Ask question

Ask question

All categories

3 years ago

12

two automobiles leave a city at the same time and travel along straight highways that differ in direction by 84 degrees. if thei

r speeds are 60 mi/hr and 45 mi/hr, approximatly how far apart are the cars at the end of 20 minutes?

1 answer:

Gelneren [198K]3 years ago

7 0

Answer:

D = 23.7 miles

Step-by-step explanation:

Given data:

$\theta=84 degrees$

Time $= 20 min = \frac{1}{3} hr$

Distance of A $= (60 mi/hr)\times \frac{1}{3} hr =20 mi$

Distance of B = $(45 mi/hr)\frac{1}{3} hr = 15 mi$

Draw a triangle.

By using cosine formula we can determine the distance between them

$D^2 = 20^2 + 15^2 - 2\times 20\times 15\times cos(84)$

$D^2 = 625 - 600 cos(84)$

$D^2 = 625 - 63.2$

$D^2 = 561.8$

Thus D = 23.7 miles

You might be interested in

Answer the following question regarding the table below. 40 41 43 44 47 47 49 40 42 43 45 47 48 49 40 42 44 45 47 48 49 41 42 44

Stels [109]

Hello there, and thank you for posting your question here on brainly.

The smallest number out of the bunch is 40.

Hope this helps!

4 0

3 years ago

Remi has $8.40 in a jar that contains only nickels and dimes. There are 108 coins in the jar. How many nickles are in the jar

Anit [1.1K]

There are 21 nickles in the jar.

4 0

3 years ago

Read 2 more answers

For each equation determine whether it is linear.

Elan Coil [88]

The only linear equation is choice d

8 0

3 years ago

Let set C = {1, 2, 3, 4, 5, 6, 7, 8} and set D = {2, 4, 6, 8}.

g100num [7]

<span>1. If it is intersection then it SHOULD be included in both the sets right? Now we know that odd numbers from 1-100 but the second set are multiples of 5 from 50-150! So we mainly need to look for common numbers which are ODD and are a MULTIPLE OF 5 BETWEEN 50 - 100!! So A={51,53,57,59,61......99} B={55,60,65,70.......95} [We stop till 100 because set A has no such element] So what is A ∩ B here? A ∩ B = {All odd numbers and multiples of 5 between 50 - 100}

</span>

6 0

3 years ago

(a) Find the equation of the normal to the hyperbola x = 4t, y = 4/t at the point (8,2).

gtnhenbr [62]

The slope of the tangent line to the curve at (8, 2) is given by the derivative $\frac{dy}{dx}$ at that point. By the chain rule,

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Differentiate the given parametric equations with respect to $t$ :

$x = 4t \implies \dfrac{dx}{dt} = 4$

$y = \dfrac4t \implies \dfrac{dy}{dt} = -\dfrac4{t^2}$

Then

$\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}$

We have $x=8$ and $y=2$ when $t=2$ , so the slope at the given point is $\frac{dy}{dx} = -\frac14$ .

The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation

$y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}$

Alternatively, we can eliminate the parameter and express $y$ explicitly in terms of $x$ :

$x = 4t \implies t = \dfrac x4 \implies y = \dfrac4t = \dfrac4{\frac x4} = \dfrac{16}x$

Then the slope of the tangent line is

$\dfrac{dy}{dx} = -\dfrac{16}{x^2}$

At $x = 8$ , the slope is again $-\frac{16}{64}=-\frac14$ , so the normal has slope +4, and so on.

5 0

2 years ago