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3 years ago

Solve this inequality: 3p – 6 > 21.

1 answer:

6 0

<span>3p - 6 > 21 <=> 3p > 21+6 <=> 3p > 27 <=> p > 27/3 <=> p > 9. So the answer is B. When solving inequalities the idea is to move the terms which contain the unknown quantity to one side and the rest of the terms to the other side. And then perform valid mathematical operations to find the interval value for the unknown quantity.</span>

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Plz answer my question quickly!!!!! First person to respond will get brainliest!!!!!

garri49 [273]

Answer:

The Correct and final Answer Is

D) Or option 4

Hope this helps!

Please mark brainliest!

Bye!

4 0

3 years ago

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What is less than 0.625

lidiya [134]

0.624 is less than 0.625

8 0

2 years ago

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Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

NEED THESE ANSWERED ASAP SO I CAN GRADUATE THIS WEEK

babunello [35]

For this case we must solve each of the equations proposed:

A) $-6 (2s-1) = 30$

We apply distributive property to the terms within parentheses:

$-12s + 6 = 30$

Subtracting 6 from both sides of the equation we have:

$-12s = 24$

Dividing between -12 on both sides of the equation:

$s = \frac {24} {- 12}s = -2$

B) $-3m = -5 (m-8)$

We apply distributive property to the terms within parentheses:

$-3m = -5m + 40$

We add 5m on both sides of the equation:

$2m = 40$

Dividing between 2 on both sides of the equation:

$m = \frac {40} {2}\\m = 20$

C) $7 (2-g) = 49$

We apply distributive property to the terms within parentheses:

$14-7g = 49$

We subtract 14 from both sides of the equation:

$-7g = 35$

Dividing between -7 on both sides of the equation:

$g = \frac {35} {- 7}\\g = -5$

D) - $7 (3w + 4) = 14$

We apply distributive property to the terms within parentheses:

$-21w-28 = 14$

We add 28 to both sides of the equation:

$-21w = 42$

Dividing between -21 on both sides of the equation:

$w = \frac {42} {- 21}\\w = -2$

Answer:

$s = -2\\m = 20\\g = -5\\w = -2$

6 0

2 years ago

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Solve for w in the proportion 32/40 = 28/w W= ?

Nana76 [90]

W= 35 this is because it is and i need 20 characters i don’t know how to explain but it is 35

8 0

2 years ago

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