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2 years ago

Select all the expressions that are equal to 12x + 36

Mathematics

1 answer:

kvv77 [185]2 years ago

8 0

Answer:

The answer are

12(x + 24) ❌

x(12 + 36) ❌

3(4x + 12) ✅

4(3x + 9) ✅

4(3x + 9x) (Again?) ✅

Step-by-step explanation:

12x + 36 equals 12(x + 3) (Keep that in mind)

12(x + 24) is not correct because the result is 12x + 288. Therefore, the first option is incorrect. ❌

x(12 + 36) isn't correct either. This is because the result is 48x. Therefore, the second option is incorrect too. ❌

3(4x + 12) is correct because the result is 12x+36, but wait! Remember how I told you to keep 12(x + 3) in mind? Well, when you solve 12x+36, the result is 12(x + 3). Therefore, the third option is correct. ✅

4(3x + 9) is correct because the result is 12x+36 just like the third option, but wait! Since it is the same as the third option, that means that the result will be 12(x + 3). Therefore, the fourth option is correct too.✅

Hope this helped! Keep strong and good luck! :)

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Let me know if you want me to help you with anything else. I'll be happy enough to help you. :>

- ✨7272033Alt✨

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Step-by-step explanation:

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3 years ago

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Kaylis [27]

Answer:

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Step-by-step explanation:

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2 years ago

Can someone please help!!

anastassius [24]

X=f
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2 years ago

Choose the best coordinate system to find the volume of the portion of the solid sphere rho <_4 that lies between the cones φ

MrRissso [65]

Answer:

So, the volume is:

$\boxed{V=\frac{128\sqrt{2}\pi}{3}}$

Step-by-step explanation:

We get the limits of integration:

$R=\left\lbrace(\rho, \varphi, \theta):\, 0\leq \rho \leq 4,\, \frac{\pi}{4}\leq \varphi\leq \frac{3\pi}{4},\, 0\leq \theta \leq 2\pi\right\rbrace$

We use the spherical coordinates and we calculate a triple integral:

$V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_0^4 \rho^2 \sin \varphi \, d\rho\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \left[\frac{\rho^3}{3}\right]_0^4\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \cdot \frac{64}{3} \, d\varphi\, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} [-\cos \varphi]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\$

we get:

$V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\V=\frac{64\sqrt{2}}{3}\cdot[\theta]_0^{2\pi}\\\\V=\frac{128\sqrt{2}\pi}{3}$

So, the volume is:

$\boxed{V=\frac{128\sqrt{2}\pi}{3}}$

4 0

2 years ago

So ya i don get is it is crazy

Furkat [3]

so ya I don't see it, it must be crazy

3 0

2 years ago

Read 2 more answers