Calculate the number of sides of a polygon with an interior angle of 60°.

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10.
Calculate the number of sides of a polygon with an interior angle of $60^\circ$ :

The formula for the interior angle of a regular polygon with $n$ sides is $\frac{(n-2) \times 180^\circ}{n}$ .
Set the formula equal to $60^\circ$ and solve for $n$ .
$\frac{(n-2) \times 180^\circ}{n} = 60^\circ$ $(n-2) \times 180 = 60n$ $180n - 360 = 60n$ $180n - 60n = 360$ $120n = 360$ $n = \frac{360}{120}$ $n = \boxed{3}$
11.
A map has a scale of $1:200,000$ :
This means $1 \text{ cm}$ on the map represents $200,000 \text{ cm}$ in actual distance.
Convert the actual distance from cm to km.
$200,000 \text{ cm} = 200,000 \times \frac{1}{100} \text{ m} = 2000 \text{ m}$ $2000 \text{ m} = 2000 \times \frac{1}{1000} \text{ km} = 2 \text{ km}$
So, $1 \text{ cm}$ on the map represents $2 \text{ km}$ actual distance.

a) Find the distance in km represented by $4 \text{ cm}$ on the map:
Multiply the equivalent distance for $1 \text{ cm}$ by 4.
$4 \text{ cm} \times 2 \text{ km/cm} = \boxed{8 \text{ km}}$
b) Find the actual area in km $^2$ represented by $8 \text{ cm}^2$ on the map:
The area scale is the square of the linear scale.
$(1 \text{ cm})^2 : (2 \text{ km})^2$ $1 \text{ cm}^2 : 4 \text{ km}^2$
So, $1 \text{ cm}^2$ on the map represents $4 \text{ km}^2$ actual area.
Multiply the equivalent area for $1 \text{ cm}^2$ by 8.
$8 \text{ cm}^2 \times 4 \text{ km}^2/\text{cm}^2 = \boxed{32 \text{ km}^2}$
12.

a) Evaluate $\frac{5}{7} - \frac{1}{3}$ :
Find a common denominator, which is $7 \times 3 = 21$ .
$\frac{5}{7} - \frac{1}{3} = \frac{5 \times 3}{7 \times 3} - \frac{1 \times 7}{3 \times 7}$
Perform the subtraction.
$= \frac{15}{21} - \frac{7}{21} = \frac{15 - 7}{21} = \boxed{\frac{8}{21}}$
b) Evaluate $5\frac{3}{3} \div \frac{1}{4}$ :
Simplify the mixed number $5\frac{3}{3}$ .
$5\frac{3}{3} = 5 + 1 = 6$
Perform the division by multiplying by the reciprocal.
$6 \div \frac{1}{4} = 6 \times 4 = \boxed{24}$
c) Express 0.025 as a fraction in its lowest term:
Write the decimal as a fraction.
$0.025 = \frac{25}{1000}$
Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 25.
$\frac{25 \div 25}{1000 \div 25} = \boxed{\frac{1}{40}}$
13.
Given the formula $\frac{1}{c} = \sqrt{b-a}$ :

a) Find $c$ if $a = -4, b = 21$ :
Substitute the given values of $a$ and $b$ into the formula.
$\frac{1}{c} = \sqrt{21 - (-4)}$ $\frac{1}{c} = \sqrt{21 + 4}$
Simplify the square root.
$\frac{1}{c} = \sqrt{25}$ $\frac{1}{c} = 5$
Solve for $c$ .
$c = \boxed{\frac{1}{5}}$
b) Make $b$ the subject of the formula:
Square both sides of the equation to eliminate the square root.
$\left(\frac{1}{c}\right)^2 = (\sqrt{b-a})^2$ $\frac{1}{c^2} = b-a$
Add $a$ to both sides to isolate $b$ .
$b = \boxed{\frac{1}{c^2} + a}$
14.
Given $A = \begin{pmatrix} 2 & 1 \\ -3 & 0 \end{pmatrix}$ :

a) Find $AB$ :
The matrix $B$ is not provided in the question. Therefore, $AB$ cannot be calculated.

