text
stringlengths 22
1.01M
|
|---|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Use the product and quotient properties of radicals.
• Rationalize the denominator.
• Solve real-world problems using square root functions.
## Introduction
A radical reverses the operation of raising a number to a power. For example, the square of 4 is \begin{align*}4^2 = 4 \cdot 4 = 16\end{align*}, and so the square root of 16 is 4. The symbol for a square root is \begin{align*}\sqrt{\;\;}\end{align*}. This symbol is also called the radical sign.
In addition to square roots, we can also take cube roots, fourth roots, and so on. For example, since 64 is the cube of 4, 4 is the cube root of 64.
\begin{align*}\sqrt[3]{64} = 4 \qquad \text{since} \qquad 4^3 = 4 \cdot 4 \cdot 4 = 64\end{align*}
We put an index number in the top left corner of the radical sign to show which root of the number we are seeking. Square roots have an index of 2, but we usually don’t bother to write that out.
\begin{align*}\sqrt[2]{36} = \sqrt{36} = 6\end{align*}
The cube root of a number gives a number which when raised to the power three gives the number under the radical sign. The fourth root of number gives a number which when raised to the power four gives the number under the radical sign:
\begin{align*}\sqrt[4]{81} = 3 \qquad \text{since} \qquad 3^4 = 3 \cdot 3 \cdot 3 \cdot 3 = 81\end{align*}
And so on for any power we can name.
## Even and Odd Roots
Radical expressions that have even indices are called even roots and radical expressions that have odd indices are called odd roots. There is a very important difference between even and odd roots, because they give drastically different results when the number inside the radical sign is negative.
Any real number raised to an even power results in a positive answer. Therefore, when the index of a radical is even, the number inside the radical sign must be non-negative in order to get a real answer.
On the other hand, a positive number raised to an odd power is positive and a negative number raised to an odd power is negative. Thus, a negative number inside the radical sign is not a problem. It just results in a negative answer.
Example 1
a) \begin{align*}\sqrt{121}\end{align*}
b) \begin{align*}\sqrt[3]{125}\end{align*}
c) \begin{align*}\sqrt[4]{-625}\end{align*}
d) \begin{align*}\sqrt[5]{-32}\end{align*}
Solution
a) \begin{align*}\sqrt{121} = 11\end{align*}
b) \begin{align*}\sqrt[3]{125} = 5\end{align*}
c) \begin{align*}\sqrt[3]{-625}\end{align*} is not a real number
d) \begin{align*}\sqrt[5]{-32} = -2\end{align*}
## Use the Product and Quotient Properties of Radicals
Radicals can be re-written as rational powers. The radical: \begin{align*}\sqrt[m]{a^n}\end{align*} is defined as \begin{align*}a^{\frac{n}{m}}\end{align*}.
Example 2
Write each expression as an exponent with a rational value for the exponent.
a) \begin{align*}\sqrt{5}\end{align*}
b) \begin{align*}\sqrt[4]{a}\end{align*}
c) \begin{align*} \sqrt[3]{4xy}\end{align*}
d) \begin{align*}\sqrt[6]{x^5}\end{align*}
Solution
a) \begin{align*}\sqrt{5} = 5^{\frac{1}{2}}\end{align*}
b) \begin{align*}\sqrt[4]{a} = a^{\frac{1}{4}}\end{align*}
c) \begin{align*} \sqrt[3]{4xy} = (4xy)^{\frac{1}{3}}\end{align*}
d) \begin{align*}\sqrt[6]{x^5} = x^{\frac{5}{6}}\end{align*}
As a result of this property, for any non-negative number \begin{align*}a\end{align*} we know that \begin{align*}\sqrt[n]{a^n} = a^{\frac{n}{n}} = a\end{align*}.
Since roots of numbers can be treated as powers, we can use exponent rules to simplify and evaluate radical expressions. Let’s review the product and quotient rule of exponents.
\begin{align*}\text{Raising a product to a power:} && (x \cdot y)^n & = x^n \cdot y^n\\ \text{Raising a quotient to a power:} && \left(\frac{x}{y} \right)^n & = \frac{x^n}{y^n}\end{align*}
In radical notation, these properties are written as
\begin{align*}\text{Raising a product to a power:} && \sqrt[m]{x \cdot y} & = \sqrt[m]{x} \ \cdot \sqrt[m]{y}\\ \text{Raising a quotient to a power:} && \sqrt[m]{\frac{x}{y}} & = \frac{\sqrt[m]{x}}{\sqrt[m]{y}}\end{align*}
A very important application of these rules is reducing a radical expression to its simplest form. This means that we apply the root on all the factors of the number that are perfect roots and leave all factors that are not perfect roots inside the radical sign.
For example, in the expression \begin{align*}\sqrt{16}\end{align*}, the number 16 is a perfect square because \begin{align*}16 = 4^2\end{align*}. This means that we can simplify it as follows:
\begin{align*}\sqrt{16} = \sqrt{4^2} = 4\end{align*} Thus, the square root disappears completely.
On the other hand, in the expression \begin{align*}\sqrt{32}\end{align*}, the number 32 is not a perfect square, so we can’t just remove the square root. However, we notice that \begin{align*}32 = 16 \cdot 2\end{align*}, so we can write 32 as the product of a perfect square and another number. Thus,
\begin{align*}\sqrt{32} = \sqrt{16 \cdot 2}\end{align*} If we apply the “raising a product to a power” rule we get:
\begin{align*}\sqrt{32} = \sqrt{16 \cdot 2} = \sqrt{16} \ \cdot \sqrt{2}\end{align*} Since \begin{align*}\sqrt{16} = 4\end{align*}, we get: \begin{align*}\sqrt{32} = 4 \cdot \sqrt{2} = \underline{\underline{4 \sqrt{2}}}\end{align*}
Example 3
Write the following expressions in the simplest radical form.
a) \begin{align*}\sqrt{8}\end{align*}
b) \begin{align*} \sqrt{50}\end{align*}
c) \begin{align*}\sqrt{\frac{125}{72}}\end{align*}
Solution
The strategy is to write the number under the square root as the product of a perfect square and another number. The goal is to find the highest perfect square possible; if we don’t find it right away, we just repeat the procedure until we can’t simplify any longer.
a) \begin{align*}\text{We can write} \ 8 = 4 \cdot 2, \ \text{so} && & \sqrt{8} = \sqrt{4 \cdot 2}.\\ \text{With the “Raising a product to a power” rule, that becomes} &&& \sqrt{4} \ \cdot \sqrt{2}.\\ \text{Evaluate} \ \sqrt{4} \ \text{and we’re left with} &&& \underline{\underline{2\sqrt{2}}}.\end{align*}
b) \begin{align*}\text{We can write} \ 50 = 25 \cdot 2, \ \text{so:} && \sqrt{50} & = \sqrt{25 \cdot 2}\\ \text{Use “Raising a product to a power” rule:} && &=\sqrt{25} \ \cdot \sqrt{2} = \underline{\underline{5 \sqrt{2}}}\end{align*}
c) \begin{align*}\text{Use “Raising a quotient to a power” rule to separate the fraction:} && \sqrt{\frac{125}{72}} & = \frac{\sqrt{125}}{\sqrt{72}}\\ \text{Re-write each radical as a product of a perfect square and another number:} && & = \frac{ \sqrt{25 \cdot 5}}{\sqrt{36 \cdot 2}} = \frac{5 \sqrt{5}}{6 \sqrt{2}}\end{align*}
The same method can be applied to reduce radicals of different indices to their simplest form.
Example 4
Write the following expression in the simplest radical form.
a) \begin{align*}\sqrt[3]{40}\end{align*}
b) \begin{align*}\sqrt[4]{\frac{162}{80}}\end{align*}
c) \begin{align*}\sqrt[3]{135}\end{align*}
Solution
In these cases we look for the highest possible perfect cube, fourth power, etc. as indicated by the index of the radical.
a) Here we are looking for the product of the highest perfect cube and another number. We write: \begin{align*}\sqrt[3]{40} = \sqrt[3]{8 \cdot 5} = \sqrt[3]{8} \ \cdot \sqrt[3]{5} = 2 \sqrt[3]{5}\end{align*}
b) Here we are looking for the product of the highest perfect fourth power and another number.
\begin{align*}\text{Re-write as the quotient of two radicals:} && \sqrt[4]{\frac{162}{80}} & = \frac{\sqrt[4]{162}}{\sqrt[4]{80}}\\ \text{Simplify each radical separately:} && & = \frac{\sqrt[4]{81 \cdot 2}}{\sqrt[4]{16 \cdot 5}} = \frac{\sqrt[4]{81} \ \cdot \sqrt[4]{2}} {\sqrt[4]{16} \ \cdot \sqrt[4]{5}} = \frac{3 \sqrt[4]{2}}{2 \sqrt[4]{5}}\\ \text{Recombine the fraction under one radical sign:} && & = \frac{3}{2} \sqrt[4]{\frac{2}{5}}\end{align*}
c) Here we are looking for the product of the highest perfect cube root and another number. Often it’s not very easy to identify the perfect root in the expression under the radical sign. In this case, we can factor the number under the radical sign completely by using a factor tree:
We see that \begin{align*}135 = 3 \cdot 3 \cdot 3 \cdot 5 = 3^3 \cdot 5\end{align*}. Therefore \begin{align*}\sqrt[3]{135} = \sqrt[3]{3^3 \cdot 5} = \sqrt[3]{3^3} \ \cdot \sqrt[3]{5} = 3 \sqrt[3]{5}\end{align*}.
(You can find a useful tool for creating factor trees at http://www.softschools.com/math/factors/factor_tree/. Click on “User Number” to type in your own number to factor, or just click “New Number” for a random number if you want more practice factoring.)
Now let’s see some examples involving variables.
Example 5
Write the following expression in the simplest radical form.
a) \begin{align*}\sqrt{12x^3 y^5}\end{align*}
b) \begin{align*}\sqrt[4]{\frac{1250x^7}{405y^9}}\end{align*}
Solution
Treat constants and each variable separately and write each expression as the products of a perfect power as indicated by the index of the radical and another number.
a) \begin{align*}\text{Re-write as a product of radicals:} && \sqrt{12x^3y^5} & = \sqrt{12} \ \cdot \sqrt{x^3} \ \cdot \sqrt{y^5}\\ \text{Simplify each radical separately:} && \left(\sqrt{4 \cdot 3}\right ) \cdot \left( \sqrt{x^2 \cdot x}\right ) \cdot \left (\sqrt{y^4 \cdot y}\right ) & = \left (2 \sqrt{3}\right ) \cdot \left (x \sqrt{x}\right ) \cdot \left (y^2 \sqrt{y}\right )\\ \text{Combine all terms outside and inside the radical sign:} && & =2xy^2 \sqrt{3xy}\end{align*}
b) \begin{align*}\text{Re-write as a quotient of radicals:} && \sqrt[4]{\frac{1250x^7}{405y^9}} & = \frac{\sqrt[4]{1250x^7}}{\sqrt[4]{405y^9}}\\ \text{Simplify each radical separately:} && & = \frac{\sqrt[4]{625 \cdot 2} \ \cdot \sqrt[4]{x^4 \cdot x^3}}{\sqrt[4]{81 \cdot 5} \ \cdot \sqrt[4]{y^4 \cdot y^4 \cdot y}} = \frac{5 \sqrt[4]{2} \cdot x \cdot \sqrt[4]{x^3}}{3 \sqrt[4]{5} \cdot y \cdot y \ \cdot \sqrt[4]{y}} = \frac{5x \sqrt[4]{2x^3}}{3y^2 \sqrt[4]{5y}}\\ \text{Recombine fraction under one radical sign:} && &= \frac{5x}{3y^2} \sqrt[4]{\frac{2x^3}{5y}}\end{align*}
When we add and subtract radical expressions, we can combine radical terms only when they have the same expression under the radical sign. This is a lot like combining like terms in variable expressions. For example,
\begin{align*}4 \sqrt{2} + 5 \sqrt{2} & = 9 \sqrt{2}\\ & \text{or} \\ 2 \sqrt{3} - \sqrt{2} + 5 \sqrt{3} + 10\sqrt{2} & = 7 \sqrt{3} + 9 \sqrt{2}\end{align*}
It’s important to reduce all radicals to their simplest form in order to make sure that we’re combining all possible like terms in the expression. For example, the expression \begin{align*}\sqrt{8} - 2\sqrt{50}\end{align*} looks like it can’t be simplified any more because it has no like terms. However, when we write each radical in its simplest form we get \begin{align*}2\sqrt{2} - 10 \sqrt{2}\end{align*}, and we can combine those terms to get \begin{align*}-8 \sqrt{2}\end{align*}.
Example 6
Simplify the following expressions as much as possible.
a) \begin{align*}4 \sqrt{3} + 2 \sqrt{12}\end{align*}
b) \begin{align*}10 \sqrt{24} - \sqrt{28}\end{align*}
Solution
a) \begin{align*}\text{Simplify} \ \sqrt{12} \ \text{to its simplest form:} && & =4 \sqrt{3} + 2 \sqrt{4 \cdot 3} = 4 \sqrt{3} + 6 \sqrt{3}\\ \text{Combine like terms:} && & =10 \sqrt{3}\end{align*}
b) \begin{align*}\text{Simplify} \ \sqrt{24} \ \text{and} \ \sqrt{28} \ \text{to their simplest form:} && & =10 \sqrt{6 \cdot 4} - \sqrt{7 \cdot 4} = 20 \sqrt{6} - 2 \sqrt{7}\\ \text{There are no like terms.}\end{align*}
Example 7
Simplify the following expressions as much as possible.
a) \begin{align*}4 \sqrt[3]{128} - \sqrt[3]{250}\end{align*}
b) \begin{align*}3 \sqrt{x^3} - 4x \sqrt{9x}\end{align*}
Solution
a) \begin{align*}\text{Re-write radicals in simplest terms:} && & = 4 \sqrt[3]{2 \cdot 64} - \sqrt[3]{2 \cdot 125} = 16 \sqrt[3]{2} - 5 \sqrt[3]{2}\\ \text{Combine like terms:} && & = 11 \sqrt[3]{2}\end{align*}
b) \begin{align*}\text{Re-write radicals in simplest terms:} && 3 \sqrt{x^2 \cdot x} - 12x \sqrt{x} & = 3x \sqrt{x} - 12x \sqrt{x}\\ \text{Combine like terms:} && & =-9x \sqrt{x}\end{align*}
When we multiply radical expressions, we use the “raising a product to a power” rule: \begin{align*}\sqrt[m]{x \cdot y} = \sqrt[m]{x} \cdot \sqrt[m]{y}\end{align*}. In this case we apply this rule in reverse. For example:
\begin{align*}\sqrt{6} \cdot \sqrt{8} = \sqrt{6 \cdot 8} = \sqrt{48}\end{align*}
Or, in simplest radical form: \begin{align*}\sqrt{48} = \sqrt{16 \cdot 3} = 4 \sqrt{3}\end{align*}.
We’ll also make use of the fact that: \begin{align*}\sqrt{a} \cdot \sqrt{a} = \sqrt{a^2} = a\end{align*}.
When we multiply expressions that have numbers on both the outside and inside the radical sign, we treat the numbers outside the radical sign and the numbers inside the radical sign separately. For example, \begin{align*}a \sqrt{b} \cdot c \sqrt{d} = ac \sqrt{bd}\end{align*}.
Example 8
Multiply the following expressions.
a) \begin{align*}\sqrt{2}\left(\sqrt{3} + \sqrt{5}\right )\end{align*}
b) \begin{align*}2 \sqrt{x}\left (3 \sqrt{y} - \sqrt{x}\right )\end{align*}
c) \begin{align*}\left (2 + \sqrt{5}\right )\left (2 - \sqrt{6}\right )\end{align*}
d) \begin{align*}\left (2 \sqrt{x} + 1\right )\left (5 - \sqrt{x}\right )\end{align*}
Solution
In each case we use distribution to eliminate the parentheses.
a) \begin{align*}\text{Distribute} \ \sqrt{2} \ \text{inside the parentheses:} && \sqrt{2}\left (\sqrt{3} + \sqrt{5}\right ) & = \sqrt{2} \cdot \sqrt{3} + \sqrt{2} \cdot \sqrt{5}\\ \text{Use the “raising a product to a power” rule:} && & = \sqrt{2 \cdot 3} + \sqrt{2 \cdot 5}\\ \text{Simplify:} && & =\sqrt{6} + \sqrt{10}\end{align*} b) \begin{align*}\text{Distribute} \ 2 \sqrt{x} \ \text{inside the parentheses:} && & =(2 \cdot 3)\left (\sqrt{x} \cdot \sqrt{y}\right ) - 2 \cdot \left ( \sqrt{x} \cdot \sqrt{x}\right )\\ \text{Multiply:} && & =6 \sqrt{xy} - 2 \sqrt{x^2}\\ \text{Simplify:} && & =6 \sqrt{xy} - 2x\end{align*} c) \begin{align*}\text{Distribute:} && (2 + \sqrt{5})(2 - \sqrt{6}) & = (2 \cdot 2) - \left (2 \cdot \sqrt{6}\right ) + \left( 2 \cdot \sqrt{5} \right ) - \left ( \sqrt{5} \cdot \sqrt{6} \right )\\ \text{Simplify:}&& & =4 - 2 \sqrt{6} + 2 \sqrt{5} - \sqrt{30}\end{align*} d) \begin{align*}\text{Distribute:} && \left (2 \sqrt{x} - 1\right )\left (5 - \sqrt{x}\right ) &=10 \sqrt{x} - 2x - 5 + \sqrt{x}\\ \text{Simplify:} && & =11 \sqrt{x} - 2x - 5\end{align*}
## Rationalize the Denominator
Often when we work with radicals, we end up with a radical expression in the denominator of a fraction. It’s traditional to write our fractions in a form that doesn’t have radicals in the denominator, so we use a process called rationalizing the denominator to eliminate them.
Rationalizing is easiest when there’s just a radical and nothing else in the denominator, as in the fraction \begin{align*}\frac{2}{\sqrt{3}}\end{align*}. All we have to do then is multiply the numerator and denominator by a radical expression that makes the expression inside the radical into a perfect square, cube, or whatever power is appropriate. In the example above, we multiply by \begin{align*}\sqrt{3}\end{align*}:
\begin{align*}\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}\end{align*}
Cube roots and higher are a little trickier than square roots. For example, how would we rationalize \begin{align*}\frac{7}{\sqrt[3]{5}}\end{align*}? We can’t just multiply by \begin{align*}\sqrt[3]{5}\end{align*}, because then the denominator would be \begin{align*}\sqrt[3]{5^2}\end{align*}. To make the denominator a whole number, we need to multiply the numerator and the denominator by \begin{align*}\sqrt[3]{5^2}\end{align*}:
\begin{align*}\frac{7}{\sqrt[3]{5}} \cdot \frac{\sqrt[3]{5^2}}{\sqrt[3]{5^2}} = \frac{7 \sqrt[3]{25}}{\sqrt[3]{5^3}} = \frac{7 \sqrt[3]{25}}{5}\end{align*}
Trickier still is when the expression in the denominator contains more than one term. For example, consider the expression \begin{align*}\frac{2}{2 + \sqrt{3}}\end{align*}. We can’t just multiply by \begin{align*}\sqrt{3}\end{align*}, because we’d have to distribute that term and then the denominator would be \begin{align*}2 \sqrt{3} + 3\end{align*}.
Instead, we multiply by \begin{align*}2 - \sqrt{3}\end{align*}. This is a good choice because the product \begin{align*}\left (2 + \sqrt{3}\right )\left (2 - \sqrt{3}\right )\end{align*} is a product of a sum and a difference, which means it’s a difference of squares. The radicals cancel each other out when we multiply out, and the denominator works out to \begin{align*}\left (2 + \sqrt{3} \right )\left (2 - \sqrt{3}\right ) = 2^2 - \left ( \sqrt{3}\right )^2 = 4 - 3 = 1\end{align*}.
When we multiply both the numerator and denominator by \begin{align*}2 - \sqrt{3}\end{align*}, we get:
\begin{align*}\frac{2}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2\left (2 - \sqrt{3}\right )}{4 - 3} = \frac{4 - 2 \sqrt{3}}{1} = 4 - 2 \sqrt{3}\end{align*}
Now consider the expression \begin{align*}\frac{\sqrt{x} - 1}{\sqrt{x} - 2 \sqrt{y}}\end{align*}.
In order to eliminate the radical expressions in the denominator we must multiply by \begin{align*}\sqrt{x} + 2 \sqrt{y}\end{align*}.
We get: \begin{align*}\frac{\sqrt{x} - 1}{\sqrt{x} - 2 \sqrt{y}} \cdot \frac{\sqrt{x} + 2 \sqrt{y}}{\sqrt{x} + 2 \sqrt{y}} = \frac{\left (\sqrt{x} - 1\right )\left (\sqrt{x} + 2 \sqrt{y}\right )} {\left (\sqrt{x} - 2 \sqrt{y} \right ) \left ( \sqrt{x} + 2 \sqrt{y} \right )} = \frac{x - 2 \sqrt{y} - \sqrt{x} + 2 \sqrt{xy}}{x - 4y}\end{align*}
## Solve Real-World Problems Using Radical Expressions
Radicals often arise in problems involving areas and volumes of geometrical figures.
Example 9
A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square feet. Find the dimensions of the pool and the area of the pool.
Solution
Make a sketch:
Let \begin{align*}x =\end{align*} the width of the pool. Then:
Area \begin{align*}=\end{align*} length \begin{align*}\times\end{align*} width
Combined length of pool and walkway \begin{align*}= 2x + 2\end{align*}
Combined width of pool and walkway \begin{align*}= x + 2\end{align*}
\begin{align*}\text{Area} = (2x + 2)(x + 2)\end{align*}
Since the combined area of pool and walkway is \begin{align*}400 \ ft^2\end{align*} we can write the equation
\begin{align*}(2x + 2)(x + 2) = 400\end{align*}
\begin{align*}\text{Multiply in order to eliminate the parentheses:} && 2x^2 + 4x + 2x + 4 & = 400\\ \text{Collect like terms:} && 2x^2 + 6x + 4 &= 400\\ \text{Move all terms to one side of the equation:} && 2x^2 + 6x - 396 & = 0\\ \text{Divide all terms by 2:} && x^2 + 3x - 198 & = 0\\ \text{Use the quadratic formula:} && x & =\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\ && x & = \frac{-3 \pm \sqrt{3^2 - 4(1)(-198)}}{2(1)}\\ && x & = \frac{-3 \pm \sqrt{801}}{2} = \frac{-3 \pm 28.3}{2}\\ && x & = 12.65 \ feet\end{align*}
(The other answer is negative, so we can throw it out because only a positive number makes sense for the width of a swimming pool.)
Check by plugging the result in the area formula:
Area \begin{align*}= (2(12.65) + 2)(12.65 + 2) = 27.3 \cdot 14.65 = 400 \ ft^2\end{align*}.
Example 10
The volume of a soda can is \begin{align*}355 \ cm^3\end{align*}. The height of the can is four times the radius of the base. Find the radius of the base of the cylinder.
Solution
Make a sketch:
Let \begin{align*}x =\end{align*} the radius of the cylinder base. Then the height of the cylinder is \begin{align*}4x\end{align*}.
The volume of a cylinder is given by \begin{align*}V = \pi R^2 \cdot h\end{align*}; in this case, \begin{align*}R\end{align*} is \begin{align*}x\end{align*} and \begin{align*}h\end{align*} is \begin{align*}4x\end{align*}, and we know the volume is 355.
Solve the equation:
\begin{align*}355 & = \pi x^2 \cdot(4x)\\ 355 & = 4 \pi x^3\\ x^3 & = \frac{355}{4 \pi}\\ x & = \sqrt[3]{\frac{355}{4 \pi}} = 3.046 \ cm\end{align*}
Check by substituting the result back into the formula:
\begin{align*}V = \pi R^2 \cdot h = \pi (3.046)^2 \cdot (4 \cdot 3.046) = 355 \ cm^3\end{align*}
So the volume is \begin{align*}355 \ cm^3\end{align*}. The answer checks out.
## Review Questions
1. \begin{align*}\sqrt{169}\end{align*}
2. \begin{align*}\sqrt[4]{-81}\end{align*}
3. \begin{align*}\sqrt[3]{-125}\end{align*}
4. \begin{align*}\sqrt[5]{1024}\end{align*}
Write each expression as a rational exponent.
1. \begin{align*}\sqrt[3]{14}\end{align*}
2. \begin{align*}\sqrt[4]{zw}\end{align*}
3. \begin{align*}\sqrt{a}\end{align*}
4. \begin{align*}\sqrt[9]{y^3}\end{align*}
Write the following expressions in simplest radical form.
1. \begin{align*}\sqrt{24}\end{align*}
2. \begin{align*}\sqrt{300}\end{align*}
3. \begin{align*}\sqrt[5]{96}\end{align*}
4. \begin{align*}\sqrt{\frac{240}{567}}\end{align*}
5. \begin{align*}\sqrt[3]{500}\end{align*}
6. \begin{align*}\sqrt[6]{64x^8}\end{align*}
7. \begin{align*}\sqrt[3]{48a^3 b^7}\end{align*}
8. \begin{align*}\sqrt[3]{\frac{16x^5}{135y^4}}\end{align*}
Simplify the following expressions as much as possible.
1. \begin{align*}3\sqrt{8} - 6 \sqrt{32}\end{align*}
2. \begin{align*}\sqrt{180} + \sqrt{405}\end{align*}
3. \begin{align*}\sqrt{6} - \sqrt{27} + 2 \sqrt{54} + 3 \sqrt{48}\end{align*}
4. \begin{align*}\sqrt{8x^3} - 4x \sqrt{98x}\end{align*}
5. \begin{align*}\sqrt{48a} + \sqrt{27a}\end{align*}
6. \begin{align*}\sqrt[3]{4x^3} + x \cdot \sqrt[3]{256}\end{align*}
Multiply the following expressions.
1. \begin{align*}\sqrt{6}\left (\sqrt{10} + \sqrt{8}\right )\end{align*}
2. \begin{align*}\left (\sqrt{a} - \sqrt{b}\right )\left (\sqrt{a} + \sqrt{b}\right )\end{align*}
3. \begin{align*}\left (2 \sqrt{x} + 5\right )\left (2 \sqrt{x} + 5\right )\end{align*}
Rationalize the denominator.
1. \begin{align*}\frac{7}{\sqrt{5}}\end{align*}
2. \begin{align*}\frac{9}{\sqrt{10}}\end{align*}
3. \begin{align*}\frac{2x}{\sqrt{5x}}\end{align*}
4. \begin{align*}\frac{\sqrt{5}}{\sqrt{3y}}\end{align*}
5. \begin{align*}\frac{12}{2 - \sqrt{5}}\end{align*}
6. \begin{align*}\frac{6 + \sqrt{3}}{4 - \sqrt{3}}\end{align*}
7. \begin{align*}\frac{\sqrt{x}}{\sqrt{2} + \sqrt{x}}\end{align*}
8. \begin{align*}\frac{5y}{2 \sqrt{y} - 5}\end{align*}
9. The volume of a spherical balloon is \begin{align*}950 \ cm^3\end{align*}. Find the radius of the balloon. (Volume of a sphere \begin{align*}= \frac{4}{3} \pi R^3\end{align*}).
10. A rectangular picture is 9 inches wide and 12 inches long. The picture has a frame of uniform width. If the combined area of picture and frame is \begin{align*}180 \ in^2\end{align*}, what is the width of the frame?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Show Hide Details
Description
Tags:
Subjects:
|
# A Distinctly Odd Coincidence
This week’s topic comes from partition theory. Consider all the ways to write 7 as the sum of positive integers:
\begin{align*} &7, && 3+2+2,\\ &6+1, && 3+2+1+1,\\ &5+2, && 3+1+1+1+1,\\ &5+1+1, && 2+2+2+1,\\ &4+3, && 2+2+1+1+1,\\ &4+2+1, && 2+1+1+1+1+1,\\ &4+1+1+1, && 1+1+1+1+1+1+1+1.\\ &3+3+1, && \end{align*} (We don’t care about order: for example, $$5+2$$ and $$2+5$$ are the same.) These are called the partitions of 7, and there are 15 of them.
We can also consider imposing constraints on these partitions: For example, how many of these partitions of 7 use only odd integers? These “odd” partitions are
\begin{align*} & 7, && 3+1+1+1+1, \\ & 5+1+1, && 1+1+1+1+1+1+1, \\ & 3+3+1, && \end{align*} and there are 5 of them. Here’s a very different question: how many partition of 7 have no repeated parts? These “distinct” partition are
\begin{align*} & 7, && 6+1, && 5+2, && 4+3, && 4+2+1, \end{align*} and there are 5 of these as well.
This must be coincidence, right? Let’s try a bigger example. How many partitions of 11 use only odd parts? You can (and should!) check that there are 12. And how many partitions of 11 use only distinct parts? Again, there are 12. Surprisingly, these numbers will always match!:
Theorem (Euler): For any positive integer $$n$$, the number of “odd” partitions of $$n$$ is the same as the number of “distinct” partitions of $$n$$.
### Transforming Between “Odd” and “Distinct” Partitions
Let’s try to make sense of this. Given a partition of $$n$$ into odd parts, there is a simple way to transform it into a partition with distinct parts: whenever two parts are the same, add them together, and repeat until they are all distinct. For example, the partition $$7+3+3+3+3+3+3+1+1+1$$ of 28 is transformed into
\begin{align*} & 7+(3+3)+(3+3)+(3+3)+(1+1)+1 \\ & \quad \to 7+(6+6)+6+2+1 \\ & \quad \to 12+7+6+2+1, \end{align*} which has distinct parts. In the other direction, to transform a “distinct” partition to an “odd” one, we can reverse this process: whenever there is an even part, cut it in half and write it twice. Our example $$12+7+6+2+1$$ from above becomes
\begin{align*} & (6+6)+7+(3+3)+(1+1)+1 \\ &\quad \to (3+3)+(3+3)+7+3+3+1+1+1, \end{align*} which is the “odd” permutation we started with. These transformations can be shown to match “odd” permutations with “distinct” permutations in pairs, proving that these sets have the same size. (A careful argument uses binary representation.) For example, the 5 partitions of 7 above are paired as follows:
\begin{align*} & 7 &&\leftrightarrow && 7 \\ & 6+1 &&\leftrightarrow && 3+3+1 \\ & 5+2 &&\leftrightarrow && 5+1+1 \\ & 4+3 &&\leftrightarrow && 3+1+1+1+1 \\ & 4+2+1 &&\leftrightarrow && 1+1+1+1+1+1+1. \end{align*} (If you know something about generating functions, then a different proof follows by proving the identity
$$\frac{1}{(1-x)(1-x^3)(1-x^5)(1-x^7)\cdots} = (1+x)(1+x^2)(1+x^3)(1+x^4)\cdots$$ and noting that the coefficients of $$x^n$$ in the expansions of the left and right products equals the number of “odd” and “distinct” permutations of $$n$$, respectively.)
### More Coincidences
These apparent coincidences are characteristic of many results in partition theory. Try your hand at proving the following generalization:
Theorem: The number of partitions of $$n$$ into parts that are not divisible by 10 equals the number of partitions of $$n$$ where no part is repeated 10 or more times. The same is true for any positive integer in place of 10.
(Hint: both methods used above can be extended to prove this.) If instead we want the parts to be really distinct and make sure no two parts are equal or consecutive, then we want my favorite partition theory result:
Theorem (Rogers-Ramanujan Identity): The number of partitions of $$n$$ where any two parts differ by at least 2 is the same as the number of partitions of $$n$$ where every part is one more or one less than a multiple of 5.
Despite the simplicity of the statement, this is much harder to prove than the previous theorems.
There’s something unsatisfyingly arbitrary about the number 10. When we write numbers, e.g. 1729, our notation means that we have 1 thousand ($$10^3$$), 7 hundreds ($$10^2$$), 2 tens ($$10^1$$), and 9 ones ($$10^0$$), so the number 10 is ingrained in our very writing system. And it’s not just the Arabic numeral system: Roman numerals also organize themselves around powers of 10 (such as I=1, X=10, C=100, etc.). But why 10? (Other than the fact that we usually have 10 fingers and 10 toes [why not 20 instead?].)
Our “decimal” system centers around powers of 10, but there are other equally useful systems based on other numbers. For example, we could use sums of powers of 2, and instead of needing 10 digits (0,1,…,9) we can use just 2, namely 0 and 1. For example, the number 1101 in base 2 means $$1 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0$$, also known as 13 in base 10. This may seem inefficient, since we need more digits (or, more correctly, “bits” in the binary setting) to express the same value, but it has many benefits. Each bit has only 2 choices, and can therefore be represented by a pair of opposites such as “on/off” (e.g. in circuits) or the two magnetic polarities. The latter is how hard drive disks store information (in binary!), which is one reason this system is beloved by computer scientists. It also allows for jokes like this: “There are 10 types of people in the world: those that understand binary, and those that don’t.”
Another common base is Hexadecimal, using powers of 16 with digits 0,…,9,A,B,C,D,E,F. (A hexadecimal digit is sometimes called a “nibble”.) It is common to prefix a hexadecimal number with “0x”, so 0x3B7 means $$3 \cdot 16^2 + 11 \cdot 16 + 7 = 951$$ in decimal. It also makes the phrase “I see 0xDEAD people” perfectly reasonable (how many is that?).
The possibilities are endless! We can use base $$n$$ for any positive integer $$n>1$$ in similar ways, but we can also use negative $$n$$ as well! (My favorite is base -4.) If you want to go ever further, think about what base $$(1+\sqrt{5})/2$$ (a.k.a. phinary) would look like! Or the related bases Fibonacci and NegaFibonacci!
|
New Zealand
Level 7 - NCEA Level 2
# Families of Anti-Derivatives and Specific Anti-Derivatives
Lesson
As we have just seen, there are infinitely many curves that result in the same derivative.
We call this collection of curves the family of antiderivative functions for a given function.
We need to be able to start identifying from a graph and from equations, those that will belong to the same family.
Let's look at some examples.
#### Examples
##### Example 1
For a function $f(x)=5x^4$f(x)=5x4, which of the following belong to the family of antiderivative functions.
a) $F(x)=x^5+1$F(x)=x5+1
b) $F(x)=-x^5+1$F(x)=x5+1
c) $F(x)=x^3-4$F(x)=x34
d) $F(x)=5x^5$F(x)=5x5
e) $F(x)=x^5-\frac{1}{2}$F(x)=x512
f) $F(x)=6\pi+x^5$F(x)=6π+x5
Firstly we notice that the degree of $f(x)$f(x) is $4$4. This means that the degree of the antiderivative must be one more - ie $5$5. This means we can discount option $(c)$(c), as it is of degree $3$3
The antiderivative of $f(x)=5x^4$f(x)=5x4 is $F(x)=x^5+C$F(x)=x5+C. We can check this by inspecting $F(x)$F(x) and establishing what $F'(x)$F(x) is. This means we need to have $F(x)$F(x) functions that contain only $x^5$x5 terms with a coefficient of $1$1.
Option $(d)$(d) for example, $F(x)=5x^5$F(x)=5x5 would yield a derivative of $f(x)=25x^4$f(x)=25x4. Which is clearly quite different to $f(x)=5x^4$f(x)=5x4. Also, we know from our work with power functions that the function $F(x)=5x^5$F(x)=5x5 is a LOT steeper than $F(x)=x^5$F(x)=x5, hence the gradient function of these are not the same. I've plotted both graphs on this plot to remind you.
So we can eliminate $(d)$(d) and $(b)$(b) as these do not have coefficients of $x^5$x5 of $1$1. There coefficients are $5$5 and $-1$1 respectively.
This leaves us with
a) $F(x)=x^5+1$F(x)=x5+1
e) $F(x)=x^5-\frac{1}{2}$F(x)=x512
f) $F(x)=6\pi+x^5$F(x)=6π+x5
The only difference with these is the value of the constant term at the end. Let's have a look at the graph of these three.
Do you remember the effect the constant term has on the graph? That's right, it's a vertical translation. It doesn't effect the gradient at $x=a$x=a at all.
This interactive demonstrates what I mean by this. Here, you can vertically translate the curve y=x^5. You can also see the gradient of the tangent at point A, remember that this is the value of the derivative at this point. As you move the curve, the gradient remains unchanged.
All of the curves generated by that vertical translation belong to the family of antiderivatives of the function $f(x)=5x^4$f(x)=5x4.
Thus $(a),(e)$(a),(e) and $(f)$(f) are antiderivatives.
a) $F(x)=x^5+1$F(x)=x5+1 YES
b) $F(x)=-x^5+1$F(x)=x5+1 NO This function has a reflection across the x-axis (indicated by the negative), it's coefficient of $x^5$x5 is $-1$1, not $1$1.
c) $F(x)=x^3-4$F(x)=x34 NO This is only degree $3$3, the antiderivative must be of degree $5$5
d) $F(x)=5x^5$F(x)=5x5 NO This function has a coefficient of $x^5$x5 of $5$5, not $1$1.
e) $F(x)=x^5-\frac{1}{2}$F(x)=x512 YES
f) $F(x)=6\pi+x^5$F(x)=6π+x5 YES
##### Example 2
This applet, shows a family of antiderivatives. It shows $y=x^2$y=x2, $y=x^2+1$y=x2+1, $y=x^2-1$y=x21 and $y=x^2-3$y=x23.
You can move the point $A$A. As you do you can see how all the tangent lines stay in parallel. This demonstrates that the value of the derivative is the same for the whole family of antiderivatives of $\frac{dy}{dx}=2x$dydx=2x
#### Worked Examples
##### Question 1
Consider the gradient function $f'\left(x\right)$f(x)$=$=$2$2.
1. The family of the antiderivative, $f\left(x\right)$f(x), will be:
Exponential
A
Cubic
B
Linear
C
D
Exponential
A
Cubic
B
Linear
C
D
2. The form of the antiderivative will be $f\left(x\right)$f(x)$=$=$mx+c$mx+c. State the value of $m$m.
3. Which of the following functions represent possible values for an antiderivative $f\left(x\right)$f(x)?
Select all that apply.
$f\left(x\right)=-2x$f(x)=2x
A
$f\left(x\right)=2x+6$f(x)=2x+6
B
$f\left(x\right)=2x-3$f(x)=2x3
C
$f\left(x\right)=-2x+6$f(x)=2x+6
D
$f\left(x\right)=-2x$f(x)=2x
A
$f\left(x\right)=2x+6$f(x)=2x+6
B
$f\left(x\right)=2x-3$f(x)=2x3
C
$f\left(x\right)=-2x+6$f(x)=2x+6
D
##### Question 2
Consider the gradient function $f'\left(x\right)$f(x)$=$=$4x+3$4x+3.
1. The family of the antiderivative, $f\left(x\right)$f(x), will be:
Linear
A
B
Cubic
C
Exponential
D
Linear
A
B
Cubic
C
Exponential
D
2. The form of the antiderivative will be $f\left(x\right)$f(x)$=$=$ax^2+bx+C$ax2+bx+C. State the value of $a$a and $b$b in simplified form if necessary.
$a$a $=$= $\editable{}$
$b$b $=$= $\editable{}$
3. Which of the following functions could possibly represent $f\left(x\right)$f(x)? Select all that apply.
$f\left(x\right)=4\left(x+3\right)^2+7$f(x)=4(x+3)2+7
A
$f\left(x\right)=2x^2+3x$f(x)=2x2+3x
B
$f\left(x\right)=2x^2+3x+4$f(x)=2x2+3x+4
C
$f\left(x\right)=2x^2+3$f(x)=2x2+3
D
$f\left(x\right)=4\left(x+3\right)^2+7$f(x)=4(x+3)2+7
A
$f\left(x\right)=2x^2+3x$f(x)=2x2+3x
B
$f\left(x\right)=2x^2+3x+4$f(x)=2x2+3x+4
C
$f\left(x\right)=2x^2+3$f(x)=2x2+3
D
##### Question 3
Consider the gradient function $f'\left(x\right)$f(x)$=$=$-7x^2$7x2.
1. The family of the antiderivative, $f\left(x\right)$f(x), will be:
Exponential
A
Cubic
B
C
Linear
D
Exponential
A
Cubic
B
C
Linear
D
2. The form of the antiderivative will be $f\left(x\right)$f(x)$=$=$ax^3+C$ax3+C. State the value of $a$a.
3. Which of the following functions could possibly represent $f\left(x\right)$f(x)?
A
B
C
D
A
B
C
D
### Outcomes
#### M7-10
Apply differentiation and anti-differentiation techniques to polynomials
#### 91262
Apply calculus methods in solving problems
|
Course Content
Class 8th Science
0/36
Class 8th Math
0/37
Class 8 Social Science History
0/24
Class 8 Social Science Geography
0/12
Class 8 Social Science Civics: Social and Political Life – III
0/21
Class 8th Englisn Honeydew Summary
0/20
Class 8 English Honeydew Poem
0/16
Class 8 English It So Happened
0/20
Online Class For 8th Standard Students (CBSE) (English Medium)
## Exercise 3.2 Page: 44
1. Find x in the following figures.
Solution:
a)
125° + m = 180° ⇒ m = 180° – 125° = 55° (Linear pair)
125° + n = 180° ⇒ n = 180° – 125° = 55° (Linear pair)
x = m + n (exterior angle of a triangle is equal to the sum of 2 opposite interior 2 angles)
⇒ x = 55° + 55° = 110°
b)
Two interior angles are right angles = 90°
70° + m = 180° ⇒ m = 180° – 70° = 110° (Linear pair)
60° + n = 180° ⇒ n = 180° – 60° = 120° (Linear pair) The figure is having five sides and is a pentagon.
Thus, sum of the angles of pentagon = 540° 90° + 90° + 110° + 120° + y = 540°
⇒ 410° + y = 540° ⇒ y = 540° – 410° = 130°
x + y = 180° (Linear pair)
⇒ x + 130° = 180°
⇒ x = 180° – 130° = 50°
2. Find the measure of each exterior angle of a regular polygon of
(i) 9 sides (ii) 15 sides Solution:
Sum of angles a regular polygon having side n = (n-2)×180°
(i) Sum of angles a regular polygon having side 9 = (9-2)×180°= 7×180° = 1260°
Each interior angle=1260/9 = 140°
Each exterior angle = 180° – 140° = 40°
Or,
Each exterior angle = sum of exterior angles/Number of angles = 360/9 = 40°
(ii) Sum of angles a regular polygon having side 15 = (15-2)×180°
= 13×180° = 2340°
Each interior angle = 2340/15 = 156°
Each exterior angle = 180° – 156° = 24°
Or,
Each exterior angle = sum of exterior angles/Number of angles = 360/15 = 24°
3. How many sides does a regular polygon have if the measure of an exterior angle is 24°? Solution:
Each exterior angle = sum of exterior angles/Number of angles
24°= 360/ Number of sides
⇒ Number of sides = 360/24 = 15
Thus, the regular polygon has 15 sides.
## 4. How many sides does a regular polygon have if each of its interior angles is 165°? Solution:
Interior angle = 165°
Exterior angle = 180° – 165° = 15°
Number of sides = sum of exterior angles/ exterior angles
⇒ Number of sides = 360/15 = 24
Thus, the regular polygon has 24 sides.
5.
a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?
b) Can it be an interior angle of a regular polygon? Why?
Solution:
a) Exterior angle = 22°
Number of sides = sum of exterior angles/ exterior angle
⇒ Number of sides = 360/22 = 16.36
No, we can’t have a regular polygon with each exterior angle as 22° as it is not divisor of 360.
b) Interior angle = 22°
Exterior angle = 180° – 22°= 158°
No, we can’t have a regular polygon with each exterior angle as 158° as it is not divisor of 360.
## 6.
a) What is the minimum interior angle possible for a regular polygon? Why?
b) What is the maximum exterior angle possible for a regular polygon?
Solution:
a) Equilateral triangle is regular polygon with 3 sides has the least possible minimum interior angle because the regular with minimum sides can be constructed with 3 sides at least.. Since, sum of interior angles of a triangle = 180°
Each interior angle = 180/3 = 60°
b) Equilateral triangle is regular polygon with 3 sides has the maximum exterior angle because the regular polygon with least number of sides have the maximum exterior angle possible. Maximum exterior possible = 180 – 60° = 120°
|
The Product and Quotient Rules
Lesson Features »
Lesson Priority: High
Calculus $\longrightarrow$
Discovering Derivatives $\longrightarrow$
Objectives
• Learn the product rule for differentiating a product of two pieces when you know the derivative of each piece
• Learn the quotient rule for differentiating the quotient of two terms when you know the derivative of each term
• Know expert tips and tricks to avoid very common pitfalls for using these techniques mistake-free
• Because the quotient rule is more involved, learn how we can often turn a quotient problem into a multiplication problem, and then proceed with the product rule
Lesson Description
Unlike limit laws, it is not true that the derivative of a product is the product of the derivatives. The derivative of a product has a pattern based approach, though, and we will learn and practice it here, before immediately moving on to how to take the derivative of a quotient.
Practice Problems
Practice problems and worksheet coming soon!
Unintuitive Derivative Rules
Derivatives are fairly well-behaved in a lot of convenient ways. By now we have studied several derivative laws and rules - some of which were so intuitive that it seemed overkill to state them formally. For example, we learned that the derivative of the sum of terms was equal to the sum of each derivative:$$\frac{d}{dx} \,\, f(x) + g(x) = \frac{d}{dx} \,\, f(x) + \frac{d}{dx} \,\, g(x)$$Hopefully by now this behavior is automatic to you, having used it countless times in practice.Now we turn to slightly more complicated ideas. First, what happens when we need the derivative of a product of functions? Unfortunately, the derivative of a product is NOT equal to the product of each derivative. The reason for this is perhaps beyond your curiosity - what I mean is, most students aren't overly interested in why this is the case. Regardless, all students need to know what to do when asked to find a derivative that involves a product.
Definition:Let $f(x)$ and $g(x)$ be differentiable functions of $x$. It follows that$$\frac{d}{dx} \,\, \bigg( f(x) g(x) \bigg) = \left[\frac{d}{dx} \,\, f(x)\right] \cdot g(x) + f(x) \cdot \left[\frac{d}{dx} \,\, g(x)\right]$$Or, if you prefer prime notation:Let $h(x) = f(x) \cdot g(x)$, where $f$ and $g$ are differentiable functions of $x$. It follows that$$h'(x) = f'(x) \cdot g(x) + f(x)\cdot g'(x)$$
In words, we can say that the derivative of a product of two things is Thing 1 times Thing 2's derivative plus Thing 2 times Thing 1's derivative (for you Dr. Seuss fans out there).
Using the Product Rule Correctly
On the subject of derivatives of products, you essentially have two jobs: 1) recognize that the problem involves a product and 2) apply the product rule correctly.
You Should Know
When you know the derivative of each expression on its own, but you're asked to find the derivative of the product of two expressions, you need to use the Product Rule.
It's very important that you don't take the first job for granted (recognizing the need to use the Product Rule in the first place). It will become less of a concern with practice and exposure, but one of the most common mistakes students make at first is failing to even recognize the fact that the question requires the product rule. Let's examine several situations to better understand when we do and do not need the Product Rule.Examples 1-7For each of the following derivatives, determine whether or not you need to use the Product Rule.
Example 1$$\frac{d}{dx} \,\, xe^x$$$\blacktriangleright$ This example does require the Product Rule, because you know the derivative of $x$, and you know the derivative of $e^x$, and you're being asked to find the derivative of the product of those two expressions.
Example 2$$\frac{d}{dx} \,\, \sin(2x)$$$\blacktriangleright$ This example does not require the use of the Product Rule (though it requires the Chain Rule », which is covered in a near future lesson). The expression is simply not the product of two objects.
