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3 years ago

what is the equation of the graph that represents f(x)=x^4 stretched vertically by 2 and shifted up 3 spaces

Mathematics

1 answer:

BabaBlast [244]3 years ago

8 0

Answer: $g(x)=2x^4+3$

Step-by-step explanation:

These are some transformations for a function f(x):

If $f(x)+k$ , then the function is shifted up "k" units.

If $mf(x)$ , and $k>1$ , then the function is stretched vertically by a factor of "m".

Knowing this transformation and knowing that the function $f(x)=x^4$ is stretched vertically by 2 and shifted up 3 spaces, then we can conclude that new function, which we can call g(x), is:

$g(x)=2(x^4)+3$

$g(x)=2x^4+3$

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Step-by-step explanation:

Hello there, to answer this question, one must know the formula of area of a circle.

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3 years ago

The strength of an electrical current x flowing through the electric circuit shown is expressed as a function of time t and sati

Elza [17]

Answer:

The current of the circuit at t = 0 is equal to 0.

If we take the limit as t approaches infinity, the current is equal to ε/R or V/R.

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]:

$\displaystyle (cu)' = cu'$

Derivative Property [Addition/Subtraction]:

$\displaystyle (u + v)' = u' + v'$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Slope Fields

Separation of Variables

Integration

Integrals

Integration Rule [Reverse Power Rule]:

$\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]:

$\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]:

$\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Method: U-Substitution

Electricity

Ohm's Law: V = IR

V is voltage (in Volts)
I is current (in Amps)
R is resistance (in Ohms)

Circuits

Circuit Symbols
Kirchhoff's Laws (Loop and Junction Rule)
Inductors

Step-by-step explanation:

*Note:

In the given equation, our variable of differentiation is x. I will rewrite this as current I for physics notation purposes.

Step 1: Define

Identify given.

$\displaystyle L \frac{dI}{dt} + RI = V$

[Assuming switch S is closed] Recall that an inductor is used in a circuit to resist change. After a long period of time, when it hits steady-state equilibrium, we expect to see the inductor act like a wire.

Step 2: Find Current Expression Pt. 1

[Kirchhoff's Law] Rewrite expression:
$\displaystyle L \frac{dI}{dt} = V - IR$
Rewrite expression by dividing R on both sides:
$\displaystyle \frac{L}{R} \frac{dI}{dt} = \frac{\mathcal E}{R} - I$

Step 3: Find Current Expression Pt. 2

Identify variables for u-substitution.

Set u:
$\displaystyle u = \frac{\mathcal E}{R} - I$
[u] Differentiation [Derivative Rules and Properties]:
$\displaystyle du = - \, dI$

Step 4: Find Current Expression Pt. 3

[Kirchhoff's Law] Apply U-Substitution:
$\displaystyle - \frac{L}{R} \frac{du}{dt} = u$
[Kirchhoff's Law] Apply Separation of Variables:
$\displaystyle \frac{1}{u} \, du = -\frac{L}{R} \, dt$

Recall that our initial condition is when t = 0, denoted as u₀, and we go to whatever position u we are trying to find. Also recall that time t always ranges from t = 0 (time can't be negative) and to whatever t we are trying to find.

[Kirchhoff's Law] Integrate both sides:
$\displaystyle \int\limits^u_{u_0} {\frac{1}{u}} \, du = \int\limits^t_0 {- \frac{R}{L}} \, dt$
[Kirchhoff's Law] Rewrite [Integration Property]:
$\displaystyle \int\limits^u_{u_0} {\frac{1}{u}} \, du = - \frac{R}{L} \int\limits^t_0 {} \, dt$
[1st Integral] Apply Logarithmic Integration:
$\displaystyle \ln | u | \bigg| \limits^u_{u_0} = - \frac{R}{L} \int\limits^t_0 {} \, dt$
[2nd Integral] Apply Integration Rule [Reverse Power Rule]:
$\displaystyle \ln | u | \bigg| \limits^u_{u_0} = - \frac{R}{L} t \bigg| \limits^t_0$
Apply Integration Rule [Fundamental Theorem of Calculus 1]:
$\displaystyle \ln | \frac{u}{u_0} | = - \frac{R}{L} t$
Apply e to both sides:
$\displaystyle e^{\ln | \frac{u}{u_0} |} = e^{- \frac{R}{L} t}$
Simplify:
$\displaystyle \frac{u}{u_0} = e^{- \frac{R}{L} t}$
Rewrite:
$\displaystyle u = u_0 e^{- \frac{R}{L} t}$

Recall that our initial condition u₀ (derived from Ohm's Law) contains only the voltage across resistor R, where voltage is supplied by the given battery. This is because the current is stopped once it reaches the inductor in the circuit since it resists change.

Back-Substitute in u and u₀:
$\displaystyle \frac{\mathcal E}{R} - I = \frac{\mathcal E}{R} e^{- \frac{R}{L} t}$
Solve for I:
$\displaystyle I = \frac{\mathcal E}{R} - \frac{\mathcal E}{R} e^{- \frac{R}{L}t}$

Step 5: Solve

If we are trying to find the strength of the electrical current I at t = 0, we simply substitute t = 0 into our current function:

$\displaystyle\begin{aligned}I(t) & = \frac{\mathcal E}{R} - \frac{\mathcal E}{R} e^{- \frac{R}{L}t} \\I(0) & = \frac{\mathcal E}{R} - \frac{\mathcal E}{R} e^{- \frac{R}{L}(0)} \\& = \boxed{\bold{0}}\end{aligned}$

If we are taking the limit as t approaches infinity of the current function I(t), we are simply just trying to find the current after a long period of time, which then would just be steady-state equilibrium:

$\displaystyle\begin{aligned}I(t) & = \frac{\mathcal E}{R} - \frac{\mathcal E}{R} e^{- \frac{R}{L}t} \\\lim_{t \to \infty} I(t) & = \frac{\mathcal E}{R} - \frac{\mathcal E}{R} e^{- \frac{R}{L}(\infty)} \\& = \boxed{\bold{\frac{\mathcal E}{R}}}\end{aligned}$

∴ we have found the current I at t = 0 and the current I after a long period of time and proved that an inductor resists current running through it in the beginning and acts like a wire when in electrical equilibrium.

---

Topic: AP Physics C - EMAG

Unit: Induction

4 0

2 years ago

Which best describes the relationship between the successive terms in the sequence shown?

timurjin [86]

Answer:

-7.2

Step-by-step explanation:

2.4 + -7.2 = -4.8

-4.8 + (-7.2 *2) = 9.6

9.6 + (-7.2 *3) = -19.2

7 0

3 years ago

Point U is on line segment \overline{TV} TV . Given UV=13UV=13 and TV=19,TV=19, determine the length \overline{TU}. TU .

Gelneren [198K]

Answer:

<h2>6 units</h2>

Step-by-step explanation:

If point U is on the line segment TV, this means that point T, U and V are collinear. The expression TU+UV = TV will represent the sum of the line segments.

Given UV = 13, TV = 19, we are to find TU.

Substitute the given values into the formula to get TU as shown;

TU + 13 = 19

subtract 13 from both sides

TU+13-13 = 19-13

TU = 6

Hence the length of TU is 6 units

7 0

4 years ago