b) Find $A^{-1}$ :
Calculate the determinant of matrix $A$ .
$\det(A) = (2)(0) - (1)(-3) = 0 - (-3) = 3$
Find the adjoint of matrix $A$ .
For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the adjoint is $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ .
$\text{adj}(A) = \begin{pmatrix} 0 & -1 \\ -(-3) & 2 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 3 & 2 \end{pmatrix}$
Calculate the inverse using the formula $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$ .
$A^{-1} = \frac{1}{3} \begin{pmatrix} 0 & -1 \\ 3 & 2 \end{pmatrix} = \boxed{\begin{pmatrix} 0 & -\frac{1}{3} \\ 1 & \frac{2}{3} \end{pmatrix}}$
15.
Given $\vec{OP} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and $\vec{Q} = (1,5)$ (assuming $\vec{OQ} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}$ ):

a) Find $|\vec{OP}|$ :
The magnitude of a vector $\begin{pmatrix} x \\ y \end{pmatrix}$ is $\sqrt{x^2 + y^2}$ .
$|\vec{OP}| = \sqrt{4^2 + 1^2}$
Calculate the magnitude.
$|\vec{OP}| = \sqrt{16 + 1} = \boxed{\sqrt{17}}$
b) Find $\vec{PQ}$ :
Use the vector subtraction formula $\vec{PQ} = \vec{OQ} - \vec{OP}$ .
$\vec{PQ} = \begin{pmatrix} 1 \\ 5 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \end{pmatrix}$
Perform the subtraction.
$\vec{PQ} = \begin{pmatrix} 1-4 \\ 5-1 \end{pmatrix} = \boxed{\begin{pmatrix} -3 \\ 4 \end{pmatrix}}$
c) If $\vec{QR} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}$ , find the coordinates of R:
Let the coordinates of R be $(x, y)$ , so $\vec{OR} = \begin{pmatrix} x \\ y \end{pmatrix}$ .
Use the vector addition formula $\vec{OR} = \vec{OQ} + \vec{QR}$ .
$\vec{OR} = \begin{pmatrix} 1 \\ 5 \end{pmatrix} + \begin{pmatrix} -2 \\ -3 \end{pmatrix}$
Perform the addition.
$\vec{OR} = \begin{pmatrix} 1 + (-2) \\ 5 + (-3) \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}$
The coordinates of R are the components of $\vec{OR}$ .
$R = \boxed{(-1, 2)}$
16.
Given a right-angled triangle ABC with $\angle B = 90^\circ$ , $AB = 4 \text{ cm}$ , $AC = 5 \text{ cm}$ . Point D is on BC extended.
Use the Pythagorean theorem to find the length of BC.
$AB^2 + BC^2 = AC^2$ $4^2 + BC^2 = 5^2$ $16 + BC^2 = 25$ $BC^2 = 25 - 16$ $BC^2 = 9$ $BC = \sqrt{9} = 3 \text{ cm}$
a) Find $\cos A\hat{C}B$ as a fraction:
In $\triangle ABC$ , $\cos(\angle ACB) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC}$ .
$\cos A\hat{C}B = \frac{3}{5}$ $\boxed{\frac{3}{5}}$
b) Find $\cos A\hat{C}D$ as a fraction:
Angles $\angle ACB$ and $\angle ACD$ form a linear pair, so they are supplementary.
$\angle ACD = 180^\circ - \angle ACB$
Use the trigonometric identity $\cos(180^\circ - \theta) = -\cos(\theta)$ .
$\cos A\hat{C}D = -\cos A\hat{C}B$ $\cos A\hat{C}D = \boxed{-\frac{3}{5}}$
17.
A trader made a profit of 5% when he sold an item for $200. Find the cost price:
Let the cost price be $CP$ .
The selling price ( $SP$ ) is the cost price plus the profit.
$SP = CP + 5\% \text{ of } CP$ $SP = CP + 0.05 \times CP$ $SP = 1.05 \times CP$
Substitute the given selling price and solve for $CP$ .
$200 = 1.05 \times CP$ $CP = \frac{200}{1.05}$
Calculate the value.
$CP = \frac{200}{\frac{105}{100}} = \frac{200 \times 100}{105} = \frac{20000}{105}$
$CP = \frac{4000}{21} \approx \boxed{\$190.48}$

18.

a) Convert 4.21 hours to hours and minutes:
Separate the integer and fractional parts of the hours.
$4.21 \text{ hours} = 4 \text{ hours} + 0.21 \text{ hours}$
Convert the fractional part of the hours to minutes by multiplying by 60.
$0.21 \text{ hours} \times 60 \text{ minutes/hour} = 12.6 \text{ minutes}$ $4.21 \text{ hours} = \boxed{4 \text{ hours and } 12.6 \text{ minutes}}$
b) Convert $18 \text{ km/hr}$ to $\text{m/s}$ :
Convert kilometers to meters ( $1 \text{ km} = 1000 \text{ m}$ ) and hours to seconds ( $1 \text{ hour} = 3600 \text{ seconds}$ ).
$18 \frac{\text{km}}{\text{hr}} = 18 \times \frac{1000 \text{ m}}{3600 \text{ s}}$
Simplify the fraction.
$= 18 \times \frac{10}{36} \frac{\text{m}}{\text{s}}$ $= 18 \times \frac{5}{18} \frac{\text{m}}{\text{s}}$ $= \boxed{5 \text{ m/s}}$

Calculate the number of sides of a polygon with an interior angle of 60°.

