both sides by 216:

$\frac {216 ^ {n-2}} {216} = (\frac {1} {36}) ^ {3n}$

To divide powers of the same base, we place the same base and subtract the exponents:

$216 ^ {n-2-1} = (\frac {1} {36}) ^ {3n}\\216 ^ {n-3} = (\frac {1} {36}) ^ {3n}$

Rewriting:

$(6 ^ 3) ^ {n-3} = (\frac {1} {6 ^ 2}) ^ {3n}\\6 ^ {3n-9} = \frac {1} {6 ^ {6n}}\\6^{ 3n-9} * 6^{ 6n} = 1$

To multiply powers of the same base, we place the same base and add the exponents:

$6^{ 3n-9 + 6n} = 1\\6^{ 9n-9} = 1$

We know that any number raised to zero is 1, $a ^ 0 = 1.$

So, for equality to be true:

$9n-9 = 0\\9n = 9\\n = \frac {9} {9}\\n = 1$

Answer:

$n = 1$

3 0

3 years ago

Read 2 more answers

PLEASE HELP ME WITH THIS QUESTION!! Thanks

Gre4nikov [31]

Answer:

20% probability

Step-by-step explanation:

First, we need to calculate the area of all the figures inside the rectangle.

Area of triangle= 1/2 BH

So, 3x2=(6)

Area of Square= L^2

3x3=(9)

Area of rectangle= LxW

=60.

60+9+6= 75.

75 is the total area.

Now, lets find the added area of only the triangle and square.

9+6= 15

The probability will be 15/75=

Convert into percent form

= 20%

Hope this helps and please mark me brainliest :)

7 0

1 year ago