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3 years ago

Simplify (5ab^2-4ab 7a^2b) (ab )^-1

Mathematics

1 answer:

algol133 years ago

7 0

$(5ab^2-4ab+7a^2b) (ab )^{-1 }=(5ab^2-4ab+7a^2b) a^{-1}b^{-1}\\\\=\boxed{5b-4+7a}$

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Use powers to rewrite these problems: Example: 5 x 5 = 52

jeyben [28]

Answer:

Problem A

5 * 5 * x * x * x

$5^2*x^3$

Problem B

8 * 8 * 8

$8^3$

Problem C

4 * 4 * 4 * x * x * x * x

$4^3*x^4$

5 0

3 years ago

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

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Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

3 years ago

What is the function for a table with a domain of 5, 10, 15, 20 and a range of 40, 80, 120, and 160

kykrilka [37]

To find the range we need to find the vertex of the parabola which is at (-b/2a , y)

-b/2a would be 2/2 = 1 ... so find the y value at x = 1

12 - 2(1) - 15 = 1-17 = -16

So the vertex is at (1,-16)

Since the parabola opens upward from that point the minimum value of the range is y = -16

The range would include the point -16 so it is actually

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3 years ago

Someone please help with solving this problem so I can use it as an example for others!!

Katarina [22]

2+3i
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3 years ago

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tatyana61 [14]

4x^2-4x-120
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answer
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3 years ago

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