You deposit 350 in an account that pays 3% annual intrest. Find the balance afer 2 years if the intrest is compounded

Answers

Answer 1

Answer:

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \n\n A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\n P=\textit{original amount deposited}\dotfill &\$350\n r=rate\to 3\%\to (3)/(100)\dotfill &0.03\n n= \begin{array}{llll} \textit{times it compounds per year}\n \textit{annually, thus once} \end{array}\dotfill &1\n t=years\dotfill &2 \end{cases} \n\n\n A=350\left(1+(0.03)/(1)\right)^(1\cdot 2)\implies A=350(1.03)^2\implies A=371.315$

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What is the indefinite integral of 6sin(3t) dt?

Answers

Answer:

$\displaystyle \int {6sin(3t)} \, dt = -2cos(3t) + C$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals
[Indefinite Integrals] Integration Constant C

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int {6sin(3t)} \, dt$

Step 2: Integrate Pt. 1

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {6sin(3t)} \, dt = 6\int {sin(3t)} \, dt$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set u: $\displaystyle u = 3t$
[u] Differentiate [Basic Power Rule, Multiplied Constant]: $\displaystyle du = 3 \ dt$

Step 4: integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {6sin(3t)} \, dt = 2\int {3sin(3t)} \, dt$
[Integral] U-Substitution: $\displaystyle \int {6sin(3t)} \, dt = 2\int {sin(u)} \, du$
[Integral] Trigonometric Integration: $\displaystyle \int {6sin(3t)} \, dt = -2cos(u) + C$
Back-Substitute: $\displaystyle \int {6sin(3t)} \, dt = -2cos(3t) + C$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Given that cos 63°≈ 0.454, enter the sine of a complementary angle.
sin

Answers

Answer:

$\cos(63^\circ)$ is the same as $\sin(27^\circ)$ by co-function identities

Step-by-step explanation:

Remember that complementary angles add up to 90°. The angle that i s complementary to 63° is 27°.

Also recall the co-function identities:

sin (90° – x) = cos x
cos (90° – x) = sin x

This means that $\cos(90^\circ-27^\circ)=\sin(27^\circ)\approx0.454$ .

A+(-b) where a=4 and b=2

Answers

Answer:

The sum of a and (-b) is 2

Step-by-step explanation:

method 1:First, we take the negate(negative) of b to get -2. Then, we take the sum of 4 and -2 to get 2.

method 2: since adding a negative is the same as subtracting a positive, we get 4-2, which is 2.

What is the angle supplementary to the angle measuring 165°12′?A. 75°12′
B. 180′
C. 14°48′
D. 60′

Answers

Two angles are supplementary if their sum is 180 degrees. One degree is made up of 60 minutes. So the angle supplementary to an angle measuring 165d12m is 180d - 165d12m, which gives us 14d48m. So the answer is C.

Answer:

Option C is correct.

$14^(\circ)48'$ is the angle supplementary to the angle measuring $165^(\circ)12'$