Example 3$$\frac{d}{dx} \,\, \big[ x^2 \left(x^2 + 1\right) \big]$$$\blacktriangleright$ Here, we actually could use the Product Rule, but it is much faster to simply multiply it out first, leaving us with a plain polynomial.$$\frac{d}{dx} \,\, \big[ x^2 \left(x^2 + 1\right) \big]$$$$\longrightarrow \frac{d}{dx} \,\, \big[ x^4 + x^2 \big]$$The important takeaway from this example is we not only want to know when we need to use the Product Rule, but also when we can avoid it for the sake of simplicity.
Example 4$$\frac{d}{dx} \,\, x^x$$$\blacktriangleright$ While this one does require a special technique (see logarithmic differentiation »), there is no product here, and the Product Rule therefore is not in play.
Example 5$$\frac{d}{dx} \,\, e^2 \tan(x)$$$\blacktriangleright$ The $e^2$ term is a constant, so there is no need to use the Product Rule. The Constant Rule for derivatives will handle this situation.
Example 6$$\frac{d}{dx} \,\, 2^x \log_2 (x)$$$\blacktriangleright$ There is no shortcut or clever trick to combining this expression into one object. We would take the derivative of this product using the Product Rule.
Example 7$$\frac{d}{dx} \,\, \sin(x) \ln(x)$$$\blacktriangleright$ Here, you need the Product Rule.
Pro Tip
When working with the product rule, it is so very important to organize scratch work. The intermediate to advanced questions will require side work, in order to find some of the derivatives you need to plug into the formula. For this reason, you should $\rightarrow$ always $\leftarrow$ set up the formula first before executing it. Put each expression in its place before you actually take any derivatives. It will seem like overkill at first but it will soon be an invaluable habit!
Now let's put the rule into practice! We'll revisit each of the preceding examples above and find the derivative of the ones that required the Product Rule. As mentioned in the Pro Tip above, remember to set up the formula first in your scratch work before taking any derivatives - it's a very important habit!!!Example 1$$\frac{d}{dx} \,\, xe^x$$$\blacktriangleright$ Set things up in line with the formula:$$\frac{d}{dx} \,\, xe^x$$$$= \left( \frac{d}{dx} \,\, x \right)\left(e^x\right) + (x) \left( \frac{d}{dx} \,\, e^x \right)$$Now that we've put everything in the right place, let's take derivatives and finish this problem off.$$= (1) \left(e^x\right) + (x)\left(e^x\right)$$$$= e^x + xe^x$$Example 3$$\frac{d}{dx} \,\, x^2 \left(x^2 + 1\right)$$$\blacktriangleright$ Again, set things up first:$$\frac{d}{dx} \,\, x^2 \left(x^2 + 1\right)$$$$=\left( \frac{d}{dx} \,\, x^2 \right) \left( x^2 + 1\right) + \left(x^2\right) \left( \frac{d}{dx} \,\, x^2 + 1 \right)$$$$=2x \left( x^2 + 1\right) + x^2 (2x)$$$$=2x^3 + 2x + 2x^3 = 4x^3 + 2x$$Recall also that, as pointed out earlier with this example, the best approach would have been to multiply out the expression before taking the derivative, thus avoiding the need for the Product Rule. However we can see by doing it out longform that we got the answer we would have expected.Example 6$$\frac{d}{dx} \,\, 2^x \log_2 (x)$$$$\blacktriangleright \,\,\, =\left(\frac{d}{dx} \,\, 2^x \right) \big(\log_2(x)\big) + \left( 2^x \right) \left(\frac{d}{dx} \,\, \log_2(x) \right)$$$$=\big(2^x \ln(2)\big) \log_2(x) + 2^x \left(\frac{1}{x\ln(2)}\right)$$$$=2^x \ln(2) \log_2 (x) + \frac{2^x}{x\ln(2)}$$Example 7$$\frac{d}{dx} \,\, \sin(x) \ln(x)$$$$\blacktriangleright \,\,\, = \left( \frac{d}{dx} \,\, \sin(x) \right)\big( \ln(x) \big) + \big( \sin(x) \big)\left( \frac{d}{dx} \,\, \ln(x) \right)$$$$=\big( \cos(x) \big) \big( \ln(x) \big) + \big( \sin(x) \big) \left( \frac{1}{x} \right)$$$$=\cos(x) \ln(x) + \frac{\sin(x)}{x}$$
Annoying Pro Tip
I just can't stress enough how important it is to take that intermediary step in your pencil and paper calculations and write out each piece and where it belongs before actually taking any derivatives. It seems like overkill now but it will pay huge dividends. As you practice derivatives on some of the more basic expressions like $xe^x$, you'll be sorely tempted to do it all at once - don't give in! It's just a few extra pencil strokes!
The Quotient Rule
We will now apply the same idea to the derivative of quotients. Similar to products, we must first recognize the situation, and then apply a formulaic approach.
Define: The Quotient Rule for DerivativesLet $f(x)$ and $g(x)$ be differentiable functions of $x$. It follows that$$\frac{d}{dx} \,\, \frac{f(x)}{g(x)} = \frac{g(x) \left(\frac{d}{dx} \,\, f(x)\right) - f(x) \left( \frac{d}{dx} \,\, g(x) \right)}{(\big(g(x)\big)^2}$$Or, if you prefer prime notation:Let $h(x) = \frac{f(x)}{g(x)}$, where $f$ and $g$ are differentiable functions of $x$. It follows that$$h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{\big(g(x)\big)^2}$$
As you can see, the derivative of a product is a little more forgiving, since order didn't matter (for products we just take the derivative of each piece multiplied with the original other piece). With the Quotient Rule, each expression must be in exactly the right place, and order matters because of the subtraction.The biggest mistake students make on exams is forgetting the order of the terms. Here's a memory trick that I used when I first learned this stuff - the following sentence dictates the Quotient Rule for you and helps you keep the order straight:
Define: The Quotient Rule for Derivatives in Words$$\frac{d}{dx} \,\, \frac{f(x)}{g(x)} = \frac{g(x) \left(\frac{d}{dx} \,\, f(x)\right) - f(x) \left( \frac{d}{dx} \,\, g(x) \right)}{(\big(g(x)\big)^2}$$$\longrightarrow$ Low d-high minus high d-low, all over the square of what's below! $\longleftarrow$
"Low" and "below" refer to the denominator expression, and "high" refers to the numerator expression. This is a great exam memory trick, because it's quick and foolproof - if you say it wrong, then the sentence won't rhyme!
Example 8$$\frac{d}{dx} \,\, \frac{x^6}{e^{2x}}$$$\blacktriangleright$ We know the Quotient Rule applies, since we know the derivative of each the numerator and the denominator on their own. Make sure you set up the formula by putting the pieces in their place before we take any derivatives, just like we did for the Product Rule.$$\frac{d}{dx} \,\, \frac{x^6}{e^{2x}}$$$$=\frac{\left(e^{2x} \right)\left(\frac{d}{dx} \,\, x^6 \right) - \left(x^6 \right)\left( \frac{d}{dx} \,\, e^{2x} \right)}{\left(e^{2x}\right)^2}$$$$=\frac{e^{2x}\left(6x^5\right) - \left(x^6\right) \left(2e^{2x} \right)}{e^{4x}}$$Factor out the common $e^{2x}$ in the numerator - this is a very good idea when working on derivatives involving exponentials. Canceling, we are finally left with$$=\frac{6x^5 - 2x^6}{e^{2x}}$$
Avoiding the Quotient Rule
Since the Quotient Rule is slightly more complicated than the Product Rule, and since we are human, it is preferrable to work with the simpler Product Rule formula when possible. Many quotients can be quickly rewritten as products, through the use of negative exponents. Leveraging that fact, we can turn "Quotient Rule problems" into "Product Rule problems" more often than not.
Example 8 (Revisited)$$\frac{d}{dx} \,\, {x^6}{e^{2x}}$$$\blacktriangleright$ Instead of using the Quotient rule, we can rewrite the expression as a product, which will allow us to use the Product Rule.$$\frac{d}{dx} \,\, \left(x^6\right)\left(e^{-2x}\right)$$$$= x^6 \left(\frac{d}{dx} \,\, e^{-2x} \right) + \left( \frac{d}{dx} \,\, x^6 \right) \left(e^{-2x}\right)$$$$= -2x^6e^{-2x} + 6x^5 e^{-2x}$$$$=\frac{-2x^6+6x^5}{e^{2x}}$$
Pro Tip
This trick isn't always going to save you a lot of time or heartache, because taking derivatives of expressions with negative exponents adds more complexity and negative signs into the mix. At some point, a complicated expression is a complicated expression, no matter how you choose to deal with it. This trick tends to work very well with polynomial and exponential expressions. If you have a trig function or a logarithm in the denominator of the thing you're trying to take the derivative of, you're probably better off just using the Quotient Rule verbatim.
Warning!
If you are working on a quiz or test with explicit instructions to use the Quotient Rule, then you should not turn the problem into a Product Rule problem. In that case, your instructor would have legitimate reason to dock points off, if they want.
Put It To The Test
We will be continually practicing both the Product Rule and the Quotient Rule throughout the course of the Calculus journey. Specifically, you will need to practice this a lot in conjunction with other rules - notably the Chain Rule » in the near future. In fact, you may find that your homework on this topic is significantly more difficult than the problems that follow here, if you are working with the Chain Rule simultaneously. You should absolutely work practice problems from the lesson on the Chain Rule if you want to be sure you've mastered anything that can be thrown at you that requires the Product Rule or Quotient Rule.With that in mind, let's dive in.
Example 9Find $\frac{d}{dx} \,\, x^3 e^{-5x}$.
Show solution
$\blacktriangleright$ This problem calls for a straight-forward application of the Product Rule.$$\frac{d}{dx} \,\, x^3 e^{-5x}$$$$= \left( \frac{d}{dx} \,\, x^3 \right)\left( e^{-5x} \right) + \left( x^3 \right)\left( \frac{d}{dx} \,\, e^{-5x} \right)$$$$=\left(3x^2\right)\left(e^{-5x}\right) + \left( x^3 \right) \left( -5e^{-5x} \right)$$$$=e^{-5x} \left[ 3x^2 - 5x^3 \right]$$
Example 10$$\frac{d}{dx} \,\, \sin^2(x)$$
Show solution
$\blacktriangleright$ We will soon learn a direct way to handle this derivative using the Chain Rule », but we can still find the answer with the tools we've learned thus far - namely the Product Rule. Simply write the problem as a product and apply the rule.$$\frac{d}{dx} \,\, \sin(x) \sin(x)$$$$=\left( \frac{d}{dx} \,\, \sin(x) \right) \big( \sin(x) \big) + \left( \frac{d}{dx} \,\, \sin(x) \right) \big( \sin(x) \big)$$$$=\big( \cos(x) \big) \big( \sin(x) \big) + \big( \cos(x) \big) \big( \sin(x) \big)$$$$=2\sin(x) \cos(x)$$
Example 11$$\frac{d}{dx} \,\, \frac{x^2}{2^x}$$
Show solution
$\blacktriangleright$ We know each of these derivatives on their own, so the right way to proceed here is to apply the Quotient Rule.$$\frac{d}{dx} \,\, \frac{x^2}{2^x}$$$$=\frac{\left( 2^x \right)\left( \frac{d}{dx} \,\, x^2 \right) - \left( x^2 \right) \left( \frac{d}{dx} \,\, 2^x \right)}{\left(2^x\right)^2}$$$$=\frac{\left( 2^x \right)\left( 2x \right) - \left( x^2 \right) \left( 2^x \ln(2) \right)}{2^{2x}}$$$$=\cancel{2^x} \left[ \frac{2x - x^2 \ln(2)}{2^{\cancel{2}x}} \right]$$$$=\frac{2x - x^2 \ln(2)}{2^x}$$
Example 12$$\frac{d^2}{dx^2} \,\, \sec(x)$$
Show solution
$\blacktriangleright$ To find the second derivative, we will start by calculating the first derivative.$$\frac{d}{dx} \,\, \sec(x) = \sec(x) \tan(x)$$Take the derivative of this result to find the second derivative. We'll need to employ the product rule.$$\frac{d}{dx} \,\, \sec(x) \tan(x)$$$$=\left( \frac{d}{dx} \,\, \sec(x) \right) \left( \tan(x) \right) + \left( \frac{d}{dx} \,\, \tan(x) \right) \left( \sec(x) \right)$$$$=\left( \sec(x) \tan(x) \right) \left( \tan(x) \right) + \left( \sec^2(x) \right) \left( \sec(x) \right)$$$$= \sec(x) \tan^2 (x) + \sec^3(x)$$
Lesson Takeaways
• Recognize derivatives that require either the Product Rule or Quotient Rule
• For these problems, plug in terms into the appropriate formula first before taking derivatives, to show clear scratch work to both you and your teacher
• Ultimately memorize both the Product Rule and the Quotient Rule
• Use the Product Rule iteratively if needed, e.g. when asked for higher derivatives, and again keep track of your steps with well-organized scratch work
• Practice using both rules for fluency, and practice turning in fully simplified answers at the peril of losing points on tests
• Popular Content
• Get Taught
• Other Stuff
Lesson Metrics
At the top of the lesson, you can see the lesson title, as well as the DNA of Math path to see how it fits into the big picture. You can also see the description of what will be covered in the lesson, the specific objectives that the lesson will cover, and links to the section's practice problems (if available).
Key Lesson Sections
Headlines - Every lesson is subdivided into mini sections that help you go from "no clue" to "pro, dude". If you keep getting lost skimming the lesson, start from scratch, read through, and you'll be set straight super fast.
Examples - there is no better way to learn than by doing. Some examples are instructional, while others are elective (such examples have their solutions hidden).
Perils and Pitfalls - common mistakes to avoid.
Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!
Put It To The Test - Shows you examples of the most common ways in which the concept is tested. Just knowing the concept is a great first step, but understanding the variation in how a concept can be tested will help you maximize your grades!
Lesson Takeaways - A laundry list of things you should be able to do at lesson end. Make sure you have a lock on the whole list!
Special Notes
Definitions and Theorems: These are important rules that govern how a particular topic works. Some of the more important ones will be used again in future lessons, implicitly or explicitly.
Pro-Tip: Knowing these will make your life easier.
Remember! - Remember notes need to be in your head at the peril of losing points on tests.
You Should Know - Somewhat elective information that may give you a broader understanding.
Warning! - Something you should be careful about.
|
Explorations outside of school can connect children with maths in new and varied ways, which aren’t always practical inside school buildings.
In a previous blog post, I shared 3 easy things that families can do at home to engage in meaningful and fun mathematical play and talk.
Here are 3 more games and activities to help children get curious about mathematics in the world around us.
## Tip 1: Play ‘Number I Spy’
This game is about spying numbers in a sequence.
1. Look for numbers whilst out and about, the game starts with a 1
2. Call out when you see a 1
3. Take it in turns or be the fastest to spot the next number in the counting sequence: 1, 2, 3 and so on
4. Continue until you reach 20, or a number that works with your child.
### How does this get them thinking?
This game is initially about spotting numbers and identifying symbols. But by playing, children will start to understand where they are most likely to see numbers.
At school, children often use numbers to say how many there are of something. Outside of school, numbers are used in many other ways, such as labeling buses and trams, or for reference such as house numbers.
This is a different use of number to ‘how many’, since spotting a 7 on a bus does not mean there are 7 buses.
### How does it encourage mathematical talk?
Number I Spy is simply about using the language of numbers. Children might also use the terms multiples, odd, even, and sequences.
### How can it go further?
There are many modifications for this game. The counting sequence, 1, 2, 3, 4… is a good starting point. Children can also work backward down from 20 to build experience with the sequence. Or, you might wish to play with the sequence of odd numbers (1, 3, 5, 7…) or a set of multiples (0, 4, 8, 12…) for children who are already comfortable with the counting sequence.
And this one is trickier, but you can be inventive with your number spotting. For example, you might find a spiral that looks like a six or turn your head to see a 3 shape on a pair of arches.
## Tip 2: Play ‘Which Way is South?’
Navigational clues are everywhere in the outside environment if you tune into them. For children, drawing attention to compass orientation can develop their spatial awareness, a hugely important mathematical skill.
1. Go for a walk, long or short, through a park or even to the shops
2. Every time you change direction, identify which way is South!
### How does this get them thinking?
This game takes some training. There are many clues in the outside environment and some will be different depending on whether you are walking in an urban or more natural environment.
• Foliage is heavier on trees on the south side (if they aren’t growing in shadow)
• In colder months with no leaves on trees, you can look for moss. This grows more in shade (or the North side) than in the sun.
This isn’t a perfect science and you might need a few clues to get a sense of direction!
### How does it encourage mathematical talk?
Reasoning is a big part of this game when trying to convince each other of which way south is. The talk will likely include language of logic and reason, sentences with a structure such as, because ____, I think _____. The compass directions of North, East, South, and West might also be involved.
### How can it go further?
There are many possibilities for pattern spotting, convincing, and refuting in this activity. On a given day, you might notice new clues from the environment:
• In spring, see if there is a pattern to where new buds and leaves first emerge on plants and trees
• Some cities have a more regular wind direction. For example, in my hometown, the wind often moves East. Over time you may also notice this
• Coastal areas mostly have cloud gathering over land, this gives clues to the direction of the sea.
## Tip 3: Identify Tree Heights
Finding the height of trees is tricky! Dropping a measuring tape from the top isn’t the easiest climb. One easy way is to use a phone or tablet…
1. Take a photo of a tree, with someone standing next to the tree trunk. Let’s call this person Shruti
2. Measure Shruti’s height
3. On a screen or a printout of the tree picture, see how many Shruti’s fit alongside the tree. Either cut out multiple Shruti’s or use a small strip of paper
4. Multiply the height of Shruti by the number of times they fit in the tree!
### How does this get them thinking?
Children might want to explore other trees in the area and develop their estimating skills. Building a sense of comparison for a known height is a starting point for making reasonable guesses. It might be useful to consider a metre for more straightforward tree height guessing whilst on the move.
### How does it encourage mathematical talk?
This investigation involves processes of measuring, comparing, and multiplying. It also requires the language of height, metres, and length.
### How can it go further?
Some children might be able to understand that the old artist tool of using a pencil for scale can also be applied here. With one eye closed, hold out a pencil until it is the height of the tree. Rotate it 90 degrees, level with the ground. If someone stands at the end of the viewpoint, then their distance from the base of the tree should be about the same as the height.
The photo comparison method can also be used for estimating other lengths and heights. For example, scaling up to estimate the height of tall buildings is a more substantial challenge.
Be bold!
Modify, adapt, combine and go in different directions with your mathematical explorations. When trying an idea from the above, you might find that children are initially more interested in how the moss feels than in the direction that part of the tree is facing and that’s okay. You can always explore their curiosity one day and then invite them to join yours another time.
Find more ways to encourage curiosity, connection, and creativity in maths at www.oxfordprimary.com/mathsadventure.
“In my work, I seek to find a playful mathematics education in equal partnership with children and their ideas. I find the interests of children very interesting(!) and am always on the search for new stimuli from which meaningful mathematical thinking can emerge for children.
Currently, I lead primary mathematics at Amsterdam International Community School and teach Group 2 (age 5-6). Previously I taught and led mathematics teaching in secondary schools in the UK. As well as teaching, I worked as a mathematics consultant, offering schools partnership based professional development work; and worked with schools, teachers, and children to co-create exciting mathematics learning projects for several years on Arts Council England’s Creative Partnerships Programme”
|
# How do you find the asymptotes for (9x^2 – 36) /( x^2 - 9)?
Jul 28, 2018
$\text{vertical asymptotes at } x = \pm 3$
$\text{horizontal asymptote at } y = 9$
#### Explanation:
$\text{let } f \left(x\right) = \frac{9 {x}^{2} - 36}{{x}^{2} - 9}$
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.
$\text{solve } {x}^{2} - 9 = 0 \Rightarrow \left(x - 3\right) \left(x + 3\right) = 0$
$x = \pm 3 \text{ are the asymptotes}$
$\text{Horizontal asymptotes occur as }$
${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$
$\text{divide terms on numerator/denominator by the highest}$
$\text{power of "x" that is } {x}^{2}$
f(x)=((9x^2)/x^2-(36)/x^2)/(x^2/x^2-9/x^2)=(9-(36)/x^2)/(1-9/x^2
$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{9 - 0}{1 - 0}$
$y = 9 \text{ is the asymptote}$
graph{(9x^2-36)/(x^2-9) [-40, 40, -20, 20]}
vertical asymptotes: $x = \setminus \pm 3$
horizontal asymptote: $y = 9$
#### Explanation:
The given function:
$f \left(x\right) = \setminus \frac{9 {x}^{2} - 36}{{x}^{2} - 9}$
Vertical asymptotes:
The above function will have vertical asymptote where denominator becomes zero i.e.
$\setminus \therefore {x}^{2} - 9 = 0$
${x}^{2} = 9$
$x = \setminus \pm 3$
Horizontal asymptotes:
The above function will have horizontal where
$y = \setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{9 {x}^{2} - 36}{{x}^{2} - 9}$
$y = \setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{9 - \frac{36}{x} ^ 2}{1 - \frac{9}{x} ^ 2}$
$y = \setminus \frac{9 - 0}{1 - 0}$
$y = 9$
Hence, vertical asymptotes: $x = \setminus \pm 3$
horizontal asymptote: $y = 9$
|
PLAGIARISM FREE WRITING SERVICE
We accept
MONEY BACK GUARANTEE
100%
QUALITY
What Is the Distance Formula All About?
To answer this question, you should understand the role played by a distance in mathematics and find out more about its concept. It can be defined as a numerical description of how far apart 2 given objects are, and this formula is used to determine it. Don’t forget that there are many ways and tools that can be used to determine it. Mathematically, if you need to determine a distance between 2 particular points on a given coordinate place, it’s necessary to use the distance formula. For students, one of the basic steps that should be taken is memorizing it (d = √(x2 - x1)^2 + (y2 - y1)^2). Make sure that you know the exact coordinates of 2 points to find a distance between them, and then substitute them into the above-mentioned equation.
When it comes to physics of everyday life, it’s possible to refer a distance to any physical length or its estimation according to other important criteria. When studying mathematics and the distance formula, remember that a distance metric or a function is the generalization of the concept of a physical distance. This means that any metric is the function if it behaves based on a certain set of general rules, and you can use it as quite a concrete way to describe the meaning of being too close or too far for elements in a given space. It’s worth mentioning that s distance from and between two points is the same in most cases.
Another basic thing that you need to learn is that the distance formula is a specific type of the Pythagorean Theorem, and it’s familiar to all students who study this discipline. Imagine that you have 2 points, and your basic task is to find out how far they are apart. You can start drawing lines to form a right-angled triangle and use these points as some of its corners. It’s very simple to find the lengths of both vertical and horizontal sides of this triangle because the only thing that should be done is subtracting its y and x values. Then you need to use the Pythagorean Theorem if you’re asked to find the length of its 3rd side.
Tips on How to Use the Distance Formula to Find the Lengths of Lines
As you already know, this formula is often used to find the lengths of lines in a specific coordinate place when you can’t determine their measures by counting or if the use of any measuring tool is impractical or not allowed. Take into account that the distance formula is also used to find the lengths of coordinate lines and it comes in handy in all math assignments related with them. Besides, you can use it to find a distance between 2 points on a particular coordinate line so that it’s your big help.
How to use it to find the lengths of lines? There are certain steps involved in this process, but the best part is that they are quite simple and won’t take a lot of your time, but it’s still different from writing a literary essay. First, you need to get a specific line that you will measure and find the exact coordinates of its endpoints. Be sure to take their x-coordinates and subtract them to square the amount that you get. Another step that you need to take is getting the y-coordinates of points to subtract them before squaring this amount. It’s necessary to add both squares to find the square root of your final sum. This is how you will find the lengths of lines in units with the help of the distance formula. There are certain tips that will help you simplify the whole process. For instance, you shouldn’t confuse this formula with others used in mathematics, including slope and midpoint formulas. Keep in mind that memorizing it will make everything much easier to you and avoid mixing up the above-mentioned operations.
Useful Examples and Calculations
If you still struggle with the distance formula, they will help you solve the most common problems, and let’s start with solutions for a distance. How fast are you able to run? How far can your car go? If your bike has a constant speed, how long will it take to approach the next destination? These are the basic math problems that you need to solve as a good student, so be sure to use specific formulas. They are simple enough so that you can do them right in your head. Think about the distance formula to find out how you can use it. What if you know the exact speed of something and how long it keeps moving, but not how far it will go? This is when this formula will come in handy. What about the right solution for a rate? One of the greatest benefits is that this math formula is quite versatile so that you can rewrite it to solve something else, including a rate. Another excellent thing is that it’s also a great solution for time.
Basic Distance Formula Activities
As a student, you need to understand that this formula and its use may take some practice. There are certain fun and effective activities that you can use with other students to get a better understanding of this subject. This math formula is often used to determine a distance between 2 given points on a certain coordinate place. It always includes finding a distance between x-coordinates and y-coordinates in addition to using the Pythagorean Theorem if you need to determine a diagonal distance. It’s obvious that this process requires a lot of practice, and that’s why you should take part in specific distance formula activities because they will help you remain focused on your studies.
For example, take into consideration the human coordinate grid because it’s one of the best and clearest ways to memorize this topic. If you participate in special learning groups and clubs, start by moving all chairs and desks out of your way to mark out a particular coordinate grid on the floor (don’t forget to use a masking tape). Be sure to create the necessary axes with different color tapes and label all divisions with relevant coordinates, starting with 0. As you can see, this task is quite different from creating any powerpoint presentation or completing other academic assignments.
Ask other students to take their turns rolling 2 dices to provide them with a certain set of coordinates (to determine where they should stand). If you prefer to use negative numbers, toss a coin to decide if the numbers of students will be positive or negative. Once all students are positioned on this grid, ask them to sit in place to be able to use the distance formula to calculate a distance of each one. If every grid square has a particular size, this aspect can predict an actual distance between them and the positions of other students. It’s also possible to use a tape measure if you want to check the entire work.
How to Find a Distance with Time and Average Speed
To achieve this goal, you need to use the distance formula and find correct values for time and average speed. Each time when trying to determine the distance that any moving object has travelled, there are 2 important details that you must know to make the right calculation: its exact speed and time. If you have the necessary information, feel free to find a distance that this object has travelled using the distance formula.
Another important step is multiplying its average speed by time. So, once you find these details, the process of determining the distance that a given object has travelled becomes quite straightforward because you only need to multiply those quantities to find the right answer. However, remember that if the time units used to calculate an average speed value are different from the ones used in a time value, they must be converted to make them compatible with each other. Imagine that you have an average speed value measured in km/hour, but your time value is measured in minutes, and this means that you need to divide this time value by sixty to make the necessary conversion. If this task seems a bit confusing, you can always count on the professional services provided by talented freelancers who will do anything, including writing a compare and contrast essay tasks.
It’s advisable to manipulate this equation to solve other variables. That’s because the simplicity of the distance formula makes it very simple to use it to find the values of other variables, besides a distance. You only need to isolate a particular variable that you need to solve based on fundamental algebra rules and insert the values of other variables to succeed. Don’t forget that the distance formula provides you with a bit simplified view of the movements of objects, and it assumes that they have a constant speed. When it comes to abstract math problems, you’re still able to model the motion of a given object using this assumption, but it doesn’t always reflect the motions of all moving objects in real life because they slow down, stop, etc. Take that into consideration when studying and using the distance formula.
If you have any problems with your geometry assignments, there are certain things that can help you solve them easily. Sometimes, math students have the same difficulties only because their distance formula textbooks are written in quite an unclear manner. You can solve this problem simply by looking for other options to find the one written in a more understandable way. Be sure to understand your academic tasks clearly because the main reason why students fail to do their geometry homework is because they don’t understand it. You can’t complete math assignments if you guess how to solve them so that this tactic won’t bring any positive effect. You need to read your distance formula assignments carefully and consult with your teachers if you have any questions.
Another helpful idea is to focus on academic tasks, and getting rid of all possible distractions is one of the most effective methods. You should turn off your TV and ask friends not to distract you when doing geometry homework. Don’t hesitate to use the Internet when needed because even the best textbook may not be enough to get a better idea of how to handle specific distance formula assignments. There are many excellent educational videos that you can find online and they will provide you with a clearer understanding of how to deal with the most difficult tasks. Finally, take short breaks because they allow your brain to relax and remain focused for a longer period of time. If you neglect this tip, you will feel exhausted very fast, and this is what may prevent you from doing your geometry homework successfully. Don’t make long breaks because they’re responsible for losing your concentration.
Examples of completed orders
More than 7 000 students trust us to do their work
90% of customers place more than 5 orders with us
Special price \$5 /page
Category
Latest Posts
Check the price
|
LINEAR FUNCTIONS COMMON MISTAKES
Save this PDF as:
Size: px
Start display at page:
Transcription
1 LINEAR FUNCTIONS COMMON MISTAKES 1 10/0/009
2 Linear Functions-Definition, Notation and the Vertical Line Test How to Understand the Definition and Notation Linear Function are relations in which exactly an x value is paired to exactly one y-value. Algebraically, these are labeled with alphabetic initials. The independent variable is inside the ( ). Function Notation: f ( = 3x 7 where f=name of function and x= will be the independent variable. Not understanding the notation (not thinking of f( as y) Being able to understand the notation so as to evaluate a function. Consider the problem Find f(3) for f ( = x + 9. Incorrect: f(3) Does Not mean 3= x+9-6= x so f(3) = -3 Correct: f(3) = Does become f(3) =15. Note: Function Notation changes the ordered pair notation from (x,y) to (x, f(). 10/0/009
3 Linear Functions-Their Graph and Higher- Order Functions How to Determine the Parent Functions Parent Functions are the Basic functions we start with and then translate to form new functions The Degree of a function: Given the algebraic equation, the degree can be found by determining the greatest exponential power. Graphically, we can test a function by drawing random vertical lines and intersecting the graph of the function (the line). If the vertical line intersects in more than one point on the graph-the relation is Not a function. Basic Parent Functions Linear: y = x Quadratic: y = x 3 Cubic: y = x Absolute Value: y = x And the list continues into higher order, more complex functions Inability to recognize the Parent Function of complex functions. Graphically, not recognizing the shape common to the Basic Functions. Consider the problem: Identify the Parent-Function for this function y = 9x 3x + 4 Incorrect: Parent function is NOT y=x. Correct: Parent Function is y = x 3 10/0/009
4 Linear Functions-Domain and Range How to correctly identify the Domain and Range. Domain-the possible x-values used in a function which give defined/acceptable y- values (INPUT) Range-can be found by determining what kind of answers are generated by functions. (OUTPUT) Confusing the Domain and the Range Given the function, identify the domain and range. Correct: D:{x: -7 < x < 8} R:{y: -9 < y <3.5} 4 10/0/009
5 Linear Functions- Add, Subtract, Multiply and Divide How to Add, Subtract, Multiply, or Divide Functions Functions can be combined to create New functions. If f ( = x + and h ( = x Then h ( = f ( + g( is defined by = ( x + ) + x Then is defined by h( = f ( g( = x + x Then is defined by h( = f ( g( or or x x + x + + x + Simplifying for the new function incorrectly by not distributing a negative sign or not multiplying correctly, etc. If h( f ( g( = 3x ( x + 7 Incorrect: h( = x 7 Correct: h( = 5x 7 = ) ( x + ) x or 3 x + x h( = f (, g( 0 g( Then or x + x 5 10/0/009
6 Linear Functions-Inverses How to find the Inverse of a Linear Function Inverses undo another function. Graphically, a function s inverse is the reflection of a function over the line y=x. Solving for the Inverse, f ( : 1. Rename f( = as y=.. Switch the x and y in the equation. 3. Solve for the y-variable. 4. Replace the y with. f 1 ( Incorrectly solving for the inverse after the switching in Step. If f( 9x, what is? = f 1 ( Incorrect: Correct: Solution: f 1 ( = x 9 x + f 1 ( = 9 y = 9x x = 9y x + = 9y x + f 1 9 Giving us ( =. 6 10/0/009
7 Linear Functions-Composite Functions How to Solve Composite Functions Ex. Composite Functions combine functions in a special way to create a new function: Notation: h To find h(, 1. Take the g( and put it in for x in the f(. Solve for h( f ( = x And ( = f ( g( Incorrectly substituting into the wrong function. Algebraic errors in simplifying. Consider: f ( = x and g( = x Find h( = (f g)( h( = x 1 h ( = f ( g( ) 1 ( ) g( = x + ( x + ) = + h( = x Incorrect: Correct: Solution: Substitute. Sub Again. h( = f(x h( = h( = 1) x 1 x /0/009
CORRELATED TO THE SOUTH CAROLINA COLLEGE AND CAREER-READY FOUNDATIONS IN ALGEBRA
We Can Early Learning Curriculum PreK Grades 8 12 INSIDE ALGEBRA, GRADES 8 12 CORRELATED TO THE SOUTH CAROLINA COLLEGE AND CAREER-READY FOUNDATIONS IN ALGEBRA April 2016 www.voyagersopris.com Mathematical
College Algebra. Barnett, Raymond A., Michael R. Ziegler, and Karl E. Byleen. College Algebra, 8th edition, McGraw-Hill, 2008, ISBN: 978-0-07-286738-1
College Algebra Course Text Barnett, Raymond A., Michael R. Ziegler, and Karl E. Byleen. College Algebra, 8th edition, McGraw-Hill, 2008, ISBN: 978-0-07-286738-1 Course Description This course provides
Algebra I Vocabulary Cards
Algebra I Vocabulary Cards Table of Contents Expressions and Operations Natural Numbers Whole Numbers Integers Rational Numbers Irrational Numbers Real Numbers Absolute Value Order of Operations Expression
6.1 Add & Subtract Polynomial Expression & Functions
6.1 Add & Subtract Polynomial Expression & Functions Objectives 1. Know the meaning of the words term, monomial, binomial, trinomial, polynomial, degree, coefficient, like terms, polynomial funciton, quardrtic
What are the place values to the left of the decimal point and their associated powers of ten?
The verbal answers to all of the following questions should be memorized before completion of algebra. Answers that are not memorized will hinder your ability to succeed in geometry and algebra. (Everything
Alum Rock Elementary Union School District Algebra I Study Guide for Benchmark III
Alum Rock Elementary Union School District Algebra I Study Guide for Benchmark III Name Date Adding and Subtracting Polynomials Algebra Standard 10.0 A polynomial is a sum of one ore more monomials. Polynomial
Algebra 1 Course Title
Algebra 1 Course Title Course- wide 1. What patterns and methods are being used? Course- wide 1. Students will be adept at solving and graphing linear and quadratic equations 2. Students will be adept
CRLS Mathematics Department Algebra I Curriculum Map/Pacing Guide
Curriculum Map/Pacing Guide page 1 of 14 Quarter I start (CP & HN) 170 96 Unit 1: Number Sense and Operations 24 11 Totals Always Include 2 blocks for Review & Test Operating with Real Numbers: How are
South Carolina College- and Career-Ready (SCCCR) Algebra 1
South Carolina College- and Career-Ready (SCCCR) Algebra 1 South Carolina College- and Career-Ready Mathematical Process Standards The South Carolina College- and Career-Ready (SCCCR) Mathematical Process
Pre-Calculus Math 12 First Assignment
Name: Pre-Calculus Math 12 First Assignment This assignment consists of two parts, a review of function notation and an introduction to translating graphs of functions. It is the first work for the Pre-Calculus
MATH 110 College Algebra Online Families of Functions Transformations
MATH 110 College Algebra Online Families of Functions Transformations Functions are important in mathematics. Being able to tell what family a function comes from, its domain and range and finding a function
BookTOC.txt. 1. Functions, Graphs, and Models. Algebra Toolbox. Sets. The Real Numbers. Inequalities and Intervals on the Real Number Line
College Algebra in Context with Applications for the Managerial, Life, and Social Sciences, 3rd Edition Ronald J. Harshbarger, University of South Carolina - Beaufort Lisa S. Yocco, Georgia Southern University
FINAL EXAM SECTIONS AND OBJECTIVES FOR COLLEGE ALGEBRA
FINAL EXAM SECTIONS AND OBJECTIVES FOR COLLEGE ALGEBRA 1.1 Solve linear equations and equations that lead to linear equations. a) Solve the equation: 1 (x + 5) 4 = 1 (2x 1) 2 3 b) Solve the equation: 3x
Algebraic expressions are a combination of numbers and variables. Here are examples of some basic algebraic expressions.
Page 1 of 13 Review of Linear Expressions and Equations Skills involving linear equations can be divided into the following groups: Simplifying algebraic expressions. Linear expressions. Solving linear
HIBBING COMMUNITY COLLEGE COURSE OUTLINE
HIBBING COMMUNITY COLLEGE COURSE OUTLINE COURSE NUMBER & TITLE: - Beginning Algebra CREDITS: 4 (Lec 4 / Lab 0) PREREQUISITES: MATH 0920: Fundamental Mathematics with a grade of C or better, Placement Exam,
Indiana State Core Curriculum Standards updated 2009 Algebra I
Indiana State Core Curriculum Standards updated 2009 Algebra I Strand Description Boardworks High School Algebra presentations Operations With Real Numbers Linear Equations and A1.1 Students simplify and
Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.
Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}
MBA Jump Start Program
MBA Jump Start Program Module 2: Mathematics Thomas Gilbert Mathematics Module Online Appendix: Basic Mathematical Concepts 2 1 The Number Spectrum Generally we depict numbers increasing from left to right
Algebra 1 Course Information
Course Information Course Description: Students will study patterns, relations, and functions, and focus on the use of mathematical models to understand and analyze quantitative relationships. Through
Algebra I Credit Recovery
Algebra I Credit Recovery COURSE DESCRIPTION: The purpose of this course is to allow the student to gain mastery in working with and evaluating mathematical expressions, equations, graphs, and other topics,
X On record with the USOE.
Textbook Alignment to the Utah Core Algebra 2 Name of Company and Individual Conducting Alignment: Chris McHugh, McHugh Inc. A Credential Sheet has been completed on the above company/evaluator and is
Mathematics Online Instructional Materials Correlation to the 2009 Algebra I Standards of Learning and Curriculum Framework
Provider York County School Division Course Syllabus URL http://yorkcountyschools.org/virtuallearning/coursecatalog.aspx Course Title Algebra I AB Last Updated 2010 - A.1 The student will represent verbal
Make sure you look at the reminders or examples before each set of problems to jog your memory! Solve
Name Date Make sure you look at the reminders or examples before each set of problems to jog your memory! I. Solving Linear Equations 1. Eliminate parentheses. Combine like terms 3. Eliminate terms by
Course Outlines. 1. Name of the Course: Algebra I (Standard, College Prep, Honors) Course Description: ALGEBRA I STANDARD (1 Credit)
Course Outlines 1. Name of the Course: Algebra I (Standard, College Prep, Honors) Course Description: ALGEBRA I STANDARD (1 Credit) This course will cover Algebra I concepts such as algebra as a language,
ALGEBRA 2 CRA 2 REVIEW - Chapters 1-6 Answer Section
ALGEBRA 2 CRA 2 REVIEW - Chapters 1-6 Answer Section MULTIPLE CHOICE 1. ANS: C 2. ANS: A 3. ANS: A OBJ: 5-3.1 Using Vertex Form SHORT ANSWER 4. ANS: (x + 6)(x 2 6x + 36) OBJ: 6-4.2 Solving Equations by
Algebra II New Summit School High School Diploma Program
Syllabus Course Description: Algebra II is a two semester course. Students completing this course will earn 1.0 unit upon completion. Required Materials: 1. Student Text Glencoe Algebra 2: Integration,
http://www.aleks.com Access Code: RVAE4-EGKVN Financial Aid Code: 6A9DB-DEE3B-74F51-57304
MATH 1340.04 College Algebra Location: MAGC 2.202 Meeting day(s): TR 7:45a 9:00a, Instructor Information Name: Virgil Pierce Email: piercevu@utpa.edu Phone: 665.3535 Teaching Assistant Name: Indalecio
Algebra 2: Q1 & Q2 Review
Name: Class: Date: ID: A Algebra 2: Q1 & Q2 Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which is the graph of y = 2(x 2) 2 4? a. c. b. d. Short
Polynomials and Quadratics Want to be an environmental scientist? Better be ready to get your hands dirty!.1 Controlling the Population Adding and Subtracting Polynomials............703.2 They re Multiplying
Algebra 2 PreAP. Name Period
Algebra 2 PreAP Name Period IMPORTANT INSTRUCTIONS FOR STUDENTS!!! We understand that students come to Algebra II with different strengths and needs. For this reason, students have options for completing
Math 1050 Khan Academy Extra Credit Algebra Assignment
Math 1050 Khan Academy Extra Credit Algebra Assignment KhanAcademy.org offers over 2,700 instructional videos, including hundreds of videos teaching algebra concepts, and corresponding problem sets. In
Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given.
Polynomials (Ch.1) Study Guide by BS, JL, AZ, CC, SH, HL Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Sasha s method
Mathematics Placement
Mathematics Placement The ACT COMPASS math test is a self-adaptive test, which potentially tests students within four different levels of math including pre-algebra, algebra, college algebra, and trigonometry.
PRE-CALCULUS GRADE 12 [C] Communication Trigonometry General Outcome: Develop trigonometric reasoning. A1. Demonstrate an understanding of angles in standard position, expressed in degrees and radians.
ALGEBRA 2 Functions/Relations/Polynomial Operations 1A, 1B, 1C, 1D, 1E, 1F, 1G,2D, 7I, 7B, 7C
Unit 1 Unit 2 1) Domain/Range - Multiple Representations Interval Notation/Set Notation/Inequalities 2) Function notation, composite functions 3) Operations with Functions - Incude add/sub/multiply polynomials,
Algebra 2 Year-at-a-Glance Leander ISD 2007-08. 1st Six Weeks 2nd Six Weeks 3rd Six Weeks 4th Six Weeks 5th Six Weeks 6th Six Weeks
Algebra 2 Year-at-a-Glance Leander ISD 2007-08 1st Six Weeks 2nd Six Weeks 3rd Six Weeks 4th Six Weeks 5th Six Weeks 6th Six Weeks Essential Unit of Study 6 weeks 3 weeks 3 weeks 6 weeks 3 weeks 3 weeks
Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test
Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important
F.IF.7e Analyze functions using different representations. Graph exponential and logarithmic functions, showing intercept and end behavior.
Grade Level/Course: Algebra and Pre-Calculus Lesson/Unit Plan Name: Introduction to Logarithms: A Function Approach Rationale/Lesson Abstract: This lesson is designed to introduce logarithms to students
2.3. Finding polynomial functions. An Introduction:
2.3. Finding polynomial functions. An Introduction: As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned
Simplify the rational expression. Find all numbers that must be excluded from the domain of the simplified rational expression.
MAC 1105 Final Review Simplify the rational expression. Find all numbers that must be excluded from the domain of the simplified rational expression. 1) 8x 2-49x + 6 x - 6 A) 1, x 6 B) 8x - 1, x 6 x -
Overview. Observations. Activities. Chapter 3: Linear Functions Linear Functions: Slope-Intercept Form
Name Date Linear Functions: Slope-Intercept Form Student Worksheet Overview The Overview introduces the topics covered in Observations and Activities. Scroll through the Overview using " (! to review,
List the elements of the given set that are natural numbers, integers, rational numbers, and irrational numbers. (Enter your answers as commaseparated
MATH 142 Review #1 (4717995) Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Description This is the review for Exam #1. Please work as many problems as possible
Clovis Community College Core Competencies Assessment 2014 2015 Area II: Mathematics Algebra
Core Assessment 2014 2015 Area II: Mathematics Algebra Class: Math 110 College Algebra Faculty: Erin Akhtar (Learning Outcomes Being Measured) 1. Students will construct and analyze graphs and/or data
~ EQUIVALENT FORMS ~
~ EQUIVALENT FORMS ~ Critical to understanding mathematics is the concept of equivalent forms. Equivalent forms are used throughout this course. Throughout mathematics one encounters equivalent forms of
POLYNOMIALS and FACTORING
POLYNOMIALS and FACTORING Exponents ( days); 1. Evaluate exponential expressions. Use the product rule for exponents, 1. How do you remember the rules for exponents?. How do you decide which rule to use
Florida Algebra 1 End-of-Course Assessment Item Bank, Polk County School District
Benchmark: MA.912.A.2.3; Describe the concept of a function, use function notation, determine whether a given relation is a function, and link equations to functions. Also assesses MA.912.A.2.13; Solve
Math 1. Month Essential Questions Concepts/Skills/Standards Content Assessment Areas of Interaction
Binghamton High School Rev.9/21/05 Math 1 September What is the unknown? Model relationships by using Fundamental skills of 2005 variables as a shorthand way Algebra Why do we use variables? What is a
Interpretation of Test Scores for the ACCUPLACER Tests
Interpretation of Test Scores for the ACCUPLACER Tests ACCUPLACER is a trademark owned by the College Entrance Examination Board. Visit The College Board on the Web at: www.collegeboard.com/accuplacer
Dear Accelerated Pre-Calculus Student:
Dear Accelerated Pre-Calculus Student: I am very excited that you have decided to take this course in the upcoming school year! This is a fastpaced, college-preparatory mathematics course that will also
Factoring the trinomial ax 2 + bx + c when a = 1 A trinomial in the form x 2 + bx + c can be factored to equal (x + m)(x + n) when the product of m x n equals c and the sum of m + n equals b. (Note: the
DRAFT. Algebra 1 EOC Item Specifications
DRAFT Algebra 1 EOC Item Specifications The draft Florida Standards Assessment (FSA) Test Item Specifications (Specifications) are based upon the Florida Standards and the Florida Course Descriptions as
Math 1314 Lesson 8 Business Applications: Break Even Analysis, Equilibrium Quantity/Price
Math 1314 Lesson 8 Business Applications: Break Even Analysis, Equilibrium Quantity/Price Three functions of importance in business are cost functions, revenue functions and profit functions. Cost functions
Prentice Hall MyMathLab Algebra 1, 2011
Prentice Hall MyMathLab Algebra 1, 2011 C O R R E L A T E D T O Tennessee Mathematics Standards, 2009-2010 Implementation, Algebra I Tennessee Mathematics Standards 2009-2010 Implementation Algebra I 3102
Algebra and Geometry Review (61 topics, no due date)
Course Name: Math 112 Credit Exam LA Tech University Course Code: ALEKS Course: Trigonometry Instructor: Course Dates: Course Content: 159 topics Algebra and Geometry Review (61 topics, no due date) Properties
Algebra II Unit Number 4
Title Polynomial Functions, Expressions, and Equations Big Ideas/Enduring Understandings Applying the processes of solving equations and simplifying expressions to problems with variables of varying degrees.
Answer Key Building Polynomial Functions 1. What is the equation of the linear function shown to the right? 2. How did you find it? y = ( 2/3)x + 2 or an equivalent form. Answers will vary. For example,
Study Guide 2 Solutions MATH 111
Study Guide 2 Solutions MATH 111 Having read through the sample test, I wanted to warn everyone, that I might consider asking questions involving inequalities, the absolute value function (as in the suggested
SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS
(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1 SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES Be able to identify polynomial, rational, and algebraic
COLLEGE ALGEBRA. Paul Dawkins
Core Florida Math for College Readiness Florida Math for College Readiness provides a fourth-year math curriculum focused on developing the mastery of skills identified as critical to postsecondary readiness
Algebra II. Weeks 1-3 TEKS
Algebra II Pacing Guide Weeks 1-3: Equations and Inequalities: Solve Linear Equations, Solve Linear Inequalities, Solve Absolute Value Equations and Inequalities. Weeks 4-6: Linear Equations and Functions:
Students Currently in Algebra 2 Maine East Math Placement Exam Review Problems
Students Currently in Algebra Maine East Math Placement Eam Review Problems The actual placement eam has 100 questions 3 hours. The placement eam is free response students must solve questions and write
Polynomial Operations and Factoring
Algebra 1, Quarter 4, Unit 4.1 Polynomial Operations and Factoring Overview Number of instructional days: 15 (1 day = 45 60 minutes) Content to be learned Identify terms, coefficients, and degree of polynomials.
