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Colt1911 [192]
2 years ago
8

Find the volume of a 2by2​

Mathematics
1 answer:
STALIN [3.7K]2 years ago
3 0

Answer:

24

Step-by-step explanation:

2+2+2 = 6*2 = 12*2 = 24

You might be interested in
Use the function F(3)<br> F(x) =(1/5)^x<br> A.)1/15<br> B.)1/125<br> C.)3/5<br> D.)3/125
nalin [4]
Here is your answer

B) 1/125

REASON:

Given,

F (x)={(1/5)}^{x}

for F (3)

Putting x=3, in above function we get

F (3)= {(1/5)}^{3}= 1/125


HOPE IT IS USEFUL
6 0
3 years ago
Jeremy bought a 5-kilogram can of peanuts for $4.50. What is the unit price?
Goryan [66]

Answer:

$0.90

Step-by-step explanation:

you would divide it by 5

$4.50/5=$0.90

hope this helps

brainliest?

6 0
3 years ago
Name all similar triangles to triangle CDE.
blondinia [14]

Answer:ceo

Step-by-step explanation:

Ceo

4 0
3 years ago
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
A translation moves point X to X' using the rule (x,y) → (x-2, y + 1). If X' is (3,-4), what was the
Dimas [21]

Answer:

X' is (5,-5)

Step-by-step explanation:

Knowing that this is just a translation, we can reverse the (polarity?) of each equation, such as changing -5 to +5 to find the original answer.

With that in mind, we can add 2 to X, and -1 from Y, which puts us at, 5 on the X axis, and -5 on the Y axis.

3 0
3 years ago
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