🧮 algebra

1. We are given the expression $5c + 2$ and the value $c = 6$. 2. Substitute $6$ for $c$ in the expression: $$5 \cdot 6 + 2$$

1. Problem: Find the domain of the functions \( f(x) = \sqrt{x+2} \) and \( g(x) = \frac{1}{x^2 - x} \). 2. For \( f(x) = \sqrt{x+2} \), the expression inside the square root must

1. Stating the problem: Simplify the expression \( \frac{3}{10} - \frac{1}{10} \). 2. Since the denominators are the same, simply subtract the numerators:

1. The problem is to add the mixed number $1 \frac{1}{2}$ and the fraction $\frac{3}{8}$. 2. First, convert the mixed number to an improper fraction. $1 \frac{1}{2} = \frac{2 \time

1. State the problem: Add the fractions $\frac{2}{3}$ and $\frac{1}{4}$. 2. Find a common denominator for the fractions. The denominators are 3 and 4, so the least common denominat

1. The problem is to evaluate the expression $\frac{6}{11} - 2 \frac{3}{4}$. 2. First, convert the mixed number $2 \frac{3}{4}$ to an improper fraction.

1. We need to subtract the fractions $\frac{3}{4}$ from $\frac{5}{6}$.\n2. Find a common denominator. The denominators are 6 and 4. The least common denominator (LCD) of 6 and 4 is

1. State the problem: Simplify the expression $5 \frac{9}{11} - \left(-\frac{6}{11}\right)$.\n\n2. Convert the mixed number to an improper fraction:\n$$5 \frac{9}{11} = 5 + \frac{9

1. **Problem:** If you have $\frac{3}{4}$ of a chocolate bar and you want to divide it into portions each of size $\frac{1}{2}$ of a chocolate bar, how many portions can you get? *

1. The problem is to evaluate the expression $-1 \frac{3}{7} - \frac{2}{3}$. 2. First, convert the mixed number $-1 \frac{3}{7}$ into an improper fraction. Since it is negative, we

1. Let's evaluate the expression $$\frac{-23 - (-14)}{-1 + 2(-1)}$$ step by step. - Compute the numerator: $$-23 - (-14) = -23 + 14 = -9$$

1. Let's clarify what "use the proper solution for division" means. Division is the operation of splitting a number into equal parts or groups. It is often expressed as $a \div b$

1. We are asked to simplify the expression $\frac{4}{9} - \frac{7}{9}$. 2. Since both fractions have the same denominator 9, subtract the numerators directly: $4 - 7 = -3$.

1. The problem is not explicitly stated, but let's demonstrate solving a simple algebraic equation: $2x + 3 = 7$. 2. Subtract 3 from both sides to isolate the term with $x$: $$2x +

1. Énonçons le problème : Calculer l'expression de $A$ donnée par

1. Evaluate \(\frac{-23 - (-14)}{-1 + 2(-1)}\). Step 1: Simplify the numerator:

1. **Problem:** Divide $\frac{3}{4}$ by $\frac{2}{5}$.\n\nStep 1: Write the problem as division of fractions: $\frac{3}{4} \div \frac{2}{5}$.\n\nStep 2: To divide by a fraction, mu

1. Problem 7: Find the domain and range of the relation $R = \{(x,y) \mid y \leq x-1 \text{ and } y \geq 2x-1\}$. - The domain is the set of all $x$ values for which there exists $

1. Énonçons le problème : Soit $a,b,c$ trois scalaires tels que $a\neq 1$ et $b\neq 1$. On considère la matrice ou expression définie par

1. **Problem:** Classify each given function. 2. **Function (a):** $f(x) = \sqrt[5]{x} = x^{1/5}$.

1. **State the problem:** We want to prove that if for some natural number $n$ and real number $x > -1$, the inequality $$ (1+x)^n \geq 1 + nx $$ holds, then it also holds for $n+1