Give example of solution to solve the equation 2x-7=5+x

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

2 years ago

Rectangle QRST is dilated with the origin as the center of dilation to create rectangle Q'R'S'T'.

scoundrel [369]

Answer:

Rule: dilation about the origin by a scale factor of 2

Step-by-step explanation:

Given

See attachment for QRST and Q'R'S'T

Required

The dilation rule

First, we calculate the scale factor (k)

From the attachment, we have:

$QR = 4$

$Q'R' = 8$

k, is then calculated as:

$k = \frac{Q'R'}{QR}$

$k = \frac{8}{4}$

$k =2$

Hence, QRST is dilated by a factor of 2

3 0

3 years ago

Find a solution x = x(t) of the equation x′ + 2x = t2 + 4t + 7 in the form of a quadratic function of t, that is, of the form x(

Temka [501]

The particular quadratic solution to the ODE is found as follows:

$x=at^2+bt+c$
$x'=2at+b$

$(2at+b)+2(at^2+bt+c)=t^2+4t+7$
$2at^2+(2a+2b)t+(b+2c)=t^2+4t+7$

$\begin{cases}2a=1\\2(a+b)=4\\b+2c=7\end{cases}\implies a=\dfrac12,b=\dfrac32,c=\dfrac{11}4$

Note that there's also the fundamental solution to account for, which is obtained from the characteristic equation for the ODE:

$x'+2x=0\implies r+2=0\implies r=-2$

so that $x_c=Ce^{-2t}$ is a characteristic solution to the ODE, and the general solution would be

$x=Ce^{-2t}+\dfrac{t^2}2+\dfrac{3t}2+\dfrac{11}4$

4 0

3 years ago

SCORPION-xisa [38]

Answer:

It's 15

i took the test

Step-by-step explanation:

5 0

2 years ago

HELP ASAPPPPP When graphing the inequality y > -4x + 6, what type of line should be graphed?

melomori [17]

Answer:

A dashed line

Step-by-step explanation:

The end points of the line for the inequality would not be included since y is only greater, not greater than or equal to.

6 0

3 years ago