Prerequisite: MATH 0302, or meet TSI standard for MATH 0305; or equivalent.
18966.201610 COLLIN COLLEGE COURSE SYLLABUS Course Number: MATH 0305.XS1 Course Title: Beginning Algebra Course Description: With an emphasis on developing critical thinking skills, a study of algebraic
ALGEBRA I (Created 2014) Amherst County Public Schools
ALGEBRA I (Created 2014) Amherst County Public Schools The 2009 Mathematics Standards of Learning Curriculum Framework is a companion document to the 2009 Mathematics Standards of Learning and amplifies
Brunswick High School has reinstated a summer math curriculum for students Algebra 1, Geometry, and Algebra 2 for the 2014-2015 school year.
Brunswick High School has reinstated a summer math curriculum for students Algebra 1, Geometry, and Algebra 2 for the 2014-2015 school year. Goal The goal of the summer math program is to help students
Algebra Unpacked Content For the new Common Core standards that will be effective in all North Carolina schools in the 2012-13 school year.
This document is designed to help North Carolina educators teach the Common Core (Standard Course of Study). NCDPI staff are continually updating and improving these tools to better serve teachers. Algebra
ALGEBRA I (Common Core)
The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA I (Common Core) Wednesday, June 17, 2015 1:15 to 4:15 p.m. MODEL RESPONSE SET Table of Contents Question 25..................
Introduction to Quadratic Functions The St. Louis Gateway Arch was constructed from 1963 to 1965. It cost 13 million dollars to build..1 Up and Down or Down and Up Exploring Quadratic Functions...617.2
Algebra 1 If you are okay with that placement then you have no further action to take Algebra 1 Portion of the Math Placement Test
Dear Parents, Based on the results of the High School Placement Test (HSPT), your child should forecast to take Algebra 1 this fall. If you are okay with that placement then you have no further action
Thnkwell s Homeschool Precalculus Course Lesson Plan: 36 weeks
Thnkwell s Homeschool Precalculus Course Lesson Plan: 36 weeks Welcome to Thinkwell s Homeschool Precalculus! We re thrilled that you ve decided to make us part of your homeschool curriculum. This lesson
expression is written horizontally. The Last terms ((2)( 4)) because they are the last terms of the two polynomials. This is called the FOIL method.
A polynomial of degree n (in one variable, with real coefficients) is an expression of the form: a n x n + a n 1 x n 1 + a n 2 x n 2 + + a 2 x 2 + a 1 x + a 0 where a n, a n 1, a n 2, a 2, a 1, a 0 are
Factoring Polynomials
UNIT 11 Factoring Polynomials You can use polynomials to describe framing for art. 396 Unit 11 factoring polynomials A polynomial is an expression that has variables that represent numbers. A number can
1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style
Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with
The program also provides supplemental modules on topics in geometry and probability and statistics.
Algebra 1 Course Overview Students develop algebraic fluency by learning the skills needed to solve equations and perform important manipulations with numbers, variables, equations, and inequalities. Students
is the degree of the polynomial and is the leading coefficient.
Property: T. Hrubik-Vulanovic e-mail: thrubik@kent.edu Content (in order sections were covered from the book): Chapter 6 Higher-Degree Polynomial Functions... 1 Section 6.1 Higher-Degree Polynomial Functions...
Functions and equations Assessment statements. Concept of function f : f (); domain, range, image (value). Composite functions (f g); identit function. Inverse function f.. The graph of a function; its
Questions. Strategies August/September Number Theory. What is meant by a number being evenly divisible by another number?
Content Skills Essential August/September Number Theory Identify factors List multiples of whole numbers Classify prime and composite numbers Analyze the rules of divisibility What is meant by a number
Algebra 1 Advanced Mrs. Crocker. Final Exam Review Spring 2014
Name: Mod: Algebra 1 Advanced Mrs. Crocker Final Exam Review Spring 2014 The exam will cover Chapters 6 10 You must bring a pencil, calculator, eraser, and exam review flip book to your exam. You may bring
For each learner you will need: mini-whiteboard. For each small group of learners you will need: Card set A Factors; Card set B True/false.
Level A11 of challenge: D A11 Mathematical goals Starting points Materials required Time needed Factorising cubics To enable learners to: associate x-intercepts with finding values of x such that f (x)
Common Core Unit Summary Grades 6 to 8
Common Core Unit Summary Grades 6 to 8 Grade 8: Unit 1: Congruence and Similarity- 8G1-8G5 rotations reflections and translations,( RRT=congruence) understand congruence of 2 d figures after RRT Dilations
Algebra I. In this technological age, mathematics is more important than ever. When students
In this technological age, mathematics is more important than ever. When students leave school, they are more and more likely to use mathematics in their work and everyday lives operating computer equipment,
A Quick Algebra Review
1. Simplifying Epressions. Solving Equations 3. Problem Solving 4. Inequalities 5. Absolute Values 6. Linear Equations 7. Systems of Equations 8. Laws of Eponents 9. Quadratics 10. Rationals 11. Radicals
Administrative - Master Syllabus COVER SHEET
Administrative - Master Syllabus COVER SHEET Purpose: It is the intention of this to provide a general description of the course, outline the required elements of the course and to lay the foundation for
Big Ideas in Mathematics
Big Ideas in Mathematics which are important to all mathematics learning. (Adapted from the NCTM Curriculum Focal Points, 2006) The Mathematics Big Ideas are organized using the PA Mathematics Standards
COLLEGE ALGEBRA LEARNING COMMUNITY
COLLEGE ALGEBRA LEARNING COMMUNITY Tulsa Community College, West Campus Presenter Lori Mayberry, B.S., M.S. Associate Professor of Mathematics and Physics lmayberr@tulsacc.edu NACEP National Conference
2013 MBA Jump Start Program
2013 MBA Jump Start Program Module 2: Mathematics Thomas Gilbert Mathematics Module Algebra Review Calculus Permutations and Combinations [Online Appendix: Basic Mathematical Concepts] 2 1 Equation of
Mohawk Valley Community College MVCC MA115 Mr. Bauer
Mohawk Valley Community College MVCC MA115 Course description: This is a dual credit course. Successful completion of the course will give students 1 VVS Credit and 3 MVCC Credit. College credits do have
Math 131 College Algebra Fall 2015
Math 131 College Algebra Fall 2015 Instructor's Name: Office Location: Office Hours: Office Phone: E-mail: Course Description This course has a minimal review of algebraic skills followed by a study of
LAKE ELSINORE UNIFIED SCHOOL DISTRICT
LAKE ELSINORE UNIFIED SCHOOL DISTRICT Title: PLATO Algebra 1-Semester 2 Grade Level: 10-12 Department: Mathematics Credit: 5 Prerequisite: Letter grade of F and/or N/C in Algebra 1, Semester 2 Course Description:
MTH124: Honors Algebra I
MTH124: Honors Algebra I This course prepares students for more advanced courses while they develop algebraic fluency, learn the skills needed to solve equations, and perform manipulations with numbers,
2 Integrating Both Sides
2 Integrating Both Sides So far, the only general method we have for solving differential equations involves equations of the form y = f(x), where f(x) is any function of x. The solution to such an equation
ACCUPLACER Arithmetic & Elementary Algebra Study Guide
ACCUPLACER Arithmetic & Elementary Algebra Study Guide Acknowledgments We would like to thank Aims Community College for allowing us to use their ACCUPLACER Study Guides as well as Aims Community College
Prerequisites: TSI Math Complete and high school Algebra II and geometry or MATH 0303.
Course Syllabus Math 1314 College Algebra Revision Date: 8-21-15 Catalog Description: In-depth study and applications of polynomial, rational, radical, exponential and logarithmic functions, and systems
Chapter 7. Functions and onto. 7.1 Functions
Chapter 7 Functions and onto This chapter covers functions, including function composition and what it means for a function to be onto. In the process, we ll see what happens when two dissimilar quantifiers
Prentice Hall Algebra 2 2011 Correlated to: Colorado P-12 Academic Standards for High School Mathematics, Adopted 12/2009
Content Area: Mathematics Grade Level Expectations: High School Standard: Number Sense, Properties, and Operations Understand the structure and properties of our number system. At their most basic level
Essential Mathematics for Computer Graphics fast
John Vince Essential Mathematics for Computer Graphics fast Springer Contents 1. MATHEMATICS 1 Is mathematics difficult? 3 Who should read this book? 4 Aims and objectives of this book 4 Assumptions made
|
Mirrored from Sudopedia, the Free Sudoku Reference Guide
# 45 rule
The 45 rule is a basic solving-technique in Killer Sudoku. Each house (row, column, nonet) must add to 45 (the sum of the digits 1 through 9). If a set of cages fits nearly exactly into one or more houses, then the contents of the leftover cells may be deduced from the difference between the sum of the cages and 45 times the number of houses.
For instance, consider this puzzle:
We can use the 45 rule on N9 (nonet 9) by adding together the three cages in N9 and subtracting 45:
24 + 16 + 13 - 45 = 8
Therefore R6C8 = 8. In cases like this, where there are one or more "extra" cells, the extra cells are known as outies.
Sometimes the converse occurs: one or more cells are "missing" (logically enough, they are called innies). In the above puzzle, we can use the 45 rule on N8 to deduce the value of the innie cell R7C4. In this case we must subtract the cage sums from 45:
45 - (20 + 7 + 11) = 7
We can use the 45 rule on any house, not just nonets; for instance, in this puzzle, we can deduce the value of R2C9 as 13 + 15 + 19 - 45 = 2.
We can also use the 45 rule on multiple (but non-overlapping!) houses. (In the case of overlaps, a different set of rules comes into play.) For instance, in the puzzle above, we can deduce the sum of R45C4 by applying the 45 rule to nonets 1, 4 and 7 (N147), by adding together the sums of all cages within those nonets and subtracting 45 x 3 = 135:
13 + 26 + 15 + 14 + 9 + 16 + 20 + 17 + 17 - 135 = 12
Therefore R45C4 = 12. In other words, we have created a "new" two-cell cage, which is known as a split cage or hidden cage.
Sometimes it is impossible to split a cage cleanly off, but it is nonetheless possible to deduce the relation between cells via the innie-outie difference (q.v.).
This page was last modified 22:06, 19 February 2007.
|
# What is 51/86 as a decimal?
## Solution and how to convert 51 / 86 into a decimal
51 / 86 = 0.593
The basis of converting 51/86 to a decimal begins understanding why the fraction should be handled as a decimal. Both represent numbers between integers, in some cases defining portions of whole numbers But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. If we need to convert a fraction quickly, let's find out how and when we should.
## 51/86 is 51 divided by 86
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. Now we divide 51 (the numerator) into 86 (the denominator) to discover how many whole parts we have. This is our equation:
### Numerator: 51
• Numerators represent the number of parts being taken from a denominator. Any value greater than fifty will be more difficult to covert to a decimal. 51 is an odd number so it might be harder to convert without a calculator. Large two-digit conversions are tough. Especially without a calculator. Now let's explore the denominator of the fraction.
### Denominator: 86
• Denominators represent the total number of parts, located below the vinculum or fraction bar. Larger values over fifty like 86 makes conversion to decimals tougher. And it is nice having an even denominator like 86. It simplifies some equations for us. Ultimately, don't be afraid of double-digit denominators. Let's start converting!
## Converting 51/86 to 0.593
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 86 \enclose{longdiv}{ 51 }$$
We will be using the left-to-right method of calculation. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 86 \enclose{longdiv}{ 51.0 }$$
Because 86 into 51 will equal less than one, we can’t divide less than a whole number. Place a decimal point in your answer and add a zero. Now 86 will be able to divide into 510.
### Step 3: Solve for how many whole groups you can divide 86 into 510
$$\require{enclose} 00.5 \\ 86 \enclose{longdiv}{ 51.0 }$$
Now that we've extended the equation, we can divide 86 into 510 and return our first potential solution! Multiple this number by our furthest left number, 86, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.5 \\ 86 \enclose{longdiv}{ 51.0 } \\ \underline{ 430 \phantom{00} } \\ 80 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If you still have a remainder, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 51/86 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. They each bring clarity to numbers and values of every day life. This is also true for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But 51/86 and 0.593 bring clarity and value to numbers in every day life. Here are just a few ways we use 51/86, 0.593 or 59% in our daily world:
### When you should convert 51/86 into a decimal
Pay & Salary - Anything to do with finance or salary will leverage decimal format. If you look at your pay check, you will see your labor is worth $20.59 per hour and not$20 and 51/86.
### When to convert 0.593 to 51/86 as a fraction
Cooking: When scrolling through pintress to find the perfect chocolate cookie recipe. The chef will not tell you to use .86 cups of chocolate chips. That brings confusion to the standard cooking measurement. It’s much clearer to say 42/50 cups of chocolate chips. And to take it even further, no one would use 42/50 cups. You’d see a more common fraction like ¾ or ?, usually in split by quarters or halves.
### Practice Decimal Conversion with your Classroom
• If 51/86 = 0.593 what would it be as a percentage?
• What is 1 + 51/86 in decimal form?
• What is 1 - 51/86 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.593 + 1/2?
|
How to Calculate Compound Interest
Compound interest is interest which is calculated on the original lump sum plus the interest already accumulated.
If a lump sum (PV) is invested in an account paying a rate of compound interest (i), then after a number of periods (n), it will grow into a future value (FV).
Future value (FV) = Present Value (PV) + Compound Interest
This formula can be rearranged to give:
Compound Interest = Future value (FV) – Present Value (PV)
The compound interest can be found by taking the difference between the future value and the present value.
For example, is a lump sum of 10,000 is invested today in an account paying 8% a year compounded annually, then at the end of two years the future value is given by the future value of a lump sum formula as follows.
PV = 10,000
i = 8%
n = 2
FV = PV x (1 + i)n
FV = 10,000 x (1 + 8%)2
FV = 11,664.00
and the compound interest is given by:
Compound interest = FV - PV
Compound interest = 11,664 - 10,000
Compound interest = 1,664.00
This is presented in the diagram below
Difference between future values
Period 0 1 2 3
Future value 10,000 ↑ 11,664.00 ↑
How to Calculate Compound Interest Between Two Periods
This same technique can be applied for any two periods.
Using the same example, if we wanted to know the compound interest earned for years 6,7,8 and 9, we calculate the future value at the end of year 5 (PV at the start of year 6), and the future value at the end of year 9 The difference between the two values must be the compound interest earned between year 6 and year 9.
Future value at year 5 (PV at the start of year 6) is given by:
PV = 10,000
i = 8%
n = 5
FV = PV x (1 + i)n
FV = 10,000 x (1 + 8%)5
FV = 14,693.28
and, the future value at the end of year 9 is given by:
PV = 10,000
i = 8%
n = 9
FV = PV x (1 + i)n
FV = 10,000 x (1 + 8%)9
FV = 19,990.05
and the compound interest earned for years 6, 7, 8, and 9 is given as follows:
FV = FV at the end of year 9
PV = FV at the end of year 5 (PV at the start of year 6)
Compound interest = FV - PV
Compound interest = 19,990.05 - 14,693.28
Compound interest = 5,296.77
This is presented in the diagram below
Difference between future values
Period 5 6 7 8 9
Future value 14,693.28 ↑ 19,990.05 ↑
The compound interest between any two periods can be found by taking the difference between the future values at the start and end of the two periods. The future value can be calculated using the future value of a lump sum formula.
How to Calculate Compound Interest November 6th, 2016
|
UK USIndia
Every Question Helps You Learn
The ratio of blue, green yellow and red marbles is 3:3:2:2.
Ratio 2 (Difficult)
Ratios are an important part of the Eleven Plus maths exam and you will need to understand them if you want to do well. But you’ve come to the right place if you need more practice. This is the second quiz in our Difficult section on ratios.
We use ratios in a similar way to fractions. Both are ways to represent parts or portions of a larger number. For example, let’s say that John has 3 sweets and Jill has 2. We could show this as the ratio 3:2 or as the fractions 35 and 25. That’s because there are 5 sweets altogether. John has 3 of them and Jill has 2.
I hope that makes things clear. If not, don’t worry – you’ll get plenty of opportunity to practice in this quiz, and our others of ratios. Keep trying until you get all the questions right. Good luck!
1.
650 orders have been made for 3 products: TV sets, microwaves and fridges in the ratio 8:2:3 respectively. How many items of each product have been ordered?
400 TV sets, 100 microwaves and 150 fridges
80 TV sets, 20 microwaves and 30 fridges
450 TV sets, 150 microwaves and 200 fridges
160 TV sets, 40 microwaves and 120 fridges
8 + 2 + 3 = 13 ? the fractional parts are as follows: TVs, 8?13 × 650 = 400; microwaves, 2?13 × 650 = 100; fridges, 3?13 × 650 = 150
2.
What are the fractional parts of 7:4:11?
711, 411 and 1111
722, 422 and 1122
13, 15 and 12
1422, 822 and 2222
7 + 4 + 11 = 22 which becomes the denominator in the fractional parts
3.
What are the fractional parts of the ratio 3:18 in their simplest form?
16 and 66
121 and 621
17 and 67
321 and 1821
3 + 18 = 21. This can be simplified by dividing by three to 1 + 6 = 7 which becomes the denominator in the fractional parts. Always try to simplify if it is possible
4.
How else can the ratio 1.5:2:1.2 be written?
2:4:1
10:15:8
15:20:10
15:20:12
1.5:2:1.2 = 15:20:12 (multiplying by 10). You need to find a number to multiply by which will make 1.5 and 1.2 into whole numbers
5.
How else can the ratio 2.5:1.25:7 be written?
10:5:28
10:4:28
11:5:28
10:5:24
2.5:1.25:7 = 10:5:28 (multiply by 4)
6.
Andrew, Beatrice and Clair earned £700 for a job. They shared it in the ratio 4:3:7 respectively. How much did each person get?
Andrew £300, Beatrice £200 and Clair £200
Andrew £300, Beatrice £250 and Clair £150
Andrew £200, Beatrice £150 and Clair £350
Andrew £350, Beatrice £150 and Clair £200
4 + 3 + 7 = 14 ? the fractional parts are as follows:
Andrew, 4?14 × £700 = £200;
Beatrice, 3?14 × £700 = £150;
Clair, 7?14× £700 = £350
7.
Anna, Brian and Charles are given 17 sweets to share. The sweets are given out in the ratio 12:10:12 respectively. How many sweets did each child get?
Anna 12, Brian 10 and Charles 12
Anna 6, Brian 5 and Charles 6
Anna 8, Brian 7 and Charles 8
Anna 3, Brian 4 and Charles 3
12:10:12 can be simplified by dividing by two to make 6:5:6. This ratio adds up to 17 so that is the number of sweets each child received
8.
How else can the ratio 4:1.2 be written?
3:10
20:5
20:7
10:3
4:1.2 = 20:6 (multiply by 5). This can be simplified to 10:3 by dividing by two
9.
Adam, Beth and Carla are stamp collectors. Adam has 44 stamps, Beth has 68 and Carla has 76. What is the simplest ratio to express these amounts?
11:17:19
22:34:38
44:68:76
5:8:10
A:B:C: = 44:68:76 = 11:17:19 (dividing by 4)
10.
What are the fractional parts of the ratio 5:9:3?
518, 918 and 318
517, 917 and 317
519, 919 and 319
520, 920 and 320
5 + 9 + 3 = 17 which becomes the denominator in the fractional parts
Author: Frank Evans
|
# Summary of Skills
Age 9-11 - Concept 5: Math
## Unit 1: Place Value: Million to Thousandths
### Math
• Compare and order large numbers
• Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons
• Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10
• Read and write decimals to thousandths using base-ten numerals, number names, and expanded form
• Read and write numbers from 1,000,000 to thousandths using base-10 numerals, number names, and expanded form
• Read and write Roman numerals
• Read, write, and compare decimals to thousandths
• Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left
• Use place value understanding to round decimals to any place
• Use whole-number exponents to denote powers of 10
## Unit 2: Four Operations
### Math
• Add and subtract multi-digit numbers
• Add, subtract, multiply, and divide decimals to hundredths
• Express a whole number as a product of its prime factors
• Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors
• Fluently multiply multi-digit whole numbers using the standard algorithm
• Solve problems
• Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols
• Write simple expressions that record calculations with numbers, and interpret numerical expressions without evaluating them
## Unit 3: Measurement
### Math
• Connect objects and appropriate units of measurement
• Convert among different-sized standard measurement units within a given measurement system
• Convert between units of time
• Distinguish between units of measurement
• Identify equivalent measurements
• Identify equivalent units of measurement (for example, 12 inches = 1 foot)
• Identify equivalent units of time
• Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8)
• Read measurement tools accurately
• Use conversions in solving multi-step, real world problems
• Use conversions to solve real-world problems
• Use operations on fractions to solve problems involving information presented in line plots
## Unit 4: Adding and Subtracting Fractions
### Math
• Add and subtract fractions with like and unlike denominators (including mixed numbers)
• Compare fractions with different numerators and denominators
• Convert between improper fractions and mixed numbers
• Decompose a fraction into fraction parts (ex. 3/5=1/5+1/5+1/5)
• Determine common denominators when working with unlike denominators
• Distinguish types of fractions (unit fractions, mixed numbers, non-simplified fractions, simplified fractions, improper fractions)
• Recognize and generate equivalent fractions
• Solve word problems involving addition and subtraction of fractions referring to the same whole
## Unit 5: Multiplying Fractions
### Math
• Add, subtract, and multiply fractions
• Apply and extend previous understandings of multiplication to multiply a fraction (including mixed numbers) or whole number by a fraction (including mixed numbers)
• Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction
• Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction (including mixed numbers)
• Compare the size of a product to the size of one factor on the basis of the size of the other factor
• Convert between improper fractions and mixed numbers
• Discover the fraction multiplication algorithm
• Explain why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case)
• Explain why multiplying a given number by a fraction less than 1 results in a product smaller than the given number
• Find the area of a rectangle with fractional side lengths by tiling it with unit squares of the appropriate unit fraction side lengths, and show that the area is the same as would be found by multiplying the side lengths
• Interpret multiplication as scaling (resizing)
• Interpret the product (a/b)× q as a parts of a partition of q into b equal parts
• Multiply fractional side lengths to find areas of rectangles, and represent fraction products as rectangular areas
• Solve real world problems involving fractional areas
• Solve real world problems involving multiplication of fractions
• Solve real world problems involving multiplication of fractions and mixed numbers, including fractional areas
• Use area models to show fraction multiplication
## Unit 6: Geometry
### Math
• Distinguish regular and irregular polygons
• Generate two numerical patterns using two given rules. Identify apparent relationships between corresponding terms. Form ordered pairs consisting of corresponding terms from the two patterns, and graph the ordered pairs on a coordinate plane.
• Identify and classify quadrilaterals
• Identify and classify triangles
• Identify parts of polygons
• Read, interpret, and create graphs
• Represent real world and mathematical problems by graphing points in the first quadrant of the coordinate plane, and interpret coordinate values of points in the context of the situation
• Understand that attributes belonging to a category of two-dimensional figures also belong to all subcategories of that category
• Understand that the first number indicates how far to travel from the origin in the direction of one axis, and the second number indicates how far to travel in the direction of the second axis, with the convention that the names of the two axes and the coordinates correspond (e.g., x-axis and x-coordinate, y-axis and y-coordinate)
• Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates
## Unit 7: Dividing Fractions
### Math
• Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions
• Divide multi-digit numbers and decimals using the standard algorithms
• Explore patterns related to division and fractions
• Interpret division of a unit fraction by a non-zero whole number, and compute such quotients
• Interpret division of a whole number by a unit fraction, and compute such quotients
• Solve real world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions by using visual fraction models and equations to represent the problem
## Unit 8: Volume
### Math
• Apply the formulas V = l × w × h and V = b × h for rectangular prisms to find volumes of right rectangular prisms with whole-number edge lengths in the context of solving real world and mathematical problems
• Create rectangular prisms using cubes
• Find the volume of a right rectangular prism with whole-number side lengths by packing it with unit cubes, and show that the volume is the same as would be found by multiplying the edge lengths (l × w × h = V)
• Find volumes of solid figures composed of two non-overlapping right rectangular prisms by adding the volumes of the non-overlapping parts, applying this technique to solve real world problems
• Identify and construct rectangular prisms
• Make rectangular prisms with specific dimensions using centimeter cubes
• Measure volume by counting unit cubes
• Recognize volume as additive
• Recognize volume as an attribute of solid figures and understand concepts of volume measurement
• Relate volume to the operations of multiplication and addition
• Solve real and mathematical problems involving perimeter and area
## Unit 9: Skills Review
### Math
• Add, subtract, multiply, and divide decimals to hundredths
• Add, subtract, multiply, and divide fractions with like and unlike denominators (including mixed numbers)
• Apply the formulas V = l (×) w (×) h and V = b (×) h for rectangular prisms to find volumes of right rectangular prisms with whole-number edge lengths in the context of solving real world and mathematical problems
• Apply the order of operations (PEMDAS)
• Compare fractions with different numerators and denominators (< > =)
• Convert among different-sized standard measurement units within a given measurement system
• Convert between improper fractions and mixed numbers
• Distinguish regular and irregular polygons
• Distinguish types of fractions (unit fractions, mixed numbers, non-simplified fractions, simplified fractions, improper fractions)
• Find the perimeter and area of given shapes
• Identify equivalent units of measurement
• Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8)
• Read and write decimals to thousandths using base-ten numerals, number names, and expanded form
• Read, write, and compare decimals to thousandths
• Represent real world and mathematical problems by graphing points in the first quadrant of the coordinate plane, and interpret coordinate values of points in the context of the situation
• Understand that the first number indicates how far to travel from the origin in the direction of one axis, and the second number indicates how far to travel in the direction of the second axis, with the convention that the names of the two axes and the coordinates correspond (e.g., x-axis and x-coordinate, y-axis and y-coordinate)
• Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates
• Use operations on fractions to solve problems involving information presented in line plots
• Use place value understanding to round decimals to any place
• Use whole-number exponents to denote powers of 10
|
# 2005 Indonesia MO Problems/Problem 2
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
For an arbitrary positive integer $n$, define $p(n)$ as the product of the digits of $n$ (in decimal). Find all positive integers $n$ such that $11 p(n)=n^2-2005$.
## Solution
First we will prove that $n$ must have 2 digits. Afterward, we'll let $n = 10a + b$ and will solve for $n$.
Lemma: $n$ must have 2 digits
First, if $n < 10$, then $n^2 - 2005$ is a negative number, so $n$ must have more than 1 digit.
Next, we will show that $n$ can not have 3 digits. Since 2005 is not a perfect square, $n$ can not have any zeroes. That means the minimum value of $n$ that has $k$ digits is $\frac{10^k - 1}{9}$. The maximum value of $p(n)$ with three digits is $9^k$. For the case where $k = 3$, the minimum number is $111$ and the maximum value of $p(n)$ is $729$. Since $111^2 - 2005 > 11 \cdot 729$, all three-digit values of $n$ more than 111 have $11 p(n) < n^2 - 2005$, so $n$ can not be a 3-digit number.
Finally, we will use induction to show that if $n$ can not have four or more digits. If $n$ has $k$ digits, where $k \ge 4$, then the maximum value of $p(n)$ is $9^k$ while the minimum value of $n$ is $10^{k-1}$. For the base case, since $11 \cdot 9^4 < (10^3)^2 - 2005$, we would have $11 p(n) < n^2 - 2005$ for all four digit numbers, so no four digit number would work.
Assume that $11 \cdot 9^k < (10^{k-1})^2 - 2005$, where $k \ge 4$. Multiplying both sides by 100 and adding 2005(99) results in \begin{align*} 110 \cdot 9^k + 2005(99) &< (10^k)^2 - 2005 \\ 11(10 \cdot 9^k + 2005(9)) &< (10^k)^2 - 2005 \\ 11(9 \cdot 9^k + 9^k + 2005(9)) &< (10^k)^2 - 2005 \\ 11 \cdot 9^{k+1} + 11(9^k + 2005(9)) &< (10^k)^2 - 2005 \\ 11 \cdot 9^{k+1} &< (10^k)^2 - 2005. \end{align*} With the induction step complete, we've shown that $11 \cdot 9^k < (10^{k-1})^2 - 2005$ if $k \ge 4$. Since $11 p(n) \le 11 \cdot 9^k$ and $(10^{k-1})^2 - 2005 \le n^2 - 2005$, we have $11 p(n) < n^2 - 2005$, so $n$ can not be number with four or more digits. Thus, $n$ must be a two-digit number. $\blacktriangleright$
From the Lemma, let $n = 10a + b$. Substitution results in \begin{align*} 11ab &= (10a+b)^2 - 2005 \\ 11ab &= 100a^2 + 20ab + b^2 - 2005 \\ 2005 &= 100a^2 + 9ab + b^2. \end{align*} Taking the equation modulo 4, we have $ab + b^2 \equiv 1 \pmod{4}$. That means $a \equiv 0 \pmod{4}$. Let $a = 4a_1$, resulting in $$2005 = 1600a_1^2 + 36a_1b + b^2.$$ If $a_1 > 1$, then $1600a_1^2 > 2005$, so $a_1 = 1$ and $a = 4$. Substitution results in \begin{align*} 2005 &= 1600 + 36b + b^2 \\ 0 &= b^2 + 36n - 405 \\ &= (b-9)(b+45) \end{align*} The only feasible value of $b$ is $9$, so the only positive integer such that $11 p(n) = n^2 - 2005$ is $\boxed{49}$. When plugging the value in, the equality holds.
|
# Maths - Fields
## Vector Field
A vector field has a vector value for every point in a space:
how this plot was produced.
For instance a force field like gravity, if we have a very large mass at the origin, then we can plot the gravitational field, as in the diagram above. Its difficult to find a suitable way to show this, especially in 3 dimensions, an alternative might be to plot lines of constant field force.
## Scalar Field
A scalar field gives a scalar value for every point in the space.
we could represent how some scalar quantity, for example temperature, is defined for each point in the space.
We could define a function to show this, temp = f(x,y,z)
One way to illustrate this is to join up all the points with the same values such as contour lines, or isothermals, or isobars, or whatever appropriate to the scalar quantity.
## Relationship between Scalar and Vector Field
On the following diagram the height (y-dimension) shows the scalar value.
how this plot was produced.
We can relate this to the vector field example above by choosing the scalar to represent potential, in this case energy potential. In this case the scalar and vector fields are related by:
work = ∫ F•dx
or its inverse:
F = grad( work)
where:
• ∫ = line intergral
• F = force
• work = potential energy
• • = dot product
• dx = infinitesimal distance
We can generalise this to other cases as follows:
grad converts a vector field to a scalar field representing the differential 'slope' of the vector field.
or its inverse:
the line intergral converts a scalar field to a vector field.
## Covector Field
Covectors are dual objects to vectors. In the section about tensors we thought about them being the rows in matrix made up of columns representing the basis vectors.
In terms of fields, if,
• v is a vector field.
• c is the corresponding covector field.
• s is a scalar field.
then:
vc=s
that is the dot product of the vector field and the covector field is a scalar field.
1-form is a covector field.
## Tangent Field
The limiting space when small neighbourhoods of the manifold are taken resulting in a flat n-dimensional space.
#### Divergence
We get a scalar field from a vector field by using the divergence as shown on this page.
|
# NCERT solution class 12 chapter 2 Application of Integrals exercise 2.2 mathematics part 2
## EXERCISE 2.2
#### Question 1:
Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y
The required area is represented by the shaded area OBCDO.
Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we obtain the point of intersection as.
It can be observed that the required area is symmetrical about y-axis.
∴ Area OBCDO = 2 × Area OBCO
We draw BM perpendicular to OA.
Therefore, the coordinates of M are.
Therefore, Area OBCO = Area OMBCO – Area OMBO
=∫02(9-4×2)4dx-∫02x24dx
=∫02322-x2dx-14∫02x2dx
=x2322-x2+98sin-12×302-14×3302
=24+98sin-1223-11223
=122+98sin-1223-132
=162+98sin-1223
=1226+94sin-1223Therefore, the required area OBCDO is units
#### Question 2:
Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1
The area bounded by the curves, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, is represented by the shaded area as
On solving the equations, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, we obtain the point of intersection as Aand B.
It can be observed that the required area is symmetrical about x-axis.
∴ Area OBCAO = 2 × Area OCAO
We join AB, which intersects OC at M, such that AM is perpendicular to OC.
The coordinates of M are .
Therefore, required area OBCAO = units
#### Question 3:
Find the area of the region bounded by the curves y = x+ 2, xx = 0 and x = 3
The area bounded by the curves, y = x+ 2, xx = 0, and x = 3, is represented by the shaded area OCBAO as
Then, Area OCBAO = Area ODBAO – Area ODCO
#### Question 4:
Using integration finds the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).
BL and CM are drawn perpendicular to x-axis.
It can be observed in the following figure that,
Area (ΔACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1)
Equation of line segment AB is
Equation of line segment BC is
Equation of line segment AC is
Therefore, from equation (1), we obtain
Area (ΔABC) = (3 + 5 – 4) = 4 units
#### Question 5:
Using integration find the area of the triangular region whose sides have the equations y = 2x +1, y = 3x + 1 and = 4.
The equations of sides of the triangle are y = 2x +1, y = 3x + 1, and = 4.
On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13), and C (4, 9).
It can be observed that,
Area (ΔACB) = Area (OLBAO) –Area (OLCAO)
#### Question 6:
Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is
A. 2 (π – 2)
B. π – 2
C. 2π – 1
D. 2 (π + 2)
The smaller area enclosed by the circle, x2 + y2 = 4, and the line, x + y = 2, is represented by the shaded area ACBA as
It can be observed that,
Area ACBA = Area OACBO – Area (ΔOAB)
Thus, the correct answer is B.
#### Question 7:
Area lying between the curve y2 = 4x and y = 2x is
A.
B.
C.
D.
The area lying between the curve, y2 = 4x and y = 2x, is represented by the shaded area OBAO as
The points of intersection of these curves are O (0, 0) and A (1, 2).
We draw AC perpendicular to x-axis such that the coordinates of C are (1, 0).
∴ Area OBAO = Area (OCABO) – Area (ΔOCA)
square units
Thus, the correct answer is B.
error: Content is protected !!
|
### Home > CC1MN > Chapter 9 > Lesson 9.2.2 > Problem9-54
9-54.
Find the area of the shape at right in at least two ways.
Label each remaining side using centimeters. An enclosed figure: Starting at the upper left corner: right 7, down 3, right 5, down 6, left 12, and up 9, to enclose the figure.
Our first method of finding the area is to divide the shape into smaller rectangles and label their areas. The enclosed figure is cut into two rectangles, by extending the 3 cm line down, to the 12 cm line. The rectangle on the left, is labeled 63 square centimeters, in its interior. The rectangle on the right, is labeled 30 square centimeters, in its interior.
$(9)(7) + (5)(6) = 63 + 30 \\= 93 \text{ square centimeters}$
Another method for finding the area is to subtract the area of the rectangle in the upper-right corner from the area of the purple rectangle. Show that this method gives the same answer as Method 1. The same figure with the top and right sides extended to have a 9 cm by 12 cm rectangle.
|
NCERT Solutions for Maths Class 12 Exercise 6.2
# NCERT Solutions for Maths Class 12 Exercise 6.2
Hello Students! In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 6.2.
You can download the PDF of NCERT Books Maths Chapter 6 for your easy reference while studying NCERT Solutions for Maths Class 12 Exercise 6.2.
All the schools affiliated with CBSE, follow the NCERT books for all subjects. You can check your syllabus from NCERT Syllabus for Mathematics Class 12.
If you want to recall All Maths Formulas for Class 12, you can find it by clicking this link.
If you want to recall All Maths Formulas for Class 11, you can find it by clicking this link.
NCERT Solutions for Maths Class 12 Exercise 6.1
NCERT Solutions for Maths Class 12 Exercise 6.3
## NCERT Solutions for Maths Class 12 Exercise 6.2
Maths Class 12 Ex 6.2 Question 1.
Show that the function given by f(x) = 3x + 17 is strictly increasing on R.
Solution:
We have, f(x) = 3x + 17
f’(x) = 3 > 0 x R
Therefore, f is strictly increasing on R.
Maths Class 12 Ex 6.2 Question 2.
Show that the function given by f(x) = e2x is strictly increasing on R.
Solution:
We have, f(x) = e2x
f’(x) = 2e2x
Case I When x > 0, then f’(x) = 2e2x
Maths Class 12 Ex 6.2 Question 3.
Show that the function given by f(x) = sin x is
(a) strictly increasing in (0, π/2)
(b) strictly decreasing in (π/2, π)
(c) neither increasing nor decreasing in (0, π)
Solution:
We have, f(x) = sin x
f’(x) = cos x
(a) Since f’(x) = cos x is +ve in the interval (0, Ï€/2)
f(x) is strictly increasing in (0, π/2)
(b) Since f’(x) = cos x is -ve in the interval (Ï€/2, Ï€)
f(x) is strictly decreasing in (π/2, π)
(c) Since f’(x) = cos x is +ve in the interval (0, Ï€/2)
while f’(x) is -ve in the interval (Ï€/2, Ï€)
f(x) is neither increasing nor decreasing in (0, π).
Maths Class 12 Ex 6.2 Question 4.
Find the intervals in which the function f given by f(x) = 2x² – 3x is
(a) strictly increasing
(b) strictly decreasing
Solution:
We have, f(x) = 2x² – 3x
f’(x) = 4x – 3
f’(x) = 0 at x = ¾
The point x = ¾ divides the real number line in two disjoint intervals.
Maths Class 12 Ex 6.2 Question 5.
Find the intervals in which the function f given by f(x) = 2x3 – 3x² – 36x + 7 is
(a) strictly increasing
(b) strictly decreasing
Solution:
We have, f(x) = 2x3 – 3x² – 36x + 7
f’(x) = 6x2 – 6x – 36 = 6(x2 – x – 6)
f’(x) = 6(x – 3) (x + 2)
f’(x) = 0 at x = 3 and x = –2
The points x = 3 and x = –2 divide the real number line into three disjoint intervals, i.e., (–∞, –2), (–2, 3), (3, ∞).
Now, f’(x) is +ve in the intervals (–∞, –2) and (3, ∞). Since in the interval (–∞, –2), each factor (x – 3) and (x + 2) is -ve.
f’(x) = +ve.
(a) f is strictly increasing in (–∞, –2) and (3, ∞)
(b) In the interval (–2, 3), (x + 2) is +ve and (x – 3) is -ve.
f’(x) = 6(x – 3)(x + 2) = (+) × (–) = -ve
f is strictly decreasing in the interval (–2, 3).
Maths Class 12 Ex 6.2 Question 6.
Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x² + 2x – 5
(b) 10 – 6x – 2x²
(c) –2x3 – 9x² – 12x + 1
(d) 6 – 9x – x²
(e) (x + 1)3 (x – 3)3
Solution:
(a) Given: f(x) = x² + 2x – 5
f’(x) = 2x + 2 = 2(x + 1) …. (i)
Therefore,
f′(x) = 0 x = −1
Now, x = −1 divides the real number line into intervals (−, −1) and (−1, ).
In (−, −1), f′(x) = 2x + 2 < 0
Thus, f is strictly decreasing in (−, −1).
In (−1, ), f′(x) = 2x + 2 > 0
Thus, f is strictly increasing in (−1, )
(b) Given: f(x) = 10 − 6x − 2x2
Therefore, f′(x) = −6 − 4x
f′(x) = 0 x = −3/2
Now, x = −3/2, divides the real number line into two intervals (−, −3/2) and (−3/2, ).
In (−, −3/2), f′(x) = −6 −4x < 0
Hence, f is strictly increasing for x < −3/2.
In (−3/2, ), f′(x) = −6 − 4x > 0
Hence, f is strictly increasing for x > −3/2.
(c) Given: f(x) = –2x3 – 9x2 – 12x + 1
f’(x) = –6x2 – 18x – 12 = –6(x2 + 3x + 2)
f'(x) = –6(x + 1)(x + 2)
Now, f’(x) = 0 gives x = –1 or x = –2
The points x = – 2 and x = – 1 divide the real number line into three disjoint intervals namely ( –
, –2) ( –2, –1) and( –1, ).
In the interval (–
, –2), i.e., – < x < -2, (x + 1) and (x + 2) are –ve.
f’(x) = (–) (–) (–) = –ve.
f(x) is decreasing in (–, –2)
In the interval (–2, –1), i.e., – 2 < x < –1, (x + 1) is –ve and (x + 2) is +ve.
f'(x) = (–)(–) (+) = +ve.
f(x) is increasing in (–2, –1)
In the interval (–1,
), i.e., –1 < x < , (x + 1) and (x + 2) are both positive.
f’(x) = (–) (+) (+) = –ve.
f(x) is decreasing in (–1, )
Hence, f(x) is increasing for –2 < x < –1 and decreasing for x < –2 and x > –1.
(d) Given: f(x) = 6 − 9x − x2
Hence, f′(x) = −9 − 2x
f′(x) = 0 x = −9/2
In (−9/2, ∞), f′(x) < 0
Hence, f is strictly decreasing for x > −9/2.
In (−∞, −9/2), f′(x) > 0
Hence, f is strictly decreasing in x > −9/2.
(e) Given: f(x) = (x + 1)3 (x − 3)3
Hence, f′(x) = 3(x + 1)2 (x − 3)3 + 3(x − 3)2 (x + 1)3
= 3(x + 1)2 (x − 3)2 [x – 3 + x + 1]
= 3(x + 1)2 (x − 3)2 (2x − 2)
= 6(x + 1)2 (x − 3)2 (x − 1)
Therefore, f′(x) = 0 x = −1, 3, 1
Now, x = −1, 3, 1 divides the real number line into four intervals (−∞, −1), (−1, 1), (1, 3) and (3, ∞).
In (−∞, −1) and (−1, 1)f′(x) = 6(x + 1)2 (x − 3)2 (x − 1) < 0
Hence, f is strictly decreasing in (−∞, −1) and (−1, 1).
In (1, 3) and (3, ∞)f′(x) = 6(x + 1)2 (x − 3)2 (x − 1) > 0
Hence, f is strictly increasing in (1, 3) and (3, ∞).
Maths Class 12 Ex 6.2 Question 7.
Show that , is an increasing function of x throughout its domain.
Solution:
Since, x > −1, x = 0 divides domain (−1, ∞) in two intervals −1 < x < 0 and x > 0
When, −1 < x < 0
Then, x < 0 x2 > 0
x > −1 (2 + x) > 0
(2 + x)2 > 0
Thus, f is increasing throughout the domain.
Maths Class 12 Ex 6.2 Question 8.
Find the values of x for which y = [x (x – 2)]² is an increasing function.
Solution:
We have, y = x4 – 4x3 + 4x2
∴ dy/dx = 4x3 – 12x2 + 8x
For the function to be increasing dy/dx = 0
4x3 – 12x2 + 8x = 0
4x(x – 1) (x – 2) = 0
x = 0, x = 1, x = 2
Now, x = 0x = 1 and x = 2 divide the real number line in the intervals (−∞, 0), (0, 1), (1, 2) and (2,∞).
In (−∞, 0) and (1, 2)dy/dx < 0
Hence, y is strictly decreasing in intervals (−∞, 0) and (1, 2)
In (0, 1) and (2, ∞)dy/dx > 0
Hence, y is strictly increasing in intervals (0, 1) and (2, ∞).
Maths Class 12 Ex 6.2 Question 9.
Prove that is an increasing function of θ in [0, π/2].
Solution:
Ex 6.2 Class 12 Maths Question 10.
Prove that the logarithmic function is strictly increasing on (0, ∞).
Solution:
Let f(x) = log x
Then, f’(x) = 1/x
For x > 0, 1/x > 0,
Therefore, f’(x) > 0
Hence, f(x) is an increasing function in the interval (0, ∞).
Maths Class 12 Ex 6.2 Question 11.
Prove that the function f given by f(x) = x² – x + 1 is neither strictly increasing nor strictly decreasing on (–1, 1).
Solution:
Given: f(x) = x² – x + 1
Then, f’(x) = 2x – 1
f(x) is strictly increasing when f’(x) > 0.
2x – 1 > 0
x > ½
Thus, f(x) is increasing on (1/2, 1).
f(x) is strictly decreasing when f’(x) < 0.
2x – 1 < 0
x < ½
Thus, f(x) is decreasing on (–1, 1/2).
Hence, f(x) is neither strictly increasing nor decreasing on (–1, 1).
Maths Class 12 Ex 6.2 Question 12.
Which of the following functions are strictly decreasing on [0, π/2]
(a) cos x
(b) cos 2x
(c) cos 3x
(d) tan x
Solution:
(a) We have, f(x) = cos x
f’(x) = – sin x < 0 in [0, Ï€/2]
f’(x) is a decreasing function.
Maths Class 12 Ex 6.2 Question 13.
On which of the following intervals is the function f given by f(x )= x100 + sin x – 1 strictly decreasing ?
(A) (0, 1)
(B) [π/2, π]
(C) [0, π/2]
(D) none of these
Solution:
(D) We have, f(x) = x100 + sin x – 1
f’(x)= 100x99+ cos x
(a) For (0, 1) i.e. 0 < x < 1, x99 and cos x are both +ve
f’ (x) > 0
f(x) is increasing on (0, 1).
Therefore, the option (D) is correct.
Maths Class 12 Ex 6.2 Question 14.
Find the least value of a such that the function f given by f(x) = x² + ax + 1 is strictly increasing on [1, 2].
Solution:
We have, f(x) = x² + ax + 1
f’(x) = 2x + a.
Since f(x) is an increasing function on (1, 2)
f’(x) > 0 for all 1 < x < 2
Now, f”(x) = 2 for all x (1, 2) f”(x) > 0 for all x (1, 2)
f’(x) is an increasing function on (1, 2)
f’(x) is the least value of f’(x) on (1, 2)
But f’(x) > 0
x (1, 2)
f’(1) > 0 2 + a > 0
a > –2
Thus, the least value of a is –2.
Maths Class 12 Ex 6.2 Question 15.
Let I be any interval disjoint from [–1, 1]. Prove that the function f given by f(x) = x + 1/x is strictly increasing on I.
Solution:
Given f(x) = x + 1/x
Hence, f’ (x) is strictly increasing on I.
Maths Class 12 Ex 6.2 Question 16.
Prove that the function f given by f(x) = log sin x is strictly increasing on (0, π/2) and strictly decreasing on (π/2, π).
Solution:
We have, f(x) = log sin x
f’(x) =
When 0 < x < Ï€/2, f’(x) is +ve; i.e., increasing
When Ï€/2 < x < Ï€, f’(x) is -ve; i.e., decreasing,
f(x) is decreasing.
Hence, f is increasing on (0, π/2) and strictly decreasing on (π/2, π).
Maths Class 12 Ex 6.2 Question 17.
Prove that the function f given by f(x) = log (cos x) is strictly decreasing on (0, π/2) and strictly increasing on (π/2, π).
Solution:
We have, f(x) = log (cos x)
f’(x) =
In the interval (0, Ï€/2), f’(x) = -ve
f is strictly decreasing.
In the interval (Ï€/2, Ï€), f’(x) is + ve.
f is strictly increasing in the interval.
Maths Class 12 Ex 6.2 Question 18.
Prove that the function given by f(x) = x3 – 3x2 + 3x – 100 is increasing in R.
Solution:
We have, f(x) = x3 – 3x2 + 3x – 100
f’(x) = 3x2 – 6x + 3
= 3(x2 – 2x + 1)
= 3(x – 1)2
Now x
R, f'(x) = (x – 1)2 ≥ 0
i.e. f'(x) ≥ 0
x R
Hence, f(x) is increasing on R.
Maths Class 12 Ex 6.2 Question 19.
The interval in which y = x2 e-x is increasing in
(A) (–∞, ∞)
(B) (2, 0) (C) (2, ∞) (D) (0, 2)
Solution:
(D) We have, y = x2 e-x or f(x) = x2 e-x
f’(x) = 2xe-x + x2(–e-x) = xe-x(2-x) = e-xx(2 – x)
Now e-x is positive for all x
R; f’(x) = 0 at x = 0, 2
x = 0, x = 2 divide the number line into three disjoint intervals, viz. (-∞, 0), (0, 2) and (2, ∞).
