Convert 0.000437 to scientific notation and calculate (3.2 × 103) + (4.5 × 102). Simplify 18/45 and express…

Question

Convert 0.000437 to scientific notation and calculate (3.2 × 103) + (4.5 × 102). Simplify 18/45 and express 45 as a percentage of 120.

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Here are the solutions to the questions provided.

Section A

1. a)

Identify the number 0.000437.
Move the decimal point to the right until there is one non-zero digit to its left.
The decimal point moves 4 places to the right.
Write the number in the form $a \times 10^n$ , where $1 \le a < 10$ and $n$ is the number of places the decimal point moved.
Since the original number is less than 1, $n$ is negative.
$0.000437 = \boxed{4.37 \times 10^{-4}}$
1. b)
Evaluate $3.2 \times 10^3$ .
$3.2 \times 10^3 = 3.2 \times 1000 = 3200$
Evaluate $4.5 \times 10^2$ .
$4.5 \times 10^2 = 4.5 \times 100 = 450$
Add the results from Step 1 and Step 2.
$3200 + 450 = \boxed{3650}$
2. a)
Simplify the fraction $\frac{18}{45}$ .
Find the greatest common divisor (GCD) of 18 and 45.
The divisors of 18 are 1, 2, 3, 6, 9, 18.
The divisors of 45 are 1, 3, 5, 9, 15, 45.
The GCD is 9.
Divide both the numerator and the denominator by the GCD.
$\frac{18 \div 9}{45 \div 9} = \frac{2}{5}$
The simplified fraction is $\boxed{\frac{2}{5}}$ .

2. b)
To express 45 as a percentage of 120, divide 45 by 120 and multiply by 100%.
$\frac{45}{120} \times 100\%$
Simplify the fraction.
$\frac{45}{120} = \frac{9 \times 5}{24 \times 5} = \frac{9}{24} = \frac{3 \times 3}{8 \times 3} = \frac{3}{8}$
Multiply by 100%.
$\frac{3}{8} \times 100\% = 0.375 \times 100\% = \boxed{37.5\%}$
3. a)
Expand the expression $(x-3)(x+5)$ using the distributive property (FOIL method).
$(x-3)(x+5) = x(x+5) - 3(x+5)$ $= x^2 + 5x - 3x - 15$
Combine like terms.
$x^2 + (5x - 3x) - 15 = \boxed{x^2 + 2x - 15}$
3. b)
Factorize the expression $6y^2 - 9y$ .
Find the greatest common factor (GCF) of $6y^2$ and $9y$ .
The GCF of 6 and 9 is 3.
The GCF of $y^2$ and $y$ is $y$ .
So, the GCF of $6y^2$ and $9y$ is $3y$ .
Factor out the GCF.
$6y^2 - 9y = 3y(2y) - 3y(3) = \boxed{3y(2y - 3)}$
3. c)
Solve the equation $4(2x-3) = 5x+6$ .
First, distribute the 4 on the left side.
$8x - 12 = 5x + 6$
Subtract $5x$ from both sides of the equation.
$8x - 5x - 12 = 6$ $3x - 12 = 6$
Add 12 to both sides of the equation.
$3x = 6 + 12$ $3x = 18$
Divide by 3.
$x = \frac{18}{3}$ $x = \boxed{6}$
4. a)
The ratio of flour to sugar is $3:2$.
This means for every 3 parts of flour, there are 2 parts of sugar.
If 480g of flour is used, this corresponds to 3 parts of the ratio.
Let $x$ be the amount of sugar needed.
$\frac{\text{Flour}}{\text{Sugar}} = \frac{3}{2}$ $\frac{480 \text{ g}}{x} = \frac{3}{2}$
Cross-multiply to solve for $x$ .
$3x = 480 \text{ g} \times 2$ $3x = 960 \text{ g}$
Divide by 3.
$x = \frac{960 \text{ g}}{3}$ $x = \boxed{320 \text{ g}}$
4. b)
The angles in a triangle are in the ratio $2:3:4$.
Let the angles be $2k$, $3k$, and $4k$.
The sum of angles in a triangle is $180^\circ$ .
$2k + 3k + 4k = 180^\circ$ $9k = 180^\circ$
Solve for $k$ .
$k = \frac{180^\circ}{9}$ $k = 20^\circ$
Calculate each angle.
First angle: $2k = 2 \times 20^\circ = 40^\circ$
Second angle: $3k = 3 \times 20^\circ = 60^\circ$
Third angle: $4k = 4 \times 20^\circ = 80^\circ$
The angles are $\boxed{40^\circ, 60^\circ, 80^\circ}$ .

