Solve the math and word problems presented in the image.

Understand the Problem

The image contains a series of math and word problems. The questions cover a variety of topics including number representation, algebraic manipulation, probability, geometry, and set theory. We will classify each question into its appropriate category and subcategory.

Answer

1. (a) Eight trillion, four hundred eighty-eight billion, four hundred fifty-six million, nineteen thousand, five hundred. (b) $\frac{727}{210}$ (c) $16$ and $7$ 2. (a) $2$ (b) $\{2,6\}$ 3. (a) $14$ (b) $0.091$ 4. (a) $x=5$ (b) (i) A solution is a value or set of values that satisfies a given equation or set of equations. (ii) $k = 0$

Steps to Solve

Convert the decimal to a fraction (1a)

The given number is 8,488,456,019,500. In words, this is eight trillion, four hundred eighty-eight billion, four hundred fifty-six million, nineteen thousand, five hundred.

Convert the repeating decimal to a fraction (1b)

Let $x = 1.0\overline{3} = 1.0333...$. Then $10x = 10.3333...$ and $100x = 103.3333...$. Subtracting these gives $90x = 93$, so $x = \frac{93}{90} = \frac{31}{30}$.

Convert the other decimal to a fraction (1b)

$2.428571 = 2 + \frac{428571}{1000000}$. Notice that $1/7 = 0.\overline{142857}$, and $3/7 = 0.\overline{428571}$, thus $2.428571 \approx 2 + \frac{3}{7} = \frac{17}{7}$

Add the two fractions (1b)

$\frac{31}{30} + \frac{17}{7} = \frac{31 \cdot 7 + 17 \cdot 30}{30 \cdot 7} = \frac{217 + 510}{210} = \frac{727}{210}$.

Set up equations based on the problem statement (1c)

Let the two numbers be $x$ and $y$, where $x > y$. We are given $x + y = 23$ and $2x = 4y + 4$.

Solve for one variable in terms of the other (1c)

From the first equation, $x = 23 - y$.

Substitute and solve for $y$ (1c)

Substitute into the second equation: $2(23 - y) = 4y + 4$, which simplifies to $46 - 2y = 4y + 4$. Then $6y = 42$, so $y = 7$.

Solve for $x$ (1c)

$x = 23 - y = 23 - 7 = 16$.

Square the expression (2a)

Let $A = \sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}$. Then $A^2 = (\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}})^2 = (3+2\sqrt{2}) - 2\sqrt{(3+2\sqrt{2})(3-2\sqrt{2})} + (3-2\sqrt{2})$ $A^2 = 6 - 2\sqrt{9 - 4(2)} = 6 - 2\sqrt{9 - 8} = 6 - 2\sqrt{1} = 6 - 2 = 4$.

Take the square root of the resulting number (2a)

$A = \sqrt{4} = 2$. Since $2$ can be written as $\frac{2}{1}$, it is a rational number.

Apply logarithm rules (2b)

We have $2\log x = \log 4 + \log (2x-3)$. Using logarithm rules, $\log x^2 = \log[4(2x-3)]$, so $x^2 = 4(2x-3) = 8x - 12$.

Solve the quadratic equation (2b)

$x^2 - 8x + 12 = 0$. Factoring, we get $(x-6)(x-2) = 0$, so $x=6$ or $x=2$.

Check the solutions to discard the invalid one (2b)

If $x=2$, then $2x-3 = 4-3 = 1 > 0$ and $x > 0$, thus $x=2$ is a valid solution. If $x=6$, then $2x-3 = 12-3 = 9 > 0$ and $x > 0$, thus $x=6$ is a valid solution.

Use a Venn diagram or set theory to solve the problem (3a)

Let $P$ be the set of students who study physics and $C$ be the set of students who study chemistry. We are given $|P| = 20$, $|P \cap C| = 12$, total students $= 30$, and students who study neither $= 8$. This means $|P \cup C| = 30 - 8 = 22$. We know that $|P \cup C| = |P| + |C| - |P \cap C|$, so $22 = 20 + |C| - 12$. Therefore, $|C| = 22 - 20 + 12 = 14$.