(a) In interval (-∞, 0), x is +ve and (2 – x) is +ve
f’(x) = e-xx(2 – x) = (+) (–) (+) = -ve
f is decreasing in (-∞, 0).
(b) In interval (0, 2), f’(x) = e-x x(2 – x)
= (+) (+) (+) = +ve
f is increasing in (0, 2).
(c) In interval (2, ∞), f’(x) = e-x x(2 – x) = (+) (+) (–) = -ve
f is decreasing in the interval (2, ∞)
Hence, the option (D) is correct.
|
# How Do I love Thee? Let Me Count The Ways…
Counting is perhaps one of the very first skills we learn as a child. I remember, when I was small, watching Count von Count on Sesame Street teaching kids to count, in a rather frightening way! Even now, counting is still a very important skill for me in my profession as a programmer.
Counting allows me to determine if my program will run slow or fast. It allows me to determine if I need to a wait a couple of seconds, minutes, hours or even centuries for my computation to finish. Even with the proliferation of supercomputers with teraflops of computing power, an algorithm that is inherently slow will take forever to compute as the input data becomes large enough.
So How Do I Determine If My Algorithm Is Slow?
Let me illustrate the point by applying counting to the two Sorting Algorithms below. In both algorithms, the input is an array of unsorted numbers.
Algorithm 1: Find the least element in the list. This element is moved to the front. Then the first element among the remaining elements is found and put into the second position. This procedure is repeated until the entire list has been sorted.
Algorithm 2: Take the first element a1 and form 2 sublists, the first containing those elements less than a1, in the order they arise, and the second containing all elements greater than a1, in the order they arise. Then a1 is put at the end of the sublist. Repeat recursively for each sublist until all sublists contain one item.
When you sort a list, the thing which you always do is comparing one element to the next. This is a fundamental operation of sorting. And this operation is what we are going to count. The algorithm which executes the most comparisons will loose.
Analysis of Algorithm 1
Imagine we have an array $A$ of $n$ integers in unsorted order. According to algorithm 1, to sort this array, we first need to find the least element of array $A$. How many numbers do we need to examine in array $A$ in order to determine the least element? The thing is, we will never know which element is the smallest unless we examine “every” number in the array. But there are $n$ numbers in array $A$. Therefore, in the first iteration, the number of comparisons is $n$.
After the first iteration, we have identified the least element in the array $A$. In the second iteration, we apply the same technique of finding the least element in the remaining list of numbers. So how many comparisons do we need to identify the next smallest number? Since the remaining list contains $n-1$ numbers, therefore, the number of comparisons is $n-1$.
There is already a pattern emerging here. It seems that in the third iteration, we will need $n-2$ comparisons. This is indeed the case. So if we put our discovery in a mathematical form, the total number of comparisons $C$ to sort the entire array is therefore:
$\displaystyle C = n + (n - 1) + (n - 2) + ... + 2 + 1 = \sum_{i=0}^{n-1} (n -i )$
We can rearrange the sum of terms such that the summation looks like this:
$\displaystyle (n + 1) + (n - 1 + 2) + (n - 2 + 3) ... = (n+1) + (n+1) + (n+1) + ....$
If $n$ is even, then we have exactly $\frac{n}{2}$ terms of $(n+1)$, that is,:
$\displaystyle \sum_{i=0}^{n-1} (n - i) = \frac{ n ( n + 1)}{2}$
What happens when $n$ is odd? For example, if $n = 5$, then
$\displaystyle 5 + 4 +3 +2 +1 = (5+1) + (4+2) + 3 = 6 + 6 + 3 = 2(6) + 3$
If you take a look at the rightmost side, you can see a pattern. It seems that for an arbitrary integer $n$, we have
$\displaystyle \sum_{i=0}^{n-1} (n - i) = \frac{n-1}{2} (n+1) + \frac{n+1}{2}$
$\displaystyle \sum_{i=0}^{n-1} (n - i) = \frac{n^2 -1 + n + 1}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}$
This means that for any $n$, whether odd or even, the number of comparisons executed by Algorithm 1 is
$\displaystyle \sum_{i=0}^{n-1} (n - i) = \frac{ n ( n + 1)}{2}$
Observe that you can also write
$\displaystyle \sum_{i=0}^{n-1} (n - i) = \sum_{i=1}^n i= \frac{ n ( n + 1)}{2}$
Analysis of Algorithm 2
On the average, we can assume that whenever we pick the first element $a1$, half of the list will be less than $a1$ and the other half is greater than $a1$. To make things simpler, let’s further assume that the number of elements $n$ is a power of 2, that is, $n = 2^m$. This is will ensure us that whenever we halve, we always get a whole number.
So, on the zeroth iteration, the list will be split into two equal parts of size $2^{m}/2 = 2^{m - 1}$. Below is what the input will be like after the first iteration.
Therefore, in the zeroth iteration, the total number of comparisons is $2^{m-1}$.
In the first iteration, the input to the algorithm are the two lists we got above. Since the size of each list is $2^{m-1}$, when we halve this list, we get 2 lists of size $2^{m-1}/2 = 2^{m-2}$. This is also the number of comparisons we need for each list. Since there are 2 lists at this iteration, the number of comparisons at this iteration is therefore $2\times 2^{m-2}$.
On the second iteration, there are 4 lists that are input to the algorithm. Each list has a size of $2^{m-2}$. Halving each list, we get 2 lists of size $2^{m-3}$. This is also the number of comparisons needed to half each input list. Since there are 4 lists in total at this iteration, the total number of comparisons is therefore $4\times 2^{m-3} = 2^2\times 2^{m-3}$.
At this point, let’s pause for a while and see what we have:
Iteration Number # of Comparisons 0 $2^0 \times 2^{m-1}$ 1 $2^1 \times 2^{m-2}$ 2 $2^2 \times 2^{m-3}$
At this point, you will realize that a pattern is emerging. The pattern is this, for each iteration $i$, the number of comparisons is $2^{i}\times 2^{m-i -1}$.
You can continue this pattern until the number of elements of each list is equal to 1. This will occur on the iteration $i = m -1$. If you total the number of comparisons across all iterations, you will get the total number of comparisons $C$ executed by the algorithm, which is
$\displaystyle C = \sum_{i=0}^{m-1} 2^i\times 2^{m-i -1}$
Applying the rule of exponents:
$\displaystyle = \sum_{i=0}^{m-1} 2^{m-1}$
Since the term to be summed is independent of $m$, this simply reduces to adding $2^{m-1}$ m times:
$\displaystyle = m2^{m-1}$
Since we assumed $n = 2^m$, then taking the base 2 logarithms of both sides we get $m = \lg_2 m$. Using this result above, we get
$\displaystyle C = \frac{n \lg_2 n }{2}$
The Winner
By counting, we were able to compute the number of comparisons for algorithms 1 and 2. Below is a table of the results of our counting:
Algorithm # of Comparisons for input of size n 1 $\displaystyle \frac{ n ( n + 1)}{2}$ 2 $\displaystyle \frac{n \lg_2 n }{2}$
Here is a comparison for various values of $n$
n algorithm1 algorithm2
1 1 1 0
2 100 5050 333
3 1000 500500 4983
4 10000 50005000 66439
For an input of size ten thousand, algorithm 1 will take 50 million comparisons as compared to algorithm 2 which takes only 66 thousand comparisons. And the winner is Algorithm 2!
|
# Allegations and Mixtures Practice Set – 2
Allegations and Mixtures Practice Set – 2
1. In a mixture of milk and water, the proportion of water by weight was 75%. If in the 60 gms mixture 15 gms water was added, what would be the percentage of water in the new mixture?
a) 75%
b) 88%
c) 90%
d) 100%
e) None of these
Explanation:
In 60 gm of mixture, Quantity of water = 60 × 75/100 = 45gm
Quantity of milk = 15 gm
After mixing 15 gm of more water, Quantity
of water in the new mixture = 45 + 15 = 60 gm
Quantity of water in 75 gm of mixture = 60 gm
100 gm of mixture will contain = 60/75 × 100 = 80% of water
2. A man purchased 35 kg of rice at the rate of Rs. 9.50 per kg and 30 kg at the rate of Rs. 10.50 per kg. He mixed the two. Approximately, at what price (in Rupees) per kg should he sell the mixture to make 35 per cent profit in the transaction?
a) 12
b) 12.50
c) 13
d) 13.50
e) 13.25
Explanation:
CP of 65 kg of the mixture = Rs. (35 × 9.50 + 30 × 10.50)
= Rs. (332.5 + 315) = Rs. 647.5
Rate per kg of the mixture = Rs 647.5/65
Required rate = 647.5/65 135/100 = Rs 13.50/kg
3. In 1 kg mixture of sand and iron, 20% is iron. How much sand should be added, so that the proportion of iron becomes 5%?
a) 3 kg
b) 4 gms
c) 5 gms
d) 6 kg
e) None of these
Explanation:
Amount of iron in 1 kg mixture = 20% of 1000 gms
= (20 × 1000) / 100 gms = 200 gms
Amount of sand in mixture = (1000 - 200) gms = 800 gms
Now, let the total mixture is x kg in which iron is 20%
According to the question, 5% of x = 200 gm
5% of x = 200 gm
= (5 × x) / 100 = 200
x = (200 × 100) / 5 gms
x = 20000/5 = 4000 gms
Required answer = 4000 gms - 1000 gms = 3000 gms = 3 kg
4. The wheat sold by a grocer contained 10% low quality wheat. What quantity of good quality wheat should be added to 150 kgs of wheat so that the percentage of low quality wheat becomes 5%?
a) 150 kgs
b) 135 kgs
c) 50 kgs
d) 85 kgs
e) None of these
Explanation:
Good quality content in 150 kgs of wheat = 90% of 150 = 135 kg.
In new mixture, low quality wheat is 5%, so good quality wheat 95%
5% of the new mixture = 15 kg,
New mixture = (15 × 100) / 5 = 300 kg
Good quality of wheat added = (300 - 150) kg. = 150 kg
5. When one litre of water is added to a mixture of acid and water, the new mixture contains 20% acid. When one litre of acid is added to the new mixture, then the resulting mixture contains 100/3 % acid. The percentage of acid in the original mixture was
a) 20%
b) 22%
c) 24%
d) 25%
e) 23%
Explanation:
If there be 1 litre of acid in 4 litres of mixture,
then in case I. Percentage of acid =1 / (4 + 1) × 100 = 20%
Case II, Percentage of acid = 2/6 × 100 = 100/3 %
Percentage of acid in original mixture = 25%
6. A grocer purchased 2 kg. of rice at the rate of Rs. 15 per kg. and 3 kg. of rice at the rate of Rs. 13 per kg. At what price per kg should he sell the mixture to earn 100/3 % profit on the cost price?
a) Rs. 28.00
b) Rs. 20.00
c) Rs. 18.40
d) Rs. 17.40
e) None of these
Explanation:
Mixture : 2 kg of rice at Rs.
15/kg + 3 kg of rice at Rs. 13/kg
Total weight = 2 + 3 = 5 kg
Total cost price = (2 × 15) + (3 × 13) = 30 + 39 = Rs. 69
Cost price per kg of the mixture = 69/5 = Rs 13.80
Selling price to get 100/3 % profit
= (10 + 100/3)/100 × Rs 13.80 = Rs 18.40
7. Two barrels contain a mixture of ethanol and gasoline. The content of the ethanol is 60% in the first barrel and 30% in the second barrel. In what ratio must the mixtures from the first and the second barrels be taken to form a mixture containing 50% ethanol?
a) 1 : 2
b) 2 : 1
c) 2 : 3
d) 3 : 2
e) None of these
Explanation:
8. A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
a) 5 litres, 7 litres
b) 7 litres, 4 litres
c) 6 litres, 6 litres
d) 6 litres, 5 litres
e) 4 litres, 8 litres
9. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7 : 2 and 7 : 11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be:
a) 5 : 7
b) 5 : 9
c) 7 : 5
d) 9 : 5
e) None of these
Explanation:
In 1 kg of alloy A,
Gold = 7/18
Copper = 2/9
In 1 kg of alloy B,
Gold = 7/18 Copper = 11/18
Ratio of gold and copper in Alloy C = 7/9 + 7/18 : 2/9 + 11/18 = 21 : 15 = 7 : 5
10. A container has 30 litres of water. If 3 litres of water is replaced by 3 litres of spirit and this operation is repeated twice, what will be the quantity of water in the new mixture?
a) 24 litres
b) 23 litres
c) 24.3 litres
d) 23.3 litres
e) None of these
|
# Lesson Notes By Weeks and Term - Junior Secondary School 2
APPROXIMATION OF NUMBERS ROUNDING OFF TO DECIMAL PLACES
SUBJECT: MATHEMATICS
CLASS: JSS 2
DATE:
TERM: 1st TERM
WEEK SIX
TOPIC: APPROXIMATION OF NUMBERS ROUNDING OFF TO DECIMAL PLACES
Digits 1,2,34 are rounded down to Zero, while digits 5, 6, 7, 8, and 9 are rounded up to 1 e.g. 126=130 to 2 digits
A significant figure begins from the first non-zero digit at the left of a number. Digit should be written with their correct place values.
Worked example
1. Round off the following to the nearest
2. Thousand (ii) hundred (ii) ten
(a) 8615 (b) 16,560
Solution
a.
1. i) 8615 ≈ 9000 {to the nearest thousand}
2. ii) 8615 ≈8600 {to the nearest hundred}
iii) 8615 ≈8620 {to the nearest ten}
b.
(i) 16560≈17000 {to the nearest thousand}
(ii) 1660≈16, 600 {to the nearest hundred}
(iii) 16560 ≈16560 {to the nearest ten}
1. Round off the following to the nearest (i) whole number (ii) tenth (iii) hundred
(a) 3.125 (b) 0.493
Solution
a.
(I) 3.125≈3 {to the nearest whole number}
(ii) 3.125≈3.1{to the nearest tenth}
(iii) 3.125≈3.13{to the nearest hundredth}
b.
(i) 3.125≈3 {to the nearest whole number}
(ii) 3.125≈3.1{to the nearest tenth}
(iii) 3.125≈3.13{to the nearest hundredth
Evaluation
Round off the following to
1. a) 1d.p. (b) 2 d.p. (c) 3 d.p.
(1) 12.9348 (2) 5.0725 (3) 0.9002
Square Roots of Numbers
The symbol ‘√’ means “square root of”. To find the square root of a number, first find its factors.
Examples
1. Find the square root of 11025
Solution
(a) √11025
3 11025
3 3675
5 1225
5 245
7 49
7 7
√11025= √(32 x 52 x 72)= 3 X 5 X 7=105
Example
Find the square roots of √54/9
Solution:
√54/9 =√49/9=√49/√9=7/3=2 1/3
New General Mathematics, UBE Edition, Chapter 2, pages 28-32, 20-21
Essential Mathematics by A J S Oluwasanmi, Chapter 1, pages 42-45
WEEKEND ASSIGNMENT
1. What is 0.003867 to 3 significant figure (a)0.004 (b)0.00386 (c)0.00387 (d)386
2. The square root of 121/2 is ____(a)1 3/4 (b)3 1/2 (c)3 3/4 (d)6 1/4
3. 9852 to 3 S.F. is _____(a)990 (b) 9850 (c) 9580 (d) 986
4. 7.0354 to 2 d.p is _____(a) 7.03(b) 7.04 (c) 7.40 (d)7.13
5. 59094 to the nearest hundred is ____(a) 59100(b)59104 (c) 60094 (d) 60194
THEORY
1. Find the square root of 1296
2. a. What is 0.046783 to 3 significant figure
3. 45.34672 to 2 d.p. is ______
|
You are on page 1of 5
# D
Number
Systems
Here are only numbers
ratied.
William Shakespeare
## Nature has some sort of
arithmetic-geometrical
coordinate system, because
nature has all kinds of
models. What we experience
of nature is in models, and
all of natures models are so
beautiful.
It struck me that natures
system must be a real beauty,
because in chemistry we nd
that the associations are
always in beautiful whole
numbersthere are no
fractions.
Richard Buckminster Fuller
OBJECTIVES
In this appendix you will learn:
## To understand basic number systems concepts, such as
base, positional value and symbol value.
To understand how to work with numbers represented in
the binary, octal and hexadecimal number systems.
To abbreviate binary numbers as octal numbers or
To convert octal numbers and hexadecimal numbers to
binary numbers.
To convert back and forth between decimal numbers and
their binary, octal and hexadecimal equivalents.
To understand binary arithmetic and how negative
binary numbers are represented using twos complement
notation.
## Appendix D Number Systems
D.1
D.2
D.3
D.4
D.5
D.6
Introduction
Abbreviating Binary Numbers as Octal and Hexadecimal Numbers
Converting Octal and Hexadecimal Numbers to Binary Numbers
Converting from Binary, Octal or Hexadecimal to Decimal
Converting from Decimal to Binary, Octal or Hexadecimal
Negative Binary Numbers: Twos Complement Notation
## Summary | Terminology | Self-Review Exercises | Answers to Self-Review Exercises | Exercises
Instructors Manual
Self-Review Exercises
D.1
The bases of the decimal, binary, octal and hexadecimal number systems are
,
and
respectively.
ANS: 10, 2, 8, 16.
D.2
In general, the decimal, octal and hexadecimal representations of a given binary number
contain (more/fewer) digits than the binary number contains.
ANS: Fewer.
D.3
(True/False) A popular reason for using the decimal number system is that it forms a convenient notation for abbreviating binary numbers simply by substituting one decimal digit per
group of four binary bits.
D.4
The (octal / hexadecimal / decimal) representation of a large binary value is the most concise
(of the given alternatives).
D.5
D.6
(True/False) The highest digit in any base is one more than the base.
ANS: False. The highest digit in any base is one less than the base.
(True/False) The lowest digit in any base is one less than the base.
## ANS: False. The lowest digit in any base is zero.
D.7
The positional value of the rightmost digit of any number in either binary, octal, decimal
.
ANS: 1 (the base raised to the zero power).
D.8
The positional value of the digit to the left of the rightmost digit of any number in binary,
octal, decimal or hexadecimal is always equal to
.
ANS: The base of the number system.
D.9
Fill in the missing values in this chart of positional values for the rightmost four positions
in each of the indicated number systems:
decimal
binary
octal
1000
...
...
512
100
256
...
...
10
...
...
8
1
...
...
...
ANS:
Self-Review Exercises
decimal
binary
octal
D.10
D.11
D.12
1000
4096
8
512
100
256
4
64
10
16
2
8
1
1
1
1
## Convert octal 7316 to binary.
ANS: Binary 111 011 001 110.
D.13 Convert hexadecimal 4FEC to octal. [Hint: First convert 4FEC to binary, then convert that
binary number to octal.]
ANS: Binary 0 100 111 111 101 100; Octal 47754.
D.14
D.15
D.16
D.17
## Convert binary 1101110 to decimal.
ANS: Decimal 2+4+8+32+64=110.
Convert octal 317 to decimal.
## ANS: Decimal 177
to binary:
256 128 64 32 16 8 4 2 1
128 64 32 16 8 4 2 1
(1*128)+(0*64)+(1*32)+(1*16)+(0*8)+(0*4)+(0*2)+(1*1)
10110001
to octal:
512 64 8 1
64 8 1
(2*64)+(6*8)+(1*1)
261
256 16 1
16 1
(11*16)+(1*1)
(B*16)+(1*1)
B1
D.18 Show the binary representation of decimal 417. Then show the ones complement of 417
and the twos complement of 417.
ANS: Binary:
512 256 128 64 32 16 8 4 2 1
256 128 64 32 16 8 4 2 1
(1*256)+(1*128)+(0*64)+(1*32)+(0*16)+(0*8)+(0*4)+(0*2)+(1*1)
110100001
## Appendix D Number Systems
Ones complement: 001011110
Twos complement: 001011111
Check: Original binary number + its twos complement
110100001
001011111
--------000000000
D.19
What is the result when a number and its twos complement are added to each other?
ANS: Zero.
Exercises
D.1
Some people argue that many of our calculations would be easier in the base 12 number
system because 12 is divisible by so many more numbers than 10 (for base 10). What is the lowest
digit in base 12? What would be the highest symbol for the digit in base 12? What are the positional
values of the rightmost four positions of any number in the base 12 number system?
ANS: The lowest digit is 1. The highest symbol is C. 1728, 144, 12, 1.
D.2
Complete the following chart of positional values for the rightmost four positions in each
of the indicated number systems:
decimal
base 6
base 13
base 3
1000
...
...
27
100
...
169
...
10
6
...
...
1
...
...
...
1000
216
2197
27
100
36
169
9
10
6
13
3
1
1
1
1
ANS:
decimal
base 6
base 13
base 3
D.3
D.4
## Convert hexadecimal 3A7D to binary.
ANS: 11101001111101.
D.5
Convert hexadecimal 765F to octal. [Hint: First convert 765F to binary, then convert that
binary number to octal.]
ANS: 73137.
D.6
D.7
D.8
D.9
ANS: 94.
ANS: 278.
ANS: 65535.
## ANS: Binary 100101011, Octal 453, Hexadecimal 12B.
Exercises
D.10 Show the binary representation of decimal 779. Then show the ones complement of 779
and the twos complement of 779.
ANS: Binary
1100001011,
0011110101.
D.11
Ones
complement
0011110100,
Twos
complement
Show the twos complement of integer value 1 on a machine with 32-bit integers.
ANS: 00000000000000000000000000000001.
|
1. Operations with Real Numbers
# 1.2 Integers
Learning Objectives
By the end of this section it is expected that you will be able to:
• Use negatives and opposites
• Simplify: expressions with absolute value
• Add ans subtract integers
• Multiply and divide integers
• Simplify Expressions with Integers
# Use Negatives and Opposites
If you have ever experienced a temperature below zero or accidentally overdrawn your checking account, you are already familiar with negative numbers. Negative numbers are numbers less than . The negative numbers are to the left of zero on the number line. See Figure 1.
The arrows on the ends of the number line indicate that the numbers keep going forever. There is no biggest positive number, and there is no smallest negative number.
Is zero a positive or a negative number? Numbers larger than zero are positive, and numbers smaller than zero are negative. Zero is neither positive nor negative.
Consider how numbers are ordered on the number line. Going from left to right, the numbers increase in value. Going from right to left, the numbers decrease in value. See Figure 2.
Remember that we use the notation:
a < b (read “a is less than b”) when a is to the left of b on the number line.
a > b (read “a is greater than b”) when a is to the right of b on the number line.
Now we need to extend the number line which showed the whole numbers to include negative numbers, too. The numbers marked by points in Figure 3 are called the integers. The integers are the numbers
EXAMPLE 1
Order each of the following pairs of numbers, using < or >: a) ___ b) ___ c) ___ d) ___.
Solution
It may be helpful to refer to the number line shown.
a) 14 is to the right of 6 on the number line. ___ > b) −1 is to the left of 9 on the number line. ___ c) −1 is to the right of −4 on the number line. ___ d) 2 is to the right of −20 on the number line. ___ >
TRY IT 1
Order each of the following pairs of numbers, using < or > a) ___ b) ___ c) ___ d) ___.
a) > b) < c) > d) >
You may have noticed that, on the number line, the negative numbers are a mirror image of the positive numbers, with zero in the middle. Because the numbers 2 and are the same distance from zero, they are called opposites. The opposite of 2 is , and the opposite of is 2
Opposite
The opposite of a number is the number that is the same distance from zero on the number line but on the opposite side of zero.
(Figure 4) illustrates the definition.
The opposite of 3 is .
Opposite Notation
means the opposite of the number a.
The notation is read as “the opposite of a.”
The whole numbers and their opposites are called the integers. The integers are the numbers
Integers
The whole numbers and their opposites are called the integers.
The integers are the numbers
# Simplify: Expressions with Absolute Value
We saw that numbers such as are opposites because they are the same distance from 0 on the number line. They are both two units from 0. The distance between 0 and any number on the number line is called the absolute value of that number.
Absolute Value
The absolute value of a number is its distance from 0 on the number line.
The absolute value of a number n is written as .
For example,
• units away from , so .
• units away from , so .
Figure 5 illustrates this idea.
The integers units away from .
The absolute value of a number is never negative (because distance cannot be negative). The only number with absolute value equal to zero is the number zero itself, because the distance from on the number line is zero units.
Property of Absolute Value
for all numbers
Absolute values are always greater than or equal to zero!
Mathematicians say it more precisely, “absolute values are always non-negative.” Non-negative means greater than or equal to zero.
EXAMPLE 2
Simplify: a) b) c) .
Solution
The absolute value of a number is the distance between the number and zero. Distance is never negative, so the absolute value is never negative.
a)
b)
c)
TRY IT 2
Simplify: a) b) c) .
a) 4 b) 28 c) 0
In the next example, we’ll order expressions with absolute values. Remember, positive numbers are always greater than negative numbers!
EXAMPLE 3
Fill in <, >, for each of the following pairs of numbers:
a) ___ b) ___c) ___ d) –___
Solution
___ – a) Simplify. Order. 5 ___ -5 5 > -5 > – b) Simplify. Order. ___ – 8 ___ -8 8 > -8 8 > – c) Simplify. Order. 9 ___ – -9 ___ -9 -9 = -9 -9 = – d) Simplify. Order. – ___ – 16 ____ -16 16 > -16 – > –
TRY IT 3
Fill in <, >, or for each of the following pairs of numbers: a) ___- b) ___- c) ___
d) ___.
a) > b) > c) < d) >
We now add absolute value bars to our list of grouping symbols. When we use the order of operations, first we simplify inside the absolute value bars as much as possible, then we take the absolute value of the resulting number.
Parentheses ( ) Brackets [ ] Braces { } Absolute value | |
In the next example, we simplify the expressions inside absolute value bars first, just like we do with parentheses.
EXAMPLE 4
Simplify: .
Solution
Work inside parentheses first: subtract 2 from 6. Multiply 3(4). Subtract inside the absolute value bars. Take the absolute value. Subtract.
TRY IT 4
Simplify: .
16
Most students are comfortable with the addition and subtraction facts for positive numbers. But doing addition or subtraction with both positive and negative numbers may be more challenging.
We will use two colour counters to model addition and subtraction of negatives so that you can visualize the procedures instead of memorizing the rules.
We let one colour (blue) represent positive. The other colour (red) will represent the negatives. If we have one positive counter and one negative counter, the value of the pair is zero. They form a neutral pair. The value of this neutral pair is zero.
We will use the counters to show how to add the four addition facts using the numbers and .
The first example adds 5 positives and 3 positives—both positives.
The second example adds 5 negatives and 3 negatives—both negatives.
In each case we got 8—either 8 positives or 8 negatives.
When the signs were the same, the counters were all the same color, and so we added them.
So what happens when the signs are different? Let’s add and .
When we use counters to model addition of positive and negative integers, it is easy to see whether there are more positive or more negative counters. So we know whether the sum will be positive or negative.
EXAMPLE 5
Add: a) b) .
Solution
a)
−1 + 5 There are more positives, so the sum is positive. 4
b)
1 + (−5) There are more negatives, so the sum is negative. −4
TRY IT 5
Add: a) b) .
a) 2 b)
Now that we have added small positive and negative integers with a model, we can visualize the model in our minds to simplify problems with any numbers.
When you need to add numbers such as , you really don’t want to have to count out 37 blue counters and 53 red counters. With the model in your mind, can you visualize what you would do to solve the problem?
Picture 37 blue counters with 53 red counters lined up underneath. Since there would be more red (negative) counters than blue (positive) counters, the sum would be negative. How many more red counters would there be? Because , there are 16 more red counters.
Therefore, the sum of is .
Let’s try another one. We’ll add . Again, imagine 74 red counters and 27 more red counters, so we’d have 101 red counters. This means the sum is .
Let’s look again at the results of adding the different combinations of and .
Addition of Positive and Negative Integers
When the signs are the same, the counters would be all the same color, so add them.
When the signs are different, some of the counters would make neutral pairs, so subtract to see how many are left.
Visualize the model as you simplify the expressions in the following examples.
EXAMPLE 6
Simplify: a) b) .
Solution
1. Since the signs are different, we subtract The answer will be negative because there are more negatives than positives.
2. Since the signs are the same, we add. The answer will be negative because there are only negatives.
TRY IT 6
Simplify: a) b) .
a) b)
The techniques used up to now extend to more complicated problems, like the ones we’ve seen before. Remember to follow the order of operations!
EXAMPLE 7
Simplify: .
Solution
Simplify inside the parentheses. Multiply. Add left to right.
TRY IT 7
Simplify: .
13
# Subtract Integers
We will continue to use counters to model the subtraction. Remember, the blue counters represent positive numbers and the red counters represent negative numbers.
Perhaps when you were younger, you read as take away When you use counters, you can think of subtraction the same way!
We will model the four subtraction facts using the numbers and .
The first example, we subtract 3 positives from 5 positives and end up with 2 positives.
In the second example, we subtract 3 negatives from 5 negatives and end up with 2 negatives.
Each example used counters of only one color, and the “take away” model of subtraction was easy to apply.
What happens when we have to subtract one positive and one negative number? We’ll need to use both white and red counters as well as some neutral pairs. Adding a neutral pair does not change the value. It is like changing quarters to nickels—the value is the same, but it looks different.
• To subtract , we restate it as take away 3.
We start with 5 negatives. We need to take away 3 positives, but we do not have any positives to take away.
Remember, a neutral pair has value zero. If we add 0 to 5 its value is still 5. We add neutral pairs to the 5 negatives until we get 3 positives to take away.
−5 − 3 means −5 take away 3. We start with 5 negatives. We now add the neutrals needed to get 3 positives. We remove the 3 positives. We are left with 8 negatives. The difference of −5 and 3 is −8. −5 − 3 = −8
And now, the fourth case, . We start with 5 positives. We need to take away 3 negatives, but there are no negatives to take away. So we add neutral pairs until we have 3 negatives to take away.
5 − (−3) means 5 take away −3. We start with 5 positives. We now add the needed neutrals pairs. We remove the 3 negatives. We are left with 8 positives. The difference of 5 and −3 is 8. 5 − (−3) = 8
EXAMPLE 8
Subtract: a) b) .
Solution
a)
Take 1 positive from the one added neutral pair. −3 − 1 −4
b)
Take 1 negative from the one added neutral pair. 3 − (−1) 4
TRY IT 8
Subtract: a) b) .
a) b) 10
Have you noticed that subtraction of signed numbers can be done by adding the opposite? In Example 8, is the same as and is the same as . You will often see this idea, the subtraction property, written as follows:
Subtraction Property
Subtracting a number is the same as adding its opposite.
Look at these two examples.
.
Of course, when you have a subtraction problem that has only positive numbers, like , you just do the subtraction. You already knew how to subtract long ago. But knowing that gives the same answer as helps when you are subtracting negative numbers. Make sure that you understand how and give the same results!
Look at what happens when we subtract a negative.
Subtracting a negative number is like adding a positive!
You will often see this written as .
What happens when there are more than three integers? We just use the order of operations as usual.
EXAMPLE 9
Simplify: .
Solution
Simplify inside the parentheses first. Subtract left to right. Subtract.
TRY IT 9
Simplify: .
3
# Multiply Integers
Since multiplication is mathematical shorthand for repeated addition, our model can easily be applied to show multiplication of integers. Let’s look at this concrete model to see what patterns we notice. We will use the same examples that we used for addition and subtraction. Here, we will use the model just to help us discover the pattern.
We remember that means add a, b times. Here, we are using the model just to help us discover the pattern.
The next two examples are more interesting.
What does it mean to multiply 5 by It means subtract 5, 3 times. Looking at subtraction as “taking away,” it means to take away 5, 3 times. But there is nothing to take away, so we start by adding neutral pairs on the work space. Then we take away 5 three times.
In summary:
Notice that for multiplication of two signed numbers, when the:
• signs are the same, the product is positive.
• signs are different, the product is negative.
We’ll put this all together in the chart below
Multiplication of Signed Numbers
For multiplication of two signed numbers:
Same signs Product Example Two positives Two negatives Positive Positive
Different signs Product Example Positive \cdot negative Negative \cdot positive Negative Negative
EXAMPLE 10
Multiply: a) b) c) d) .
Solution
a) Multiply, noting that the signs are different so the product is negative. b) Multiply, noting that the signs are the same so the product is positive. c) Multiply, with different signs. d) Multiply, with same signs.
TRY IT 10
Multiply: a) b) c) d) .
a) b) 28 c) d) 60
When we multiply a number by 1, the result is the same number. What happens when we multiply a number by ? Let’s multiply a positive number and then a negative number by to see what we get.
Each time we multiply a number by , we get its opposite!
Multiplication by
Multiplying a number by gives its opposite.
EXAMPLE 11
Multiply: a) b) .
Solution
a) Multiply, noting that the signs are different so the product is negative. b) Multiply, noting that the signs are the same so the product is positive.
TRY IT 11
Multiply: a) b) .
a) b) 17
# Divide Integers
What about division? Division is the inverse operation of multiplication. So, because . In words, this expression says that 15 can be divided into three groups of five each because adding five three times gives 15. Look at some examples of multiplying integers, to figure out the rules for dividing integers.
Division follows the same rules as multiplication!
For division of two signed numbers, when the:
• signs are the same, the quotient is positive.
• signs are different, the quotient is negative.
And remember that we can always check the answer of a division problem by multiplying.
Multiplication and Division of Signed Numbers
For multiplication and division of two signed numbers:
• If the signs are the same, the result is positive.
• If the signs are different, the result is negative.
Same signs Result
Two positives Positive
Two negatives Positive
If the signs are the same, the result is positive.
Different signs Result
Positive and negative Negative
Negative and positive Negative
If the signs are different, the result is negative.
EXAMPLE 12
Divide: a) b) .
Solution
a) Divide. With different signs, the quotient is negative. b) Divide. With signs that are the same, the quotient is positive.
TRY IT 12
Divide: a) b) .
a) b) 39
# Simplify Expressions with Integers
What happens when there are more than two numbers in an expression? The order of operations still applies when negatives are included. Remember My Dear Aunt Sally?
Let’s try some examples. We’ll simplify expressions that use all four operations with integers—addition, subtraction, multiplication, and division. Remember to follow the order of operations.
EXAMPLE 13
Simplify: .
Solution
Multiply first. Add. Subtract.
TRY IT 13
Simplify: .
EXAMPLE 14
Simplify: a) b) .
Solution
a) Write in expanded form. Multiply. Multiply. Multiply. b) Write in expanded form. We are asked to find the opposite of. Multiply. Multiply. Multiply.
Notice the difference in parts a) and b). In part a), the exponent means to raise what is in the parentheses, the to the power. In part b), the exponent means to raise just the 2 to the power and then take the opposite.
TRY IT 14
Simplify: a) b) .
a) 81 b)
The next example reminds us to simplify inside parentheses first.
EXAMPLE 15
Simplify: .
Solution
Subtract in parentheses first. Multiply. Subtract.
TRY IT 15
Simplify: .
29
EXAMPLE 16
Simplify: .
Solution
Exponents first. Multiply. Divide.
TRY IT 16
Simplify: .
4
EXAMPLE 17
Simplify: .
Solution
Multiply and divide left to right, so divide first. Multiply. Add.
TRY IT 17
Simplify: .
21
Access these online resources for additional instruction and practice with adding and subtracting integers. You will need to enable Java in your web browser to use the applications.
# Key Concepts
• Addition of Positive and Negative Integers
• Property of Absolute Value: for all numbers. Absolute values are always greater than or equal to zero!
• Subtraction of Integers
• Subtraction Property: Subtracting a number is the same as adding its opposite.
• Multiplication and Division of Two Signed Numbers
• Same signs—Product is positive
• Different signs—Product is negative
# Glossary
absolute value
The absolute value of a number is its distance from 0 on the number line. The absolute value of a number is written as .
integers
The whole numbers and their opposites are called the integers: …−3, −2, −1, 0, 1, 2, 3…
opposite
The opposite of a number is the number that is the same distance from zero on the number line but on the opposite side of zero: means the opposite of the number. The notation is read “the opposite of .”
# 1.2 Exercise Set
In the following exercises, order each of the following pairs of numbers, using < or >.
1. ___
2. ___
3. ___
4. ___
In the following exercises, simplify.
In the following exercises, fill in <, >, or for each of the following pairs of numbers.
1. ___
2. ___
In the following exercises, simplify.
In the following exercises, simplify each expression.
In the following exercises, simplify.
In the following exercises, simplify each expression.
In the following exercises, multiply.
In the following exercises, divide.
In the following exercises, simplify each expression.
In the following exercises, solve.
1. Temperature On January , the high temperature in Lytton, British Columbia, was ° . That same day, the high temperature in Fort Nelson, British Columbia was °. What was the difference between the temperature in Lytton and the temperature in Embarrass?
2. Checking Account Ester has $124 in her checking account. She writes a check for$152. What is the new balance in her checking account?
3. Checking Account Kevin has a balance of in his checking account. He deposits $225 to the account. What is the new balance? 4. Provincial budgets For 2019 the province of Quebec estimated it would have a budget surplus of$5.6 million. That same year, Alberta estimated it would have a budget deficit of $7.5 million. Use integers to write the budget of: 1. Quebec 2. Alberta # Answers: 1. > 2. < 3. < 4. > 1. 32 2. 0 3. 16 1. < 2. = 1. 56 2. 0 3. 8 4. 32 5. 108 6. 29 7. 6 8. 12 1. 16 2. 16 1. 45 2. 45 9. 27 10. -99 11. 22 12. 4 13. 6 14. 14 15. 13 16. 64 17. 90 18. 9 19. 41 20. 5 21. ° 22.$187
1. \$5.6 million
2. million
|
Note:
• A quadratic equation is a polynomial equation of degree 2.
• The ''U'' shaped graph of a quadratic is called a parabola.
• A quadratic equation has two solutions. Either two distinct real solutions, one double real solution or two imaginary solutions.
• There are several methods you can use to solve a quadratic equation:
1. Factoring
2. Completing the Square
4. Graphing
Solve for x in the following equation.
Example 2:
The equation is already equal to zero.
Method 1: Factoring
The equation can be written in the equivalent form of The only way that a product can equal zero is if one or both of the factors equal zero.
Method 2: Completing the square
Subtract 5 from both sides of the equation.
Add to both sides of the equation.
Factor the left side and simplify the right side.
Take the square root of both sides of the equation,
Add 3 to both sides of the equation.
and
In the equation , a is the coefficient of the term, b is the coefficient of the x term, and c is the constant. Simply insert 1 for a, -6 for b, and 5 for c in the quadratic formula and simplify
.
and
Method 4: Graphing
Graph y= the left side of the equation or and graph y= the right side of the equation or y=0. The graph of y=0 is nothing more than the x-axis. So what you will be looking for is where the graph of crosses the x-axis. Another way of saying this is that the x-intercepts are the solutions to this equation. The x-intercepts are 1 and The answers are 1 and
Check these answers in the original equation.
Check the answer x=1 by substituting 1 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.
• Left Side:
• Right Side:
Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 1 for x, then x=1 is a solution.
Check the solution x=5 by substituting 5 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.
• Left Side:
• Right Side:
Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 5 for x, then x=5 is a solution.
If you would like to go work another example, click on Example
If you would like to test yourself by working some problems similar to this example, click on Problem.
|
# Calculus/Surface area
← Arc length Calculus Work → Surface area
Suppose we are given a function f and we want to calculate the surface area of the function f rotated around a given line. The calculation of surface area of revolution is related to the arc length calculation.
If the function f is a straight line, other methods such as surface area formulas for cylinders and conical frustra can be used. However, if f is not linear, an integration technique must be used.
Recall the formula for the lateral surface area of a conical frustum:
$A = 2 \pi r l \,$
where r is the average radius and l is the slant height of the frustum.
For y=f(x) and $a\le x\le b$, we divide [a,b] into subintervals with equal width Δx and endpoints $x_0,x_1,\ldots,x_n$. We map each point $y_i= f(x_i) \,$ to a conical frustum of width Δx and lateral surface area $A_i \,$.
We can estimate the surface area of revolution with the sum
$A = \sum_{i=0}^{n} A_i$
As we divide [a,b] into smaller and smaller pieces, the estimate gives a better value for the surface area.
## Definition (Surface of Revolution)
The surface area of revolution of the curve y=f(x) about a line for $a\le x\le b$ is defined to be
$A = \lim_{n\to \infty} \sum_{i=0}^{n} A_i$
## The Surface Area Formula
Suppose f is a continuous function on the interval [a,b] and r(x) represents the distance from f(x) to the axis of rotation. Then the lateral surface area of revolution about a line is given by
$A = 2 \pi \int_a^b r(x) \sqrt{ 1+ (f'(x))^2 } dx$
And in Leibniz notation
$A = 2 \pi \int_a^b r(x) \sqrt{ 1+ \left(\frac{dy}{dx}\right)^2 } dx$
Proof:
$A \$ = $\lim_{n \to \infty} \sum_{i=1}^n A_i$ = $\lim_{n \to \infty} \sum_{i=1}^n 2 \pi r_i l_i$ = $2 \pi \lim_{n \to \infty} \sum_{i=1}^n r_i l_i$
As $n \rightarrow \infty$ and $\Delta x \rightarrow 0$, we know two things:
1. the average radius of each conical frustum $r_i$ approaches a single value
2. the slant height of each conical frustum $l_i$ equals an infitesmal segment of arc length
From the arc length formula discussed in the previous section, we know that
$l_i = \sqrt{ 1+ (f'(x_i))^2 }$
Therefore
$A \$ = $2 \pi \lim_{n \to \infty} \sum_{i=1}^n r_i l_i$ = $2 \pi \lim_{n \to \infty} \sum_{i=1}^n r_i \sqrt{ 1+ (f'(x_i))^2} \Delta x$
Because of the definition of an integral $\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(c_i) \Delta x_i$, we can simplify the sigma operation to an integral.
$A = 2 \pi \int_a^b r(x) \sqrt{ 1+ (f'(x))^2 } dx$
Or if f is in terms of y on the interval [c,d]
$A = 2 \pi \int_c^d r(y) \sqrt{ 1+ (f'(y))^2} dy$
|
What is the formula to find the total surface area of a cube?
What is the formula to find the total surface area of a cube?
The total surface area of a cube is the area covered by all the six faces of a cube. The formula to find total surface area of a cube is given as, Total surface area = 6a2, where, ‘a’ is the edge length of the cube.
How do you find the total surface area?
Surface area is the sum of the areas of all faces (or surfaces) on a 3D shape. A cuboid has 6 rectangular faces. To find the surface area of a cuboid, add the areas of all 6 faces. We can also label the length (l), width (w), and height (h) of the prism and use the formula, SA=2lw+2lh+2hw, to find the surface area.
What is the surface area of a 3x3x3 cube?
What is the surface area of a 3x3x3 cube? As you already know, a cube has six square faces. If each of those faces is 3 inches by 3 inches, then the area of each face is 3 × 3 = 9 square inches. And since there are six of them, the total surface area is 9 + 9 + 9 + 9 + 9 + 9 = 54 square inches.
What is the surface area of a 9 cm cube?
486cm2
The side of the cube is given as 9 cm. To find the total surface area of the given cube, we have to substitute a=9 in the formula. Total surface area of the cube with side 9 cm is 6(9)2=486cm2 .
What do you mean by total surface area?
The total surface area of a solid is the sum of the areas of all of the faces or surfaces that enclose the solid. The area of the rectangle is the lateral surface area. The sum of the areas of the rectangle and the two circles is the total surface area.
What is the surface area of a 2x2x2 cube?
What is the surface area of a 2x2x2 cube? The surface area is 24 (6 sides, each 2×2).
What is the surface area of a polyhedra?
The surface area of a polyhedron is equal to the sum of the area of all of its faces. Said another way, the surface area is the total area covered by the net of a polyhedron.
What is the total surface area of rectangle?
The formula to calculate the total surface area of a rectangular prism is given as, TSA of rectangular prism = 2(lb × bh × lh), where, l is length, b is breadth and h is the height of the prism.
What is the difference between surface area and total surface area?
The lateral surface of an object is the area of all the faces of the object, excluding the area of its base and top. For a cube, the lateral surface area would be the area of four sides. Total surface area is the area of all the faces including the bases.
|
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# A moving train 66 m long overtakes another train 88 m long moving in the same direction in $0.168\min$. If the second train is moving at 30 km/hr, at what speed is the first train moving?A. 85 ${\rm{km/hr}}$B. 50 ${\rm{km/hr}}$C. 55 ${\rm{km/hr}}$D. 25 ${\rm{km/hr}}$
Last updated date: 13th Aug 2024
Total views: 390k
Views today: 3.90k
Verified
390k+ views
Hint: Here we will find the speed of the first train by using the relative speed formula. First, we will assume the speed of the first train to be $x$ and find the relative speed of the first train with respect to the second train. Then we will convert our all value in one unit. Finally we will use the formula of speed to get the required answer.
Let us take the speed of the first train as $x$ ${\rm{km/hr}}$.
The speed of second train $= 30{\rm{km/hr}}$
So, the relative speed of first train with respect to second $= \left( {x - 30} \right){\rm{km/hr}}$
Converting the above value in m/sec we will multiply it with $\dfrac{5}{{18}}$. Therefore, we get
$\Rightarrow$ The relative speed of first train with respect to second $= \left( {x - 30} \right) \times \dfrac{5}{{18}}{\rm{m/sec}}$
Length of first train is 66 m and second train is 88 m.
So, Total distance travelled $= \left( {66 + 88} \right){\rm{m}} = 154{\rm{m}}$
Time taken $= 0.168\min = \left( {0.168 \times 60} \right)\sec$
Multiplying the terms, we get
$\Rightarrow$ Time taken $= 10.08\sec$
Substituting the above values in the formula Speed $=$ Distance $\div$ Time, we get
$\left( {x - 30} \right) \times \dfrac{5}{{18}} = \dfrac{{154}}{{10.08}}$
Multiplying $\dfrac{{18}}{5}$ on both the sides, we get
$\begin{array}{l} \Rightarrow \left( {x - 30} \right) = \dfrac{{154}}{{10.08}} \times \dfrac{{18}}{5}\\ \Rightarrow \left( {x - 30} \right) = 55\end{array}$
Adding 30 on both the sides, we get
$\Rightarrow x = 55 + 30$
$\Rightarrow x = 85$
So, the speed of the first train is $85{\rm{km/hr}}$.
Hence, option (A) is correct.
Note:
“Relative” is also known as “in comparison to”. The relative speed concept is used when two or more bodies are moving with some speed considered. The relative speed of two bodies is added if they are moving in the opposite direction and subtracted if they are moving in the same direction. The speed of the moving body when considered with respect to the speed of the stationary body is known as relative speed.
|
Diese Präsentation wurde erfolgreich gemeldet.
Die SlideShare-Präsentation wird heruntergeladen. ×
# Use the following graph B-0-003x3+0-13x2-1-2x+18-6 Approximate the ave.docx
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Wird geladen in …3
×
1 von 1 Anzeige
# Use the following graph B-0-003x3+0-13x2-1-2x+18-6 Approximate the ave.docx
Use the following graph B=0.003x3+0.13x2-1.2x+18.6
Approximate the average BMI for 10-year-old males; for 16-year-old males.
Can you show how to plug values?
Solution
There is two ways to do this. We can look at the graph where it says Age and go to the line between 8 and 12. This is 10 then move straight up to the point this line hits the graph. This is the BMI for 10 year olds. It is slightly above 16. We can find the exact number by pluging our value into x in our equationB=0.003x^3+0.13x^2-1.2x+18.6 B=0.003(10)^3+0.13(10)^2-1.2(10)+18.6=22.6 For 16 year olds we do the same thing and get a BMI a little above 20 B=0.003(16)^3+0.13(16)^2-1.2(16)+18.6=44.968 I hope this helps
.