5. a)
Given the scores
$3, 9, 7, 4, 7, 6, 7$.
The mode is the value that appears most frequently in a data set.
Count the occurrences of each score:
3: 1 time
4: 1 time
6: 1 time
7: 3 times
9: 1 time
The score 7 appears most frequently.
The mode is $\boxed{7}$ .

5. b)
To find the median, first arrange the scores in ascending order.
Scores: $3, 9, 7, 4, 7, 6, 7$
Ordered scores: $3, 4, 6, 7, 7, 7, 9$
The number of scores is $n=7$ .
Since $n$ is odd, the median is the middle value, which is the $\frac{n+1}{2}$ -th term.
Median position = $\frac{7+1}{2} = \frac{8}{2} = 4$ -th term.
The 4th term in the ordered list is 7.
The median is $\boxed{7}$ .

6. a)
A fence has length 18m and width 12m.
This describes a rectangular shape.
The perimeter of a rectangle is given by the formula $P = 2(l+w)$ , where $l$ is length and $w$ is width.
$P = 2(18 \text{ m} + 12 \text{ m})$ $P = 2(30 \text{ m})$ $P = \boxed{60 \text{ m}}$
6. b)
The area of a rectangle is given by the formula $A = l \times w$ .
$A = 18 \text{ m} \times 12 \text{ m}$ $A = \boxed{216 \text{ m}^2}$
7. a)
Given the function $f(x) = 3x - 4$ .
To find $f(2)$ , substitute $x=2$ into the function.
$f(2) = 3(2) - 4$ $f(2) = 6 - 4$ $f(2) = \boxed{2}$
7. b)
To determine the inverse function $f^{-1}(x)$ , let $y = f(x)$ .
$y = 3x - 4$
Swap $x$ and $y$ .
$x = 3y - 4$
Solve for $y$ .
Add 4 to both sides.
$x + 4 = 3y$
Divide by 3.
$y = \frac{x+4}{3}$
So, $f^{-1}(x) = \boxed{\frac{x+4}{3}}$ .

8.
Let the amount received by the first student be $S_1$ and the amount received by the second student be $S_2$ .
The total amount received is 500 FCFA.
$S_1 + S_2 = 500$
One student has 50 FCFA more than the other.
Assume $S_1$ is the student with more money.
$S_1 = S_2 + 50$
Substitute the second equation into the first equation.
$(S_2 + 50) + S_2 = 500$ $2S_2 + 50 = 500$
Subtract 50 from both sides.
$2S_2 = 500 - 50$ $2S_2 = 450$
Divide by 2 to find $S_2$ .
$S_2 = \frac{450}{2}$ $S_2 = 225 \text{ FCFA}$
Find $S_1$ using $S_1 = S_2 + 50$ .
$S_1 = 225 + 50$ $S_1 = 275 \text{ FCFA}$
Each of them received $\boxed{275 \text{ FCFA and } 225 \text{ FCFA}}$ .

9. a)
Given vectors $\vec{r} = 3\vec{i} - \vec{j}$ and $\vec{s} = 2\vec{i} + \vec{j}$ .
Find $\vec{r} + 2\vec{s}$ .
First, calculate $2\vec{s}$ .
$2\vec{s} = 2(2\vec{i} + \vec{j}) = 4\vec{i} + 2\vec{j}$
Add $\vec{r}$ and $2\vec{s}$ .
$\vec{r} + 2\vec{s} = (3\vec{i} - \vec{j}) + (4\vec{i} + 2\vec{j})$ $= (3+4)\vec{i} + (-1+2)\vec{j}$ $= \boxed{7\vec{i} + \vec{j}}$
9. b)
Find $|\vec{r} - \vec{s}|$ .
First, calculate $\vec{r} - \vec{s}$ .
$\vec{r} - \vec{s} = (3\vec{i} - \vec{j}) - (2\vec{i} + \vec{j})$ $= (3-2)\vec{i} + (-1-1)\vec{j}$ $= \vec{i} - 2\vec{j}$
Calculate the magnitude of $\vec{i} - 2\vec{j}$ .
The magnitude of a vector $a\vec{i} + b\vec{j}$ is $\sqrt{a^2 + b^2}$ .
$|\vec{i} - 2\vec{j}| = \sqrt{(1)^2 + (-2)^2}$ $= \sqrt{1 + 4}$ $= \boxed{\sqrt{5}}$
10. a)
Express 126 as a product of its prime factors.
$126 = 2 \times 63$ $= 2 \times 3 \times 21$ $= 2 \times 3 \times 3 \times 7$ $= \boxed{2 \times 3^2 \times 7}$
Express 84 as a product of its prime factors.
$84 = 2 \times 42$ $= 2 \times 2 \times 21$ $= 2 \times 2 \times 3 \times 7$ $= \boxed{2^2 \times 3 \times 7}$
10. b)
To find the Highest Common Factor (HCF) of 126 and 84, use their prime factorizations.
$126 = 2^1 \times 3^2 \times 7^1$ $84 = 2^2 \times 3^1 \times 7^1$
For each common prime factor, take the lowest power.
Common prime factors are 2, 3, and 7.
Lowest power of 2: $2^1$
Lowest power of 3: $3^1$
Lowest power of 7: $7^1$
Multiply these lowest powers together.
$\text{HCF}(126, 84) = 2^1 \times 3^1 \times 7^1$ $= 2 \times 3 \times 7$ $= \boxed{42}$
Section B

