All categories

3 years ago

. The area of a rectangle is 45.5 square inches. The base of the rectangle is 7 inches.

Mathematics

2 answers:

VLD [36.1K]3 years ago

8 0

Answer:

beug

Step-by-step explanation:

wdqd

Ugo [173]3 years ago

7 0

What the question then

You might be interested in

Lexi has 24 students in her class 2/3 of the students are girls.how many girls are in lexis class?

Anuta_ua [19.1K]

If you divide 24 by 3 it’s 8 so 8 is 1/3 so if you double it, it will be 16 which equal 2/3

Btw the answer is 16

3 0

4 years ago

IF ANYONE HELPS ME WITH THIS I WILL GIVE U BRAINLIEST AND 100 POINTS BTW ITS INTEGRALS FOR CALCULUS

dlinn [17]

Answer:

$\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{e^\bigg{\frac{1}{4}}}{8} - \frac{e^\bigg{\frac{1}{9}}}{8}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

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

Calculus

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

eˣ Integration: $\displaystyle \int {e^u} \, dx = e^u + C$

Step-by-step explanation:

Step 1: Define

$\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set: $\displaystyle u = 4x^{-2}$
[u] Differentiate [Derivative Rule - Basic Power Rule]: $\displaystyle du = -8x^{-3} \ dx$
[du] Rewrite [Exponential Rule - Rewrite]: $\displaystyle du = \frac{-8}{x^3} \ dx$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^6_4 {\frac{-8}{x^3}e^{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{1}{9}}_{\frac{1}{4}} {e^u} \, dx$
[Integral] eˣ Integration: $\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8}(e^u) \bigg| \limits^{\frac{1}{9}}_{\frac{1}{4}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8} \bigg[ -e^\bigg{\frac{1}{9}} \bigg( e^\bigg{\frac{5}{36}} - 1 \bigg) \bigg]$
Simplify: $\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{e^\bigg{\frac{1}{4}}}{8} - \frac{e^\bigg{\frac{1}{9}}}{8}$