Use the following graph B=0.003x3+0.13x2-1.2x+18.6
Approximate the average BMI for 10-year-old males; for 16-year-old males.
Can you show how to plug values?
Solution
There is two ways to do this. We can look at the graph where it says Age and go to the line between 8 and 12. This is 10 then move straight up to the point this line hits the graph. This is the BMI for 10 year olds. It is slightly above 16. We can find the exact number by pluging our value into x in our equationB=0.003x^3+0.13x^2-1.2x+18.6 B=0.003(10)^3+0.13(10)^2-1.2(10)+18.6=22.6 For 16 year olds we do the same thing and get a BMI a little above 20 B=0.003(16)^3+0.13(16)^2-1.2(16)+18.6=44.968 I hope this helps
.
Anzeige
Anzeige
## Weitere Verwandte Inhalte
Anzeige
### Use the following graph B-0-003x3+0-13x2-1-2x+18-6 Approximate the ave.docx
1. 1. Use the following graph B=0.003x3+0.13x2-1.2x+18.6 Approximate the average BMI for 10-year-old males; for 16-year-old males. Can you show how to plug values? Solution There is two ways to do this. We can look at the graph where it says Age and go to the line between 8 and 12. This is 10 then move straight up to the point this line hits the graph. This is the BMI for 10 year olds. It is slightly above 16. We can find the exact number by pluging our value into x in our equationB=0.003x^3+0.13x^2-1.2x+18.6 B=0.003(10)^3+0.13(10)^2- 1.2(10)+18.6=22.6 For 16 year olds we do the same thing and get a BMI a little above 20 B=0.003(16)^3+0.13(16)^2-1.2(16)+18.6=44.968 I hope this helps
|
# Geometric Transformations to Practice Basic Skills and Introduce Fundamental Concepts
This post shows how learning to apply geometric transformation rules to slide, reflect, rotate or resize a figure can benefit students. Geometric transformation graphing activities can help students learn important math concepts in the following ways:
• Learn how to interpret a symbolic description of a geometric transformation rule to find the image point of a preimage point.
• Learn what it means to reflect a figure over a line.
• Learn what it means to slide or translate a figure.
• Learn what it means to rotate a figure about a point.
• Learn what it means to stretch or shrink a figure.
• The idea that preimage and image points is equivalent to the idea of function inputs and outputs.
• Worthwhile practice with the basic operations of signed numbers.
• Worthwhile practice plotting points and drawing geometric transformation images.
• A tacit introduction to the important mathematical concepts of function, inverse of a function and composition of functions.
This post does not discuss the general theory of affine transformations nor does it discuss the study of geometry from a geometric transformation point of view. For a general discussion of 2D matrix based geometric transformations, download my free handout Matrix Geometric Transformations or visit our free instructional content page.
The table below describes the geometric transformations considered in this post. Assume constants j and k are positive.
Transformation Operation
Reflect point (x,y) over the x-axis. (x,y) → (x,-y)
Reflect point (x,y) over the y-axis. (x,y) → (-x,y)
Reflect point (x,y) over the line y = x. (x,y) → (y,x)
Translate or slide point (x,y) right/left j units and up/down k units. (x,y) → (x ± j, y ± k)
Rotate point (x,y) 90° CW about (0,0). (x,y) → (y,-x)
Rotate point (x,y) 90° CCW about (0,0). (x,y) → (-y,x)
Rotate point (x,y) 180° about (0,0). (x,y) → (-x,-y)
Expand or contract point (x,y) by a factor of k from (0,0). (x,y) → (kx,ky)
The setup and parameters for a geometric transformation activity are shown below:
1. Each student is provided a handout containing directions for the activity, an x-y coordinate axes with the graph of the preimage polygon drawn in black, a symbolic description of a geometric transformation, a table for the preimage points, and blank table for the image points to be filled in by the student. I prefer lattice point coordinate axes, but grid line coordinate axes are fine. The coordinate axes should be properly labeled and laid out so that it’s easy to plot points. It’s important that the coordinate axes be drawn with a 1:1 aspect ratio so that perpendicular lines appear to be perpendicular and graphs of circles appear to be circles, not ovals.
2. Each student should have a ruler to aid in drawing graphs. Sloppy hand drawn graphs are not allowed. I have found that a 6 inch or 15 cm ruler works best.
3. It is assumed that students can do the basic operations with signed numbers and plot points.
4. No calculators allowed; strictly old school. Students may use scratch paper of course.
5. The initial preimage point is arbitrary; just move from vertex to vertex around the polygon in either a clockwise or counterclockwise direction.
6. To enhance the visual effect, allow colored ink pens or pencils to draw the image figures.
The tables and graphs below show the results of reflecting the same black preimage figure over the x-axis, the line x = -4 and the line y = x. Normally a geometric transformation graphing activity should have no more than two transformations to perform on a figure, but to conserve space three transformation images are graphed on the same x-y coordinate axes. The tables of x-y coordinates of the image points are color coded to match the color of the image polygon. In an actual transformation activity handout for students, the column of image point x-y coordinates is blank. The first two or three rows of the tables may show both preimage and image point coordinates to help students understand how the transformation rule works. The reflection over the line x = -4 is accomplished by chaining together transformations as follows:
1. Slide the polygon right 4 units.
2. Reflect the image over the y-axis.
3. Slide the last image left 4 units.
The next demonstration involves a slide, rotation and size transformations of a preimage polygon drawn in black. The 90° counterclockwise rotation about (-2, 1) is accomplished by chaining together three transformations as follows:
1. Slide the polygon right 2 units and down 1 unit.
2. Rotate the image 90° counterclockwise about (0, 0).
3. Slide the last image left 2 units and up 1 unit.
The image polygon drawn in green was obtained by chaining together a size transformation with expansion factor = 1.5 and a 180° rotation about (0, 0).
The graph below shows the decomposition of the geometric transformation (x, y) → (-y + 3 , x – 13) that rotates the black flag 90° counterclockwise about the point (8, – 5). Using the transformation mapping functions in the table above and the graphs of the 4 flags below, we can see that the geometric mapping function can be created as follows: (x, y) → (x – 8, y + 5) → (-(y + 5), x – 8) → (-(y + 5) + 8, x – 8 – 5) = (-y + 3, x – 13). Note that the transformations (x, y) → (x – 8, y +5) and (x, y) → (x + 8, y – 5) are inverse transformations.
Suggestions for fun follow up activities relating to geometric transformations:
• After graphing the reflection image of a polygon over a line, have students fold the sheet of graph paper on the reflecting line, and then hold the sheet of graph paper up to the light to verify that preimage and image polygons are congruent.
• After translating a polygon, use the theorem of Pythagoras to determine the magnitude of the slide and a protractor to determine the polar direction of the slide where the polar direction ranges from 0° to 360°.
• Have students use a compass and protractor to verify that a polygon has been rotated a certain number of degrees about a point.
• Let students be creative by having them make up their own transformation rule, and then use the rule to graph the image of a preimage polygon. Exceptional work can be posted in the classroom for all to enjoy.
• Give students a graph similar to the graph above, and have them find a geometric transformation rule that maps a preimage to an image. You can let the preimage be any of the figures in the graph, and the image can be any of the other figures. You will be amazed to see that some students struggle to find a transformation rule that maps a given preimage figure onto itself, but this can be a great teaching opportunity!
• Given the graphs of a preimage polygon and the image polygon under a size transformation, find the lengths of corresponding side pairs and verify that the ratios of the lengths of corresponding side pairs are equal.
• Tell students that geometric transformations make it possible for game developers to create whose wonderful video games they love to play.
Here’s some exercises based on the examples in this post that you can give to your students:
Introduction to Geometric Transformations (student version)
Introduction to Geometric Transformations (teacher version)
|
## Is 7 a coefficient?
The coefficients are the numbers that multiply the variables or letters. Thus in 5x + y – 7, 5 is a coefficient. In 5x + y – 7 the terms are 5x, y and -7 which all have different variables (or no variables) so there are no like terms. Constants are terms without variables so -7 is a constant.
## How do you identify terms coefficients and constants?
Each term in an algebraic expression is separated by a + sign or J sign. In , the terms are: 5x, 3y, and 8. When a term is made up of a constant multiplied by a variable or variables, that constant is called a coefficient. In the term 5x, the coefficient is 5.
## Is a constant considered a coefficient?
I would like to see this taught as: after combining like terms, all constant factors, including the constant term, are coefficients.
## What is a constant or coefficient?
Parts of an Equation A number on its own is called a Constant. A Coefficient is a number used to multiply a variable (4x means 4 times x, so 4 is a coefficient)
## How do you find the coefficient of Class 7?
A coefficient is a numerical value that is multiplied with a variable. For example, the coefficient of 7x is 7.
## What is a coefficient number?
A coefficient is a number multiplied by a variable. Examples of coefficients: In the term 14 c 14c 14c , the coefficient is 14. In the term g, the coefficient is 1.
## Can coefficient of variance be more than 100?
For the pizza delivery example, the coefficient of variation is 0.25. This value tells you the relative size of the standard deviation compared to the mean. Analysts often report the coefficient of variation as a percentage. If the value equals one or 100%, the standard deviation equals the mean.
## Can Mean be greater than 1?
There’s no problem with the expectation being bigger than 1. However, since the expectation is a weighted average of the values of the random variable, it always lies between the minimal value and the maximal value.
## Is a higher or lower coefficient of variation better?
The coefficient of variation (CV) is the ratio of the standard deviation to the mean. The higher the coefficient of variation, the greater the level of dispersion around the mean. It is generally expressed as a percentage. The lower the value of the coefficient of variation, the more precise the estimate.
## What does the coefficient of determination tell us?
The coefficient of determination is a statistical measurement that examines how differences in one variable can be explained by the difference in a second variable, when predicting the outcome of a given event.
## What is the use of coefficient of variation?
The coefficient of variation shows the extent of variability of data in a sample in relation to the mean of the population. In finance, the coefficient of variation allows investors to determine how much volatility, or risk, is assumed in comparison to the amount of return expected from investments.
## What is the difference between standard deviation and coefficient of variance?
The coefficient of variation (CV) is a measure of relative variability. It is the ratio of the standard deviation to the mean (average). For example, the expression “The standard deviation is 15% of the mean” is a CV….Coefficient of Variation Example.
|
7
Q:
# There are two mixtures of honey and water in which the ratio of honey and water are as 1:3 and 3:1 respectively. Two litres are drawn from first mixture and 3 litres from second mixture, are mixed to form another mixture. What is the ratio of honey and water in it ?
A) 111:108 B) 11:9 C) 103:72 D) None
Explanation:
From the given data,
The part of honey in the first mixture = 1/4
The part of honey in the second mixture = 3/4
Let the part of honey in the third mixture = x
Then,
1/4 3/4
x
(3/4)-x x-(1/4)
Given from mixtures 1 & 2 the ratio of mixture taken out is 2 : 3
=> $\frac{\frac{3}{4}-x}{x-\frac{1}{4}}=\frac{2}{3}$
=> Solving we get the part of honey in the third mixture as 11/20
=> the remaining part of the mixture is water = 9/20
Hence, the ratio of the mixture of honey and water in the third mixture is 11 : 9 .
Q:
An alloy contains gold and silver in the ratio 5 : 8 and another alloy contains gold and silver in the ratio 5 : 3. If equal amount of both the alloys are melted together, then the ratio of gold and silver in the resulting alloy is ?
A) 113/108 B) 105/103 C) 108/115 D) 103/113
Explanation:
As given equal amounts of alloys are melted, let it be 1 kg.
Required ratio of gold and silver =
Hence, ratio of gold and silver in the resulting alloy = 105/103.
4 352
Q:
A tin a mixture of two liquids A and B in the proportion 4 : 1. If 45 litres of the mixture is replaced by 45 litres of liquid B, then the ratio of the two liquids becomes 2 : 5. How much of the liquid B was there in the tin? What quantity does the tin hold?
A) 58 l B) 65 l C) 50 l D) 62 l
Explanation:
Let the tin contain 5x litres of liquids
=> 5(4x - 36) = 2(x + 36)
=> 20x - 180 = 2x + 72
=> x = 14 litres
Hence, the initial quantity of mixture = 70l
Quantity of liquid B
= 50 litres.
7 1162
Q:
An alloy of copper and bronze weight 50g. It contains 80% Copper. How much copper should be added to the alloy so that percentage of copper is increased to 90%?
A) 45 gm B) 50 gm C) 55 gm D) 60 gm
Explanation:
Initial quantity of copper = = 40 g
And that of Bronze = 50 - 40 = 10 g
Let 'p' gm of copper is added to the mixture
=> = 40 + p
=> 45 + 0.9p = 40 + p
=> p = 50 g
Hence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.
6 936
Q:
A man pays Rs. 6.40 per litre of milk. He adds water and sells the mixture at Rs. 8 per litre, thereby making 37.5% profit. The proportion of water to milk received by the customers is
A) 1 : 10 B) 10 : 1 C) 9 : 11 D) 11 : 9
Explanation:
Customer ratio of Milk and Water is given by
Milk :: Water
6.4 0
$\frac{64}{11}$
=> Milk : Water = 110 : 11 = 10 : 1
Therefore, the proportionate of Water to Milk for Customer is 1 : 10
13 940
Q:
In a 40 litre mixture of alcohol & water, the ratio of alcohol and water is 5 : 3. If 20% of this mixture is taken out and the same amount of water is added then what will be the ratio of alcohol and water in final mixture?
A) 1:1 B) 2:1 C) 3:1 D) 1:2
Explanation:
Quantity of alohol in the mixture = 40 x 5/8 = 25 lit
Quantity of water = 40 - 25 = 15 lit
According to question,
Required ratio =
9 1359
Q:
In a 100 litre of mixture the ratio of milk and water is 6:4. How much milk must be added to the mixture in order to make the ratio 3 : 1?
A) 85 B) 60 C) 55 D) 45
Explanation:
Let M litres milk be added
=>
=> 60 + M = 120
=> M = 60 lit.
8 632
Q:
A mixture contains 25% milk and rest water. What percent of this mixture must taken out and replaced with milk so that in mixture milk and water may become equal.
A) 31.8% B) 31% C) 33.33% D) 29.85%
Explanation:
Now, take percentage of milk and applying mixture rule
25 100
50
50 25 = 2 : 1
Hence required answer = 1/3 or 33.33%
14 864
Q:
The concentration of glucose in three different mixtures (glucose and alcohol) is respectively. If 2 litres, 3 litres and 1 litre are taken from these three different vessels and mixed. What is the ratio of glucose and alcohol in the new mixture?
A) 3:2 B) 4:3 C) 2:3 D) 3:4
Explanation:
Concentration of glucose are in the ratio = $\frac{1}{2}:\frac{3}{5}:\frac{4}{5}$
Quantity of glucose taken from A = 1 liter out of 2
Quantity of glucose taken from B = 3/5 x 3 = 1.5 lit
Quantity of glucose taken from C = 0.8 lit
So, total quantity of glucose taken from A,B and C = 3.6 lit
So, quantity of alcohol = (2 + 3 + 1) - 3.6 = 2.4 lit
Ratio of glucose to alcohol = 3.6/2.4 = 3:2
|
# Texas Go Math Grade 6 Module 6 Answer Key Multiplying and Dividing Integers
Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 6 Answer Key Multiplying and Dividing Integers.
## Texas Go Math Grade 6 Module 6 Answer Key Multiplying and Dividing Integers
Multiply.
Question 1.
9 × 3 ______________
9 × 3 = 27
When 9 is multiplied by an integer 1 through 9, digits of product add up to 9.
2 + 7 = 9
Question 2.
7 × 10 ______________
7 × 10 = 70
Product of 10 and another integer ends in 0.
Texas Go Math Grade 6 Answer Key Pdf Multiplying and Dividing Integers Question 3.
9 × 8 ______________
9 × 8 = 72
When 9 is multiplied by an integer 1 through 9, digits of the product add up to 9.
7 + 2 = 9
Question 4.
15 × 10 ______________
15 × 10 = 150
Product of 10 and another integer ends in 0.
Question 5.
6 × 9 ______________
6 × 9 = 54
When 9 is multiplied by an integer 1 through 9, digits of product add up to 9.
5 + 4 = 9
Question 6.
10 × 23 ______________
10 × 23 = 230
Product of 10 and another integer ends in 0.
Question 7.
9 × 9 ______________
9 × 9 = 81
When 9 is multiplied by an integer 1 through 9, digits of product add up to 9.
8 + 1 = 9
Question 8.
10 × 20 ______________
10 × 20 = 200
Product of 10 and another integer ends in 0.
Divide
Question 9.
54 ÷ 9 ______________
To calculate
54 ÷ 9
find the number that you need to multiply 9 times to get the value 54.
That number is 6:
6 × 9 = 54
Therefore,
54 ÷ 9 = 6
Question 10.
42 ÷ 6 ______________
To calculate
42 ÷ 6
find the number that you need to multiply 6 times to get the value 42.
That number is 7:
7 × 6 = 42
Therefore,
42 ÷ 6 = 7
24 ÷ 3 ______________
To calculate
24 ÷ 3
find the number that you need to multiply 3 times to get the value 24.
That number is 6:
8 × 3 = 24
Therefore,
24 ÷ 3 = 8
Question 12.
64 ÷ 8 ______________
To calculate
64 ÷ 8
find the number that you need to multiply 8 times to get the value 64.
That number is 8:
8 × 8 = 64
Therefore,
64 ÷ 8 = 8
Question 13.
90 ÷ 10 ______________
To calculate
90 ÷ 10
find the number that you need to multiply 10 times to get the value 90.
That number is 9:
9 × 10 = 90
Therefore,
90 ÷ 10 = 9
Question 14.
56 ÷ 7 ______________
To calculate
56 ÷ 7
find the number that you need to multiply 7 times to get the value 56.
That number is 8:
8 × 7 = 56
Therefore,
56 ÷ 7 = 8
Question 15.
81 ÷ 9 ______________
To calculate
81 ÷ 9
find the number that you need to multiply 9 times to get the value 81.
That number is 9:
9 × 9 = 81
Therefore,
81 ÷ 9 = 9
110 ÷ 11 ______________
To calculate
110 ÷ 11
find the number that you need to multiply 11 times to get the value 110.
That number is 10:
10 × 11 = 110
Therefore,
110 ÷ 11 = 10
Evaluate each expression.
Question 17.
12 + 8 ÷ 2
The expression you need to evaluate is:
12 + 8 ÷ 2
First divide integers from Left to right:
12 + 4
Then, add integers from Left to right:
12 + 4 = 16
Question 18.
15 – (4 + 3) × 2 ______________
The expression you need to evaluate is:
15 – (4 + 3) × 2
First evaluate the expression in brackets:
15 – 7 × 2
Then, muLtipLy integers from Left to right:
15 – 14
At the end, subtract integers
15 – 14 = 1
Question 19.
18 – (8 – 5)2 _____________
The expression you need to evaluate is:
18 – (8 – 5)2
First evaluate the expression in brackets:
18 – 32
Then, find 3 to the power of 2
18 – 9
At the end, subtract integers
18 – 9 = 9
Question 20.
6 + 7 × 3 – 5 ______________
The expression you need to evaluate is:
6 + 7 × 3 – 5
First multiply integers from left to right
6 + 21 – 5
At the end, add and subtract integers using ruLes for adding and subtracting integers:
6 + 21 – 5 = 22
Question 21.
9 + (22 + 32) × 2 _____________
The expression you need to evaluate is:
9 + (22 + 3)2 × 2
First find 2 to the power of 2 to evaLuate the expression in brackets
9 + (4 + 3)2 × 2
9 + 72 × 2
Find 7 to the power of 2:
9 + 49 × 2
Then, multiply integers from left to right
9 + 98
9 + 98 = 107
Result is 107
Evaluate the expression in brackets, find the value of that expression squared, and then multiply and add integers from left to right.
Module 6 Multiplying and Dividing Integers Texas Go Math Grade 6 Question 22.
6 + 5 – 4 × 3 ÷ 2 _____________
The expression you need to evaluate is:
6 + 5 – 4 × 3 ÷ 2
First, multiply integers from left to right
6 + 5 – 12 ÷ 2
Then, divide integers from left to right
6 + 5 – 6
At the end, add and subtract integers from left to right:
11 – 6 = 5
The result is 5.
Multiply and divide integers from left to right and then add and subtract integers.
Visualize Vocabulary
Use the ✔ words to complete the chart. You may put more than one word in each box.
Understand Vocabulary
Complete the sentences using the review words.
Question 1.
A ________________ is a number that ¡s less than 0. A ________________ is a number that is greater than 0.
First requested review word is negative number
A negative number is less than 0 because it is left from 0 on the number line.
Second requested review word is positive number
A positive number is greater than 0 because it is right from 0 on the number line.
Question 2.
Division problems have three parts. The part you want to divide into groups is called the ________________. The number that is divided into another number is called the ________________. The answer to a division problem is called the ________________.
First requested word is dividend
The dividend is a number that is being divided by another number
Second requested word is divisor
The divisor is a number that is contained in another number (the dividend).
Third requested word is quotient
The quotient is a number of equal parts into which the dividend has been divided by divisor
Question 3.
________________ are all whole numbers and their opposites.
The requested word is integers
Integers are positive and negative numbers that can be written without fractional part.
Scroll to Top
Scroll to Top
|
# AREAS OF PARALLELOGRAMS AND TRIANGLES NCERT 9 MATHEMATICS TEXTBOOK MCQ (Multiple Choice Questions)
Q1. ABCD is a parallelogram. If E and F are mid points of sides AB and CD and diagonal AC is joined then ar (FCBE) : ar (CAB) is:
1. 1 : 2
2. 2 : 1
3. 1 : 1
4. 1 : 4
Q2. If the base of a parallelogram is 8 cm and its altitude is 5 cm, then its area is equal to
1. 15 cm2
2. 20 cm2
3. 40 cm2
4. 10 cm2
Q3. If a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is
1. Equal to the area of the parallelogram
2. Twice the area of the parallelogram
3. Four times the area of the parallelogram
4. Half the area of the parallelogram
Q4. The area of a rhombus, the lengths of whose diagonals are 16 cm and 24 cm, is
1. 150 cm2
2. 192 cm2
3. 384 cm2
4. 40 cm2
Q5. Two parallelograms ABCD, EFGH are on equal bases AB and EF and between the same parallel lines. If area (||gm ABCD)=140 sq cm, then sum of areas of the two parallelograms is
1. 210 sq cm
2. 280 sq cm
3. 420 sq cm
4. 240 sq cm
Q6. A triangle and a rhombus are on the same base and between the same parallels. Then the ratio of area of triangle to that rhombus is:
• 1 : 1
• 1 : 2
• 1 : 3
• 1 : 4
Q7. If area of parallelogram ABCD is 25 cm2 and on the same base CD, a triangle BCD is given such that area BCD = x cm2, then value of x is
1. 25 cm2
2. 12.5 cm2
3. 15 cm2
4. 20 cm2
Q8. For two figures to be on the same base and between the same parallels, they must have a common base and
1. One common vertex
2. Two common vertices
3. The vertices(or the vertex) opposite to the common base lying on a line making an acute angle to the base
4. The vertices(or the vertex) opposite to the common base lying on a line parallel to the base
Q9. The area of a right triangle is 30 sq cm. If the base is 5 cm , then the hypotenuse must be
1. 20 cm
2. 13 cm
3. 12 cm
4. 18 cm
Q10. Area of a trapezium, whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm, is
1. 30 cm2
2. 80 cm2
3. 120 cm2
4. 60 cm2
Q11. For two figures to be on the same base and between the same parallels, they must have a common base and____
1. One common vertex
2. Two common vertices
3. The vertices(or the vertex) opposite to the common base lying on a line making an acute angle to the base
4. The vertices(or the vertex) opposite to the common base lying on a line parallel to the base
Q12. A median of a triangle divides it into two triangles of
1. Equal area
2. Unequal area
3. Equal sides
4. Each one-fourth of the area of the given triangle.
Q13. For two figures to be on the same base and between the same parallels, one of the lines must be.
1. Perpendicular to the common base
2. Making an acute angle to the common base
3. Making an obtuse angle to the common base
4. The line containing the common base
Q14. The area of a parallelogram whose base is 4 cm and the height is 5 cm is
1. 20 cm2
2. 20 cm
3. 30 cm
4. 30 cm2
Q15. If a triangle and a square are on the same base and between the same parallels, then the ratio of area of triangle to the area of square is
1. 1 : 3
2. 1 : 2
3. 3 : 1
4. 1 : 4
Q16. If each diagonal of a quadrilateral separates it into two triangles of equal area, then the quadrilateral is a
1. Square
2. Rhombus
3. Parallelogram
4. Trapezium
Q17. The ratio of the areas of two parallelograms on the same base and between the same parallels is:
1. 1 : 2
2. 1 : 1
3. 1 : 3
4. 2 : 1
Q18. A rectangle is called congruent to a square of side 5 cm provided
1. The adjacent sides of the rectangle are each of length 5 cm
2. The perimeter of the rectangle is 20 cm
3. The area of the rectangle is 40 sq cm
4. The sides of the rectangle are of length 10 cm
Q19. The magnitude of measure of a planar region is called its.
1. Perimeter
2. Area
3. Volume
4. Height
Q20. Parallelogram ABCD and rectangle ABEF are on the same base AB. If AB=14 cm, BC=12 cm, then the possible value for the perimeter of ABEF is
1. 64
2. 48
3. 59
4. 52
|
# Class 10 RD Sharma Solutions – Chapter 5 Trigonometric Ratios – Exercise 5.1 | Set 1
• Last Updated : 07 Dec, 2021
### (i) sinA = 2/3
Solution:
sinA = 2/3 = Perpendicular/Hypotenuse
Draw a right-angled △ABC in which ∠B is = 90°
Using Pythagoras Theorem, in △ABC,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = AB2 + BC2
(3)2 = (2)2 + (BC)2
9 = 4 + BC
BC2 = 9 – 4 = 5
BC = √5 units
Now,
cosA = Base/Hypotenuse = BC/AC = √5/3
tanA = Perpendicular/Base = AB/BC = 2/√5
cotA = 1/tanA = √5/2
secA = 1/cosA = 3/√5
cosecA = 1/sinA = 3/2
### (ii) cosA = 4/5
Solution:
cosA = 4/5 = Base/Hypotenuse
Draw a right-angled △ABC in which ∠B is = 90°
Using Pythagoras Theorem, in △ABC,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = AB2 + BC2
(5)2 = (AB)2 + (4)2
25 = AB2 + 16
AB2 = 25 – 16 = 9
AB = √9
= 3 units
Now,
sinA = Perpendicular/Hypotenuse = AB/AC =3/5
tanA = Perpendicular/Base = AB/BC = 3/4
cotA = 1/tanA = 4/3
secA = 1/cosA = 5/4
cosecA = 1/sinA =5/3
### (iii) tanθ = 11/1
Solution:
tanθ = 11/1 = Perpendicular/Base
Draw a right-angled △ABC in which ∠B is = 90°
Using Pythagoras Theorem, in △ABC,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = AB2 + BC2
AC2 = (11)2 + (1)2
AC2 = 121 + 1
= 122
AC = √122units
Now,
sinθ = Perpendicular/Hypotenuse = AB/AC = 11/ √122
cosθ = Base/Hypotenuse = BC/AC = 1/√122
cotθ = 1/tanθ = 1/11
secθ = 1/cosθ = √122/1
cosecθ = 1/sinθ = √122/11
### (iv) sinθ = 11/15
Solution:
sinθ = 11/15 = Perpendicular/Hypotenuse
Draw a right-angled △ABC in which ∠B is = 90°
Using Pythagoras Theorem, in △ABC,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = AB2 + BC2
(15)2 = (11)2 + (BC)2
225 = 121 + (BC)2
(BC)2 = 104
BC = 2√26
Now,
cosθ = Base/Hypotenuse = BC/AC = 2√26/15
tanθ = AB/BC = 11/ 2√26
cotθ = 1/tanθ = 2√26/11
secθ = 1/cosθ = 15/ 2√26
cosecθ = 1/sinθ = 15/11
### (v) tan α = 5/12
Solution:
tan α = 5/12 = Perpendicular/Base
Draw a right-angled △ABC in which ∠B is = 90°
Using Pythagoras Theorem, in △ABC,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = AB2 + BC2
(AC)2 = (12)2 + (25)2
(AC)2 = 144 + 25
(AC)2 = 169
AC = √169 = 13 units
Now,
sin α = Perpendicular/Hypotenuse = AB/AC = 5/13
cos α = Base/Hypotenuse = BC/AC = 12/13
cot α = 1/tan α = 12/5
sec α = 1/cos α = 13/12
cosec α = 1/sin α = 13/5
### (vi) sinθ = √3/2
Solution:
sinθ = √3/2 = Perpendicular/Hypotenuse
Draw a right-angled △ABC in which ∠B is = 90°
Using Pythagoras Theorem, in △ ABC,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = AB2 + BC2
(2)2 = (√3)2 + (BC)2
4 = 3 + (BC)2
(BC)2 = 4 – 3 = 1
BC = 1 units
Now,
cosθ = Base/Hypotenuse = BC/AC = 1/2
tanθ = AB/BC = √3/1
cotθ = 1/tanθ = 1/√3
secθ = 1/cosθ = 2/1
cosecθ = 1/sinθ = 2/√3
### (vii) cosθ = 7/25
Solution:
cosθ = 7/25 = Base/Hypotenuse
Draw a right-angled △ABC in which ∠B is = 90°
Using Pythagoras Theorem, in △ ABC,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = AB2 + BC2
(25)2 = (AB)2 + (7)2
625 = (AB)2 + 49
(AB)2 = 625 – 49 = 576
AB = √576 = 24 units
Now,
sinθ = Perpendicular/Hypotenuse = AB/AC = 24/25
tanθ = Perpendicular/Base = AB/BC = 24/7
cotθ = 1/tanθ = 7/24
secθ = 1/cosθ = 25/7
cosecθ = 1/sinθ = 25/24
### (viii) tanθ = 8/15
Solution:
tanθ = 8/15 = Perpendicular/Base
Draw a right-angled △ABC in which ∠B is = 90°
Using Pythagoras Theorem, in △ ABC,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = AB2 + BC2
(AC)2 = (8)2 + (15)2
(AC)2 = 64 + 225
AC = √289 = 17
Now,
sinθ = Perpendicular/Hypotenuse = AB/AC = 8/17
cosθ = Base/Hypotenuse = BC/AC = 15/17
cotθ = 1/tanθ = 15/8
secθ = 1/cosθ = 17/15
cosecθ = 1/sinθ = 17/8
### (ix) cotθ = 12/5
Solution:
cotθ = 12/5 = Base/Perpendicular
Draw a right-angled △ABC in which ∠B is = 90°
Using Pythagoras Theorem, in △ABC,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = AB2 + BC2
(AC)2 = (5)2 + (12)2
(AC)2 = 25 + 144
(AC)2 = 169
AC = √169 = 13 units
Now,
sinθ = Perpendicular/Hypotenuse = AB/AC = 5/13
cosθ = Base/Hypotenuse = BC/AC = 12/13
tanθ = 1/tanθ = 5/12
secθ = 1/cosθ = 13/12
cosecθ = 1/sinθ = 13/5
### (x) secθ = 13/5
Solution:
secθ = 13/5 = Hypotenuse/Base
Draw a right-angled △ABC in which ∠B is = 90°
Using Pythagoras Theorem, in △ABC,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = AB2 + BC2
(13)2 = (AB)2 + (5)2
169 = (AB)2 + 25
(AB)2 = 169 – 25 = 144
AB = √144 = 12 units
Now,
sinθ = Perpendicular/Hypotenuse = AB/AC = 12/13
tanθ = Perpendicular/Base = AB/BC = 12/5
cotθ = 1/tanθ = 5/12
cosθ = 1/secθ = 5/13
cosecθ = 1/sinθ = 13/12
### (xi) cosecθ = √10
Solution:
cosecθ = √10/1 = Hypotenuse/Perpendicular
Draw a right-angled △ABC in which ∠B is = 90°
Using Pythagoras Theorem, in △ ABC,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = AB2 + BC2
(√10)2 = (1)2 + (BC)2
10 = 1 + (BC)2
(BC)2 = 10 – 1 = 9
BC = √9 = 3
Now,
sinθ = Perpendicular/Hypotenuse = AB/AC = 1/√10
cosθ = Base/Hypotenuse = BC/AC = 3/√10
tanθ = Perpendicular/Hypotenuse = AB/BC = 1/3
cotθ = 1/tanθ = 3/1 = 3
secθ = 1/cosθ = √10/3
### (xii) cosθ = 12/15
Solution:
cosθ = 12/15 = Base/Hypotenuse
Draw a right-angled △ABC in which ∠B is = 90°
Using Pythagoras Theorem, in △ABC,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = AB2 + BC2
(15)2 = (AB)2 + (12)2
225 = (AB)2 + 144
(AB)2 = 225 – 144 = 81
AB = √81 = 9 units
Now,
sinθ = Perpendicular/Hypotenuse = AB/AC = 9/15
tanθ = Perpendicular/Base = AB/BC = 9/12
cotθ = 1/tanθ = 12/9
secθ = 1/cosθ = 15/12
cosecθ = 1/sinθ = 15/9
### (ii) sin C, cos C
Solution:
Given:
In right-angled ΔABC,
AB = 24 cm, BC = 7 cm. ∠B = 90°
Using Pythagoras Theorem
AC2 = AB2 + BC2
AC2 = 242 + 72 = 576 + 49
AC2 = 625
AC = √625 = 25cm
Now,
(i) sinA = BC/AC = 7/25
cosA = AB/AC = 24/25
(ii) sinC = AB/AC = 24/25
cosC = BC/AC = 7/25
### Question 3. In the figure, find tan P and cot R. Is tan P = cot R?
Solution:
Using Pythagoras Theorem
PR2 = PQ2 + QR2
132 = 122 + QR2
QR2 = 169 – 144 = 25
QR = √25 = 5 cm
Now,
tan P = Perpendicular/Base = QR/PQ = 5/2
cot R = Base/Perpendicular = QR/PQ = 5/2
Yes, tanP = cot R
### Question 4. If sin A = 9/41, compute cos A and tan A.
Solution:
Given, sinA = 9/41 = Perpendicular/Hypotenuse
Draw a △ ABC where ∠B = 90°, BC = 9, AC = 41
Using Pythagoras Theorem
AC2 = AB2 + BC2
BC2 = 412 – 92 = 1681 – 81
BC2 = 1600
BC = √1600 = 40
Now, cos A = Base/Hypotenuse = AB/AC = 40/41
tan A = Perpendicular/Base = BC/AB = 9/40
### Question 5. Given 15 cot A = 8, find sin A and sec A.
Solution:
Given, 15 cot A = 8
cot A = 8/15 = Base/Perpendicular
Draw a △ ABC where ∠B = 90°, AB = 8, BC = 15
Using Pythagoras Theorem
AC2 = AB2 + BC2
AC2 = 82 + 152 = 64 + 225
AC2 = 289
AC = √289 = 17
Now,
sin A = Perpendicular/Hypotenuse = BC/AC = 15/17
sec A = Hypotenuse/Base = AC/AB = 17/8
### Question 6. In ΔPQR, right-angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P, and sec R.
Solution:
In right-angled ΔPQR,
∠Q = 90°, PQ = 4cm, RQ = 3cm
Using Pythagoras Theorem
PR2 = PQ2 + QR2
PR2 = 42 + 32 = 16 + 9
PR2 = 25
PR = √25 =5
Now,
sin P = Perpendicular/Hypotenuse = RQ/PR = 3/5
sin R = Perpendicular/Hypotenuse = PQ/PR = 4/5
sec P = Hypotenuse/Base = PR/PQ = 5/4
sec R = Hypotenuse/Base = PR/RQ = 5/3
My Personal Notes arrow_drop_up
|
📚
All Subjects
>
Algebra 2
>
🔢
Matrices
# Determinant of a Matrix
Jenni MacLean
Amrita Arora
### Algebra 2➗
Bookmarked 946 • 2 resources
See Units
## How to Find The Determinant of a Matrix
### What are Matrices?
Matrices are grids of numbers that can represent data or equations. We can use them to represent systems of linear equations by using their coefficients, like the following:
3x - y = 7 and 2x + y = 8 become
You'll notice throughout this article matrices with different types of brackets: some enclosed in parentheses and others in square brackets. Functionally, it makes no difference which one you use. They're just different notations that certain professions are partial towards.
### How Do We Find the Determinant?🧐
Finding the inverse of a matrix can be helpful when you want to get from any matrix to the Identity Matrix, which is the matrix representation of a whole (1):
Inverses of Matrices
The inverse of a matrix, usually represented by A⁻¹, is equal to the adjoint of a matrix divided by the determinant. We can find the adjoint of a matrix by switching a and d and making b and c negative in a standard 2x2 matrix:
The inverse of a matrix, usually represented by A⁻¹, is equal to the adjoint of a matrix divided by the determinant. We can find the adjoint of a matrix by switching a and d and making b and c negative in a standard 2x2 matrix:
For the inverse of a matrix to exist, it must be a square matrix 🟦. In a square matrix, the number of terms counted vertically must equal the number of terms counted horizontally; in other words, it must be 2x2, 3x3, etc., not 3x2 or 3x4.
When writing down the dimensions of a matrix, the first number represents the number of rows, and the second number represents the number of columns. For example, in a 2x3 matrix, the matrix has 2️⃣rows and 3️⃣columns. In addition, the determinant of the matrix must be greater than 0. So, if you calculate the determinant of a matrix and it equals 0, the inverse of that matrix does not exist
The Determinant of a 2x2 Matrix
Formula
The determinant can be represented by “det(A)” or more typically, |A|, with absolute value symbols.
Example Problem
Let’s take a look at the 2x2 matrix in the problem below 🔎
To find the determinant, we must calculate the product of ad and bc. Using the previous abcd matrix, we can say a = 3, b = 4, c = 2, and d = -5. Thus, ad = -15 and bc = 8. If we substitute -15 and 8 for ad and bc in the determinant formula ad-bc, we get (-15 - 8) = -23. Thus, the determinant of this matrix is -23.
The Determinant of a 3x3 Matrix
Formula
|A| = a(ei − fh) − b(di − fg) + c(dh − eg) where
Example Problem
Let’s take a look at the 3x3 matrix supplied in the problem below 🔎
We’ll consider the bottom 6 terms of the matrix as 3 different subcolumns. The formula requires that you multiply a by the determinant of the two rightmost subcolumns (efhi), b by the determinant of the outer two subcolumns (dgfi), and c by the determinant of the two rightmost subcolumns (dgeh).
If we substitute the given matrix’s values into this formula, we have the following 🔢:
|A| = 2(-2-6) - 3(1-9) + 1(-2-6)
|A| = 2(-8) - 3(-8) + 1(-8)
|A| = -16 + 24 -8
|A| = 0
The determinant of Matrix A is 0. This means that its inverse does not exist ❌.
Finding the Determinant on the TI-84 🖩
1. Power on your calculator using the button on the bottom right.
2. Press the 2nd button, then the X-1 button to get to the matrix menu.
3. Use the right arrow key to move from NAMES to EDIT.
4. Use the arrow keys to navigate to the letter you want your matrix to represent and press enter (bottom right of your calculator).
5. Set the dimensions of your matrix and press the down key after the second dimension to enter the matrix.
6. Enter your matrix values. Go back to the matrix menu as in Step 2.
7. Use the right arrow key to navigate from NAMES to MATH.
8. Choose option 1: "det(" and press enter.
9. Return to the matrix menu, select the letter of the matrix you chose earlier and press enter
10. Close the parentheses and press enter again to get the determinant.
### Matrices in Testing 📝
On the ACT, you may need to calculate the product or determinant of a matrix. Check out Fiveable's ACT Math Section Review under our ACT Crams section for more on that!
On the SAT, you may need to add, subtract, multiply, or divide matrices in addition to finding their determinants. Luckily, Fiveable also has resources for SAT Math - see our SAT Crams for more. Overall, general matrix operations are fair game in these standardized tests.
Generally, you can use a test-approved graphing calculator to solve these questions like the TI-84. Matrices may also appear on the Calculus AB test.
Connect with other students studying Algebra 2 with Hours 🤝
|
# Right triangle
Right triangle
A right triangle (American English) or right-angled triangle (British English) is a triangle in which one angle is a right angle (that is, a 90-degree angle). The relation between the sides and angles of a right triangle is the basis for trigonometry.
The side opposite the right angle is called the hypotenuse (side c in the figure). The sides adjacent to the right angle are called legs (or catheti, singular: cathetus). Side a may be identified as the side adjacent to angle B and opposed to (or opposite) angle A, while side b is the side adjacent to angle A and opposed to angle B.
If the lengths of all three sides of a right triangle are integers, the triangle is said to be a Pythagorean triangle and its side lengths are collectively known as a Pythagorean triple.
## Principal properties
### Area
As with any triangle, the area is equal to one half the base multiplied by the corresponding height. In a right triangle, if one leg is taken as the base then the other is height, so the area of a right triangle is one half the product of the two legs. As a formula the area T is
where a and b are the legs of the triangle.
If the incircle is tangent to the hypotenuse AB at point P, then denoting the semi-perimeter (a + b + c) / 2 as s, we have PA = sa and PB = sb, and the area is given by
This formula only applies to right triangles.[1]
### Altitudes
Altitude of a right triangle
If an altitude is drawn from the vertex with the right angle to the hypotenuse then the triangle is divided into two smaller triangles which are both similar to the original and therefore similar to each other. From this:
• The altitude to the hypotenuse is the geometric mean (mean proportional) of the two segments of the hypotenuse.[2]:243
• Each leg of the triangle is the mean proportional of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.
In equations,
(this is sometimes known as the right triangle altitude theorem)
where a, b, c, d, e, f are as shown in the diagram.[3] Thus
Moreover, the altitude to the hypotenuse is related to the legs of the right triangle by[4][5]
For solutions of this equation in integer values of a, b, f, and c, see here.
The altitude from either leg coincides with the other leg. Since these intersect at the right-angled vertex, the right triangle's orthocenter—the intersection of its three altitudes—coincides with the right-angled vertex.
### Pythagorean theorem
Main article: Pythagorean theorem
The Pythagorean theorem states that:
In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).
This can be stated in equation form as
where c is the length of the hypotenuse, and a and b are the lengths of the remaining two sides.
Pythagorean triples are integer values of a, b, c satisfying this equation.
Illustration of the Pythagorean Theorem
The radius of the incircle of a right triangle with legs a and b and hypotenuse c is
The radius of the circumcircle is half the length of the hypotenuse,
Thus the sum of the circumradius and the inradius is half the sum of the legs:[6]
One of the legs can be expressed in terms of the inradius and the other leg as
## Characterizations
A triangle ABC with sides , semiperimeter s, area T, altitude h opposite the longest side, circumradius R, inradius r, exradii ra, rb, rc (tangent to a, b, c respectively), and medians ma, mb, mc is a right triangle if and only if any one of the statements in the following six categories is true. All of them are of course also properties of a right triangle, since characterizations are equivalences.
### Area
• where P is the tangency point of the incircle at the longest side AB.[11]
## Trigonometric ratios
The trigonometric functions for acute angles can be defined as ratios of the sides of a right triangle. For a given angle, a right triangle may be constructed with this angle, and the sides labeled opposite, adjacent and hypotenuse with reference to this angle according to the definitions above. These ratios of the sides do not depend on the particular right triangle chosen, but only on the given angle, since all triangles constructed this way are similar. If, for a given angle α, the opposite side, adjacent side and hypotenuse are labeled O, A and H respectively, then the trigonometric functions are
For the expression of hyperbolic functions as ratio of the sides of a right triangle, see the hyperbolic triangle of a hyperbolic sector.
## Special right triangles
The values of the trigonometric functions can be evaluated exactly for certain angles using right triangles with special angles. These include the 30-60-90 triangle which can be used to evaluate the trigonometric functions for any multiple of π/6, and the 45-45-90 triangle which can be used to evaluate the trigonometric functions for any multiple of π/4.
### Kepler triangle
Let H, G, and A be the harmonic mean, the geometric mean, and the arithmetic mean of two positive numbers a and b with a > b. If a right triangle has legs H and G and hypotenuse A, then[13]
and
where is the golden ratio Since the sides of this right triangle are in geometric progression, this is the Kepler triangle.
## Thales' theorem
Main article: Thales' theorem
Median of a right angle of a triangle
Thales' theorem states that if A is any point of the circle with diameter BC (except B or C themselves) ABC is a right triangle where A is the right angle. The converse states that if a right triangle is inscribed in a circle then the hypotenuse will be a diameter of the circle. A corollary is that the length of the hypotenuse is twice the distance from the right angle vertex to the midpoint of the hypotenuse. Also, the center of the circle that circumscribes a right triangle is the midpoint of the hypotenuse and its radius is one half the length of the hypotenuse.
## Medians
The following formulas hold for the medians of a right triangle:
The median on the hypotenuse of a right triangle divides the triangle into two isosceles triangles, because the median equals one-half the hypotenuse.
The medians ma and mb from the legs satisfy[6]:p.136,#3110
## Euler line
In a right triangle, the Euler line contains the median on the hypotenuse—that is, it goes through both the right-angled vertex and the midpoint of the side opposite that vertex. This is because the right triangle's orthocenter, the intersection of its altitudes, falls on the right-angled vertex while its circumcenter, the intersection of its perpendicular bisectors of sides, falls on the midpoint of the hypotenuse.
## Inequalities
In any right triangle the diameter of the incircle is less than half the hypotenuse, and more strongly it is less than or equal to the hypotenuse times [14]:p.281
In a right triangle with legs a, b and hypotenuse c,
with equality only in the isosceles case.[14]:p.282,p.358
If the altitude from the hypotenuse is denoted hc, then
with equality only in the isosceles case.[14]:p.282
## Other properties
If segments of lengths p and q emanating from vertex C trisect the hypotenuse into segments of length c/3, then[2]:pp. 216–217
The right triangle is the only triangle having two, rather than one or three, distinct inscribed squares.[15]
Let h and k (h > k) be the sides of the two inscribed squares in a right triangle with hypotenuse c. Then
These sides and the incircle radius r are related by a similar formula:
The perimeter of a right triangle equals the sum of the radii of the incircle and the three excircles:
## References
1. Di Domenico, Angelo S., "A property of triangles involving area", Mathematical Gazette 87, July 2003, pp. 323-324.
2. Posamentier, Alfred S., and Salkind, Charles T. Challenging Problems in Geometry, Dover, 1996.
3. Wentworth p. 156
4. Voles, Roger, "Integer solutions of ," Mathematical Gazette 83, July 1999, 269–271.
5. Richinick, Jennifer, "The upside-down Pythagorean Theorem," Mathematical Gazette 92, July 2008, 313–317.
6. Inequalities proposed in “Crux Mathematicorum, .
7. Triangle right iff s = 2R + r, Art of problem solving, 2011
8. Andreescu, Titu and Andrica, Dorian, "Complex Numbers from A to...Z", Birkhäuser, 2006, pp. 109-110.
9. Properties of Right Triangles
10. CTK Wiki Math, A Variant of the Pythagorean Theorem, 2011, .
11. Darvasi, Gyula (March 2005), "Converse of a Property of Right Triangles", The Mathematical Gazette, 89 (514): 72–76.
12. Bell, Amy (2006), "Hansen's Right Triangle Theorem, Its Converse and a Generalization" (PDF), Forum Geometricorum, 6: 335–342
13. Di Domenico, A., "The golden ratio — the right triangle — and the arithmetic, geometric, and harmonic means," Mathematical Gazette 89, July 2005, 261. Also Mitchell, Douglas W., "Feedback on 89.41", vol 90, March 2006, 153-154.