1. (i) a)
Given two lines
$y = 2x + 1$ and $3x - y = 5$ .
To find the intersection point, substitute the expression for $y$ from the first equation into the second equation.
$3x - (2x + 1) = 5$
Solve for $x$ .
$3x - 2x - 1 = 5$ $x - 1 = 5$ $x = 5 + 1$ $x = 6$
Substitute the value of $x$ back into the first equation to find $y$ .
$y = 2(6) + 1$ $y = 12 + 1$ $y = 13$
The intersection point is $\boxed{(6, 13)}$ .

1. (i) b)
Given point A $(-3, 4)$ and point B $(1, 7)$ .
Use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .
Let $(x_1, y_1) = (-3, 4)$ and $(x_2, y_2) = (1, 7)$ .
$d = \sqrt{(1 - (-3))^2 + (7 - 4)^2}$ $d = \sqrt{(1 + 3)^2 + (3)^2}$ $d = \sqrt{(4)^2 + (3)^2}$ $d = \sqrt{16 + 9}$ $d = \sqrt{25}$ $d = \boxed{5 \text{ units}}$
1. (i) c)
Given points L $(5, 6)$ and N $(1, 2)$.
Use the midpoint formula $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ .
Let $(x_1, y_1) = (5, 6)$ and $(x_2, y_2) = (1, 2)$ .
$M = \left(\frac{5 + 1}{2}, \frac{6 + 2}{2}\right)$ $M = \left(\frac{6}{2}, \frac{8}{2}\right)$ $M = \boxed{(3, 4)}$
1. (ii) a)
Given the quadratic function $y = x^2 - 4x + 3$ and the linear function $y = x - 1$ .
Complete the table for $y = x^2 - 4x + 3 = 0$ .
For $x = -1$ : $y = (-1)^2 - 4(-1) + 3 = 1 + 4 + 3 = 8$
For $

Convert 0.000437 to scientific notation and calculate (3.2 x 103) + (4.5 x 102). Simplify 18/45 and express 45 as a percentage of 120.

Step-by-step solution

Identify the number 0.000437.

Move the decimal point to the right until there is one non-zero digit to its left.

Write the number in the form a×10na \times 10^na×10n, where 1≤a<101 \le a < 101≤a<10 and nnn is the number of places the decimal point moved.

Evaluate 3.2×1033.2 \times 10^33.2×103.

Evaluate 4.5×1024.5 \times 10^24.5×102.

Add the results from Step 1 and Step 2.

Simplify the fraction 1845\frac{18}{45}4518​.

Divide both the numerator and the denominator by the GCD.

To express 45 as a percentage of 120, divide 45 by 120 and multiply by 100%.

Simplify the fraction.

Multiply by 100%.

Expand the expression (x−3)(x+5)(x-3)(x+5)(x−3)(x+5) using the distributive property (FOIL method).

Combine like terms.

Factorize the expression 6y2−9y6y^2 - 9y6y2−9y.

Find the greatest common factor (GCF) of 6y26y^26y2 and 9y9y9y.

Factor out the GCF.

Solve the equation 4(2x−3)=5x+64(2x-3) = 5x+64(2x−3)=5x+6.

Subtract $5x$ from both sides of the equation.

Add 12 to both sides of the equation.

Divide by 3.

The ratio of flour to sugar is $3:2$.

If 480g of flour is used, this corresponds to 3 parts of the ratio.

Cross-multiply to solve for xxx.

Divide by 3.

The angles in a triangle are in the ratio $2:3:4$.

The sum of angles in a triangle is 180∘180^\circ180∘.

Solve for kkk.

Calculate each angle.

Given the scores

The mode is the value that appears most frequently in a data set.

To find the median, first arrange the scores in ascending order.

The number of scores is n=7n=7n=7.

The 4th term in the ordered list is 7.

A fence has length 18m and width 12m.

The perimeter of a rectangle is given by the formula P=2(l+w)P = 2(l+w)P=2(l+w), where lll is length and www is width.

