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4 years ago

The answer plx i need this real bad

Mathematics

2 answers:

muminat4 years ago

8 0

the equation you need to do is (2/5)x15

the answer is 6

So B is the correct answer

bezimeni [28]4 years ago

6 0

15/5=3 3*2=6 The answer is B

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Craig wants to prove that if quadrilateral ABCD has diagnols that biscet each other then it is a parallelogram

Novosadov [1.4K]

Solution :

Consider quadrilateral ABCD is a parallelogram. The parallelogram have diagonals AC and DB.

So in the given quadrilateral ABCD, let the diagonal AC and diagonal DB intersects at a point E.

Thus in the quadrilateral ABCD we see that :

1. AC and DB are the diagonals of quadrilateral ABCD.

2. Angle DCE is congruent to angle BAE and angle CDE is congruent to angle ABE. (they are alternate interior angles)

3. Line DC is congruent to line AB. (opposites sides are congruent in a parallelogram )

4. Angle ABE is congruent to angle CDE. (Angle side angle)

5. Line AE is congruent to line EC. And line DE is congruent to line EB. (CPCTC)

Thus we see that if the diagonals of a $\text{quadrilateral bisects each other}$ , then the quadrilateral is a parallelogram.

7 0

3 years ago

Which function has an inverse that is also a function?

UNO [17]

The answer is C or the third one

5 0

3 years ago

what is the distance around a triangle that has sides measuring 2 and one eight feet , 3 one half feet, and 2 one half feet?

Ymorist [56]

Perimeter = 2 1/8 + 3 1/2 + 2 1/2 = 8 1/8

4 0

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

1/4 + 5/12 i rlly need this

loris [4]

8/12 is the answer bc 1/4 can be converted to 3/12 and with a common denominator you can now add and you’ll get 8/12

4 0

3 years ago