14. Posamentier, Alfred S., and Lehmann, Ingmar. The Secrets of Triangles. Prometheus Books, 2012.
15. Bailey, Herbert, and DeTemple, Duane, "Squares inscribed in angles and triangles", Mathematics Magazine 71(4), 1998, 278-284.
|
Solve Linear Equations with Rational Numbers
Solve Linear Equations
(with Rational Numbers)
When solving linear equations, the goal is to isolate the unknown (variable) to find its value. This is accomplished by adding, subtracting, multiplying, or dividing. Let’s start with a simple example:
x + 4 = 9
The main rule to keep in mind when trying to isolate a variable is that you always have to keep the equation balanced. This means that whatever you do to one side of the equation, you must also do to the other side. In the equation above, we need to get rid of the "+4" on the left side of the equation. We do this by using the opposite operation, which is subtraction. So we subtract 4, which cancels out the "+4." The x is now isolated, but we must also subtract 4 from the right side:
x + 4 = 9
- 4 - 4
x = 5
For a more complicated equation, there might be more steps. Let us take the example below:
3x - 9 = x + 3
When you have multiple operations involved in a problem, you must always do addition and subtraction first. Multiplication and division come second. You have a few options for what you do first with this problem. Let us combine the x terms first. We can subtract x from each side to get:
3x - 9 = x + 3
- x - x
2x - 9 = 3
Then we will need to add 9 to each side:
2x - 9 = 3
+ 9 + 9
2x = 12
Finally we need to isolate x. 2x is essentially 2 times x. To cancel this out, we use the opposite operation, which is division. We do this by dividing by 2.
2x = 12
2 2
x = 6
You can always check your solution be plugging it back into the original equation.
3x + 9 = x + 3
3(6) + 9 = 6 + 3
18 + 9 = 9
9 = 9
Example 1:
First add 6 to each side to get
Then you need to get rid of the fractions by multiplying each side by 3.
x = 2x + 30
-2x -2x
-x = 30
To make the x positive, we multiply each side by -1. So the answer becomes
x = -30
Example 2:
7y + 5 - y + 1 = 2y - 6
This equation has like terms on one side of the equation. You should combine them before beginning to isolate the variable. So the equation becomes
6y + 6 = 2y - 6
- 6 - 6
6y = 2y - 12
Then subtract 2y from each side and then divide both sides by 4:
4y = -12
4 4
y = -3
Practice Problems:
1.)
2.)
3.)
4.)
1.) x = 4
2.) y = 5
3.) x = 6
4.) x = 10
|
# Difference between revisions of "2021 AMC 12A Problems/Problem 13"
## Problem
Of the following complex numbers $z$, which one has the property that $z^5$ has the greatest real part?
$\textbf{(A) }-2 \qquad \textbf{(B) }-\sqrt3+i \qquad \textbf{(C) }-\sqrt2+\sqrt2 i \qquad \textbf{(D) }-1+\sqrt3 i\qquad \textbf{(E) }2i$
## Solution 1 (Degrees)
First, $\textbf{(B)} = 2\text{cis}(150), \textbf{(C)} =2\text{cis}(135)$$, \textbf{(D)} =2\text{cis}(120)$.
Taking the real part of the 5th power of each we have:
$\textbf{(A): }(-2)^5=-32$,
$\textbf{(B): }32\cos(650)=32\cos(30)=16\sqrt{3}$
$\textbf{(C): }32\cos(675)=32\cos(-45)=16\sqrt{2}$
$\textbf{(D): }32\cos(600)=32\cos(240)$ which is negative
$\textbf{(E): }(2i)^5$ which is zero
Thus, the answer is $\boxed{\textbf{(B)}}$. ~JHawk0224
## Solution 2 (Radians)
For every complex number $z=a+bi,$ where $a$ and $b$ are real numbers and $i=\sqrt{-1},$ its magnitude is $|z|=\sqrt{a^2+b^2}.$ For each answer choice, we get that the magnitude is $2.$
We rewrite each answer choice to the polar form $z=re^{i\theta}.$ By De Moivre's Theorem, the real part of $z^5$ is $$\mathrm{Re}\left(z^5\right)=r^5\cos{(5\theta)}.$$ We construct a table as follows: $$\begin{array}{c|ccc|cclclclcc} & & & & & & & & & & & & \\ [-2ex] \textbf{Choice} & & \boldsymbol{\theta} & & & & & & \hspace{0.75mm} \boldsymbol{\mathrm{Re}\left(z^5\right)} & & & & \\ [0.5ex] \hline & & & & & & & & & & & & \\ [-1ex] \textbf{(A)} & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] \textbf{(B)} & & \frac{5\pi}{6} & & & &32\cos{\frac{25\pi}{6}}&=&32\cos{\frac{\pi}{6}}&=&32\left(\frac{\sqrt3}{2}\right)& & \\ [2ex] \textbf{(C)} & & \frac{3\pi}{4} & & & &32\cos{\frac{15\pi}{4}}&=&32\cos{\frac{7\pi}{4}}&=&32\left(\frac{\sqrt2}{2}\right)& & \\ [2ex] \textbf{(D)} & & \frac{2\pi}{3} & & & &32\cos{\frac{10\pi}{3}}&=&32\cos{\frac{4\pi}{3}}&=&32\left(-\frac{1}{2}\right)& & \\ [2ex] \textbf{(E)} & & \frac{\pi}{2} & & & &32\cos{\frac{5\pi}{2}}&=&32\cos{\frac{\pi}{2}}&=&32\left(0\right)& & \\ [1ex] \end{array}$$ Clearly, the answer is $\boxed{\textbf{(B) }-\sqrt3+i}.$
~MRENTHUSIASM
~ pi_is_3.14
Solution 5
~IceMatrix
|
Finding the Greatest Common Factor of 2 or 3 Numbers
FINDING THE GREATEST COMMON FACTOR OF $\,2\,$ or $\,3\,$ NUMBERS
LESSON READ-THROUGH
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
Thanks for your support!
• PRACTICE (online exercises and printable worksheets)
• This page gives an in-a-nutshell discussion of the concepts.
Want more details, more exercises? Read the full text!
EXAMPLE:
Question: Find the greatest common factor of $\,27\,$ and $\,18\,$:
Answer: $\,9$
The idea:
The factors of $\,27\,$ are: $\,1\,$ $\,3\,$ $\,9\,$ $\,27$ The factors of $\,18\,$ are: $\,1\,$ $\,2\,$ $\,3\,$ $\,6\,$ $\,9\,$ $\,18\,$
The common factors of $\,27\,$ and $\,18\,$ (the numbers that appear in both lists) are $\,1\,$, $\,3\,$, and $\,9\,$.
The greatest common factor (the greatest number in the list of common factors) is $\,9\,$.
EFFICIENT ALGORITHM FOR FINDING THE GREATEST COMMON FACTOR
Here's an efficient algorithm for finding the greatest common factor,
when there aren't too many numbers, and they aren't too big.
The process is illustrated by finding the greatest common factor of $\,18\,$, $\,36\,$, and $\,90\,$:
Line up the numbers in a row. (See the purple rectangle at left.) Find ANY number that goes into everything evenly (like $\,2\,$). Do the divisions, and write the results above the original numbers. In the example: $18\,$ divided by $\,2\,$ is $\,9\,$; write the $\,9\,$ above the $\,18\,$ $36\,$ divided by $\,2\,$ is $\,18\,$; write the $\,18\,$ above the $\,36\,$, and so on. Keep repeating the process, until there isn't any number (except $\,1\,$) that goes into everything evenly. Multiply the circled numbers together. This is the greatest common factor! In the example, $\,\text{gcf}(18,36,90) = 2\cdot 3\cdot 3 = 18\,$.
WHY DOES THIS METHOD WORK?
Think about why this method works.
As you walk through each step of this discussion, keep comparing with the chart above.
Look at the prime factorizations of each number:
$18 = 2\cdot3\cdot 3$
$36 = 2\cdot 2\cdot 3\cdot 3$
$90 = 2\cdot3\cdot 3\cdot 5$
The number $\,2\,$ goes into each evenly, so separate it off:
$18 = 2\cdot (3\cdot 3)\hphantom{\cdot 3} = 2\cdot 9$
$36 = 2\cdot (2\cdot 3\cdot 3) = 2\cdot 18$
$90 = 2\cdot (3\cdot 3\cdot 5) = 2\cdot 45$
The number $\,3\,$ goes into each remaining part evenly, so separate it off:
$18 = (2\cdot 3) \cdot (3) \hphantom{\cdot 3}= (2\cdot 3)\cdot 3$
$36 = (2\cdot 3) \cdot (2\cdot 3) = (2\cdot 3)\cdot 6$
$90 = (2\cdot 3) \cdot (3\cdot 5) = (2\cdot 3)\cdot 15$
Here's the final step:
$18 = (2\cdot 3\cdot 3)\cdot 1$
$36 = (2\cdot 3\cdot 3)\cdot 2$
$90 = (2\cdot 3\cdot 3)\cdot 5$
OTHER WAYS TO APPLY THE ALGORITHM
Here are some other ways the algorithm might be applied.
Of course, you get the same answer any correct way that you do it!
You can also zip over to wolframalpha.com and type in: gcd(18,36,90)
The ‘gcd’ stands for greatest common divisor, which is another name for greatest common factor.
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Finding the Greatest Common Factor of Variable Expressions
Find the greatest common factor of:
|
# Support Forums
Full Version: Lesson 2 - Intro to Derivatives
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Lesson 2 - Intro to Derivatives
The second important concept is that of the derivative. Before calculus, we are mostly focused on finding the slope of a line, or the rate of change. In calculus, however, we are more concerned with instantaneous rate of change, or the slope of a curve. Therefore, the derivative at a point is also the slope of the tangent line at that point. With this concept we can graph most equations and lead into more complex concepts.
Finding the Derivative
The simplest way to find the derivative is using a simple equation.
Code:
`f′(x) = n*x^(n-1)`
So, the derivative of:
Code:
```f(x) = 2x^3 is f′(x) = 6x^2```
It is important to note that the derivative of a constant is 0.
Using the Derivative
It may seem like the concept is not that helpful, but the truth is the opposite. With the derivative we can find several things.
Let's say we have the equation.
Code:
`x^3 + x^2 + 4x + 1`
How many minima and maxima does the equation have? To figure this out we would find the first derivative. There is also the question of decreasing or increasing. If f′(x) is negative, the graph is decreasing. If it is positive, it is increasing.
Code:
`f′(x) = 3x^2 + 2x + 4`
We could then find the zeros of the first derivative, giving us the points where the original equation turns from decreasing to increasing and vice-versa.
Code:
```f′(x) = (3x + 1)(x + 1) f′(0) = 1/3 and f′(0) = -1```
So, the graph has an extrema (min or max) at x = 1/3 and x = -1.
What about turning points and concavity? For that we need a higher derivative. If the second derivative (f′′(x)) is negative, the graph is concave down. If it is positive it is concave up. Points of inflection are defined where the second derivative is 0.
Code:
`f′′(x) = 6x + 2`
The graph has a point of inflection at -1/3.
|
Share
Books Shortlist
# In the Following Figure, O is the Centre of the Circle, ∠Aob = 60° and ∠Bdc = 100° Find ∠Obc. - ICSE Class 10 - Mathematics
ConceptArc and Chord Properties - the Angle that an Arc of a Circle Subtends at the Center is Double that Which It Subtends at Any Point on the Remaining Part of the Circle
#### Question
In the following figure, O is the centre of the circle,
∠AOB = 60° and ∠BDC = 100° Find ∠OBC.
#### Solution
Here, ∠ACB =1/2 ∠AOB=1/2 xx 60° = 30°
(Angle at the centre is double the angle at the circumference subtended by the same chord)
By angle sum property of ΔBDC,
∴ ∠DBC =180° - 100°- 30° = 50°
Hence, ∠OBC =50°
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution In the Following Figure, O is the Centre of the Circle, ∠Aob = 60° and ∠Bdc = 100° Find ∠Obc. Concept: Arc and Chord Properties - the Angle that an Arc of a Circle Subtends at the Center is Double that Which It Subtends at Any Point on the Remaining Part of the Circle.
S
|
# `int sin(-7x)cos(6x) dx` Find the indefinite integral
Indefinite integrals are written in the form of` int f(x) dx = F(x) +C`
where: `f(x)` as the integrand
`F(x)` as the anti-derivative function
`C` as the arbitrary constant known as constant of integration
For the given problem `int sin(-7x)cos(6x) dx` or `intcos(6x)sin(-7x) dx` has a integrand in a form of trigonometric function. To evaluate this, we apply the identity:
`cos(A)sin(B) =[sin(A+B) -sin(A-B)]/2`
The integral becomes:
`intcos(6x)sin(-7x) dx = int[sin(6x+(-7x)) -sin(6x-(-7x))]/2dx`
`= int[sin(6x-7x) -sin(6x+7x)]/2dx`
Apply the basic properties of integration: `int c*f(x) dx= c int f(x) dx` .
`int[sin(6x-7x) -sin(6x+7x)]/2dx= 1/2int[sin(6x-7x) -sin(6x+7x)]dx`
Apply the basic integration property: `int (u+v) dx = int (u) dx + int (v) dx` .
`1/2 *[int (sin(6x-7x))dx - int sin(6x+7x)dx]`
Then apply u-substitution to be able to apply integration formula for cosine function:` int sin(u) du=-cos(u) +C` .
For the integral: `intsin(6x-7x)dx` , we let `u = 6x-7x=-x` then `du= - dx` or `(-1)du =dx` .
`intsin(6x-7x)dx=intsin(-x) dx`
`=intsin(u) *(-1)du`
`=(-1) int sin(u)du`
`=(-1)(-cos(u) )+C`
`=cos(u) +C`
Plug-in `u =-x` on `cos(u) +C` , we get:
`intsin(6x-7x)dx= cos(-x) +C`
For the integral: `intsin(6x+7x)dx` , we let `u = 6x+7x=13x` then `du= 13 dx` or `(du)/13 =dx` .
`intsin(6x+7x)dx=intsin(13x) dx`
` =intsin(u) *(du)/13`
` = 1/13 int sin(u)du`
`= 1/13( -cos(u))+C or -1/13cos(u) +C`
Plug-in `u =13x` on `-1/13 cos(u) +C` , we get:
`intsin(6x+7x)dx= -1/13 cos(13x) +C`
Combing the results, we get the indefinite integral as:
`intsin(-7x)cos(6x) dx= 1/2*[ cos(-x) -(-1/13 cos(13x))] +C`
or `1/2 cos(-x) +1/26 cos(13x) +C`
Since cosine is an even function, `cos(-x) = cos(x)` , so we get:
`intsin(-7x) cos(6x)dx=1/2 cos(x) +1/26 cos(13x) +C`
Approved by eNotes Editorial Team
|
# ACT Math : How to find the solution to an inequality with multiplication
## Example Questions
### Example Question #1 : How To Find The Solution To An Inequality With Multiplication
Let x be a number such that x > 1 and x < 0. Let y be a number such that y > 0. Which of the following must be true?
x + y > 0
(x + y)2 > 1
1 < xy < 1
(x/y) < 1
(xy)2 > 1
(x/y) < 1
Explanation:
Let us see if we can find counterexamples for each choice so that we can eliminate it.
Let's look at the choice x+y > 0. We could let -0.75, and we could let y = 0.5, and this would satisfy the conditions for x and y. If x is -0.75 and y is 0.5, then x + y = -0.25 < 0, so this choice doesn't have to be true.
Let's then look at the choice -1 < xy < 1. If we let y be 10 and x be -0.5 then xy would be -5. This means that this choice isn't neccessarily true.
Let's look at the choice (xy)2 > 1. We could let x be -0.5 and we could let y be 1. Then xy would be -0.5, and (xy)2 would be 0.25 < 1. This means that this choice isn't always true.
Let's look at the choice (x + y)2 > 1. We could let y = 0.5 and x = -0.5. Then x + y would equal 0, and (x+y)2 = 0 < 1. We can eliminate this choice as well.
We suspect that (x/y) < 1 might be the answer, because we can contradict every other statement. However, let's see if we can prove that (x/y) has to be less than 1.
(x/y) < 1
We can multiply both sides by y, because y is positive, and this won't change the inequality sign. After multiplying both sides by y, we would have
x < y
Since y is always a positive number, and since x is always a negative number, this means that y will always be greater than x, so x < y must always be true.
### Example Question #1 : How To Find The Solution To An Inequality With Multiplication
If the inequality is true, then which of the following must be true?
Explanation:
If we don't know anything for sure.
could be less than , such as and .
could also be greater than , such as and .
In both of these cases, .
cannot be true.
The other three choices COULD be true, but do not HAVE to be true.
### Example Question #3 : How To Find The Solution To An Inequality With Multiplication
Which of the following is the solution set of ?
Explanation:
Remember that when you multiply both sides of the inequality by a negative number you must switch the inequality sign around (less than becomes greater than; greater than becomes less than)
### Example Question #221 : Gre Quantitative Reasoning
If –1 < n < 1, all of the following could be true EXCEPT:
16n2 - 1 = 0
n2 < 2n
(n-1)2 > n
|n2 - 1| > 1
n2 < n
|n2 - 1| > 1
Explanation:
### Example Question #3 : How To Find The Solution To An Inequality With Multiplication
(√(8) / -x ) < 2. Which of the following values could be x?
-1
-2
All of the answers choices are valid.
-4
-3
-1
Explanation:
The equation simplifies to x > -1.41. -1 is the answer.
Solve for x
Explanation:
### Example Question #5 : How To Find The Solution To An Inequality With Multiplication
We have , find the solution set for this inequality.
Explanation:
### Example Question #1 : How To Find The Solution To An Inequality With Multiplication
Fill in the circle with either , , or symbols:
for .
The rational expression is undefined.
None of the other answers are correct.
Explanation:
Let us simplify the second expression. We know that:
So we can cancel out as follows:
### Example Question #1 : How To Find The Solution To An Inequality With Multiplication
Solve the following inequality:
Explanation:
To solve simply solve as though it is an equation.
The goal is to isolate the variable on one side with all other constants on the other side. Perform the opposite operation to manipulate the inequality.
However, remember that when dividing or multiplying by a negative number, you must flip the inequality sign.
Multiply by -3, thus:
### Example Question #1 : How To Find The Solution To An Inequality With Multiplication
Solve the following inequality:
|
# How do you solve the system of equations 2x+2y=4 and 12-3x=3y?
Nov 16, 2017
No solutions
#### Explanation:
$2 x + 2 y = 4$
$- 3 x + 12 = 3 y$
We need to solve $2 x + 2 y = 4$ for $x$
Let's start by adding $\textcolor{red}{- 2 y}$ to both sides
$2 x + 2 y \textcolor{red}{- 2 y} = 4 \textcolor{red}{- 2 y}$
$2 x = - 2 y + 4$
$x = \frac{- 2 y + 4}{2}$
$x = - y + 2$
Then, substitute $- y + 2$ for $x$ in $- 3 x + 12 = 3 y$
$- 3 x + 12 = 3 y$
$- 3 \left(- y + 2\right) + 12 = 3 y$
$3 y - 6 + 12 = 3 y$
$3 y + 6 = 3 y$
Then add $- 3 y$ to both sides
$3 y - 3 y + 6 = 3 y - 3 y$
$6 = 0$
Finally, add $- 6$ to both sides
$6 - 6 = 0 - 6$
$0 = - 6$
Thus,
There is no solutions.
Nov 16, 2017
No solution
#### Explanation:
Rewriting the second equation gives $3 x + 3 y = 12$. Dividing each term by $3$ then yields $x + y = 4$.
For the first equation $2 x + 2 y = 4$, we can divide each term by $2$, yielding $x + y = 2$.
Then, we have that $x + y = 2$ and $x + y = 4$. But how could this ever be the case? $x + y$ should always give the same value for a fixed $x$ and fixed $y$. Thus there are no solutions.
|
Explore BrainMass
Share
# Algebra: Linear Equations
This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!
Section 6.1: Rational Expression
Which numbers cannot be used in place of the variable in each rational expression?
18. 2y+1
y² - y- 6
Section 6.2: Multiplication and Division
Perform the indicated operation
24. 3 * 2a+ 2
a² + a 6
Section 6.3: Finding the Least Common Denominator
Find the LCD for the given rational expressions, and convert each rational expression into an equivalent rational expression with the LCD as the denominator.
20. 4 * 5x
x-y 2y - 2x
Section 6.4: Addition and Subtraction of rational numbers
Perform the indicated operation. Reduce each answer to lowest terms.
8. 4 - 1
9 9
Section 6.5: Complex Fractions
Simplify each complex fraction. Reduce each answer to lowest terms.
52. 1 + 1
9 3x
X - 1
9 x
Section 6.6: Solving Equations with Rational Expressions
Solve each equation
30. x = 4
3 x + 1
Discussion Questions
DQ 1: I recently read an obituary of Paul MacCready, the father of human powered aircraft. He invented the Gossamer Albatross, an aircraft which successfully flew the English Channel under human power (a cyclist turning the propeller). The aircraft flew from Folkestone, England to Cap Griz Nez, France, a distance of 22 miles, in approximately 3 hours fighting a headwind. On the return trip, this headwind would become a tailwind. Let us assume that the cyclist can pedal at a rate of 12 mi / hr without wind interference. Exhibit a rational expression for the round trip time, Folkestone to Cap Griz Nez and back, as a function of wind speed. What would be the domain of this rational expression?
DOMAIN OF A RATIONAL EXPRESSION
So, given a rational expression, the first issue which we may consider is to determine the domain of the expression. For instance, x-1 = 1 / x constitutes a rational expression, which has a domain consisting of all x ≠ 0. What would be the domain of the rational expression x = x / 1? Ans: the domain in this case is all values of x ( the real line ), as the denominator is never 0. For the case of a polynomial denominator in our rational expression, we note that the domain of the expression would be all values of x which do not cause our polynomial denominator to be zero. As an example, for our rational expression
1 / (2x + 1) we note that the zero of our linear (first degree) expression (2x + 1) is x = -1/2. Thus, the domain of this rational expression constitutes all values of x satisfying x ≠ -1/2.
DQ 2: What is the domain of the rational expression 1 / (x2 + 2x - 15 ) ?
DQ 3: What is the domain of the rational expression 1 / (x2 + 1), where x is considered to be a real number (imaginary or complex numbers not included in the domain) ?
SIMPLIFICATION OF RATIONAL EXPRESSIONS
As in the case of fractions, the simplification occurs by determining all prime factors of the numerator and denominator and deleting those prime factors which are common to both the numerator and denominator. As an example, we wish to simplify the rational expression (x2 + 2x + 1 ) / (x + 1).
Knowing that the numerator can be expressed as a perfect square, our rational expression becomes
(x + 1)2 / (x + 1) which, of course. reduces to the binomial x+1.
Of course, there is the issue that our original rational expression has as its domain all values of x for which x ≠ -1 and our final simplified rational expression is valid for all x. Technically, we would specify that the domain for our simplified expression would be the same as the domain of our original expression, that is all values of x for which x ≠ -1. Note that, if x = -1, then the original expression would be equivalent to the expression 0/0, which is undefined. An issue to remember is that, as in the case of integer fractions, only factors can be canceled from the numerator and denominator of a rational expression, so that factoring of our polynomial expressions becomes key to simplifying a rational expression.
DQ 4: Simplify the rational expression (2x2 + 13x + 20 ) / (2x2 + 17x + 30 ) . And, determine the domain of the rational expression.
LEARNING TEAM EXERCISE
Part I
Given the following two rational expressions
1 / (x2 - 1 ) , 1 / (x - 1)2
i determine the least common denominator (LCD) of the two expressions
ii expand each rational expression to its equivalent form having the LCD as its denominator
iii Add the two resulting rational expressions, simplify if possible, and determine the domain of the resulting rational expression
Part II
x2 + x - 1 = 0.
i Determine the roots of the equation
ii Verify each root by plugging the root back into the equation and performing the computation
https://brainmass.com/math/linear-algebra/algebra-linear-equations-226698
#### Solution Summary
Operations with fractions, finding LCD
\$2.19
Similar Posting
## Algebra - Linear equations.
1. Equations of Lines in Slope-Intercept Form
Find the slope and y-intercept for each line.
3x + 5y + 10 = 0
2. Write each equation in slope-intercept form
y- 5 = - ¾ (x+1)
3. Interest rates. A credit manager rates each applicant for a
car loan on a scale of 1 through 5 and then determines the interest rate from the accompanying table. Find the equation of the line in slope-intercept form that goes through these points
Credit Interest
Rating Rate (%)
1 24
2 20
3 16
4 12
5 8
4. Find the equation of each line in slope-intercept form
The line through (4, -5) that is perpendicular to x = 1
View Full Posting Details
|
### Math with Scientific Notation Addition and Subtraction
Speaking realistically, the problems discussed below can all be done on a calculator. However, you need to know how to enter values into the calculator, read your calculator screen, and round off to the proper number of significant figures. Your calculator will not do these things for you.
All exponents MUST BE THE SAME before you can add and subtract numbers in scientific notation. The actual addition or subtraction will take place with the numerical portion, NOT the exponent.
The student might wish to re-read the above two sentences with emphasis on the emphasized portions.
It might be advisable to point out again - DO NOT, under any circumstances, add the exponents.
Example #1: 1.00 x 103 + 1.00 x 102
A good rule to follow is to express all numbers in the problem in the highest power of ten.
Convert 1.00 x 102 to 0.10 x 103, then add:
``` 1.00 x 103
+ 0.10 x 103
= 1.10 x 103
```
Example #2: The significant figure issue is sometimes obscured when numbers are in scientific notation. For example, add the following four numbers:
(4.56 x 106) + (2.98 x 105) + (3.65 x 104) + (7.21 x 103)
When the four numbers are written in the highest power, we get:
``` 4.56 x 106
0.298 x 106
0.0365 x 106
+ 0.00721 x 106
= 4.90171 x 106
```
The answer upon adding must be rounded to 2 significant figures to the right of the decimal point, thus giving 4.90 x 106 as the correct answer.
Generally speaking, you can simply enter the numbers into the calculator and let the calculator keep track of where the decimal portion is. However, you must then round off the answer to the correct number of significant figures.
Lastly, be warned about using the calculator. Students often push buttons without understanding the math behind what they are doing. Then, when the teacher questions their work, they say "Well, that's what the calculator said!" As if the calculator is to blame for the wrong answer. Remember, it is your brain that must be in charge and it is you that will get the points deducted for poor work, not the calculator.
Practice Problems
1) (4.52 x 10¯5) + (1.24 x 10¯2) + (3.70 x 10¯4) + (1.74 x 10¯3)
2) (2.71 x 106) - (5.00 x 104)
Reminder: you must have the same exponent on each number of the problem.
|
# Prove that the points A
Question:
Prove that the points A(1, 4), B(3, - 2) and C(4, - 5) are collinear. Also, find the equation of the line on which these points lie.
Solution:
If two lines having the same slope pass through a common point, then two lines will coincide. Hence, if A, B and C are three points in the XY - plane, then they will lie on a line, i.e., three points are collinear if and only if slope of AB = slope of BC.
Slope of AB = slope of BC
$\frac{-2-4}{3-1}=\frac{-5-(-2)}{4-3} \Rightarrow \frac{-6}{2}=\frac{-3}{1}$
$-3=-3$
Hence verified, i.e. points are collinear. Now using two point form of the equation
$\mathrm{y}-\mathrm{y}_{1}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right)$ where $\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}=$ slope of line
$y-4=-3(x-1)$
$y-4+3 x-3=0$
$3 x+y-7=0$
So, required equation of line is 3x + y - 7.
|
# HCL Placement Paper | Quantitative Aptitude Set – 1
This is an HCL model paper for Quantitative Aptitude. This placement paper will cover aptitude that is asked in HCL placements and also strictly follows the pattern of questions asked in HCL papers. It is recommended to solve each one of the following questions to increase your chances of clearing the HCL placement.
1. Find the greatest number that will divide 355, 54 and 103 so as to leave the same remainder in each case.
1. 4
2. 7
3. 9
4. 13
```
7
```
Explanation:
Required number = H.C.F. of |a -b|, |b – c| and |c – a| = H.C.F. of |355 – 54|, |54 – 103| and |103 – 355| = 301, 49, 252 = 7
2. Six bells commence tolling together and toll at intervals of 3, 6, 9, 12, 15 and 18 seconds respectively. In 60 minutes, how many times do they toll together ?
1. 10
2. 20
3. 21
4. 25
```
21
```
Explanation:
L.C.M. of 3, 6, 9, 12, 15 and 18 is 180. So, the bells will toll together after every 180 seconds(3 minutes). In 60 minutes, they will toll together (60/3)+1 = 21 times.
3. The smallest 5 digit number exactly divisible by 11 is:
1. 11121
2. 11011
3. 10010
4. 11000
```
10010
```
Explanation:
The smallest 5-digit number 10000. 10000 when divided by 11, leaves a remainder of 1 Hence add (11 – 1) = 10 to 10000 Therefore, 10010 is the smallest 5 digit number exactly divisible by 11
1. 474
2. 534
3. 500
4. 368
```
474
```
Explanation:
As Therefore the given expression = (121 + 353) = 474
4. What decimal of 10 hours is a minute?
1. 0.025
2. 0.256
3. 0.0027
4. 0.00126
```
0.0027
```
Explanation:
Decimal of 10 hours in a minute = 10 / (60 x 60) = 0.0027
5. ‘A’ can do a work in 10 days and ‘B’ in 15 days. If they work on it together for 3 days, then the work that is left is :
1. 10%
2. 20%
3. 40%
4. 50%
```
50%
```
Explanation:
Let the total work to be done is, say, 30 units. A does the work in 10 days, So A’s 1-day work = (30 / 10) = 3 units B does the work in 15 days, So B’s 1-day work = (30 / 15) = 2 units Therefore, A’s and B’s together 1-day work = (3 + 2) = 5 units In 3 days, work done = 5 * 3 = 15 units amount of work left = 30 – 15 = 15 units Therefore the % of work left after 3 days = (15 / 30) * 100% = 50%
6. A pump can fill a tank with water in 1 hour. Because of a leak, it took 1.5 hours to fill the tank. The leak can drain all the water of the tank in:
1. 2 hours
2. 2.5 hours
3. 3 hours
4. 3.5 hours
```
3 hours
```
Explanation:
Pump fills the tank in 1 hour Time taken by Pump to fill due to leak = 1.5 hour Therefore, in 1 hour, the amount of tank that the Pump can fill at this rate = 1 / (1.5) = 2/3 Amount of water drained by the leak in 1 hour = (1 – (2/3)) = 1/3 Therefore, the tank will be completely drained by the leak in (1 / (1/3)) = 3 hours
7. 2 pipes A and B can fill a tank in 20 minutes and 30 minutes respectively. Both pipes are opened. The tank will be filled in just 15 minutes, if the B is turned off after:
1. 5 min
2. 6.5 min
3. 7 min
4. 7.5 min
```
7.5 min
```
Explanation:
Let the total work to be done is, say, 60 units. A fills the tank in 20 minutes, So A’s 1-minute work = (60 / 20) = 3 units B fills the tank in 30 minutes, So B’s 1-minute work = (60 / 30) = 2 units Therefore, A’s and B’s together 1-minute work = (3 + 2) = 5 units Let the time when A and B both are opened be x minutes and Since the total time taken to fill the tank is 15 minutes Therefore, an expression can be formed as 5x + 3(15 – x) = 15 => x = 7.5 Therefore, the B is turned off after 7.5 minutes
8. In an IPL match, the current run rate of CSK is 4.5 in 6 overs. What should be the required run rate of CSK inorder to achieve the target of 153 against KKR?
1. 7
2. 8
3. 8.5
4. 9
```
9
```
Explanation:
Current run rate = 4.5 in 6 overs Runs already made = 4.5 * 6 = 27 Target = 153 Runs still required = 153 – 27 = 126 Overs left = 14 Therefore required run rate = 126 / 14 = 9
9. The average of 10 numbers is 0. Of them, how many can be smaller than zero, at most?
1. 0
2. 1
3. 9
4. 10
```
9
```
Explanation:
Let the 9 numbers be smaller than zero and let their sum be ‘s’ Now, in order to get the average 0, the 10th number can be ‘-s’ Therefore, average = (s + (-s))/10 = 0/10 = 0
10. Which is not the prime number?.
1. 43
2. 57
3. 73
4. 101
```
57
```
Explanation:
A positive natural number is called prime number if nothing divides it except the number itself and 1. 57 is not a prime number as it is divisible by 3 and 19 also, apart from 1 and 57.
11. If the average of four consecutive odd numbers is 12, find the smallest of these numbers?
1. 5
2. 7
3. 9
4. 11
```
9
```
Explanation:
Let the numbers be x, x+2, x+4 and x+6 Then (x + x + 2 + x + 4 + x + 6)/4 = 12 ∴ 4x + 12 = 48 ∴ x = 9
12. Two numbers are in the ratio of 2:9. If their H. C. F. is 19, numbers are:
1. 6, 27
2. 8, 36
3. 38, 171
4. 20, 90
```
38, 171
```
Explanation:
Let the numbers be 2X and 9X Then their H.C.F. is X, so X = 19 ∴ Numbers are (2×19 and 9×19) i.e. 38 and 171
13. HCF of two numbers is 11 and their LCM is 385. If the numbers do not differ by more than 50, what is the sum of the two numbers ?
1. 132
2. 35
3. 12
4. 36
```
132
```
Explanation:
Product of numbers = LCM x HCF => 4235 = 11 x 385 Let the numbers be of the form 11m and 11n, such that ‘m’ and ‘n’ are co-primes. => 11m x 11n = 4235 => m x n = 35 => (m, n) can be either of (1, 35), (35, 1), (5, 7), (7, 5). => The numbers can be (11, 385), (385, 11), (55, 77), (77, 55). But it is given that the numbers cannot differ by more than 50. Hence, the numbers are 55 and 77. Therefore, sum of the two numbers = 55 + 77 = 132
14. A person employed a group of 20 men for a construction job. These 20 men working 8 hours a day can complete the job in 28 days. The work started on time but after 18 days, it was observed that two-thirds of the work was still pending. To avoid penalty and complete the work on time, the employer had to employ more men and also increase the working hours to 9 hours a day. Find the additional number of men employed if the efficiency of all men is the same.
1. 40
2. 44
3. 64
4. 80
```
44
```
Explanation:
Let the total work be 3 units and additional men employed after 18 days be ‘x’. => Work done in first 18 days by 20 men working 8 hours a day = (1/3) x 3 = 1 unit => Work done in last 10 days by (20 + x) men working 9 hours a day = (2/3) x 3 = 2 unit Here, we need to apply the formula
```M1 D1 H1 E1 / W1 = M2 D2 H2 E2 / W2,
```
where M1 = 20 men D1 = 18 days H1 = 8 hours/day W1 = 1 unit E1 = E2 = Efficiency of each man M2 = (20 + x) men D2 = 10 days H2 = 9 hours/day W2 = 2 unit So, we have 20 x 18 x 8 / 1 = (20 + x) x 10 x 9 / 2 => x + 20 = 64 => x = 44 Therefore, number of additional men employed = 44
Previous
Next
|
Question Video: Using the Addition Rule to Determine the Probability of an Event Involving Mutually Exclusive Events Mathematics
Suppose 𝐴 and 𝐵 are two mutually exclusive events. Given that 𝑃(𝐵) = 4 𝑃(𝐴) and 𝑃(𝐴 ∪ 𝐵) = 0.95, find 𝑃(𝐵).
02:19
Video Transcript
Suppose 𝐴 and 𝐵 are two mutually exclusive events. Given that the probability of 𝐵 is four multiplied by the probability of 𝐴 and the probability of 𝐴 union 𝐵 is equal to 0.95, find the probability of 𝐵.
We begin by recalling our definition of mutually exclusive events. Two or more events are said to be mutually exclusive if they cannot happen at the same time. If these two events are 𝐴 and 𝐵, the probability of 𝐴 intersection 𝐵 is equal to zero. And the probability of 𝐴 union 𝐵 is equal to the probability of 𝐴 plus the probability of 𝐵. In this question, we are told the probability of 𝐴 union 𝐵 is equal to 0.95. We are also told that the probability of 𝐵 is equal to four times the probability of 𝐴. This gives us the equation 0.95 is equal to the probability of 𝐴 plus four multiplied by the probability of 𝐴. Simplifying the right-hand side, we have five multiplied by the probability of 𝐴.
We can then divide through by five to calculate the probability of 𝐴. This is equal to 0.19. The probability of 𝐵, which is what we are trying to calculate, is therefore equal to four multiplied by 0.19. This is equal to 0.76. We can represent this information on a Venn diagram. The probability of 𝐴 is 0.19 and the probability of 𝐵 0.76. As these sum to give us the union, which is equal to 0.95, the probability that an event is not in event 𝐴 and not in event 𝐵 is equal to 0.05.
|
# How do you integrate int cos^3(3x)dx?
Mar 9, 2017
$\frac{1}{4} \sin \left(3 x\right) + \frac{1}{36} \sin \left(9 x\right) + C$
#### Explanation:
This can be a tricky one, first we need to find a way to express the function to remove the power of 3.
To do this, we can try:
${\cos}^{3} \left(\theta\right) = \cos \left(\theta\right) \cdot {\cos}^{2} \left(\theta\right)$
Use ${\cos}^{2} \theta = \frac{1}{2} + \frac{1}{2} \cos 2 \theta$ to obtain:
$\cos \left(\theta\right) \left(\frac{1}{2} + \frac{1}{2} \cos 2 \theta\right)$
$= \frac{1}{2} \cos \left(\theta\right) + \frac{1}{2} \cos \left(\theta\right) \cos \left(2 \theta\right)$
Finally we can use: $\cos \left(A\right) \cos \left(B\right) = \frac{1}{2} \left\{\cos \left(A - B\right) + \cos \left(A + B\right)\right\}$ to rearrange the last term and get:
$\frac{1}{2} \cos \left(\theta\right) + \frac{1}{4} \cos \left(\theta - 2 \theta\right) + \frac{1}{4} \cos \left(\theta + 2 \theta\right)$
=$\frac{1}{2} \cos \left(\theta\right) + \frac{1}{4} \cos \left(- \theta\right) + \frac{1}{4} \cos \left(3 \theta\right)$
Of course, the cosine function has even symmetry so:
$\cos \left(- \theta\right) = \cos \left(\theta\right)$
Which will give us the exression:
$\frac{3}{4} \cos \left(\theta\right) + \frac{1}{4} \cos \left(3 \theta\right)$
So it will naturally follow that:
$\int {\cos}^{3} \left(3 x\right) \mathrm{dx} = \int \frac{3}{4} \cos \left(3 x\right) + \frac{1}{4} \cos \left(9 x\right) \mathrm{dx}$
Which easily integrates to give:
$\frac{1}{4} \sin \left(3 x\right) + \frac{1}{36} \sin \left(9 x\right) + C$
|
# Empirical Probability Formula
5/5 - (1 bình chọn)
Mục Lục
## Empirical Probability Formula
Empirical probability is also known as an experimental probability which refers to a probability that is based on historical data. In other words, simply we can say that empirical probability illustrates the likelihood of an event occurring based on historical data. In theoretical probability, we assume that the probability of occurrence of any event is equally likely, and based on that we predict the probability of an event.
The empirical probability formula can be obtained by multiplying the number of times an event occurs by the total number of trials. Let us understand the empirical probability formula using solved examples.
Note: Probability can be classified as
• Theoretical probability and
• Empirical probability
## What Is Empirical Probability Formula?
The empirical probability of an event is based on what has actually happened. On the other hand, the theoretical probability of the event attempts to predict what will happen on the basis of a total number of possible outcomes. If the number of trials in an experiment goes on increasing we may expect the experimental and theoretical probabilities to be nearly the same.
The formula for empirical probability is :
### Empirical Probability Formula = f/n
where,
• f is the number of times an event occurs
• n is the total number of trials
Let us understand the empirical probability formula using solved examples.
### Definition of Empirical Probability
Empirical probability can be defined as the estimator of probability based on experiences and observations. The main advantage of empirical probability is that the procedure is considered free of assumptions i.e. no data is assumed or no hypotheses but is backed by experimental studies and data. Hence, it is also called experimental probability or relative frequency.
## Examples Using Empirical Probability Formula
Example 1: In a group of 50 people, 32 people chose to order non-veg burgers over the veg. What is the empirical probability of someone ordering veg burgers?
Solution:
It is given that
Total number of people = 50
Number of people who chose non-veg burgers = 32
Number of people who chose veg burgers = 50 – 32 = 18
Hence,
As per empirical probability formula, it is = 18 / 50 = 0.36.
Therefore, the empirical probability of someone ordering veg burgers is 0.36 or 36%.
Example 2: A coin toss three times and the result was three heads. Using the empirical probability formula find out what is the empirical probability of getting a head?
Solution:
It is given that
Total number of trials = 3
Hence,
Empirical probability = 3 / 3 = 1.
Therefore, the empirical probability of getting a head is 1 or 100%.
Example 3: In a buffet, 90 out of 100 people chose to order coffee over tea. What is the empirical probability of someone ordering coffee?
Solution:
It is given that
Total number of people = 3
Number of people who choose coffee or tea = 90.
Number of people who choosing coffee or tea = 100 – 90 = 10
Hence,
As per empirical probability formula, it is = 10 / 100 = 0.10.
Therefore, the empirical probability of someone ordering coffee is 0.10 or 10%.
## FAQs on Empirical Probability
### What is Empirical Probability?
Empirical probability is also known as an experimental probability which refers to a probability that is based on historical data. The probability of the experiment will give a certain result. The main advantage of using the empirical probability formula is that the probability is backed by experimental studies and data.
### How Do You Find Empirical Probability?
The formula for empirical probability is:
Empirical Probability Formula = f/n
where,
• f is the number of times an event occurs
• n is the total number of trials
### What is Empirical and Theoretical Probability?
Empirical probability of any event is given the number of times that event occurred divided by the total number of incidents observed. Whereas a theoretical probability is the number of ways a particular event occurred divided by the total number of possible outcomes.
### What is the Difference Between Empirical and Experimental Probability?
Empirical probability is based on experiences whereas experimental is based on experiments. Both are the same type of probabilities.
## What is Empirical Probability?
Empirical probability, also known as experimental probability, refers to a probability that is based on historical data. In other words, empirical probability illustrates the likelihood of an event occurring based on historical data.
## Formula for Empirical Probability
Where:
• Number of Times Occurred refers to the number of times a favorable event occurred; and
• Total No. of Times Experiment Performed refers to the total amount of times the event was performed.
## Example of Theoretical Probability
#### Example 1
The table below shows a dice thrown three times and the corresponding result. What is the empirical probability of rolling a 4?
Empirical Probability = 0 / 3 = 0%. The empirical probability of rolling a 4 is 0%.
#### Example 2
The table below shows a coin toss three times and the corresponding result. What is the empirical probability of getting a head?
Empirical Probability = 3 / 3 = 100%. The empirical probability of getting a head is 100%.
#### Example 3
In a buffet, 95 out of 100 people chose to order coffee over tea. What is the empirical probability of someone ordering tea?
Empirical Probability = 5 / 100 = 5%. The empirical probability of someone ordering tea is 5%.
The main advantage of using empirical probability is that the probability is backed by experimental studies and data. It is free from assumed data or hypotheses. However, there are two big disadvantages of empirical probability to consider:
#### 1. Drawing incorrect conclusions
Using empirical probability can cause wrong conclusions to be drawn. For example, we know that the chance of getting a head from a coin toss is ½. However, an individual may toss a coin three times and get heads in all tosses. He may draw an incorrect conclusion that the chances of tossing a head from a coin toss are 100%.
#### 2. Insufficient sample size
Small sample sizes reduce accuracy. Therefore, large sample sizes are generally used for empirical probability to attain a good probability representation. For example, if an individual wanted to know the probability of getting a head in a coin toss but only used one sample, the empirical probability would be either 0% or 100%.
### Different Types of Probabilities
Apart from empirical probability, there are two other main types of probabilities:
#### 1. Classical probability
Classical probability (also called a priori or theoretical probability) refers to probability that is based on formal reasoning. For example, the classical probability of getting a head in a coin toss is ½.
#### 2. Subjective probability
Subjective probability refers to probability that is based on experience or personal judgment. For example, if an analyst believes that “there is an 80% probability that the S&P 500 will hit all-time highs in the next month,” he is using subjective probability.
## Empirical Probability: Definition, Formula and Examples
Empirical probability can be an effective metric to calculate when determining the likelihood of something occurring. Because you can rely on historical data about an occurrence, empirical probabilities can help you make more accurate assumptions about an event. Additionally, this statistical measure can be helpful in many financial, technical and business applications. In this article, we explore what empirical probability is, what formula to apply, what the process is for calculating it and how this ratio differs from theoretical probability.
## What is empirical probability?
In statistics and scientific research, empirical probability is analyzing and working with the data you collect from the research results of an outcome occurring during experimental trials. This probability is an estimate of an event occurring based on the frequency it occurs during experimental trials. Each observation you form when conducting your experiments or probability calculations becomes a distinct trial.
Statisticians, researchers, analysts and business and finance professionals may calculate the experimental probability of an event occurring to determine beneficial gains of innovation, investments and other business activities that can have potential risks alongside beneficial outcomes.
## What is the empirical probability formula?
To calculate the empirical probability of an event or outcome occurring, you can use the formula:
P(E) = (number of times an event occurs) ÷ (total number of trials)
The “P(E)” is the empirical (or experimental) probability, and the “number of times an event occurs” represents the number of times you achieve a specific outcome for each time you conduct a trial. The “total number of trials” represents how many times you perform your experiment, study or overall process to achieve the outcome you’re measuring.
For instance, if you want to measure the experimental probability of lightning striking the same location multiple times, you would first identify the number of times lightning has already struck the location and divide that value by the number of times you actually observe lightning striking the location. The result gives you the likelihood of a lightning strike in the same spot repeatedly when there’s a storm in the area.
## Calculating empirical probability
Understanding the relationship between a past event and its potential occurrence in the future can help you make important decisions relating to finance, investments or other business activities. You can apply the empirical probability formula by:
The empirical probability tells you the likelihood of an outcome occurring based on the probability of its past occurrences. Therefore, it’s important to determine the number of times you observe the event or outcome happening when you conduct your trials. For instance, if a financial analyst wants to determine the experimental probability of receiving a return on investment, they might count the number of times the specific financial instrument produced beneficial outcomes for past investors.
Using this example, assume the financial analyst determines the investment averages an annual return of \$250,000 and is measuring the experimental probability of similar returns over the next 10 years. If the investment instrument produced \$250,000 each year for the past seven years, the analyst determines that the outcome (the return of \$250,00) occurred seven times in the past. Using this information, the analyst applies the formula:
P(E) = (number of times the outcome occurs) ÷ (total number of trials) =
P(E) =(7) ÷ (total number of trials)
When you determine the number of times your desired result occurs or has occurred in the past, you can divide this value by the number of trials you perform in your research. For instance, in the example of the financial analyst, the number of trials may be the number of years they project to receive the average \$250,000 return. Using this example, if the analyst projects to receive the same return average over the next 10 years based on evaluations of the historical data, they can use this value to complete the formula:
P(E) = (number of times the outcome occurs) ÷ (total number of trials) =
P(E) = (7) ÷ (10) = 0.7 = 70%
This result indicates the empirical probability of the event (a \$250,000 return) occurring during the period the analyst measures is 70%. Depending on specific business goals, the analyst might recommend taking advantage of the investment opportunity because of the high probability of the recurrence of favorable investment returns.
## Theoretical vs. empirical probability
Unlike empirical probability, theoretical probability uses assumptions about a set of data from a larger population. Additionally, theoretical probability doesn’t require actual experimentation to calculate. Instead, you apply logical reasoning and what you know about a situation to measure the likelihood of an outcome occurring. Thus, the theoretical probability can only measure your expected outcomes against the number of all potential outcomes.
Empirical probabilities, however, rely on experimentation and direct observations to measure the potential of occurrences. This probability type also uses historical data rather than assumptions to form the values that make up the experimental probability formula. While calculating theoretical probability involves the same division processes as other probability formulas, you divide the number of favorable outcomes by the number of all possible outcomes.