The area of a rectangle is given by the formula A=l×wA = l \times wA=l×w.

Given the function f(x)=3x−4f(x) = 3x - 4f(x)=3x−4.

To find f(2)f(2)f(2), substitute x=2x=2x=2 into the function.

To determine the inverse function f−1(x)f^{-1}(x)f−1(x), let y=f(x)y = f(x)y=f(x).

Swap xxx and yyy.

Solve for yyy.

Divide by 3.

Let the amount received by the first student be S1S_1S1​ and the amount received by the second student be S2S_2S2​.

The total amount received is 500 FCFA.

One student has 50 FCFA more than the other.

Substitute the second equation into the first equation.

Subtract 50 from both sides.

Divide by 2 to find S2S_2S2​.

Find S1S_1S1​ using S1=S2+50S_1 = S_2 + 50S1​=S2​+50.

Given vectors r⃗=3i⃗−j⃗\vec{r} = 3\vec{i} - \vec{j}r=3i−j​ and s⃗=2i⃗+j⃗\vec{s} = 2\vec{i} + \vec{j}s=2i+j​.

Find r⃗+2s⃗\vec{r} + 2\vec{s}r+2s.

Add r⃗\vec{r}r and 2s⃗2\vec{s}2s.

Find ∣r⃗−s⃗∣|\vec{r} - \vec{s}|∣r−s∣.

Calculate the magnitude of i⃗−2j⃗\vec{i} - 2\vec{j}i−2j​.

Express 126 as a product of its prime factors.

Express 84 as a product of its prime factors.

To find the Highest Common Factor (HCF) of 126 and 84, use their prime factorizations.

For each common prime factor, take the lowest power.

Multiply these lowest powers together.

Given two lines

To find the intersection point, substitute the expression for yyy from the first equation into the second equation.

Solve for xxx.

Substitute the value of xxx back into the first equation to find yyy.

Given point A (−3,4)(-3, 4)(−3,4) and point B (1,7)(1, 7)(1,7).

Use the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}d=(x2​−x1​)2+(y2​−y1​)2​.

Given points L $(5, 6)$ and N $(1, 2)$.

Use the midpoint formula M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)M=(2x1​+x2​​,2y1​+y2​​).

Given the quadratic function y=x2−4x+3y = x^2 - 4x + 3y=x2−4x+3 and the linear function y=x−1y = x - 1y=x−1.

Complete the table for y=x2−4x+3=0y = x^2 - 4x + 3 = 0y=x2−4x+3=0.

Still stuck on this one?

More Mathematics Questions

Write the number in the form $a \times 10^n$ , where $1 \le a < 10$ and $n$ is the number of places the decimal point moved.

Evaluate $3.2 \times 10^3$ .

Evaluate $4.5 \times 10^2$ .

Simplify the fraction $\frac{18}{45}$ .

Expand the expression $(x-3)(x+5)$ using the distributive property (FOIL method).

Factorize the expression $6y^2 - 9y$ .

Find the greatest common factor (GCF) of $6y^2$ and $9y$ .

Solve the equation $4(2x-3) = 5x+6$ .

Cross-multiply to solve for $x$ .

The sum of angles in a triangle is $180^\circ$ .

Solve for $k$ .

The number of scores is $n=7$ .

The perimeter of a rectangle is given by the formula $P = 2(l+w)$ , where $l$ is length and $w$ is width.

The area of a rectangle is given by the formula $A = l \times w$ .

Given the function $f(x) = 3x - 4$ .

To find $f(2)$ , substitute $x=2$ into the function.

To determine the inverse function $f^{-1}(x)$ , let $y = f(x)$ .

Swap $x$ and $y$ .

Solve for $y$ .

Let the amount received by the first student be $S_1$ and the amount received by the second student be $S_2$ .

Divide by 2 to find $S_2$ .

Find $S_1$ using $S_1 = S_2 + 50$ .

Given vectors $\vec{r} = 3\vec{i} - \vec{j}$ and $\vec{s} = 2\vec{i} + \vec{j}$ .

Find $\vec{r} + 2\vec{s}$ .

Add $\vec{r}$ and $2\vec{s}$ .

Find $|\vec{r} - \vec{s}|$ .

Calculate the magnitude of $\vec{i} - 2\vec{j}$ .

To find the intersection point, substitute the expression for $y$ from the first equation into the second equation.

Solve for $x$ .

Substitute the value of $x$ back into the first equation to find $y$ .

Given point A $(-3, 4)$ and point B $(1, 7)$ .

Use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .

Use the midpoint formula $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ .

Given the quadratic function $y = x^2 - 4x + 3$ and the linear function $y = x - 1$ .

Complete the table for $y = x^2 - 4x + 3 = 0$ .