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3 years ago

[x^2+2y]^3 expand it

2 answers:

8 0

$(a + b) {}^{3} = a {}^{3} + 3a {}^{2} b + 3ab {}^{2} + b {}^{3} \\ \\ x {}^{6} + 6x {}^{4} y + 12x {}^{2} y {}^{2} + 8y {}^{3}$

Aleksandr-060686 [28]3 years ago

3 0

Answer:

x^6 + 6x^4 y + 12x^2 y^2 + 8y^3.

Step-by-step explanation:

Using the Binomial Theorem:

(x^2 + 2y)^3 = (x^2)^3 + 3C1 (x^2)^2 (2y) + 3C2 x^2 (2y)^2 + (2y)^3

= x^6 + 3*x^4*2y + 3*x^2*4y^2 + 8y^3

= x^6 + 6x^4 y + 12x^2 y^2 + 8y^3.

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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

Find the value of c so that the polynomial p(×) is divisible by (x-4). p(×)=cx^3-15x-68

Dvinal [7]

Answer:

Step-by-step explanation:

x-4=0

x=4

p(x)=cx^3-15x-68=0

p(x)=c(4)^3-15(4)-68=0

p(x)=64c-60+68=0

p(x)=64c+8=0

p(x)=64c=-8

p(x)=c=-8/64

p(x)=c=-1/8

7 0

3 years ago

The math clubs car can travel 9 meters in a second. The race track for the soap box derby is 63 meters long. How many seconds wi

zhenek [66]

Answer: it will take 7 seconds

Step-by-step explanation: 9 x 7 = 63

5 0

3 years ago

There are five different cards numbered 1 through 5. Assume the cards are randomly chosen. If the first card picked was a 2 and

alexandr402 [8]

Answer:

Step-by-step explanation:

3 0

3 years ago

The table contains four statements. For which function below are four statements true?

Elena L [17]

I pretty sure the best answer would be f(x)=cbrt(27-x)

cbrt means cube root by the way.

8 0

3 years ago

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