According to the distributive property, 6 (a + b) =

A certain forest covers an area of 2300 km2. Suppose that each year this area decreases by 5.25%. What will the area be after 13

nordsb [41]

Answer:

The area after 13 years will be 1141 km2

Step-by-step explanation:

This problem can be modeled with an exponencial equation:

A = Ao * (1+r)^t

Where A is the Area after t years, Ao is the inicial area and r is the rate that the area changes.

For our problem, we have that Ao = 2300, r = -5,25% = -0.0525 and t = 13.

Then, we can find the area after 13 years:

A = 2300 * (1 - 0.0525)^13

A = 2300 * (0.9475)^13 = 1140.93 km2

Rounding to the nearest square kilometer, we have A = 1141 km2

8 0

4 years ago

Help! Giving brainliest and 10+ points, A ball is dropped from the top of a 400-ft tall tower. The relationship between the time

Ad libitum [116K]

Answer:

3

Step-by-step explanation:

4 0

3 years ago

Jack took a test and got 13 out of 17 points. He took a make-up test and got 12 out of 15 points.

andrezito [222]

The answer is choice A, a 3.5% increase.
12/15 = 80%
13/17 = 76.47%

80% - 76.47% = 3.5%

6 0

3 years ago

Read 2 more answers

Calculus 2. Please help

Anarel [89]

Answer:

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}}} \, dx = \infty$

General Formulas and Concepts:

Algebra I

Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integrals

Definite Integrals

Integration Constant C

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

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

U-Substitution

U-Solve

Improper Integrals

Exponential Integral Function: $\displaystyle \int {\frac{e^x}{x}} \, dx = Ei(x) + C$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx$

Step 2: Integrate Pt. 1

[Integral] Rewrite [Exponential Rule - Rewrite]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \int\limits^1_0 {\frac{e^{-x^2}}{x} \, dx$
[Integral] Rewrite [Improper Integral]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \int\limits^1_a {\frac{e^{-x^2}}{x} \, dx$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set: $\displaystyle u = -x^2$
Differentiate [Basic Power Rule]: $\displaystyle \frac{du}{dx} = -2x$
[Derivative] Rewrite: $\displaystyle du = -2x \ dx$

Rewrite u-substitution to format u-solve.

Rewrite du: $\displaystyle dx = \frac{-1}{2x} \ dx$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {-\frac{e^{-x^2}}{x} \, dx$
[Integral] Substitute in variables: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {\frac{e^{u}}{-2u} \, du$
[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}\int\limits^1_a {\frac{e^{u}}{u} \, du$
[Integral] Substitute [Exponential Integral Function]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(u)] \bigg| \limits^1_a$
Back-Substitute: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(-x^2)] \bigg| \limits^1_a$
Evaluate [Integration Rule - FTC 1]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(-1) - Ei(a)]$
Simplify: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{Ei(-1) - Ei(a)}{2}$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \infty$