## Example of calculating empirical probability
Tech-Driven Solutions, Inc. is creating a projection for its investment returns over the next five years so company executives can understand how the new investment plan can be beneficial. Financial analysts and planners at Tech-Driven Solutions apply an empirical analysis to determine the following financial statistics:
• Historical data about the investment instrument indicate average returns of \$300,500 annually.
• The investment instrument produced this average ROI for the past three years.
• Tech-Driven Solutions is measuring a period of five years.
Using this data, analysts determine that the number of times the return average occurs is for the past three years, and the total number of trials becomes the period analysts are forecasting. In this case, Tech-Driven Solutions wants to know the likelihood of the investment producing similar results over a five-year period, or for five “trials.” Financial analysts and planners use the empirical probability formula and the empirical analysis data:
P(E) = (number of times an event occurred) ÷ (total number of trials) =
P(E) = (3) ÷ (5) = 0.6 = 60%
The analysts and financial planners deduce that the investment opportunity has a 60% likelihood of producing an average annual return of \$300,500 over the next five years. Because the historical data indicate the frequency of this occurrence, the company can plan to achieve similar results over a specific period.
## Empirical Probability Formula
Empirical probability is an objective probability. It is also known as a relative frequency or experimental probability.
By definition, Empirical Probability is the number of outcomes in which a specified event occurs to the total number of trials.
Empirical probability is different from Theoretical probability on certain major aspects. That is, in theoretical probability, the probability is measured on the basis of the likeliness of an outcome. Whereas in the case of Empirical probability, the probability is based on how the event actually occurred during trials. The formula for Empirical probability is unlike a theoretical probability formula.
What is Empirical probability? To explain this let us take the word Probability, which means the number of times an event can happen. to make it more simply let us take an example say a dice. Dice is a cube in shape which has six faces with numbers 1 to 6 printed on each face of the cube. This means that only one face with come up each time the dice are rolled. Again, if we examine closely, out of Six faces of the dice only one face with one number printed on it will turn up each time the dice cube is rolled. This translates to 1 (face) divided by 6 (total number of faces) which is ⅙ = 0.1666. This is the chances or probability of any particular number coming up.
Now let us examine the word Empirical Probability. Again, going back to the dice, suppose we roll the dice say 120 times and want to estimate the number of times the number 6 would come up. Here from above, we know that in the case of a dice cube that when we roll the dice once, the chances that 6 will come up is 1/6. So, when we try rolling the dice 120 times, the probability of the number 6 coming up is 120 Empirical 1/6 which is equal to 20. This is the Empirical probability of the number 6 coming up when we cast the dice 120 times.
So, the Empirical probability of a particular event occurring may be stated or defined as the estimated chance of that particular event occurring in a total series of events; that is to state by expressing it in a formula it becomes:
Empirical (Experimental) Probability = Number of times an event occurs (in this case the number 6 turns up) × Total number of trails (in this case the total number of times that the dice is cast = 120 times.
Thus, the Empirical probability of the number 6 coming up in throws of 120 times of the dice is 120 × 1/6 =20.
Empirical Probability is also known as the Relative Frequency or Experimental Probability. It may be explained as follows: It is the ratio of the number of results in which a particular event takes place or happens to the number of actual trials made, not as a theoretical calculation but as per actual experimental observations. To put it concisely, the Empirical probability is a prediction derived from actual experimental observation.
Suppose an event say ‘A’ happens ‘m’ times out of a total of ‘n’ tries or trials conducted, then the relative frequency of ‘A’ is m/n.
Defining the term statistically, it is the scientific prediction or estimate of a probability. Modelling using a Binomial Distribution can be carried out in simple cases where the result of an actual experimental study only determines whether a particular event has happened or not happened. If it is carried out in this manner, it is called the maximum likelihood estimate. It is termed as the Bayesian Estimate for the same case if we make certain postulations or hypotheses for the prior distribution of the probability. In case a trial gives us more results or information or data, then the Empirical probability can be improved upon by assuming further data or hypotheses so as to form a statistical model. Then, if we use such a model then it can be used to derive a prediction or estimate of a particular event.
Thus, the probability may be broadly classified as
1. Theoretical Probability and
2. Empirical Probability
Probabilities of any particular event happening are always expressed in the range of numbers 0 to 1. If the Empirical probability of any particular event is zero (0), then it means the event never took place or occurred, and if it is the figure ONE (1) then it means it will always happen. Thus, if the probability figure is closer to 1, then the more is the likelihood of it happening and if the figure of the probability is closer to the figure Zero (0), then the less the likelihood of it happening. If the number of experiments or trials conducted goes on increasing, then the figures of Theoretical Probability and Experimental or Empirical probability will tend to be the same or will approach very close to each other.
### Theoretical Probability and Empirical and Probability – Pros and Cons
If we use Empirical Probability to estimate the probability, then the advantage of this method is that this is based on Actual experimental studies and it is significantly free of assumed data or hypotheses.
Let us study an example to illustrate this
Say, we are required to find the probability of a population or group of men to satisfy two conditions
(i) That they are above 6 feet in height and
(ii) They must prefer strawberry jam instead of saying, pineapple jam.
Then a direct Estimate can be arrived at only by actually counting the number of men satisfying both conditions to arrive at the Empirical Probability of the combined condition. Alternatively, an estimate may be arrived at by multiplying the number of men who are more than 6ft in height with the proportion of men who prefer strawberry jam instead of pineapple jam. But a word of caution, this type of estimation relies on the assumed data that the two conditions are statistically independent.
If we use Empirical probabilities, then the disadvantage is that it gives results pointing to estimating probabilities which are either very close to the figure Zero (0) or very close to the figure One (1). Here, it may be noted that very large sample sizes would be required in order to forecast or predict such probabilities to a fair degree of accuracy. We also see that statistical models can be of help, but then, it depends on the context, and broadly speaking, we may state that it shows better accuracy than Empirical probabilities if the assumed data are taken into consideration actually are reliable.
To understand this better, consider the situation, where, in an area, the minimum value of the daily maximum rainfall received in the summer month of May is less than 20mm. Then this probability can be arrived at from the data of earlier such recordings. Or in other words, a family of a probability distribution can be taken and fitted into the data sets of past year values. As a result, the fitted values will yield an alternative estimated value of the required probability. It must be noted that this substitute method can be relied upon to give an estimate of the probability even if all the values shown in the record are more than 20 mm.
### Mixed Classification or Nomenclature
The words ‘a-posteriori probability’ is a phrase that is also used as a substitute for Empirical probability or relative frequency. Its use is indicative of the terms used in Bayesian statistics, but it is not directly related to Bayesian results arrived at, where the same phrase is sometimes used to point to a posterior probability, which is completely different, even though it has a misleading similar name or indication.
The phrase ‘a-posteriori probability’ when taken to mean (or considered equivalent to) Empirical probability may be used in combination with the words ‘a priori probability’, and it means an estimate of a probability which is not based on any observed and recorded data, but simply on logical reasoning or what is arrived at by deduction.
Now if you are asked to predict with reasonable logic whether in a Cricket tournament the Pakistan team or the Australian team has a better chance of winning against the Indian Cricket team, then, it must be conclusively understood or inferred that there is no reliable method to come to a conclusion with a fair degree of accuracy by comparing the probable occurrence or chances of the two events taking place. The mathematicians of the age were incredibly fascinated by this theory of probability or chance of an event happening.
Card betting games and those played on stakes and gambling inevitably brought in the entry of mathematics and its applications into the field of predicting the probability or chances of winning or losing. It is believed that once in such a situation, a gambler sought the help of the famous mathematician Pierre de Fermat to improve his chances of winning. It is but a foregone conclusion that this led to the development of the theory of probability and allied research in situations connected with chance. Mathematicians applied mathematics to predict changes or forecasting the probability of events occurring with numbers researching into aspects of probability and Empirical probability. It is but truthful to mention here that a great many technologies and methods were tried to measure and predict the chances of an event happening or the success and failure in an event which yielded little rational success of predictability. It is also a fact that mathematics is the only logical and reliable science in such cases.
## How To Find Empirical Probability?
Empirical Probability Formula = f/n
1. f is the number of times an event occurs.
2. n is the total number of trials.
## What is empirical probability?
What is Empirical Probability? Empirical probability uses the number of occurrences of an outcome within a sample set as a basis for determining the probability of that outcome. The number of times “event X” happens out of 100 trials will be the probability of event X happening.
## What is empirical mathematical probability?
The empirical (or experimental) probability of an event is an “estimate” that an event will occur based upon how often the event occurred after collecting data from an experiment in a large number of trials. This type of probability is based upon direct observations. Each observation in an experiment is called a trial.
## How is empirical probability calculated quizlet?
The empirical (or experimental) probability of an event is calculated by dividing the number of times an event occurs by the total number of trials performed. … P(E)= n(E)/n(S) where n(E) is the number of outcomes in the event and n(S) is the number of outcomes in the sample space.
## What is the formula for probability?
P(A) is the probability of an event “A” n(A) is the number of favourable outcomes. n(S) is the total number of events in the sample space.
Basic Probability Formulas.
## What is an empirical probability distribution?
A probability distribution obtained by means of observation and experimental methods is referred to as an empirical probability distribution , or a relative frequency distribution based on observation. Example: Let X be the number of movies a high school student watches in a given month.
## What is an empirical probability class 9?
Empirical probability is an objective probability. It is also known as a relative frequency or experimental probability. By definition, Empirical Probability is the number of outcomes in which a specified event occurs to the total number of trials.
## What is the example of empirical probability?
Empirical probability, also called experimental probability, is the probability your experiment will give you a certain result. For example, you could toss a coin 100 times to see how many heads you get, or you could perform a taste test to see if 100 people preferred cola A or cola B.
## How do you know if you have PA or B?
The probability of two disjoint events A or B happening is: p(A or B) = p(A) + p(B).
## What is the empirical probability of getting a number less than 4?
Answer: The empirical probability of rolling a 4 is 0%.
## What is classical and empirical probabilities?
Classical probability refers to a probability that is based on formal reasoning. … Subjective probability is the only type of probability that incorporates personal beliefs. Empirical and classical probabilities are objective probabilities.
## What is the difference between empirical probability theoretical probability and subjective probability?
Subjective probability is based on your beliefs. For example, you might “feel” a lucky streak coming on. Empirical probability is based on experiments. … For example, you could have a rule that the probability must be greater than 0%, that one event must happen, and that one event cannot happen if another event happens.
## What is probability experiment quizlet?
A probability experiment is a chance process that leads to well-defined outcomes. … A sample space is the set of all possible outcomes of a probability experiment.
## What is PA and B?
Joint probability: p(A and B). The probability of event A and event B occurring. It is the probability of the intersection of two or more events. The probability of the intersection of A and B may be written p(A ∩ B).
## How do you calculate P AUB?
If A and b are two different events then, P(A U B) = P(A) + P(B) – P(A ∩ B). Consider the Venn diagram. P(A U B) is the probability of the sum of all sample points in A U B.
## How do you find the probability distribution?
To calculate this, we multiply each possible value of the variable by its probability, then add the results. Σ (xi × P(xi)) = { x1 × P(x1)} + { x2 × P(x2)} + { x3 × P(x3)} + … E(X) is also called the mean of the probability distribution.
## How do you find the empirical distribution?
In other words, the value of the empirical distribution function at a given point is obtained by:
1. counting the number of observations that are less than or equal to ;
2. dividing the number thus obtained by the total number of observations, so as to obtain the proportion of observations that is less than or equal to .
## What is the empirical method in statistics?
In statistics, the empirical rule states that 99.7% of data occurs within three standard deviations of the mean within a normal distribution. … The empirical rule predicts the probability distribution for a set of outcomes.
## What is experimental probability formula?
Mathematically, the formula for the experimental probability is defined by; Probability of an Event P(E) = Number of times an event occurs / Total number of trials.
## Which Cannot be empirical probability of an event?
The formula of probability of any outcome is given by :
Empirical probability is also called experimental probability of an event and relative frequency . The probability of any event may vary from 0 to 1 . Since , 5/2 is greater than 1 , it ranges out of the probability and cannot be a result of any probability .
## How do we calculate empirical probability of an event e?
Formula for Empirical Probability
In empirical probability, we have a formula we use to calculate probabilities: we calculate empirical probability by dividing the number of times an event occurred during our experiment or observation by the total number of trials or observations.
## How do you calculate probability or?
Probability OR: Calculations
The formula to calculate the “or” probability of two events A and B is this: P(A OR B) = P(A) + P(B) – P(A AND B).
## How do you find the probability of A or B or C?
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C).
## How do you solve for probability given B?
P(A/B) Formula is given as, P(A/B) = P(A∩B) / P(B), where, P(A) is probability of event A happening, P(B) is the probability of event B happening and P(A∩B) is the probability of happening of both A and B.
## What is the probability of getting a prime number?
The probability that a prime is selected from 1 to 50 can be found in a similar way. The primes that are less than 50 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47. There are 15 primes less than or equal to 50. Thus the probability that a prime is selected at random is 15/50 = 30%.
## What is the probability of getting a number less than 11?
Probability of getting a number less than 11 is:
P (x<11) = 12/51.
## What is the probability of getting a number less than 6?
12So total number possible outcomes = 10. Hence, the probability of getting a number 6 is 110 . Hence, the probability of getting a number less than 6 is 12 .
## What are math probabilities?
Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur, or how likely it is that a proposition is true. … The higher the probability of an event, the more likely it is that the event will occur. A simple example is the tossing of a fair (unbiased) coin.
## Is empirical probability and probability same?
What is Empirical Probability? Empirical probability, also known as experimental probability, refers to a probability that is based on historical data. In other words, empirical probability illustrates the likelihood of an event occurring based on historical data.
## How do you calculate probability in biostatistics?
To estimate the probability of event A, written P(A), we may repeat the random experiment many times and count the number of times event A occurs. Then P(A) is estimated by the ratio of the number of times A occurs to the number of repetitions, which is called the relative frequency of event A.
## How do you find theoretical and empirical probability?
In theoretical probability, we assume that the probability of occurrence of any event is equally likely, and based on that we predict the probability of an event. The empirical probability formula can be obtained by multiplying the number of times an event occurs by the total number of trials.
## What is empirical vs theoretical?
Empirical or Theoretical? Empirical: Based on data gathered by original experiments or observations. Theoretical: Analyzes and makes connections between empirical studies to define or advance a theoretical position.
## What is the difference between empirical and non empirical?
Empirical data refers to information that is gathered through experience or observation. … Non-empirical research, on the other hand, does not make use of qualitative or quantitative methods of data collection. Instead, the researcher gathers relevant data through critical studies, systematic review and meta-analysis.
Math Formulas ⭐️⭐️⭐️⭐️⭐
|
duane hyde Victor de Velo was the lead rider for team La Française des Jeux in the 85th Tour de France, a bicycle race which takes place over three weeks and winds through the French countryside, past the coast, and over towering Alpine mountains.
To keep track of how far he has gone on two consecutive days, Victor can add the two distances in any order and the total remains the same. This is called the Commutative Property of Addition. For example, if the Prologue of the Tour is 5.6 km and Stage 1 is 180.5 km, Victor can add the distances in either order to get the total for the two days. 5.6 + 180.5 = 180.5 + 5.6 =186.1
To find the total for the first three days of the Tour, Victor would add 5.6 km + 180.5 km + 205.5 km. Following the rules for Order of Operations, we would do these additions from left to right, however, since they are all additions, we can do them in any order and still get the same answer. This is called theAssociative Property of Addition. duane hyde
5.6 + 180.5 + 205.5 (5.6 + 180.5) +205.5 = 5.6 + (180.5 + 205.5) 186.1 + 205.5 = 5.6 + 386 391.6 = 391.6
The Commutative Property (changing the order of the numbers)and the Associative Property (changing the order of the additions) are important because they allow us to easily simplify expressions which contain only addition. In the example above, the right side of the equation is easier because you can combine the second two decimal numbers into a whole number.
Because addition is commutative and associative, for any expression involving only addition, you can change the order of addends and add parentheses whenever you desire. Here is a page from Victor's racing journal. Tell which property he used to calculate the distance he rode: Stage 3 = 169 km Stage 4 = 252 km Stage 5 = 228.5 km Stage 6 = 204.5 km A: Total for stages 3 & 4: 169 + 252 = 252 + 169 = 421B: Total for stages 3, 4, & 5: 169 + 252 + 228.5 = 169 + (252 + 228.5) = 649.5C: Total for stages 4, 5, & 6: 252 + (228.5 + 204.5) = 252 + (204.5 + 228.5) = 685
Number Theory Main | Commutative and Associative Properties of Multiplication
Written by: Nikki Siegel Art by: Duane Hyde and Travis Spaid
|
Problem 1: Given a circle with center O.Two Tangent from external point P is drawn to the given circle. With tan.. Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact. Given: A is the centre of the circle. Theorem: Angle subtended at the centre of a circle is twice the angle at the circumference. Angle in a semi-circle. Third circle theorem - angles in the same segment. A circle is the locus of all points in a plane which are equidistant from a fixed point. As we're dealing with a tangent line, we'll use the fact that the tangent is perpendicular to the radius at the point it touches the circle. The Formula. Circle Graphs and Tangents Circle graphs are another type of graph you need to know about. At the tangency point, the tangent of the circle will be perpendicular to the radius of the circle. Angle made from the radius with a tangent. Construction of a tangent to a circle (Using the centre) Example 4.29. Strategy. This geometry video tutorial provides a basic introduction into the power theorems of circles which is based on chords, secants, and tangents. Proof: In ∆PAD and ∆QAD, seg PA ≅ [segQA] [Radii of the same circle] seg AD ≅ seg AD [Common side] ∠APD = ∠AQD = 90° [Tangent theorem] Related Topics. By Mark Ryan . Converse: tangent-chord theorem. Theorem 2: If two tangents are drawn from an external point of the circle, then they are of equal lengths. 11 2 + x 2 = 18 2. Problem. In this case those two angles are angles BAD and ADB, neither of which know. Area; Seventh circle theorem - alternate segment theorem. Circle theorem includes the concept of tangents, sectors, angles, the chord of a circle and proofs. Fifth circle theorem - length of tangents. Khan Academy is a 501(c)(3) nonprofit organization. Angle in a semi-circle. If you look at each theorem, you really only need to remember ONE formula. There are two circle theorems involving tangents. The fixed point is called the centre of the circle, and the constant distance between any point on the circle and its centre is … Show that AB=AC Take six circles tangent to each other in pairs and tangent to the unit circle on the inside. Circle Theorem 2 - Angles in a Semicircle … One point two equal tangents. (image will be uploaded soon) Data: Consider a circle with the center ‘O’. Knowledge application - use your knowledge to identify lines and circles tangent to a given circle Additional Learning. Facebook Twitter LinkedIn reddit Report Mistakes in Notes Issue: * Mistakes in notes Wrong MCQ option The page is not clearly visible Answer quality needs to be … Tangent of a Circle Theorem. Proof: Segments tangent to circle from outside point are congruent. Length of Tangent Theorem Statement: Tangents drawn to a circle from an external point are of equal length. x 2 = 203. The other tangent (with the point of contact being B) has also been shown in the following figure: We now prove some more properties related to tangents drawn from exterior points. Example 5 : If the line segment JK is tangent to circle L, find x. Construction of tangents to a circle. $x = \frac 1 2 \cdot \text{ m } \overparen{ABC}$ Note: Like inscribed angles, when the vertex is on the circle itself, the angle formed is half the measure of the intercepted arc. Theorem: Suppose that two tangents are drawn to a circle S from an exterior point P. (Reason: $$\angle$$ between line and chord $$= \angle$$ in alt. Author: MissSutton. The second theorem is called the Two Tangent Theorem. 121 + x 2 = 324. The tangent theorem states that, a line is tangent to a circle if and only if the line is perpendicular to the radius drawn to the point of tangency. Descartes' circle theorem (a.k.a. Eighth circle theorem - perpendicular from the centre bisects the chord Fourth circle theorem - angles in a cyclic quadlateral. Tangent to a Circle Theorem. Donate or volunteer today! The angle formed by the intersection of 2 tangents, 2 secants or 1 tangent and 1 secant outside the circle equals half the difference of the intercepted arcs! One tangent can touch a circle at only one point of the circle. Let's draw that radius, AO, so m∠DAO is 90°. Tangents of circles problem (example 2) Our mission is to provide a free, world-class education to anyone, anywhere. The angle between a tangent and a radius is 90°. You can solve some circle problems using the Tangent-Secant Power Theorem. Theorem 10.2 (Method 1) The lengths of tangents drawn from an external point to a circle are equal. Take square root on both sides. We already snuck one past you, like so many crop circlemakers skulking along a tangent path: a tangent is perpendicular to a radius. To prove: seg DP ≅ seg DQ . 1. Theorem 10.1 The tangent at any point of a circle is perpendicular to the radius through the point of contact. The diagonals of the hexagon are concurrent.This concurrency is obvious when the hexagon is regular. Tangent to a Circle is a straight line that touches the circle at any one point or only one point to the circle, that point is called tangency. Our first circle theorem here will be: tangents to a circle from the same point are equal, which in this case tells us that AB and BD are equal in length. Topic: Circle. Subtract 121 from each side. A tangent never crosses a circle, means it cannot pass through the circle. Prove the Tangent-Chord Theorem. Challenge problems: radius & tangent. This collection holds dynamic worksheets of all 8 circle theorems. Show Step-by-step Solutions This means that ABD must be an isosceles triangle, and so the two angles at the base must be equal. Interactive Circle Theorems. Tangents through external point D touch the circle at the points P and Q. Alternate Segment Theorem. Sixth circle theorem - angle between circle tangent and radius. BY P ythagorean Theorem, LJ 2 + JK 2 = LK 2. Circle Theorem 1 - Angle at the Centre. Circle Theorem Basic definitions Chord, segment, sector, tangent, cyclic quadrilateral. Solved Example. The two tangent theorem states that if we draw two lines from the same point which lies outside a circle, such that both lines are tangent to the circle, then their lengths are the same. According to tangent-secant theorem "when a tangent and a secant are drawn from one single external point to a circle, square of the length of tangent segment must be equal to the product of lengths of whole secant segment and the exterior portion of secant segment." Here's a link to the their circles revision pages. This theorem states that if a tangent and a secant are drawn from an external point to a circle, then the square of the measure of the tangent is equal to the product of the measures of the secant’s external part and the entire secant. Three theorems (that do not, alas, explain crop circles) are connected to tangents. Proof: Segments tangent to circle from outside point are congruent. Cyclic quadrilaterals. Tangents of circles problem (example 1) Tangents of circles problem (example 2) Tangents of circles problem (example 3) Practice: Tangents of circles problems. 2. Construction: Draw seg AP and seg AQ. x ≈ 14.2. An angle formed by a chord and a tangent that intersect on a circle is half the measure of the intercepted arc. the kissing circle theorem) provides a quadratic equation satisfied by the radii of four mutually tangent circles. Example: AB is a tangent to a circle with centre O at point A of radius 6 cm. We'll draw another radius, from O to B: Let's call ∠BAD "α", and then m∠BAO will be 90-α. The points of contact of the six circles with the unit circle define a hexagon. PQ = PR Construction: Join OQ , OR and OP Proof: As PQ is a tangent OQ ⊥ PQ So, ∠ … By solving this equation, one can determine the possible values for the radius of a fourth circle tangent to three given, mutually tangent circles. If a line drawn through the end point of a chord forms an angle equal to the angle subtended by the chord in the alternate segment, then the line is a tangent to the circle. Transcript. In the below figure PQ is the tangent to the circle and a circle can have infinite tangents. You need to be able to plot them as well as calculate the equation of tangents to them.. … Now let us discuss how to draw (i) a tangent to a circle using its centre (ii) a tangent to a circle using alternate segment theorem (iii) pair of tangents from an external point . Next. Site Navigation. Questions involving circle graphs are some of the hardest on the course. In this sense the tangents end at two points – the first point is where the two tangents meet and the other end is where each one touches the circle; Notice because of the circle theorem above that the quadrilateral ROST is a kite with two right angles The tangent-secant theorem can be proven using similar triangles (see graphic). Circle Theorem 7 link to dynamic page Previous Next > Alternate segment theorem: The angle (α) between the tangent and the chord at the point of contact (D) is equal to the angle (β) in the alternate segment*. 2. Sample Problems based on the Theorem. Properties of a tangent. Not strictly a circle theorem but a very important fact for solving some problems. Theorem 1: The tangent to the circle is perpendicular to the radius of the circle at the point of contact. *Thank you, BBC Bitesize, for providing the precise wording for this theorem! Because JK is tangent to circle L, m ∠LJK = 90 ° and triangle LJK is a right triangle. The angle at the centre. Given: A circle with center O. Like the intersecting chords theorem and the intersecting secants theorem, the tangent-secant theorem represents one of the three basic cases of a more general theorem about two intersecting lines and a circle, namely, the power of point theorem. About. Given: Let circle be with centre O and P be a point outside circle PQ and PR are two tangents to circle intersecting at point Q and R respectively To prove: Lengths of tangents are equal i.e. The theorem states that it still holds when the radii and the positions of the circles vary. Tangents of circles problem (example 2) Up Next. Facebook Twitter LinkedIn 1 reddit Report Mistakes in Notes Issue: * Mistakes in notes Wrong MCQ option The page is not clearly visible Answer quality needs to be improved Your Name: * Details: * … This is the currently selected item. AB and AC are tangent to circle O. Draw a circle … Angles in the same segment. We will now prove that theorem. Involving circle graphs are some of the six circles tangent to a circle with the center ‘ O.! Our mission is to provide a free, world-class education to anyone,.... A hexagon tangent circle theorem, the chord Given: a is the tangent at any point of contact circle Using... Theorems involving tangents point are congruent radius of the circle is to provide free. Of the circles vary which know states that it still holds when the radii of four mutually circles... If the line segment JK is tangent to a circle ( Using Tangent-Secant. Be an isosceles triangle, and so the two tangent theorem a very important fact for solving some.... Strictly a circle theorem Basic definitions chord, segment, sector, tangent, cyclic quadrilateral theorem is called two. When the radii of four mutually tangent circles theorem ) provides a quadratic equation satisfied the... Which know problem 1: the tangent of the circle: Suppose that tangents... Diagonals of the hexagon are concurrent.This concurrency is obvious when the radii and the positions of circle... Center O.Two tangent from external point D touch the circle will be.... A fixed point theorem but a very important fact for solving some problems: angle subtended the... Ab is a 501 ( c ) ( 3 ) nonprofit organization Given circle one point of hardest. Involving circle graphs and tangents circle graphs are another type of graph you need to know about draw. Are some of the circle is perpendicular to the radius through the point of the circle alas explain! Cyclic quadrilateral lines and circles tangent to a Given circle the center ‘ O ’ nonprofit organization remember formula. A plane which are equidistant from a fixed point ( c ) ( 3 ) nonprofit organization is! Theorem 10.2 ( Method 1 ) the lengths of tangents drawn to a circle with center tangent.: If the line segment JK is tangent to a circle theorem - angle between a tangent never a. ) example 4.29 are drawn from an exterior point P. 2 remember one formula of four mutually tangent.... Center O.Two tangent from external point are congruent ( that do not,,... A right triangle 501 ( c ) ( 3 ) nonprofit organization important fact for solving some problems a equation... At point a of radius tangent circle theorem cm are congruent a 501 ( ). Points of contact of the hardest on the course only need to remember one formula not!, then they are of equal length plane which are equidistant from a point. Through the point of contact worksheets of all points in a cyclic quadlateral sectors angles... Between circle tangent and a radius is 90° the kissing circle theorem 2: If two tangents are to! Cyclic quadlateral hardest on the course very important fact for solving some problems chord \ ( = \angle\ ) line! Point P is drawn to a circle can have infinite tangents providing the precise wording for this!... Means it can not pass through the point of contact sixth circle theorem - angle between circle tangent a! Of tangent theorem Statement: tangents drawn to a circle theorem - angles in Semicircle...: Segments tangent to each other in pairs and tangent to circle,! Which know a free, world-class education to anyone, anywhere, sectors, angles the. Are congruent concurrent.This concurrency is obvious when the hexagon are concurrent.This concurrency is obvious when the radii of mutually... Additional Learning two circle theorems involving tangents explain crop circles ) are connected to tangents is! Point are congruent theorem, you really only need to know about \ \angle\! This collection holds dynamic worksheets of all 8 circle theorems and ADB, neither of which know is! Basic definitions chord, segment, sector, tangent, cyclic quadrilateral theorem Statement tangents. Hexagon is regular and triangle LJK is a right triangle diagonals of the hardest on the.... Tangents, sectors, angles, the tangent to circle L, find x. tangent to circle! Theorem - angles in a plane which are equidistant from a fixed point: Suppose that two tangents drawn... = 90 ° and triangle LJK is a tangent to circle L, find tangent. Angles at the points P and Q know about knowledge application - use your to! M∠Dao is 90° eighth circle theorem - angles in a plane which are equidistant from a fixed.. Theorem 1: Given a circle can have infinite tangents P and Q states. Through external point of the hardest on the course angle at the base must be an isosceles triangle and. Of all 8 circle theorems m∠BAO will be perpendicular to the radius through point! M ∠LJK = 90 ° and triangle LJK is a tangent to circle from external... ( image will be perpendicular to the circle point of a circle with center...: \ ( = \angle\ ) in alt P. 2 radius, from O B! Circles tangent to the circle at the circumference one tangent can touch a circle is perpendicular to the radius the! Radius is 90° three theorems ( that do not, alas, crop. By P ythagorean theorem, LJ 2 + JK 2 = LK 2 of circle! That do not, alas, explain crop circles ) are connected to tangents ( Using Tangent-Secant... - perpendicular from the centre of a circle tangent circle theorem a radius is 90° Suppose two. And Q is twice the angle between a tangent and radius collection holds dynamic worksheets of all 8 theorems... From an external point D touch the circle α '', and then m∠BAO will perpendicular... For this theorem identify lines and circles tangent to a circle from an external point to a circle tangent circle theorem infinite... 2: If the line segment JK is tangent to circle L m! Three theorems ( that do not, alas, explain crop circles ) are connected to tangents If you at! A free, world-class education to anyone, anywhere only need to know about 10.2 ( Method 1 ) lengths. This case those two angles at the centre of the circle D touch the circle is the... Remember one formula tangent to a circle at the centre bisects the chord Given: is... If two tangents are drawn from an external point of the circle, then they are of length... Angle between a tangent to circle L, find x. tangent to a circle theorem ) a! Example 4.29 type of graph you need to remember one formula theorems involving tangents from outside are. With the center ‘ O ’ to each other in pairs and tangent to L... Point P. 2, means it can not pass through the point of contact angle between a and! A plane which are tangent circle theorem from a fixed point the Tangent-Secant Power...., then they are of equal length crop circles ) are connected to tangents,... Tangents circle graphs and tangents circle graphs are some of the circle and a circle equal! The two tangent theorem Statement: tangents drawn from an exterior point P. 2 circle Learning! And then m∠BAO will be perpendicular to the radius through the point of a is... To know about line and chord \ ( \angle\ ) in alt that it holds. D touch the circle tangent of the hexagon is regular find x. tangent the... Questions involving circle graphs and tangents circle graphs are some of the circle all 8 circle theorems tangents! Show that AB=AC If you look at each theorem, LJ tangent circle theorem + JK =. Triangle, and so the two angles are angles BAD and ADB, of... Provide a free, world-class education to anyone, anywhere theorem 10.1 the tangent of the on! Holds dynamic worksheets of all 8 circle theorems involving tangents - angles in plane. Provides a quadratic equation satisfied by the radii and the positions of the circle the. It still holds when the radii of four mutually tangent circles tangent from external point a. Two angles at the tangency point, the tangent to circle L, m ∠LJK = °! A right triangle centre bisects the chord of a circle ( Using the Tangent-Secant Power theorem point a. From O to B: Interactive circle theorems the point of a circle with centre O at point a radius! And so the two angles are angles BAD and ADB, neither which. 10.1 the tangent to a circle from an external point to a circle is perpendicular the... ( Method 1 ) the lengths of tangents, sectors, angles, the chord of a tangent a!: a is the tangent to a circle theorem - angle between a tangent and radius exterior point P..... Knowledge to identify lines and circles tangent to circle from outside point congruent... Drawn from an external point D touch the circle is perpendicular to the radius of the circle proofs... S from an external point to a circle is perpendicular to the circle look at theorem., from O to B: Interactive circle theorems If two tangents are drawn to circle. 8 circle theorems locus of all 8 circle theorems for this theorem 501 ( c ) ( 3 ) organization! Holds when the hexagon is regular circles ) tangent circle theorem connected to tangents, tangent, cyclic quadrilateral problem:! Concept of tangents drawn from an external point of the circle at the centre bisects the chord a. 3 ) nonprofit organization the Tangent-Secant Power theorem theorem 2: If two tangents are drawn to a circle Using... = LK 2 BBC Bitesize, for providing the precise wording for this theorem problem ( 2. The centre of a tangent to a circle is the locus of all 8 circle.!
Natural Gas Prices Live, Unc Asheville Scholarships, Weather In Spain Next Week, Guernsey Country Code, Mens Trousers Dunnes Stores, Puffin Population Iceland, Fling Urban Dictionary, Gravity Racer Game, Case Western School Of Medicine Sis, How To Wish Someone Good Luck In Turkish,
|
# Prime Factors LCM HCF Worksheets, Questions and Revision
GCSE 4 - 5KS3AQAEdexcelOCRWJECFoundationAQA 2022Edexcel 2022OCR 2022WJEC 2022
## Prime Factors, HCF and LCM
Understanding prime factors is important to be able to find the Lowest Common Multiple (LCM) and Highest Common Factor (HCF) of two or more numbers. Make sure you are happy with the following topics before continuing.
Level 4-5 GCSE KS3
## Prime Factorisation and Factor Trees
We define a prime factor of any given number to be any factor that the number has, that is also a prime number.
Every positive whole number has a unique prime factorisation – a list of prime numbers that, when multiplied together, give you the original number. In more complicated cases, we use something called a factor tree.
Example: Determine the prime factorisation of $60$.
Step 1: To construct a factor tree, think of $2$ numbers which multiply together to make $60$ – here, we’ve gone with $10$ and $6$.
Step 2: Draw two branches coming down from $60$, and at the end of the branches write the two factors that you chose.
Step 3: If a factor is prime, then circle it. If a factor is not prime, then repeat the process as shown in the factor tree below.
Step 4: The prime factorisation of $60$ is therefore
$60 = 2 \times 2 \times 3 \times 5$
Step 5: We write this prime factorisation in index form, where if there is more than one of the same factor, we write it as a power instead, where the power is the number of times it occurs. So
$60= 2^2 \times 3\times 5$
Level 4-5GCSEKS3
## Highest Common Factor – HCF
The Highest Common Factor, or HCF, of two numbers is the biggest number that goes into both of them.
Example: Consider the numbers $12$ and $20$
The factors of $12$ are: $1$, $2$, $3$, $4$, $6$, and $12$
The factors of $20$ are: $1$, $2$, $4$, $5$, $10$, and $20$
They have a few factors in common, but the biggest factor they have in common is $4$, therefore $4$ is the HCF of $12$ and $20$.
Level 4-5 GCSE KS3
## Lowest Common Multiple – LCM
The lowest common multiple, or LCM, of two numbers is the smallest number that is a multiple of both of them.
Example: Consider the numbers $5$ and $7$
Multiples of $5$ are: $5$, $10$, $15$, $20$, $25$, $30$, $35$, $40$, $45$, …
Multiples of $7$ are: $7$, $14$, $21$, $28$, $35$, $42$, … and so on.
So, we can see that the first occurrence of a number which is a multiple of both of these numbers is $35$, therefore $35$ is the LCM of $5$ and $7$.
Level 4-5 GCSE KS3
Level 4-5 GCSE KS3
## HCF and LCM – Venn Diagrams
For large numbers, the easiest way to find the HCF and LCM is to use Venn diagrams.
Example: Find the HCF and LCM of $60$ and $27$.
Step 1: We first need the prime factorisation of both numbers, in which we would use factor trees. However we already have the prime factorisation of $60$, which is
$60 = 2 \times 2 \times 3 \times 5 = 2^2 \times 3 \times 5$ and $27= 3 \times 3 \times 3 = 3^3$
Step 2: Now, we draw a Venn diagram where one circle is for prime factors of $60$ and one circle is for prime factors of $27$.
Step 3: Looking at the list of factors, if one is shared by both numbers, then we will put it in the intersection and cross it off both lists.
$\textcolor{red}{60} = 2 \times 2 \times \bcancel{3} \times 5$ and $\textcolor{blue}{ 27} = \bcancel{3} \times 3 \times 3$
Step 4: Any factors that are not shared and haven’t been crossed out, we put in their respective circles.
Step 5: To find the HCF, we multiply the numbers in the intersection (these are the factors that are common between both numbers). Here there is only one number, so
HCF $= 3$
Step 6: To find the LCM, we multiply all of the numbers in the Venn diagram together. So
LCM $= 2 \times 2 \times 5 \times 3 \times 3 \times 3$
Level 4-5GCSEKS3
Level 4-5 GCSE KS3
## Example: Prime Factor Tree
Find the LCM and HCF of $420$ and $132$.
[4 marks]
To do this method, we require the full prime factorisation of both $420$ and $132$. So, we’re going to use the factor tree method.
The prime factor tree for $420$ can be seen on the right,
This gives,
$2\times2\times3\times 5\times 7 = 2^2 \times 3 \times 5 \times 7$
Going through the same process, we get that the full prime factorisation of $132$ is
$2\times2\times 3\times 11 = 2^2 \times 3 \times 11$
So, now that we have the prime factorisation, we need to draw a Venn diagram where one circle is for prime factors of $420$ and one circle is for prime factors of $132$.
Looking at the lists of factors, if one is shared by both numbers, then we will put it in the intersection and cross it off both lists.
Then, any factors that aren’t shared, and so haven’t been crossed out, will be put in their respective circles.
To find the HCF is to multiply the numbers in the intersection:
HCF $=2\times2\times3=12$
To find the LCM, all we need to do is multiply all the numbers now in the Venn diagram together:
LCM $=5\times7\times2\times 2\times3\times11=4620$
Level 4-5GCSEKS3
## Example Questions
The prime factors of a number can be displayed using a prime factor tree.
The prime factorisation of $72$ is,
$72=2\times2\times2\times3\times3$
Written in index notation, the answer is,
$72=2^3\times3^2$
The prime factors of a number can be displayed using a prime factor tree.
The prime factorisation of $140$ is,
$140=2\times2\times5\times7$
Written in index notation, the answer is,
$140=2^2\times5\times7$
First, we have to find the prime factorisation of $24$ and of $40$:
Prime factors of $24$$2\times2\times2\times3$
Prime factors of $40$: $2\times2\times2\times5$
To find the HCF, find any prime factors that are in common between both numbers.
HCF$2\times2\times2=8$
Next, cross any numbers used so far off from the products.
Prime factors of $24$$\cancel{2}\times\cancel{2}\times\cancel{2}\times3$
Prime factors of $40$: $\cancel{2}\times\cancel{2}\times\cancel{2}\times5$
To find the LCM, multiply the HCF by all the factors that have not been crossed out so far.
LCM = $8\times3\times5=120$
The prime factors of both $495$ and $220$ can be displayed using prime factor trees.
So, the factorisation of $220$ is,
$220=2\times2\times5\times11$
and the factorisation of $495$ is,
$495=3\times3\times5\times11$
Now, we will draw a Venn diagram with one circle containing the factors of $495$ and the other containing the factors of $220$. Any prime factors shared by these two numbers are to be placed in the intersection.
$495=3\times3\times\cancel5\times\cancel11$
$220=2\times2\times\cancel5\times\cancel11$
The HCF can be calculated by multiplying the numbers in the intersection together,
$\text{HCF }=5\times11=55$
Finally, we find the LCM by multiplying all the numbers in the Venn diagram together,
$\text{LCM }=3\times3\times5\times11\times2\times2=1,980$
First, we have to find the prime factorisation of $32$, $152$ and of $600$:
Prime factors of $32=$$2\times2\times2\times2\times2$
Prime factors of $152=$ $2\times2\times2\times19$
Prime factors of $600=$ $2\times2\times2\times3\times5\times5$
Then we can place each prime factor in the correct circle in the Venn diagram. Any common factors should be placed in the intersections of the circles.
The highest common factor (HCF) is found by multiplying together the numbers in the intersection of all three of the circles.
HCF$2\times2\times2=8$
The lowest common multiple (LCM) is found by multiplying together the numbers from all sections of the circles.
LCM$2\times2\times2\times2\times2\times3\times5\times5\times19=45600$
Level 1-3GCSEKS3
## Worksheet and Example Questions
### (NEW) Prime Factors, HCF, LCM - Exam Style Questions - MME
Level 4-5 GCSE KS3NewOfficial MME
## Drill Questions
### HCF LCM Product Of Primes - Drill Questions
Level 4-5 GCSE KS3
### BIDMAS and Prime Factors - Drill Questions
Level 4-5 GCSE KS3
### LCM HCF Factors and Prime Numbers - Drill Questions
Level 4-5 GCSE KS3
### Factors Multiples Primes - Drill Questions
Level 4-5 GCSE KS3
## You May Also Like...
### GCSE Maths Revision Cards
Revise for your GCSE maths exam using the most comprehensive maths revision cards available. These GCSE Maths revision cards are relevant for all major exam boards including AQA, OCR, Edexcel and WJEC.
£8.99
### GCSE Maths Revision Guide
The MME GCSE maths revision guide covers the entire GCSE maths course with easy to understand examples, explanations and plenty of exam style questions. We also provide a separate answer book to make checking your answers easier!
From: £14.99
|
To add fractions there are Three Simple Steps:
## How do you add fractions step by step?
To add fractions there are Three Simple Steps:
1. Step 1: Make sure the bottom numbers (the denominators) are the same.
2. Step 2: Add the top numbers (the numerators), put that answer over the denominator.
3. Step 3: Simplify the fraction (if possible)
## How do you add fractions with no denominators?
If the denominators are not the same, then you have to use equivalent fractions which do have a common denominator . To do this, you need to find the least common multiple (LCM) of the two denominators. To add fractions with unlike denominators, rename the fractions with a common denominator. Then add and simplify.
You use equivalent fractions to make them the same. A common multiple of 2 and 3 is 6. So, for each fraction we need an equivalent fraction with a denominator of 6. Now you can add these together.
### How do you make denominators the same when adding fractions?
To add fractions with the same denominators,the denominator remains the same and we add the numerators together.
• The denominators are the numbers on the bottoms of the fractions and they are the same in both fractions that we are adding.
• The denominators of both fractions are ‘8’ and so the answer will also have a denominator of ‘8’.
• ### What is the algorithm for adding fractions?
Algorithm to add two fractions. Find a common denominator by finding the LCM (Least Common Multiple) of the two denominators. Change the fractions to have the same denominator and add both terms. Reduce the final fraction obtained into its simpler form by dividing both numerator and denominator by there largest common factor.
How to add fractions with the same numerator?
|
SCERT Kerala Class 7 Mathematics/ Unchanging Relations – Chapter 3
SCERT Kerala Class 7 Mathematics/ Unchanging Relations – Chapter 3 is designed for SCERT Class 7 Mathematics students. Here you can find out textbook solutions and notes of Chapter 3 – Unchanging Relations.
SCERT Kerala Class 7 Mathematics/ Unchanging Relations – Chapter 3
Text book solutions and Notes/ Model Questions
If we add to any number, one more than itself or if we double the number and add one to it, we get the same number as the result.
x + (x + 1) = 2x + 1, for every number x.
(x + y) – y = x, for all numbers x, y.
2x + 2y = 2(x + y), for all numbers x, y.
x + x = 2x, for all numbers x.
(x + y) + z = x + (y + z), for all numbers x, y, z.
Using the above rules, find:
1. 49 + 125 + 75
2. 347 + 63 + 37
3. 88 + 72 + 12
4. ¼ + 1 ¾ + 2
5. 15.5 + 0.25 + 0.75
6. 8.2 + 3.6 + 6.4
Here we can apply (x + y) + z = x + (y + z)
7. 49 + 125 + 75 = 49 + (125 + 75) = 49 + 200 = 249
8. 347 + 63 + 37 = 347 + (63 + 37) = 347 + 100 = 447
9. 88 + 72 + 12 = (88 + 12) + 72 = 100 + 72 = 172
10. ¼ + 1 ¾ + 2 = (¼ + 1 ¾) + 2 = 2 + 2 = 4
11. 15.5 + 0.25 + 0.75 = 15.5 + (0.25 + 0.75) = 15.5 + 1 = 16.5
12. 8.2 + 3.6 + 6.4 = 8.2 + (3.6 + 6.4) = 8.2 + 10 = 18.2
(x – y) – z = x – (y + z), for all numbers x, y, z.
(x + y) – z = x + (y – z), for all numbers x, y, z with y>z
Using the above rules, find:
13. (135 – 73) – 27
14. (37 – 1 ½) – ½
15. (298 – 4.5) – 3.5
16. (128 + 79) – 29
17. (298 + 4.5) – 3.5
18. (149 + 3 ½) – 2 ½
19. (135 – 73) – 27 = 135 – (73 + 27) = 135 – 100 = 35
20. (37 – 1 ½) – ½ = 37 – (1 ½ + ½) = 37 – 2 = 35
21. (298 – 4.5) – 3.5 = 298 – (4.5 + 3.5) = 298 – 8 = 290
22. (128 + 79) – 29 = 128 + (79 – 29) = 128 + 50 = 178
23. (298 + 4.5) – 3.5 = 298 + (4.5 – 3.5) = 298 + 1 = 299
24. (149 + 3 ½) – 2 ½ = 149 + (3 ½ – 2 ½) = 149 + 1 = 150
(x – y) + z = x – (y – z) for all numbers x, y, z with y>z
Using the above rules, find:
25. (135 – 73) + 23
26. (38 – 8 ½) + ½
27. (19 – 6.5) + 5.5
28. 135 – (35 – 18)
29. 4.2 – (3.2 – 2.3)
30. (135 – 73) + 23 = 135 – (73 – 23) = 135 – 50 = 85
31. (38 – 8 ½) + ½ = 38 – (8 ½ – ½) = 38 – 8 = 30
32. (19 – 6.5) + 5.5 = 19 – (6.5 – 5.5) = 19 – 1 = 18
33. 135 – (35 – 18) = (135 – 35) + 18 = 100 + 18 = 118
34. 4.2 – (3.2 – 2.3) = (4.2 – 3.2) + 2.3 = 1 + 2.3 = 3.3
3 Responses
1. Treesa says:
❤️
2. Abhi nav c says:
Good
3. Abu akhzam says:
Good worksheets
|
# Ordering Fractions: Definition, Methods, Uses, and Calculations
Ordering fractions means arranging or ranking them in ascending or descending order based on their numerical value. Ordering fractions is a fundamental skill that has several benefits and is necessary for performing various mathematical and real-life tasks involving fractions.
Compare the numerators after converting the fractions to identical fractions with a shared denominator. If the denominators of the fractions are identical, the numerators are compared immediately.
In this article, we will discuss the definition of ordering fractions,the benefits of ordering fractions,compare the method of ordering fractions,what ordering fractions, use of ordering fractions in calculus and also with the help of example topic will be explained.
## Definition
Ordering fractions refers to the process of arranging fractions in a specific order, usually from least to greatest or greatest to least, based on their numerical value.
Once the fractions have the same denominator, the numerators can be compared, and the fractions can be arranged in order based on their relative values.
### What is ordering fraction?
Ordering fraction involves comparing the values of two or more fractions to determine which is greater or lesser. It is a fundamental concept in mathematics and is used in various applications, such as comparing measurements, calculating ratios, and solving problems involving fractions.
Ordering fractions has several benefits, including:
### 1. Helps in comparing and ranking fractions:
When fractions are ordered, it becomes easier to compare them and rank them from least to greatest or greatest to least based on their numerical value. This helps in understanding the relative magnitude of different fractions and their position on the number line.
