1 Between which two integers does sqrt(11) lie? Classify the following numbers: −1.5252... and sqrt(−25) and…

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1 Between which two integers does sqrt(11) lie? Classify the following numbers: −1.5252... and sqrt(−25) and write 0.7 as a common fraction.

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QUESTION 1

1.1 To find between which two integers $\sqrt{11}$ lies, we find the perfect squares closest to 11.

Identify perfect squares around 11.
$3^2 = 9$ $4^2 = 16$
Place $\sqrt{11}$ between the square roots of these perfect squares.
$\sqrt{9} < \sqrt{11} < \sqrt{16}$ $3 < \sqrt{11} < 4$
Thus, $\sqrt{11}$ lies between the integers 3 and 4.
$\boxed{\text{3 and 4}}$
1.2 We classify the numbers using the given symbols: N (Natural numbers), Q (Rational numbers), Q' (Irrational numbers), Z (Integers), R (Real numbers). We will assume R' refers to non-real numbers in this context.

1.2.1 $-1.525252...$
This is a repeating decimal, which can be expressed as a fraction.
Therefore, it is a Rational number (Q) and also a Real number (R).
$\boxed{\text{Q, R}}$
1.2.2 $\sqrt{-25}$
The square root of a negative number is an imaginary number. Imaginary numbers are not real numbers.
Therefore, $\sqrt{-25}$ is a non-real number (R').
$\boxed{\text{R'}}$
1.3 To write 0,7 as a common fraction, we assume 0,7 means $0.\overline{7}$ (0.7 repeating).
Let $x$ be the repeating decimal.
$x = 0.\overline{7}$
Multiply by 10 to shift the decimal.
$10x = 7.\overline{7}$
Subtract the first equation from the second.
$10x - x = 7.\overline{7} - 0.\overline{7}$ $9x = 7$
Solve for $x$ .
$x = \frac{7}{9}$ $\boxed{\frac{7}{9}}$
QUESTION 2

2.1 Simplify the following:

2.1.1 $(x - 2)(x^2 + 5x - 1)$
Distribute each term from the first parenthesis to the second.
$x(x^2 + 5x - 1) - 2(x^2 + 5x - 1)$
Multiply the terms.
$x^3 + 5x^2 - x - 2x^2 - 10x + 2$
Combine like terms.
$x^3 + (5x^2 - 2x^2) + (-x - 10x) + 2$ $x^3 + 3x^2 - 11x + 2$ $\boxed{x^3 + 3x^2 - 11x + 2}$
2.1.2 $\frac{3^x - 2 \cdot 3^{x-1}}{3^{x+2} + 2 \cdot 3^{x+1}}$
Factor out the lowest power of 3 from the numerator and denominator.
Numerator: $3^x - 2 \cdot 3^{x-1} = 3^{x-1}(3^1 - 2)$
Denominator: $3^{x+2} + 2 \cdot 3^{x+1} = 3^{x+1}(3^1 + 2)$
Simplify the terms in the parentheses.
Numerator: $3^{x-1}(3 - 2) = 3^{x-1}(1) = 3^{x-1}$
Denominator: $3^{x+1}(3 + 2) = 3^{x+1}(5) = 5 \cdot 3^{x+1}$
Rewrite the fraction and simplify using exponent rules ( $a^m/a^n = a^{m-n}$ ).
$\frac{3^{x-1}}{5 \cdot 3^{x+1}} = \frac{1}{5} \cdot 3^{(x-1) - (x+1)}$ $\frac{1}{5} \cdot 3^{x-1-x-1} = \frac{1}{5} \cdot 3^{-2}$
Evaluate $3^{-2}$ .
$\frac{1}{5} \cdot \frac{1}{3^2} = \frac{1}{5} \cdot \frac{1}{9}$ $\frac{1}{45}$ $\boxed{\frac{1}{45}}$
2.2 Factorize the following expressions fully:

2.2.1 $2x^2 - x - 6$
Find two numbers that multiply to $2 \times (-6) = -12$ and add to $-1$ .
These numbers are $-4$ and 3.
Rewrite the middle term using these numbers.
$2x^2 - 4x + 3x - 6$
Factor by grouping.
$2x(x - 2) + 3(x - 2)$
Factor out the common binomial factor $(x - 2)$ .
$(x - 2)(2x + 3)$ $\boxed{(x - 2)(2x + 3)}$
2.2.2 $a^2 - 2ab + b^2 - 100c^2$
Recognize the first three terms as a perfect square trinomial.
$(a^2 - 2ab + b^2) - 100c^2$ $(a - b)^2 - 100c^2$
Recognize the expression as a difference of squares, $X^2 - Y^2 = (X - Y)(X + Y)$ , where $X = (a - b)$ and $Y = 10c$ .
$(a - b)^2 - (10c)^2$
Apply the difference of squares formula.
$((a - b) - 10c)((a - b) + 10c)$ $(a - b - 10c)(a - b + 10c)$ $\boxed{(a - b - 10c)(a - b + 10c)}$
2.3 Solve the following equations simultaneously for $a$ and $b$ :
Equation 1: $a + b = 12$
Equation 2: $4a + 2b = 44$
From Equation 1, express $a$ in terms of $b$ .
$a = 12 - b \quad (\text{Equation 3})$
Substitute Equation 3 into Equation 2.
$4(12 - b) + 2b = 44$
Distribute and solve for $b$ .
$48 - 4b + 2b = 44$ $48 - 2b = 44$ $-2b = 44 - 48$ $-2b = -4$ $b = \frac{-4}{-2}$ $b = 2$
Substitute the value of $b$ back into Equation 3 to find $a$ .
$a = 12 - 2$ $a = 10$
State the solution.
$a = 10, b = 2$ $\boxed{a = 10, b = 2}$

1 Between which two integers does sqrt(11) lie? Classify the following numbers: -1.5252... and sqrt(-25) and write 0.7 as a common fraction.

Step-by-step solution

Identify perfect squares around 11.

Place $\sqrt{11}$ between the square roots of these perfect squares.

Let $x$ be the repeating decimal.

Multiply by 10 to shift the decimal.

Subtract the first equation from the second.

Solve for $x$ .

Distribute each term from the first parenthesis to the second.

Multiply the terms.

Combine like terms.

Factor out the lowest power of 3 from the numerator and denominator.

Simplify the terms in the parentheses.

Rewrite the fraction and simplify using exponent rules ( $a^m/a^n = a^{m-n}$ ).

Evaluate $3^{-2}$ .

Find two numbers that multiply to $2 \times (-6) = -12$ and add to $-1$ .

Rewrite the middle term using these numbers.

Factor by grouping.

Factor out the common binomial factor $(x - 2)$ .

Recognize the first three terms as a perfect square trinomial.

Recognize the expression as a difference of squares, $X^2 - Y^2 = (X - Y)(X + Y)$ , where $X = (a - b)$ and $Y = 10c$ .

Apply the difference of squares formula.

From Equation 1, express $a$ in terms of $b$ .

Substitute Equation 3 into Equation 2.

Distribute and solve for $b$ .

Substitute the value of $b$ back into Equation 3 to find $a$ .

State the solution.

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