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3 years ago

Please help right away

Mathematics

1 answer:

garik1379 [7]3 years ago

3 0

Answer:

177 balls

Step-by-step explanation:

Let's start with the space we need for each sphere. If we approximate the value of the area to that of a cube that has the same dimensions of the diameter

[tex]V=H*W*D=0.8*0.8*0.8= 0.51 inches3

I can calculate the volume of the box as

[tex]V=H*W*D=3*4*7.5= 90 inches3

Finally I divide the volume of the box between the volume I need for each sphere Vt/Vs= 90/0.51 ≅ 177 balls

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3 years ago

Evaluate the interval (Calculus 2)

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$2 \tan (6x)+2 \sec (6x)+\text{C}$

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<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

$\implies 2 \tan (6x)+2 \sec (6x)+\text{C}$

Learn more about indefinite integration here:

brainly.com/question/27805589

brainly.com/question/28155016

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4 years ago

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natima [27]

Answer

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B: 15 years

Step-by-step explanation:

A: Since $25,000 is deposited in the bank and you earn 4% interest annually over 3 years you would earn 12% interest. 12% of 25,000 is 3,000. The sum of 25,000 and 3,000 is 28,000 having you earn $28,000 over 3 years.

B: Since every year you earn 4% annual interest and you have $25,000 deposited in the bank you have to find what percentage of 25,000 equals 15,000. Since 25,000 isn't exactly easy to find multiply it by 4. That leaves you with 100,000 which we know you don't have in the bank but it will make it easier to find the percentage. You will also have to multiply 15,000 by 4 giving you 60,000. Now divide the 60,000 by 1,000 which equals 60. What's the point of the number 60? Well, that's the percentage it takes. Now divide 60 by the annual interest which is 4 and that gives you 15. 15 is the number of years it will take for the balance to reach $40,000.

Hope this helps!

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3 years ago