Step-by-step solution

The formula for the interior angle of a regular polygon with nnn sides is (n−2)×180∘n\frac{(n-2) \times 180^\circ}{n}n(n−2)×180∘​.

Set the formula equal to 60∘60^\circ60∘ and solve for nnn.

Convert the actual distance from cm to km.

Multiply the equivalent distance for 1 cm1 \text{ cm}1 cm by 4.

The area scale is the square of the linear scale.

Multiply the equivalent area for 1 cm21 \text{ cm}^21 cm2 by 8.

Find a common denominator, which is 7×3=217 \times 3 = 217×3=21.

Perform the subtraction.

Simplify the mixed number 5335\frac{3}{3}533​.

Perform the division by multiplying by the reciprocal.

Write the decimal as a fraction.

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 25.

Substitute the given values of aaa and bbb into the formula.

Simplify the square root.

Solve for ccc.

Square both sides of the equation to eliminate the square root.

Add aaa to both sides to isolate bbb.

Calculate the determinant of matrix AAA.

Find the adjoint of matrix AAA.

Calculate the inverse using the formula A−1=1det⁡(A)adj(A)A^{-1} = \frac{1}{\det(A)} \text{adj}(A)A−1=det(A)1​adj(A).

The magnitude of a vector (xy)\begin{pmatrix} x \\ y \end{pmatrix}(xy​) is x2+y2\sqrt{x^2 + y^2}x2+y2​.

Calculate the magnitude.

Use the vector subtraction formula PQ⃗=OQ⃗−OP⃗\vec{PQ} = \vec{OQ} - \vec{OP}PQ​=OQ​−OP.

Perform the subtraction.

Let the coordinates of R be (x,y)(x, y)(x,y), so OR⃗=(xy)\vec{OR} = \begin{pmatrix} x \\ y \end{pmatrix}OR=(xy​).

Perform the addition.

The coordinates of R are the components of OR⃗\vec{OR}OR.

Use the Pythagorean theorem to find the length of BC.

In △ABC\triangle ABC△ABC, cos⁡(∠ACB)=AdjacentHypotenuse=BCAC\cos(\angle ACB) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC}cos(∠ACB)=HypotenuseAdjacent​=ACBC​.

Angles ∠ACB\angle ACB∠ACB and ∠ACD\angle ACD∠ACD form a linear pair, so they are supplementary.

Use the trigonometric identity cos⁡(180∘−θ)=−cos⁡(θ)\cos(180^\circ - \theta) = -\cos(\theta)cos(180∘−θ)=−cos(θ).

Let the cost price be CPCPCP.

Substitute the given selling price and solve for CPCPCP.

Calculate the value.

Separate the integer and fractional parts of the hours.

Convert the fractional part of the hours to minutes by multiplying by 60.

Convert kilometers to meters (1 km=1000 m1 \text{ km} = 1000 \text{ m}1 km=1000 m) and hours to seconds (1 hour=3600 seconds1 \text{ hour} = 3600 \text{ seconds}1 hour=3600 seconds).

Simplify the fraction.

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More Mathematics Questions

The formula for the interior angle of a regular polygon with $n$ sides is $\frac{(n-2) \times 180^\circ}{n}$ .

Set the formula equal to $60^\circ$ and solve for $n$ .

Multiply the equivalent distance for $1 \text{ cm}$ by 4.

Multiply the equivalent area for $1 \text{ cm}^2$ by 8.

Find a common denominator, which is $7 \times 3 = 21$ .

Simplify the mixed number $5\frac{3}{3}$ .

Substitute the given values of $a$ and $b$ into the formula.

Solve for $c$ .

Add $a$ to both sides to isolate $b$ .

Calculate the determinant of matrix $A$ .

Find the adjoint of matrix $A$ .

Calculate the inverse using the formula $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$ .

The magnitude of a vector $\begin{pmatrix} x \\ y \end{pmatrix}$ is $\sqrt{x^2 + y^2}$ .

Use the vector subtraction formula $\vec{PQ} = \vec{OQ} - \vec{OP}$ .

Let the coordinates of R be $(x, y)$ , so $\vec{OR} = \begin{pmatrix} x \\ y \end{pmatrix}$ .

The coordinates of R are the components of $\vec{OR}$ .

In $\triangle ABC$ , $\cos(\angle ACB) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC}$ .

Angles $\angle ACB$ and $\angle ACD$ form a linear pair, so they are supplementary.

Use the trigonometric identity $\cos(180^\circ - \theta) = -\cos(\theta)$ .

Let the cost price be $CP$ .

Substitute the given selling price and solve for $CP$ .

Convert kilometers to meters ( $1 \text{ km} = 1000 \text{ m}$ ) and hours to seconds ( $1 \text{ hour} = 3600 \text{ seconds}$ ).