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3 years ago

Function or not a function?

Mathematics

2 answers:

den301095 [7]3 years ago

6 0

Answer:

Function

Step-by-step explanation:

Its a function because it passes the vertical line test

Dvinal [7]3 years ago

4 0

Not a function. Dont multiply its doesnt make sence

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After the closing of the mill, the town of Sawyerville experienced a decline in its population.

vfiekz [6]

Step-by-step explanation:

P(t) = 12,000 (2)^(-t/15)

9,000 = 12,000 (2)^(-t/15)

0.75 = 2^(-t/15)

ln(0.75) = ln(2^(-t/15))

ln(0.75) = (-t/15) ln(2)

-15 ln(0.75) / ln(2) = t

t = 6.23

4 0

3 years ago

Convert -18° to radians.

aev [14]

The answer is 0.314 radian

8 0

3 years ago

Read 2 more answers

An astronaut was exploring the moon of a distant planet, and found some glowing goo at the bottom of a very deep crater. She bro

Soloha48 [4]

Answer:

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8 0

3 years ago

Subtract 8/9 - 1/3 5/6 5/9 7/9 7/12

Mars2501 [29]

Answer:

3/9

Step-by-step explanation:

Common denominator: 27

24/27 - 9/27 = 15/27 = 3/9

6 0

2 years ago

Suppose Upper F Superscript prime Baseline left-parenthesis x right-parenthesis equals 3 x Superscript 2 Baseline plus 7 and Upp

Sedaia [141]

It looks like you're given

F'(x) = 3x² + 7

and

F (0) = 5

and you're asked to find F(b) for the values of b in the list {0, 0.1, 0.2, 0.5, 2.0}.

The first is done for you, F (0) = 5.

For the remaining b, you can solve for F(x) exactly by using the fundamental theorem of calculus:

$F(x)=F(0)+\displaystyle\int_0^x F'(t)\,\mathrm dt$

$F(x)=5+\displaystyle\int_0^x(3t^2+7)\,\mathrm dt$

$F(x)=5+(t^3+7t)\bigg|_0^x$

$F(x)=5+x^3+7x$

Then F (0.1) = 5.701, F (0.2) = 6.408, F (0.5) = 8.625, and F (2.0) = 27.

On the other hand, if you're expected to approximate F at the given b, you can use the linear approximation to F(x) around x = 0, which is

F(x) ≈ L(x) = F (0) + F' (0) (x - 0) = 5 + 7x

Then F (0) = 5, F (0.1) ≈ 5.7, F (0.2) ≈ 6.4, F (0.5) ≈ 8.5, and F (2.0) ≈ 19. Notice how the error gets larger the further away b gets from 0.

A better numerical method would be Euler's method. Given F'(x), we iteratively use the linear approximation at successive points to get closer approximations to the actual values of F(x).

Let y(x) = F(x). Starting with x₀ = 0 and y₀ = F(x₀) = 5, we have

x₁ = x₀ + 0.1 = 0.1

y₁ = y₀ + F'(x₀) (x₁ - x₀) = 5 + 7 (0.1 - 0) → F (0.1) ≈ 5.7

x₂ = x₁ + 0.1 = 0.2

y₂ = y₁ + F'(x₁) (x₂ - x₁) = 5.7 + 7.03 (0.2 - 0.1) → F (0.2) ≈ 6.403

x₃ = x₂ + 0.3 = 0.5

y₃ = y₂ + F'(x₂) (x₃ - x₂) = 6.403 + 7.12 (0.5 - 0.2) → F (0.5) ≈ 8.539

x₄ = x₃ + 1.5 = 2.0

y₄ = y₃ + F'(x₃) (x₄ - x₃) = 8.539 + 7.75 (2.0 - 0.5) → F (2.0) ≈ 20.164

4 0

3 years ago