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3 years ago

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The side of a square flower bed is 1m 80cm. It is enlarged by digging a strip 20cm wide all around it. find. a) the area of the

enlarged flower bed b) the increase in the area of the flower bed.

1 answer:

andreyandreev [35.5K]3 years ago

3 0

Answer:a)40,000 b) 7600cm2

Step-by-step explanation:

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ICE Princess25 [194]

Answer:

7x² + 5x

Dtep-by-step explanation:

(9x² + 8x) - (2x² + 3x)

9x² + 8x - 2x²- 3x

7x² + 5x

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3 years ago

Please help. will grant brainliest !

Igoryamba

Answer:

45

Step-by-step explanation:

Substitute i = 1, 2, 3, 4, 5 into the expression and sum the terms

3(1) + 3(2) + 3(3) + 3(4) + 3(5)

= 3 + 6 + 9 + 12 + 15

= 45

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3 years ago

If f(x) = 7x − 1, what does f(12) represent?

Rasek [7]

Hello :
f(12) represent : <span>B. The value of (7x − 1) when x = 12</span>

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4 years ago

Read 2 more answers

14 ponds equals how many onces

Vlada [557]

One pound is equivalent to 16 ounces.

Thus, multiply 14 pounds with 16 to get the number of ounces.

$14 \times 16 = 224$

14 pounds equals 224 ounces. Let me know if you need any clarifications, thanks!

~ Padoru

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3 years ago

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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

3 years ago