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2 years ago

Write the equation of the arithmetic and geometricsequences given 15, 5, ...

Mathematics

1 answer:

ivann1987 [24]2 years ago

7 0

Step-by-step explanation:

Arithmetic:

a(n) = a +(n-1)d

a=15

d=5-15=-10

°.° a(n) = 15 +(n–1)-10

Geometric:

a(n)=a*r^(n-1)

a=15

r=5/15=1/3

a(n)=15*(1/3)^(n-1)

hope this helps

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2. Write a quadratic function that has -intercepts (- 3, 0) and (4, 0) and passes through the point (5, 8)

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Answer:

$f(x)=x^2-x-12$

Step-by-step explanation:

<u>Quadratic Function</u>

Standard Form of Quadratic Function

The standard representation of a quadratic function is:

$f(x)=ax^2+bx+c$

where a,b, and c are constants.

When the zeros of f (x1 and x2) are given, it can be written as:

f(x)=a(x-x1)(x-x2)

Where a is a constant called the leading coefficient.

We are given the two roots of f: x1=-3 and x2=4, thus:

f(x)=a(x+3)(x-4)

We also know that f(5)=8, thus:

f(5)=a(5+3)(5-4)=8

Operating:

a(8)(1)=8

Solving:

a=1

The function is:

f(x)=1(x+3)(x-4)

Operating:

$\boxed{f(x)=x^2-x-12}$

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2 years ago

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

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

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

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2 years ago