Show answer in lowest terms. 25/75+ 25/125=

1 answer:

5 0

Hi Jujub! To find the answer in lowest terms, first you have to find the lowest common denominator. To find this, find the lowest common multiple of 75 and 125, which is 375. Now, to make the common denominator, you have to make equivalent fractions. 375 ÷ 75 = 5, and 5 × 25 = 125. So the new fraction is 125/375. Now the same thing for 25/125. 375 ÷ 125 = 3, and 3 × 25 is 75. So the new fraction is 75/375. Now add the new fractions:
125/375 + 75/375 = 200/375

This answer can be simplified to 8/15.

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$\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -ln(2)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Trig Derivatives

Integration

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]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Trig Integration

Logarithmic Integration

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 1 + cos(x)$
[u] Differentiate [Trig Derivative]: $\displaystyle du = -sin(x) \ dx$
[Bounds of Integration] Change: $\displaystyle [\frac{1}{2}, 1]$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -\int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{-sin \ x}{1 + cos \ x}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -\int\limits^{1}_{\frac{1}{2}} {\frac{1}{u}} \, du$
[Integral] Logarithmic Integration: $\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -(ln|u|) \bigg| \limits^{1}_{\frac{1}{2}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -ln(2)$