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4 years ago

14

A cab company charges three dollars per cab ride plus an additional two dollar per mile driven how long is a cab ride that cost

$9? Right and solve an equation to find the answer

1 answer:

Sav [38]4 years ago

6 0

If the ride cost $9 then the miles will be 3 miles.

Step-by-step explanation:

In this question we are given:

Cost of one cab ride R = $3

Additional Cost per mile M = $2

So, Total cost of Cab Ride = $3 + $2M

If cost of cab ride is $9 then we need to find the miles

$3 + $2M = $9

Solving

$2M = $9 - $3

$2M = $6

=> M = $6/$2

M = 3 miles

So, if the ride cost $9 then the miles will be 3 miles.

Learn more:

You can find more information at:

brainly.com/question/11207748

brainly.com/question/1600376

Keywords: algebraic expressions, solve an equation

#learnwithBrainly

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Trouble finding arclength calc 2

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Answer:

$S\approx1.1953$

Step-by-step explanation:

So we have the function:

$y=3-x^2$

And we want to find the arc-length from:

$0\leq x\leq \sqrt3/2$

By differentiating and substituting into the arc-length formula, we will acquire:

$\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+4x^2} \, dx$

To evaluate, we can use trigonometric substitution. First, notice that:

$\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+(2x)^2} \, dx$

Let's let y=2x. So:

$y=2x\\dy=2\,dx\\\frac{1}{2}\,dy=dx$

We also need to rewrite our bounds. So:

$y=2(\sqrt3/2)=\sqrt3\\y=2(0)=0$

So, substitute. Our integral is now:

$\displaystyle S=\frac{1}{2}\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy$

Let's multiply both sides by 2. So, our length S is:

$\displaystyle 2S=\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy$

Now, we can use trigonometric substitution.

Note that this is in the form a²+x². So, we will let:

$y=a\tan(\theta)$

Substitute 1 for a. So:

$y=\tan(\theta)$

Differentiate:

$y=\sec^2(\theta)\, d\theta$

Of course, we also need to change our bounds. So:

$\sqrt3=\tan(\theta), \theta=\pi/3\\0=\tan(\theta), \theta=0$

Substitute:

$\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{1+\tan^2(\theta)}\sec^2(\theta) \, d\theta$

The expression within the square root is equivalent to (Pythagorean Identity):

$\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{\sec^2(\theta)}\sec^2(\theta) \, d\theta$

Simplify:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta$

Now, we have to evaluate this integral. To do this, we can use integration by parts. So, let's let u=sec(θ) and dv=sec²(θ). Therefore:

$u=\sec(\theta)\\du=\sec(\theta)\tan(\theta)\, d\theta$

And:

$dv=\sec^2(\theta)\, d\theta\\v=\tan(\theta)$

Integration by parts:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\tan^2(\theta)\sec(\theta)} \, d\theta)$

Again, let's using the Pythagorean Identity, we can rewrite tan²(θ) as:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^2(\theta)-1)\sec(\theta)} \, d\theta)$

Distribute:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^3(\theta)-\sec(\theta)} \, d\theta)$

Now, let's make the single integral into two integrals. So:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta-\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)$

Distribute the negative:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta$

Notice that the integral in the first equation and the second integral in the second equation is the same. In other words, we can add the second integral in the second equation to the integral in the first equation. So:

$\displaystyle 2S= 2\int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta$

Divide the second and third equation by 2. So: $\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)$

Now, evaluate the integral in the second equation. This is a common integral, so I won't integrate it here. Namely, it is:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta))$

Therefore, our arc length will be equivalent to:

$\displaystyle 2S=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}$

Divide both sides by 2:

$\displaystyle S=\frac{1}{4}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}$

Evaluate:

$S=\frac{1}{4}((\sec(\pi/3)\tan(\pi/3)+\ln(\tan(\pi/3)+\sec(\pi/3))-(\sec(0)\tan(0)+\ln(\tan(0)+\sec(0))$

Evaluate:

$S=\frac{1}{4}((2\sqrt3+\ln(\sqrt3+2))-((1)(0)+\ln(0+1))$

Simplify:

$S=\frac{1}{4}(2\sqrt 3+\ln(\sqrt3+2)}$

Use a calculator:

$S\approx1.1953$

And we're done!

7 0

3 years ago