All categories

2 years ago

Find the exact value of the expressions

Mathematics

1 answer:

timama [110]2 years ago

4 0

Answer:

here is your correct answer and listen question no four option are incorrect

You might be interested in

375 = 3 x 5 x 5 x 5 =

Fofino [41]

375=3×5×5×5 = 15 × 25

8 0

3 years ago

Jazzy goes to bed 3 h after dinner.

makkiz [27]

Answer:

It was easy

Step-by-step explanation:

11 x 2 = 22

3 x 2 = 6

22 x 6 = 132

3 0

3 years ago

how much would $100 invested at 8% interest compounded annually be worth after 15 years? Round your answer to the nearest cent.

svet-max [94.6K]

To calculate problems abound about compounding interest use the equation A = P (1 + r/n)^(nt), where A is the future price, P is the principal amount, r is the interest rate, n is the number of times the interest is compounded per year and t for the total years. To solve, A = 100 (1 + 0.08/1)^(1 x 15) = 317.22.

5 0

3 years ago

Read 2 more answers

Han has 5 pencils and Andre has 8

Tatiana [17]

Answer:

That's all I could think that you might need. Finish your question please

Step-by-step explanation:

Together: 13

Andre has 3 more

4 0

3 years ago

Find the derivative.

krek1111 [17]

Answer:

$\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Expanding/Factoring

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \frac{\sqrt{x}}{e^x}$

Step 2: Differentiate

Derivative Rule [Quotient Rule]: $\displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}$
Basic Power Rule: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}$
Exponential Differentiation: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}$
Simplify: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}$
Rewrite: $\displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}$
Factor: $\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

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

Unit: Differentiation

7 0

2 years ago