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3 years ago

10 times as much as 80

Mathematics

1 answer:

stealth61 [152]3 years ago

4 0

Times = x (multiply)

what you want to do it multiply it

10 x 80 = 800

80 x 10 = 800

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Trouble finding arclength calc 2

kiruha [24]

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

Solve for the variables x and y in the matrix

sdas [7]

plz mark me brainliest.

i have spent much time in it.

6 0

3 years ago

Two positive improper fractions are multiplied. Is the product sometimes, always or never less than 1. explain

melisa1 [442]

<span>Given situation : 2 improper fraction is multiplied. Now is the product always more than than 1?
Yes, the product is always more than 1. Because take note that in improper fraction, the numerator is always higher compare to the denominator of the given fraction. That means when you divide the numerator from the denominator, the answer would always be more than 1.
Example:
=> 5 / 4 x 4 / 2
=> 20 / 8 or equals to 5/2
now divide
=> 5 / 2
=> 2 1/2

</span>

6 0

3 years ago

Assume f(x) = -2x+8 and g(x) = 5x. What is the value of (f o g)(3)?

makkiz [27]

Answer: C

Step-by-step explanation:

(fog)(3) is the same as finding f(g(3)). To solve, we first need to solve g(3). then, we solve for f(g(3)).

g(3)=5(3) [multiply]

g(3)=15

Now, we plug in 15 into f(x).

f(g(3))=-2(15)+8 [multiply]

f(g(3))=-30+8 [add]

f(g(3))=-22

Now, we know that C is the correct answer.

4 0

3 years ago

This is a full 30 points from a overdue quiz and I’m failing this class plzz help

erastova [34]

Answer: Just use a protractor lol...

Step-by-step explanation:

6 0

3 years ago

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