Calculate the probability of Yanga winning (3b)

The probability of Simba winning is 0.74. Since it's a knockout match, the probability of Yanga winning is $1 - 0.74 = 0.26$.

Calculate the probability of Barcelona winning (3b)

The probability of Real Madrid winning is 0.65. The probability of Barcelona winning is $1 - 0.65 = 0.35$.

Calculate the probability of both Yanga and Barcelona winning (3b)

Since the matches are independent, the probability that both Yanga and Barcelona win is $0.26 \cdot 0.35 = 0.091$.

Find the slope of line CD (4a)

The line $\overleftrightarrow{CD}$ passes through $C(2,1)$ and $D(6,1)$. The slope $m_{CD} = \frac{1-1}{6-2} = \frac{0}{4} = 0$. This means the line is horizontal.

Find the slope of the perpendicular line AB (4a)

Since $\overleftrightarrow{AB}$ is perpendicular to $\overleftrightarrow{CD}$, the slope of $\overleftrightarrow{AB}$ is undefined (vertical line). Undefined slope means a vertical line $x = c$ for some constant $c$.

Find the equation of the line AB (4a)

Since $\overleftrightarrow{AB}$ passes through $B(5,3)$ and the line must be vertical, the equation of the line is $x = 5$.

Define "solution" (4b)(i)

A solution is a value or set of values that satisfies a given equation or set of equations.

Rewrite the first equation (4b)(ii)

$\frac{\frac{y}{5}-1}{\frac{x}{15}}=x \implies \frac{y}{5} - 1 = \frac{x^2}{15} \implies \frac{y}{5} = \frac{x^2}{15} + 1 \implies y = \frac{x^2}{3} + 5$

Rewrite the second equation (4b)(ii)

$\frac{\frac{k}{40}x - \frac{y}{8}}{1}=1 \implies \frac{k}{40}x - \frac{y}{8} = 1 \implies \frac{y}{8} = \frac{k}{40}x - 1 \implies y = \frac{k}{5}x - 8$

Set the two equations as parallel (4b)(ii)

For no solution, the lines must be parallel with different y-intercepts. This means $\frac{k}{5} = 0$, $k=0$

$\frac{x^2}{3} + 5 = \frac{k}{5}x - 8$. For the system to have no solution, the lines must have the same slope but different y-intercepts, but observe that the first equation isn't a line. For these equations to have no solution, the lines should be parallel, therefore. The two equations are then: $$y = \frac{x^2}{3} + 5$$ $$y = \frac{k}{5}x - 8$$ Parallel line should means that these line should have slope. Hence, by comparison, k/5 should be 0. Hence, $k = 0$

(a) Eight trillion, four hundred eighty-eight billion, four hundred fifty-six million, nineteen thousand, five hundred. (b) $\frac{727}{210}$ (c) $16$ and $7$
(a) $2$ which is a rational number (b) ${2,6}$
(a) $14$ (b) $0.091$
(a) $x=5$ (b) (i) A solution is a value or set of values that satisfies a given equation or set of equations. (ii) $k = 0$

More Information

Here's some additional information on the answers:

In 1(a), it's crucial to correctly identify the place values when writing out large numbers in words.
In 1(b), repeating decimals can be converted to fractions using algebraic manipulation. Also, recognizing common decimal representations of fractions like $\frac{1}{7}$ simplifies the calculation.
In 2(a), squaring the given expression is a common technique to rationalize and simplify expressions involving square roots.
In 3(a), Venn diagrams are a great tool to visualize set relationships and solve related problems.
In 4(a), remember that perpendicular lines have slopes that are negative reciprocals of each other.

Tips

For 1(a), students may miscount the place values, leading to incorrect naming of the number.
For 1(b), errors may occur converting the repeating decimal to a fraction, or during the arithmetic of fractions.
For 1(c), students may incorrectly set up the equations. Paying close attention to the wording is important.
For 2(b), forgetting to check the domain of the logarithmic function when determining valid solutions. Logarithms are only defined for positive arguments.
For 3(b), failing to recognize that the probability of an opponent winning is 1 minus the probability of the team winning.
For 4(a), confusing slopes of perpendicular and parallel lines.
For 4(b)(ii), making mistakes in algebraic manipulation especially with fractions.

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