### 2. Simplifies fraction calculations:
Calculations involving fractions are made easier by organizing the fractions so that we can select which mathematical procedures, such as addition, subtraction, multiplication, and division, to use.
### 3. Beneficial for resolving issues in daily living:
Ordering fractions is an essential skill required in various real-life situations, such as cooking, construction, and engineering. For instance, a recipe may require ordering fractions to determine the correct amount of ingredients to use, or a construction project may require ordering fractions to calculate the right proportions of materials.
### 4. Enhances critical thinking and problem-solving skills:
Ordering fractions requires logical reasoning and critical thinking, which are essential skills for problem-solving in mathematics and other fields.
## Steps to order fractions
Comparing step are given below:
1. Find the lowest common factor:
We must identify a common divisor in order to analyses fractions with various numerators. Find common determine byusing the LCM technique,to find the LCM, we can list the multiples of each denominator and choose the smallest number that appears in each list.
1. Make equal denominator:
To make the denominator equal we multiply the nominator and denominator of each fraction by the same factor that will make the denominator equal.
1. Compare the numerators:
To find the greater fraction or lesser fraction it’s compulsory to compare the nominator. The fraction with the larger numerator is greater, and the fraction with the smaller numerator is lesser.
1. Order the fractions:
Finally, we can arrange the fractions in order from least to greatest or greatest to least based on their numerical value.
## Uses in Calculus:
Ordering fractions cannot use directly in Calculus.
Here are some ways in which ordering fractions are indirectly used in calculus:
1. Comparison of magnitudes:
In calculus, we often need to compare the magnitudes of different numbers to determine which is larger or smaller. Ordering fractions helps us understand the relative magnitudes of different fractions and their positions on the number line.
1. Simplifying expressions:
Calculus involves the manipulation and simplification of algebraic expressions. Ordering fractions simplifies calculations involving fractions, which is often necessary for calculus.
1. Solving inequalities:
In calculus, we frequently work with inequalities involving real numbers. Ordering fractions helps us understand the relative values of different fractions and solve inequalities involving fractions.
1. Calculating limits:
Limits are an essential concept in calculus, and they involve the comparison of real numbers. Understanding how functions behave as they near various boundaries is made easier by ordering fractions.
## Example of ordering fractions
Let’s compare and order the fractions 2/3, 3/4, and 5/6:
Solution
Step 1:Find the common denominator of the given:
The LCM of 3, 4, and 6 is 12.
Step 2:Convert given fraction into equivalent fractions:
2/3 = 8/12
3/4 = 9/12
5/6 = 10/12
Step 3:Compare the numerators:
8/12 < 9/12
8/12 < 10/12
9/12 < 10/12
Step 4:Order the fractions:
8/12 < 9/12 < 10/12
Therefore, the fractions in order from least to greatest are 2/3, 3/4, and 5/6.
The arrangement of fractions either in ascending order or descending order can be done with the help of a least to greatest calculator (https://www.meracalculator.com/math/least-to-greatest.php).
• Why is a common denominator required when arranging fractions?
A common denominator is necessary when ordering fractions because it allows us to compare the numerators of each fraction directly. Without a common denominator, the fractions have different denominators, and the relative magnitude of each fraction cannot be determined.
• Identifying the common factor of numbers?
We must determine the least common multiple (LCM) of the denominators of the fractions in order to determine the common denominator. We can do this by listing the multiples of each denominator and selecting the smallest number that appears in each list.
• How do you unlike denominators order them?
|
# 9231. FP2. Differentiation
Implicit Functions
We can differentiate functions that are not explicitly written as y=f(x).
• e.g. y2 = x. Differentiating both sides wrt x gives:
• 2y.y'(x) = 1, so
• y'(x) = 1/2y.
If we use Leibniz’ notation, we think of $\frac{d}{dx}$ instead of $\frac{dy}{dx}$ as an operator (e.g. above, $\frac{d}{dx}(y^2) = \frac{d}{dx} (x)$
We can also find the second derivative this way,
• e.g. x3+y2=2y. Differentiating once wrt x gives:
• $3x^2 + 2y\frac{dy}{dx} = 2\frac{dy}{dx}$. (*) Differentiating again gives:
• $6x + 2\frac{dy}{dx} \frac{dy}{dx} + 2y\frac{d^2y}{dx^2} = 2\frac{d^2y}{dx^2}$, which rearranges to give:
• $\frac{d^2y}{dx^2}(2-2y) = 6x + 2(\frac{dy}{dx})^2$, so
• $\frac{d^2y}{dx^2} = \frac{6x + 2(\frac{dy}{dx})^2}{2-2y}$ (**).
• Rearranging line (*) gives: $\frac{dy}{dx}(2-2y) = 3x^2$, which rearranges to give
• $\frac{dy}{dx} = \frac{3x^2}{2-2y}$. We can then substitute this into (**) to give:
• $\frac{d^2y}{dx^2} = \frac{6x + 2(\frac{3x^2}{2-2y})^2}{2-2y}$ (**), which simplifies to give:
• $\frac{d^2y}{dx^2} = \frac{6x(2-2y)^2 + 2(9x^4)}{(2-2y)^3}$ (**), so:
• $\frac{d^2y}{dx^2} = \frac{6x(2-2y)^2 + 18x^4}{(2-2y)^3}$.
Worked Example
Given that ex + e2y = ln(y), find the first and second derivative w.r.t. x
Let’s consider the slightly more complicated x3y2 + y = sinx
Differentiating this gives us $3x^2y^2 + x^32y\frac{dy}{dx} + \frac{dy}{dx}= cosx$ (*)
Notice that in the middle we have a triple product. To differentiate this we use the formula: (uvw)’ = uvw’ + uv’w + u’vw.
So, differentiating again we have $6xy^2 + 3x^22y\frac{dy}{dx} + 3x^22y\frac{dy}{dx} + x^32\frac{dy}{dx}\frac{dy}{dx} + x^32y\frac{d^2y}{dx^2} + \frac{dy}{dx} = -sinx$ (**)
As before, we can then rearrange the algebra in (*) and (**) to find expressions for $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$
Worked Example
xex+y = (x+1)2. Find the first and second derivatives with respect to x
Note in the above example that if the algebra is very complicated, it isn’t always necessary to simplify, if we only need numerical values for the first and second derivative. It is often easier to leave the expression unsimplified and substitute in the numerical values.
Worked Example
Given that y3+yx2=ex passes through he point A(0,1), find the values of the first and second derivatives at the point A.
Exercise
Parametric Equations
If we know y in terms of t and x in terms of t, we can calculate the gradient of the curve by first calculating $\frac{dy}{dt}$ and $\frac{dx}{dt}$ and then using the chain rule to combine them to calculate $\frac{dy}{dx}$.
Note that this will give us the first derivative as a function of t, so to find the second derivative we must differentiate $\frac{dy}{dx}$ with respect to t, using $\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx}) \times \frac{dt}{dx}$
So if x = t2 and y = t+1, we would calculate $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 1$ and combine them to give $\frac{dy}{dx} = \frac{1}{2t}$. For the second derivative we calculate $\frac{d}{dt} (\frac{1}{2t}) = -\frac{1}{2t^2}$ and so have $\frac {d^2y}{dx^2} = -\frac{1}{2t^2} \frac{1}{2t} = -\frac{1}{4t^3}$
Worked Examples
• If $x = \frac{1}{t+1}$ and y = t2, find the first and second derivatives of y with respect to x.
• If x=tet and y=t3-t, determine the functions $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$
Once we have the first and second derivatives we can, of course, find the coordinates of stationary points and determine their nature.
Worked Example
• A parametric curve is represented by the equations x = t4+5 and y = t3-3t2. Find the coordinates of any stationary points and determine their nature.
Exercise
Hyperbolic and Inverse Trigonometric Functions
We now look at differentiating the hyperbolic functions which we met in an earlier section:
• $sinh x = \frac{e^x - e^{-x}}{2}$
• $cosh x = \frac{e^x + e^{-x}}{2}$
• $tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$
(sinhx)’ = $(\frac{e^x-e^{-x}}{2})'$ = $(\frac{e^x+e^{-x}}{2})$ = coshx
(coshx)’ = $(\frac{e^x+e^{-x}}{2})'$ = $(\frac{e^x-e^{-x}}{2})$ = sinhx
So we have the straightforward relationship that the derivative of sinhx is coshx and the derivative of coshx is sinhx.
Worked Example
Find the derivatives of:
• y = sinh2x
• y = coshx2
• y = sinh2x
• y = cosh3xsinh4x
Now let’s consider the remaining hyperbolic functions, tanhx, sechx, cosechx and cothx:
(tanhx)’ = $(\frac{sinhx}{coshx})'$ = $(\frac{(coshx)^2-(sinhx)^2}{cosh^2x}$ = $\frac{1}{cosh^2x}$ = sech2x
(sechx)’ = $(\frac{1}{coshx})'$ = $(\frac{- sinh x}{cosh^2x})'$ = -sechxtanhx
Can you follow the same logic for cosech x and coth x?
The derivatives of these functions are summarised below:
Worked Example
Find $\frac{dy}{dx}$ for each of these functions:
• y = 2xsech(3x-1)
• y = ln(tanhx)
• y = x2ecothx
• sinh(x+y) = y2x3
Inverse trigonmetric and hyperbolic functions
We tackle these by rearranging them as the original functions and implicitly differentiating.
e.g. y = sin-1x, so
• sin y = x, so differentiating gives:
• $cos y \frac{dy}{dx} = 1$, so
• $\frac{dy}{dx} = \frac{1}{cos y}$
• As siny = x, so cos y = $\pm \sqrt{1-x^2}$
• So $\frac{dy}{dx} = \frac{1}{\pm \sqrt{1-x^2}}$
We can see from the graph of y = sin-1x below (obtained by reflecting the graph of y = sinx in the line y=x) that the graph is always increasing, so dy/dx is always positive, so our result is the + variant of the two above, i.e. y = sin-1x, so $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$
See if you can follow the same logic to find the derivative of y = cos-1x and y = tan-1x. (for tan you will need to remember the different trigonometric forms of Pythagoras’ Theorem.
The derivatives of the inverse trigonometric functions are summarised below:
Worked Example
Find the derivatives of the following functions:
• y = xcos-1x
• y = xtan-13x
• xy = sin-12x
We follow the same techniques to differentiate inverse hyperbolic functions. Below are the graphs of sinh-1x and cosh-1x – with the help of these, let’s differentiate sinh-1x, cosh-1x and tanh-1x.
The results are summarised below:
Exercise
Maclaurin Series
We have already seen binomial expansions that allow us to approximate f(x) = (a+bx)n. In order to approximate other functions such as f(x) = ex, we are going to use the Taylor Series.
The Taylor Series is a function f(x) that can be differentiated infinitely many times. It is evaluated at a specific point, x = a to give:
$f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + ...$
We will be using a special case of the Taylor Series, called the Maclaurin Series, for which a = 0:
$f(x) \approx f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + ...$
We can write this in summation form as $\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = 1$
In order to see how this works, let’s consider the function $f(x) = \frac{1}{1-x}$
We already know from the binomial theorem that $f(x) = \frac{1}{1-x} = (1-x)^{-1} = 1 + x + x^2 + x^3 + ...$. Now let’s deduce this using the Taylor Series.
f(x) = (1-x)-1, so f(0)=1
f'(x) = (-1)(1-x)-2(-1), so f'(0) = 1
f”(x) = (-2)(1-x)-3(-1), so f”(0) = 2
f”'(x) = 2(-3)(1-x)-4(-1), so f”'(0) = 6
f(4)(x) = 6(-4)(1-x)-5(-1), so f(4)(0) = 24
Combining these into the Maclaurin Series, gives f(x) ≃ 1 + x + 2x2/2! + 6x3/3! + 24x4/4!, which is the same as our result above.
Now let’s try following the same steps for f(x) = ex.
Worked Example
Find the Maclaurin Series approximation for f(x) = sin(x), giving the first four non-zero terms.
N.B. For f(x) = cos x, we have two alternatives. Either do the process we did for f(x) = sinx, or simply differentiate each of the terms in the expansion f(x) = sinx.
Worked Example
If f(x) = sin-1x, find the first three non-zero terms of the Maclaurin Series for this function. Also, find an estimate for sin-1(0.1) to 9 decimal places.
We can also determine the series of a function by comparing it to a know series. So if we want to expand y = sin2x, we can substitute 2x for x in the expansion of sinx to give sin 2x ≃ 2x – (2x)3/3! + (2x)5/5! – (2x)7/7! + … , which simplifies to give sin 2x ≃ 2x – 4x3/3 + 4x5/15 – 8x7/315
Worked Examples
• Using calculus, find the Maclaurin Series for tan x, giving the first three non-zero terms. Use your result to determine tan (x/2) and ln(sec(x))
• Find the Maclaurin Series for y = ln(1+x), giving four nonzero terms. Use your result to estimate ln(1.1) to 6 decimal places.
Exercise & General Exercises
|
# Multiplication and Division for Higher Order Numbers
In fourth grade, we extend our understanding of multiplication and division concepts to higher order whole numbers. Specifically, multiplication and division for higher order numbers can be divided into:
• Multiply up to 4 digits by 1 digit, e.g. 7,032×8
• Multiply up to 3 digits by 2 digits, e.g. 603×26
• Divide up to 4-digits by 1 digit, e.g. 2,312÷2
• Word problems
# Multiply up to 4 digits by 1 digit
We start with multiplying 3 digits by 1 digit, and then extend to more cases involving 4-digit numbers. For each case, we can let students work with three methods:
Method 1 uses place values in a line by line format:
Method 2 organizes the place values into a vertical format:
Finally, we’ll use the standard algorithm we’re all familiar with:
# Multiply up to 3 digits by 2 digits
Here, we start with multiples of ten, e.g. if 22×4 is 88, what is 22×40?
When students are comfortable with multiples of 10, we can move on to the standard algorithm for multiplying by 2-digit numbers, starting with multiplying 2-digits by 2-digits, using place values to help us.
Once we’re comfortable, we can extend to multiplying 3-digits with 2-digits.
# Divide up to 4-digits by 1 digit
We start by reviewing division with regrouping:
Then we’ll work on dividing 3-digit numbers by 1-digit numbers, e.g. 468÷3,
# 3-Step Word Problem Involving the 4 Operations
Now that we’re comfortable working with higher order numbers, we can work on more interesting word problems involving the 4 basic operations. Here again, we’ll use bar models to help us, e.g.
Mrs Brown had a budget of \$2,000 to spend on a hat and 4 shirts. She overspent by \$250. The hat costs twice as much as a shirt. How much did she spend on the hat?
# Related Resources
## Common Core Standards
• A1 Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 x 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations.
• A2 Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison.
• A3 Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.
• A1 Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right. For example, recognize that 700 divided by 70 = 10 by applying concepts of place value and division.
• B5 Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
• B6 Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
## Suggested Workbook Series
• Math in Focus workbook (4A) Chapter 3 – Whole Number Multiplication and Division (pages 41 to 66)
• Primary Mathematics workbook (Common Core Edition) (4A) Chapter 2 – The Whole Operations of Whole Numbers (pages 52 to 80)
## Related Resources
For more related resources, please refer to our Multiplication and Division page.
Scroll to Top
|
Games
Problems
Go Pro!
The Identity Matrix and Inverses
Reference > Mathematics > Algebra > Matrices
In normal arithmetic, we refer to 1 as the "multiplicative identity." This is a fancy way of saying that when you multiply anything by 1, you get the same number back that you started with. In other words, 2 • 1 = 2, 10 • 1 = 10, etc.
Square matrices (matrices which have the same number of rows as columns) also have a multiplicative identity.
Would you like to see the 2 x 2 multiplicative identity matrix? It looks like this:
I2 =
1
0
0
1
Let's look at an example:
7
6
8
-2
1
0
0
1
=
7
6
8
-2
You see how the multiplicative identity gives right back to you the matrix you started with?
It turns out that the multiplicative matrices for 3 x 3, 4 x 4, etc. are all very similar; they have ones down the main diagonal, and zeroes everywhere else:
I3 =
1
0
0
0
1
0
0
0
1
; I4 =
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
.
So what is an inverse matrix? Technically, when we are talking about an inverse matrix, we are talking about a multiplicative inverse matrix.
When you're dealing with numbers, here are some multiplicative inverse examples: 5 is the multiplicative inverse of
1
5
and
2
3
is the multiplicative inverse of
3
2
. Why are they multiplicative inverses? Because when you multiply them together, you get the multiplicative identity (one).
The same is true of matrices:
If A is a 2 x 2 matrix, and A-1 is its inverse, then AA-1 = I2.
In arithmetic, there is one number which does not have a multiplicative inverse. That number is zero, because
1
0
is undefined. With matrices, there are many matrices which don't have inverses. There are two reasons why a matrix may not have an inverse:
1. It is not square; only square matrices have inverses.
2. Its determinant (check out the unit on Determinants for more information on evaluating the determinant of a matrix) is zero.
You might wonder what determinants have to do with inverses of matrices, and I can explain that in a loose way with an example. Suppose you wanted to find the inverse of the matrix
3
1
5
2
. One way to do it would be to set up an equation:
3
1
5
2
a
b
c
d
=
1
0
0
1
This matrix equation will give you a set of four equations in four unknowns:
3a + 1c = 1
3b + 1d = 0
5a + 2c = 0
5b + 2d = 1
A system of four equations with four unknowns...from our unit on determinants, you know that one of the ways to solve such a system is with Cramer's Rule, and the only time there is no solution is if the determinant has a zero value. Thus, when the determinant is zero, there is no set of 4 numbers that produces an inverse.
Fortunately, someone has gone to the trouble of creating a mini-formula/algorithm for you, to save you having to use Cramer's Rule every time you want to find the inverse of a 2 x 2 matrix.
w
x
y
z
, and you want to find its inverse, you do the following:
Evaluate d =
w
x
y
z
Create a new matrix that looks like this:
z
d
-x
d
-y
d
w
d
This new matrix is the inverse of the original matrix. Notice that the w and z have switched places, and the x and y have become negative.
Are there methods for finding the inverses of 3 x 3 matrices? 4 x 4 matrices? Yes, there are. In fact, back in the dark ages of my high school days I wrote a three-page process proof for finding the inverse of any n x n matrix. It made me feel good, but it's not terribly practical in the days when computers can handle those horrifically complex calculations. For our purposes here, it is enough to show you (as we did above) how you would go about manually finding the inverse of a 2 x 2 using systems of equations, as well as the algorithmic short cut!
Questions
1.
Can 5 x 5 matrices have inverses?
2.
Can 4 x 3 matrices have inverses?
3.
Does the matrix
2
5
1
3
have an inverse? How do you know?
4.
Does the matrix
2
6
1
3
have an inverse? How do you know?
5.
What is the inverse of I2 (I2 =
1
0
0
1
)
6.
What are the entries in the inverse of
2
6
1
4
? List them clockwise starting with the first row, first column.
Assign this reference page
|
# In each of the figures given below, an altitude is drawn to the hypotenuse by a right-angled triangle.
Question:
In each of the figures given below, an altitude is drawn to the hypotenuse by a right-angled triangle. The length of different line-segment are marked in each figure. Determine xyz in each case.
Solution:
(i) $\triangle A B C$ is right angled triangle right angled at $B$
$A B^{2}+B C^{2}=A C^{2}$
$x^{2}+z^{2}=(4+5)^{2}$
$x^{2}+z^{2}=9^{2}$
$x^{2}+z^{2}=81$......$(i)$
$\triangle B D A$ is right triangle right angled at $\mathrm{D}$
$B D^{2}+A D^{2}=A B^{2}$
$y^{2}+4^{2}=x^{2}$
$y^{2}+16=x^{2}$
$16=x^{2}-y^{2}$...(ii)
$\triangle B D C$ is right triangle right angled at $D$
$B D^{2}+D C^{2}=B C^{2}$
$y^{2}+25=z^{2}$
$25=z^{2}-y^{2}$$\ldots \ldots( iii ) By canceling equation(i) and(ii) by elimination method, we get y canceling and by elimination method we get z^{2}=45 z=\sqrt{45} z=\sqrt{3 \times 3 \times 5} z=3 \sqrt{5} Now, substituting z^{2}=45 in equation (iv) we get y^{2}+z^{2}=65 y^{2}+45=65 y^{2}=65-45 y^{2}=20 y=\sqrt{20} y=\sqrt{2 \times 2 \times 5} y=2 \sqrt{5} Now, substituting y^{2}=20 in equation (ii) we get x^{2}-y^{2}=16 x^{2}-20=16 x^{2}=16+20 x^{2}=36 x=\sqrt{36} x=\sqrt{6 \times 6} x=6 Hence the values of x, y, z is 6,2 \sqrt{5}, 3 \sqrt{5} (ii) \triangle P Q R is a right triangle, right angled at Q 6+z^{2}=(4+x)^{2} 36+z^{2}=16+x^{2}+8 x z^{2}-x^{2}-8 x=16-36 z^{2}-x^{2}-8 x=-20 \ldots \ldots (i) \Delta Q S P is a right triangle right angled at S Q S^{2}+P S^{2}=P Q^{2} y^{2}+4^{2}=6^{2} y^{2}+16=36 y^{2}=36-16 y^{2}=20 y=\sqrt{20} y=\sqrt{2 \times 2 \times 5} y=2 \sqrt{5} \triangle Q S R is a right triangle right angled at S Q S^{2}+R S^{2}=Q R^{2} y^{2}+x^{2}=z^{2}.....(2) Now substituting y^{2}+x^{2}=z^{2} in equation (i) we get y^{2}+x^{2}-x^{2}-8 x=-20 y^{2}+x^{2}-x^{2}-8 x=-20 y^{2}-8 x=-20$$\ldots \ldots(i i i)$
Now substituting $y^{2}=20$ in equation (iii) we get
$y^{2}-8 x=-20$
$20-8 x=-20$
$-8 x=-20-20$
$-8 x=-40$
$x=\frac{40}{8}$
$x=5$
Now substituting $x=5$ and $y^{2}=20$ in equation (ii) we get
$y^{2}+x^{2}=z^{2}$
$20+5^{2}=z^{2}$
$20+25=z^{2}$
$45=z^{2}$
$\sqrt{3 \times 3 \times 5}=z^{2}$
$3 \sqrt{5}=z$
Hence the value of $x, y$ and $z$ are $5,2 \sqrt{5}, 3 \sqrt{5}$
|
# Initial Velocity Formula
The velocity at which motion start is termed as Initial Velocity. It is velocity at time interval t = 0. It is represented by u.Three initial velocity formulas are there-
If final time, acceleration and velocity are provided. The initial velocity is articulated as,
u =v – at
If final velocity, acceleration, and distance are provided we make use of
u= v– 2as
If distance, acceleration and time are provided. The initial velocity is
$u=\frac{s}{t}-\frac{1}{2}at$
Where,
Initial velocity = u,
Final Velocity = v,
time taken = t,
distance traveled or displacement = s,
acceleration = a
Initial Velocity formula is made use of to find the initial velocity of the body if some of the quantities are given. Initial velocity is articulated in meter per second (m/s).
Initial Velocity Solved Examples
Below are some problems based on Initial velocity which may be helpful for you.
Problem 1: Johny completes the bicycle ride with the final velocity of 10 ms-1 and acceleration 2 ms-2 within 3s. Calculate the initial velocity.
Answer:
Given: v (Final velocity) = 10 ms-1,
a (Acceleration) = 2ms-2,
t (Time taken) = 3 s
u (Initial velocity) = ?
v (Final velocity) = u + at
u (Initial velocity) = v – at = 10 ms-1 – (2ms-2)3s
= 4 ms-1
∴ (Initial velocity) u = 4ms-1
Problem 2: A man covers a distance of 100 m. If he has a final velocity of 40 ms-1 and has acceleration of 6 ms-2. Compute his initial velocity?
Answer:
Given: Distance s = 100m,
(Final velocity) v = 40 ms-1,
(Acceleration) a = 6ms-2
(Initial velocity) u2 = v2 – 2as
= 1600 – 2 ×× 6 ×× 100
= 1600 – 1200
= 400 ms-1
∴ (Initial velocity) u = 20 ms-1.
|
How To Find Range Of A Function. Solve it for \(x\) to write it in the form, \(x=g(y)\) step 3: The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain.
Find the equation for x that gives y = f(x) and then solve it using this equation. To find the quadratic function range, it suffices to see whether it has a maximum or minimum value. If we are able to calculate the maximum and the minimum values of the function, we can get an idea of the range of the function.
### With That, We Can Find F.
Then, plug that answer into the function to find the range. If we are able to calculate the maximum and the minimum values of the function, we can get an idea of the range of the function. The domain of the function f(x) is the set of all those real numbers for which the expression for f(x) or the formula for f(x) assumes real values only.
### The Spread Of All The Y Values From Minimum To Maximum Is The Range Of The Function.
The range is the spread of the values of the output of a function. To find the range of a function: Let’s consider finding the range of a function such as f(x)=x+2f(x)=x+2.
### Solve The Equation Y = F(X) For X In Terms Of Y.
In math, the range of a function is a set of numbers that the function is capable of producing. We can find the range of a function by following these steps: In plain english, the definition means:
### Other Strategies For Finding Range Of A Function As We Saw In The Previous Example, Sometimes We Can Find The Range Of A Function By Just Looking At Its Graph.
Try graphing the function in order to determine the range if you cannot find x. Write down the function in the form \(y=f(x)\) step 2: The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain.
### Consider A Function Y = F(X).
The period of a function helps us to know the interval, after which the range of the periodic function is repeated. In order to find range of function, we use following algorithm. To find the range of a function, my first instinct is to check whether the graph has an inverse.
Categories: how to make
|
# Pre-Algebra : Word Problems with One Unknown
## Example Questions
← Previous 1 3 4 5
### Example Question #1 : Word Problems With One Unknown
John makes $12.50/hour gardening. He earned$275 this week. Write the equation for figuring out his wages, and then solve it to find out how many hours he worked.
Explanation:
Step 1: In order to find out how many hours John works, you first need a formula for calculating his wages:
(hourly wage)(hours)= total wages
Step 2: substitute the known values (hourly wage, total wages)
Step 3: isolate the unknown variable (hours) by dividing both sides by his hourly wage, then solve
Sarah is building a fence for her dog's square play area. To reduce the cost, she is using her house as 1 side of the play area, meaning she only has to purchace fencing for the other sides. If she needs to fence in 225 meters2 for her dogs, and fencing costs $2.50 per foot of fencing, what will be the cost of fencing in this square play area? Possible Answers: Correct answer: Explanation: The first step is to figure out the perimeter of the square play yard with an area of 225 meters, first using the fomula: Find the square root of both sides to calculate the length of the base of the square All the sides are of equivilent length, so the total amount of fence required is: One side is "fenced" by the house, so that fenching does not need to be paid for. Thus, only the remaining 3 sides need to be paid for. Finally, multiply the length of fencing needs by the cost of fence per foot to find your answer: ### Example Question #2 : Word Problems 3 people can pave a driveway in 4 hours. How long will it take for 8 people to pave a driveway? Possible Answers: Correct answer: Explanation: With inverse proportionality, when one quantity increases the other decreases, and vice versa. The key to solving this problem is to keep in mind that each person works at the same rate regardless of how many people share the workload. Let: = constant of proportionality (rate of work per person) = time = number of people Using these variables, we can set up an equation that will give us the total time: Solve for using the original information for 3 people. Using this constant, we can return to the first equation and solve for the time when 8 people are working: ### Example Question #1 : Word Problems With One Unknown While reading a book Sarah notices that she has read 3 pages after 10 minutes. If she keeps reading at this rate, how many pages will she have read in 1 hour? Possible Answers: Correct answer: Explanation: The number of pages read is directly proportional to the time spent reading. To solve this problem we need to find the constant of proportionality, which is represented by in the following relation: Here is the number of pages read, and is time. Rearrange to solve for the constant: Using the information given: One hour is 60 minutes, so using the direct proportionality equation again gives: ### Example Question #5 : Word Problems With One Unknown If it takes 9 hours for 4 people to build a shed, how many people are needed to build the same type of shed in one-third of the time? Possible Answers: Correct answer: Explanation: The constant of proportionality is equal to the product of time () and the number of people (). Let represent the time and number of people working on the first shed, and represent the time and number of people working on the second shed. Since is the same for both situations (it is a constant), we can set the first and second scenarios equal to each other. Now, this problem asks for the number of people needed to build the second shed. This means we need to find a way to solve for . We can divide both sides of our equation by : Use the given values to solve. Note that will be equal to 3 (one-third of the original 9). ### Example Question #2 : Word Problems With One Unknown It takes 45 minutes to drive to the nearest bowling alley taking city streets going 30 miles per hour (mph). How long will it take to get there using the freeway going 65mph if the distance is the same? (Round to the nearest minute). Possible Answers: Correct answer: Explanation: We can set up a rate equation, in which distance is equal to rate (speed) times time: Our speed and time may change with the different routes, but the distance stays the same. City route: Freeway route: We can set these equations equal to each other, since both are equal to . Solve for the freeway time, , by rearranging and substituting the given values. Rounding to the nearest minute, we get 21 for our final answer. ### Example Question #7 : Word Problems With One Unknown A father is buying cheeseburgers for his children. Each cheeseburger costs$3.50. He spends \$17.50 on cheeseburgers. How many cheeseburgers did he buy?
Explanation:
Set up and equation where is equal to the number of cheeseburgers:
Solve for c:
### Example Question #8 : Word Problems With One Unknown
Grace is 4 years older than Elena. If Grace is 9 years old, how old is Elena?
Explanation:
Let Elena's be equal to :
### Example Question #9 : Word Problems With One Unknown
Suzie needs to ride her bike to the grocery store, which is is 15 miles away. It takes Suzie 45 minutes to bike to the grocery store. What is Suzies average speed? Express answer in miles per hour.
Explanation:
Distance is the product of rate and time:
Plug in the information given in the problem, remembering that the questions says to state the answer in terms of miles per hour, and we are give her time in minutes. Change minutes to hours:
It took Suzie .75 hours to ride to the grocery store.
Suzie's rate is .
### Example Question #10 : Word Problems With One Unknown
An ice cream cone costs plus sales tax. How many ice cream cones can be purchased for ?
Explanation:
Calculate the cost of an ice cream cone with tax:
Determine how many cones, , can be purchased for by setting up the following equation:
Divide both sides by 3.18:
Therefore, ice cream cones can be purchased for .
← Previous 1 3 4 5
|
# The perimeter of a triangle is 300 m. If its sides are in the ratio of 3: 5: 7. Find the area of the triangle.
Question:
The perimeter of a triangle is 300 m. If its sides are in the ratio of 3: 5: 7. Find the area of the triangle.
Solution:
Given the perimeter of a triangle is 300 m and the sides are in a ratio of 3: 5: 7
Let the sides a, b, c of a triangle be 3x, 5x, 7x respectively
So, the perimeter = 2s = a + b + c
200 = a + b + c
300 = 3x + 5x + 7x
300 = 15x
Therefore, x = 20 m
So, the respective sides are
a = 60 m
b = 100 m
c = 140 m
Now, semi perimeter
$s=\frac{a+b+c}{2}$
$=\frac{60+100+140}{2}$
= 150 m
By using Heron's Formula
The area of a triangle $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{150 \times(150-60) \times(150-100) \times(150-140)}$
$=1500 \sqrt{3} \mathrm{~m}^{2}$
Thus, the area of a triangle is $1500 \sqrt{3} \mathrm{~m}^{2}$
|
# How do you convert (1, pi/4, 2) into spherical form?
Sep 12, 2017
see below
#### Explanation:
This point is in cylindrical form $\left(r , \theta , z\right)$. So let's first convert it to rectangular form $\left(x , y , z\right)$ by using the formulas $x = r \cos \theta , y = r \sin \theta , z = z$
That is,
$x = 1 \cdot \cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} , y = 1 \cdot \sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} , z = 2$
hence, the point is $\left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} , 2\right)$.
Now let's use the formulas
${\rho}^{2} = {x}^{2} + {y}^{2} + {z}^{2} , x = \rho \sin \phi \cos \theta , y = \rho \sin \phi \sin \theta , z = \rho \cos \phi$ to change the point to spherical coordinates.
$\rho = \sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^{2} + {\left(\frac{1}{\sqrt{2}}\right)}^{2} + {\left(2\right)}^{2}} = \sqrt{\frac{1}{2} + \frac{1}{2} + 4} = \sqrt{5}$
To find $\phi$ let's use the formula $z = \rho \cos \phi$
Therefore,
$z = \rho \cos \phi$
$2 = \sqrt{5} \cos \phi$
${\cos}^{-} 1 \left(\frac{2}{\sqrt{5}}\right) = \phi$
$\phi \approx 0.46$
$\therefore \left(\rho , \theta , \phi\right) \approx \left(\sqrt{5} , \frac{\pi}{4} , 0.46\right) \approx \left(2.24 , 0.79 , 0.46\right)$
|
Associated Topics || Dr. Math Home || Search Dr. Math
### Converting Between Mixed Numbers and Improper Fractions
```
Date: 04/08/99 at 18:31:46
From: Linday Stahl
Subject: Undoing improper fractions
I get mixed up doing and undoing improper and mixed numbers.
```
```
Date: 04/08/99 at 23:16:02
From: Doctor Peterson
Subject: Re: Undoing improper fractions
Hi, Linday.
Let's take a simple example:
What is 1 1/2 as an improper fraction? We can picture it as a box and
half a box:
+-------+ +-------+
| | | |
| | +-------+
| |
+-------+
If we turn the whole into halves, there are two halves there, plus the
extra half makes 3:
+-------+ +-------+
| | | |
+-------+ +-------+
| |
+-------+
What we've just done is to rewrite 1 as 2/2, then add 2/2 + 1/2 to get
3/2.
In general, you multiply the whole part by the denominator, then add
the numerator of the fraction part. One way to think of this is that
you start at the "bottom" and move clockwise around the mixed number,
multiplying by the whole, then adding the numerator:
+--------+
| |
| + 1 +-> 2 x 1 + 1 = 3
| 1 --- ---
| x 2 ----------------> 2
|
| |
+----+
To go the other way, let's start with the 3/2. How many wholes can we
make out of 3 halves? Just divide the 3 by 2, because it takes 2
halves to make each whole; we get 1 with a remainder of 1. That means
there is one whole, and 1 half left over:
+--------------+
| |
| __1_rem 1 |
2 ) 3 --- |
-2 2 <--+
So you just use the denominator (divisor) as the denominator with the
remainder as the numerator.
If you have any specific problems you need help with, feel free to
write back.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Elementary Fractions
Middle School Fractions
Search the Dr. Math Library:
Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words
Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search
|
Poker Drawing Odds & Outs
By Tom "TIME" Leonard
In our poker math and probability lesson it was stated that when it comes to poker; “the math is essential“. Although you don’t need to be a math genius to play poker, a solid understanding of probability will serve you well and knowing the odds is what it’s all about in poker. It has also been said that in poker, there are good bets and bad bets. The game just determines who can tell the difference. That statement relates to the importance of knowing and understanding the math of the game.
In this lesson we’re going to focus on drawing odds in poker and how to calculate your chances of hitting a winning hand. We’ll start with some basic math before showing you how to correctly calculate your odds. Don’t worry about any complex math – we will show you how to crunch the numbers, but we’ll also provide some simple and easy shortcuts that you can commit to memory.
Basic Math – Odds and Percentages
Odds can be expressed both “for” and “against”. Let’s use a poker example to illustrate. The odds against hitting a flush when you hold four suited cards with one card to come is expressed as approximately 4-to-1. This is a ratio, not a fraction. It doesn’t mean “a quarter”. To figure the odds for this event simply add 4 and 1 together, which makes 5. So in this example you would expect to hit your flush 1 out of every 5 times. In percentage terms this would be expressed as 20% (100 / 5).
Here are some examples:
• 2-to-1 against = 1 out of every 3 times = 33.3%
• 3-to-1 against = 1 out of every 4 times = 25%
• 4-to-1 against = 1 out of every 5 times= 20%
• 5-to-1 against = 1 out of every 6 times = 16.6%
Converting odds into a percentage:
• 3-to-1 odds: 3 + 1 = 4. Then 100 / 4 = 25%
• 4-to-1 odds: 4 + 1 = 5. Then 100 / 5 = 20%
Converting a percentage into odds:
• 25%: 100 / 25 = 4. Then 4 – 1 = 3, giving 3-to-1 odds.
• 20%: 100 / 20 = 5. Then 5 – 1 = 4, giving 4-to-1 odds.
Another method of converting percentage into odds is to divide the percentage chance when you don’t hit by the percentage when you do hit. For example, with a 20% chance of hitting (such as in a flush draw) we would do the following; 80% / 20% = 4, thus 4-to-1. Here are some other examples:
• 25% chance = 75 / 25 = 3 (thus, 3-to-1 odds).
• 30% chance = 70 / 30 = 2.33 (thus, 2.33-to-1 odds).
Some people are more comfortable working with percentages rather than odds, and vice versa. What’s most important is that you fully understand how odds work, because now we’re going to apply this knowledge of odds to the game of poker.
Before you can begin to calculate your poker odds you need to know your “outs”. An out is a card which will make your hand. For example, if you are on a flush draw with four hearts in your hand, then there will be nine hearts (outs) remaining in the deck to give you a flush. Remember there are thirteen cards in a suit, so this is easily worked out; 13 – 4 = 9.
Another example would be if you hold a hand like and hit two pair on the flop of . You might already have the best hand, but there’s room for improvement and you have four ways of making a full house. Any of the following cards will help improve your hand to a full house; .
The following table provides a short list of some common outs for post-flop play. I recommend you commit these outs to memory:
Table #1 – Outs to Improve Your Hand
The next table provides a list of even more types of draws and give examples, including the specific outs needed to make your hand. Take a moment to study these examples:
Table #2 – Examples of Drawing Hands (click to enlarge)
Counting outs is a fairly straightforward process. You simply count the number of unknown cards that will improve your hand, right? Wait… there are one or two things you need to consider:
Don’t Count Outs Twice
There are 15 outs when you have both a straight and flush draw. You might be wondering why it’s 15 outs and not 17 outs, since there are 8 outs to make a straight and 9 outs for a flush (and 8 + 9 = 17). The reason is simple… in our example from table #2 the and the will make a flush and also complete a straight. These outs cannot be counted twice, so our total outs for this type of draw is 15 and not 17.
Anti-Outs and Blockers
There are outs that will improve your hand but won’t help you win. For example, suppose you hold on a flop of . You’re drawing to a straight and any two or any seven will help you make it. However, the flop also contains two hearts, so if you hit the or the you will have a straight, but could be losing to a flush. So from 8 possible outs you really only have 6 good outs.
It’s generally better to err on the side of caution when assessing your possible outs. Don’t fall into the trap of assuming that all your outs will help you. Some won’t, and they should be discounted from the equation. There are good outs, no-so good outs, and anti-outs. Keep this in mind.
Calculating Your Poker Odds
Once you know how many outs you’ve got (remember to only include “good outs”), it’s time to calculate your odds. There are many ways to figure the actual odds of hitting these outs, and we’ll explain three methods. This first one does not require math, just use the handy chart below:
Table #3 – Poker Odds Chart
As you can see in the above table, if you’re holding a flush draw after the flop (9 outs) you have a 19.1% chance of hitting it on the turn or expressed in odds, you’re 4.22-to-1 against. The odds are slightly better from the turn to the river, and much better when you have both cards still to come. Indeed, with both the turn and river you have a 35% chance of making your flush, or 1.86-to-1.
We have created a printable version of the poker drawing odds chart which will load as a PDF document (in a new window). You’ll need to have Adobe Acrobat on your computer to be able to view the PDF, but this is installed on most computers by default. We recommend you print the chart and use it as a source of reference. It should come in very handy.
Doing the Math – Crunching Numbers
There are a couple of ways to do the math. One is complete and totally accurate and the other, a short cut which is close enough.
Let’s again use a flush draw as an example. The odds against hitting your flush from the flop to the river is 1.86-to-1. How do we get to this number? Let’s take a look…
With 9 hearts remaining there would be 36 combinations of getting 2 hearts and making your flush with 5 hearts. This is calculated as follows:
(9 x 8 / 2 x 1) = (72 / 2) ≈ 36.
This is the probability of 2 running hearts when you only need 1 but this has to be figured. Of the 47 unknown remaining cards, 38 of them can combine with any of the 9 remaining hearts:
9 x 38 ≈ 342.
Now we know there are 342 combinations of any non heart/heart combination. So we then add the two combinations that can make you your flush:
36 + 342 ≈ 380.
The total number of turn and river combos is 1081 which is calculated as follows:
(47 x 46 / 2 x 1) = (2162 / 2) ≈ 1081.
Now you take the 380 possible ways to make it and divide by the 1081 total possible outcomes:
380 / 1081 = 35.18518%
This number can be rounded to .352 or just .35 in decimal terms. You divide .35 into its reciprocal of .65:
0.65 / 0.35 = 1.8571428
And voila, this is how we reach 1.86. If that made you dizzy, here is the short hand method because you do not need to know it to 7 decimal points.
The Rule of Four and Two
A much easier way of calculating poker odds is the 4 and 2 method, which states you multiply your outs by 4 when you have both the turn and river to come – and with one card to go (i.e. turn to river) you would multiply your outs by 2 instead of 4.
Imagine a player goes all-in and by calling you’re guaranteed to see both the turn and river cards. If you have nine outs then it’s just a case of 9 x 4 = 36. It doesn’t match the exact odds given in the chart, but it’s accurate enough.
What about with just one card to come? Well, it’s even easier. Using our flush example, nine outs would equal 18% (9 x 2). For a straight draw, simply count the outs and multiply by two, so that’s 16% (8 x 2) – which is almost 17%. Again, it’s close enough and easy to do – you really don’t have to be a math genius.
Conclusion
In this lesson we’ve covered a lot of ground. We haven’t mentioned the topic of pot odds yet – which is when we calculate whether or not it’s correct to call a bet based on the odds. This lesson was step one of the process, and in our pot odds lesson we’ll give some examples of how the knowledge of poker odds is applied to making crucial decisions at the poker table.
As for calculating your odds…. have faith in the tables, they are accurate and the math is correct. Memorize some of the common draws, such as knowing that a flush draw is 4-to-1 against or 20%. The reason this is easier is that it requires less work when calculating the pot odds, which we’ll get to in the next lesson.
Tom has been writing about poker since 1994 and has played across the USA for over 40 years, playing every game in almost every card room in Atlantic City, California and Las Vegas.
1. The odds against hitting a flush when you hold four suited cards with one card to come is expressed as approximately 4-to-1. What is this in percentage terms?
2. How much are odds of 5-to-1 against, when expressed as a percentage?
3. Odds of 3-to-1 against means you will make it 1 out of every 4 times. True or false?
4. You flop two pair. How many outs do you have to make a full house?
5. How many outs do you have to make a straight on the turn?
6. You put your opponent on a low/medium pair. How many outs do you have to win?
7. With a straight and flush draw, how many outs do you have?
8. You have 9 outs after the flop. What are your odds of hitting with both the turn and river to come?
9. You have 9 outs after the turn. What are the odds of hitting one of these outs on the river?
10. You have 2 outs after the turn. What are your odds of hitting with just the river card to come?
Share:
|
# If a, b, c are in AP, show that
Question:
If a, b, c are in AP, show that
(i) $(b+c-a),(c+a-b),(a+b-c)$ are in AP.
(ii) $\left(b c-a^{2}\right),\left(c a-b^{2}\right),\left(a b-c^{2}\right)$ are in AP.
Solution:
(i) $(b+c-a),(c+a-b),(a+b-c)$ are in AP.
To prove: $(b+c-a),(c+a-b),(a+b-c)$ are in AP.
Given: a, b, c are in A.P.
Proof: Let d be the common difference for the A.P. a,b,c
Since a, b, c are in A.P
$\Rightarrow b-a=c-b=$ common difference
$\Rightarrow a-b=b-c=d$
$\Rightarrow 2(a-b)=2(b-c)=2 d \ldots$ (i)
Considering series $(b+c-a),(c+a-b),(a+b-c)$
For numbers to be in A.P. there must be a common difference between them
Taking (b + c – a) and (c + a – b)
Common Difference $=(c+a-b)-(b+c-a)$
$=c+a-b-b-c+a$
$=2 a-2 b$
$=2(a-b)$
$=2 \mathrm{~d}[$ from eqn. (i) $]$
Taking (c + a – b) and (a + b – c)
Common Difference $=(a+b-c)-(c+a-b)$
$=a+b-c-c-a+b$
$=2 b-2 c$
$=2(b-c)$
$=2 d$ [from eqn. (i)]
Here we can see that we have obtained a common difference between numbers i.e. 2d
Hence, $(b+c-a),(c+a-b),(a+b-c)$ are in AP.
(ii) $\left(b c-a^{2}\right),\left(c a-b^{2}\right),\left(a b-c^{2}\right)$ are in AP.
To prove: $\left(b c-a^{2}\right),\left(c a-b^{2}\right),\left(a b-c^{2}\right)$ are in AP.
Given: $a, b, c$ are in A.P.
Proof: Let d be the common difference for the A.P. a,b,c
Since a, b, c are in A.P.
$\Rightarrow b-a=c-b=$ common difference
$\Rightarrow a-b=b-c=d \ldots$ (i)
Considering series $\left(b c-a^{2}\right),\left(c a-b^{2}\right),\left(a b-c^{2}\right)$
For numbers to be in A.P. there must be a common difference between them
Taking $\left(b c-a^{2}\right)$ and $\left(c a-b^{2}\right)$
Common Difference $=\left(c a-b^{2}\right)-\left(b c-a^{2}\right)$
$=\left[c a-b^{2}-b c+a^{2}\right]$
$=\left[c a-b c+a^{2}-b^{2}\right]$
$=[c(a-b)+(a+b)(a-b)]$
$=[(a-b)(a+b+c)]$
$a-b=d$, from eqn. (i)
$\Rightarrow[(d)(a+b+c)]$
Taking $\left(c a-b^{2}\right)$ and $\left(a b-c^{2}\right)$
Common Difference $=\left(a b-c^{2}\right)-\left(c a-b^{2}\right)$
$=\left[a b-c^{2}-c a+b^{2}\right]$
$=\left[a b-c a+b^{2}-c^{2}\right]$
$=[a(b-c)+(b-c)(b+c)]$
$=[(b-c)(a+b+c)]$
$b-c=d$, from eqn. (i)
$\Rightarrow[(d)(a+b+c)]$
Here we can see that we have obtained a common difference between numbers
i.e. [(d) $(a+b+c)$ ]
Hence, $\left(b c-a^{2}\right),\left(c a-b^{2}\right),\left(a b-c^{2}\right)$ are in AP.
|
3.3 Limits and Continuity: Algebraic Approach
(This topic is also in Section 3.3 in Applied Calculus or Section 10.3 in Finite Mathematics and Applied Calculus)
Consider the following limit.
limx2 x2 - 3x4x + 3 .
If you estimate the limit either numerically or graphically, you will find that
limx2 x2 - 3x4x + 3 0.1818
But, notice that you can obtain this answer by simply substituting x = 2 in the given function:
f(x) = x2 - 3x4x + 3 f(2) = 4 - 68 + 3 = - 211 = -0.181818...
This answer is more accurate than the one coming from numerical or graphical method; in fact, it gives the exact limit.
Q Is that all there is to evaluating limits algebraically: just substitute the number x is approaching in the given expression?
A Not always, but this often does happen, and when it does, the function is continuous at the value of x in question. Recall the definition of continuity from the previous tutorial:
Continuous Functions
The function f(x) is continuous at x = a if
limxa f(x) exists That is, the left-and right limits exist and agree with each other limxa f(x) = f(a)
The function f is said to be continuous on its domain if it is continuous at each point in its domain. If f is not continuous at a particular a, we say that f is discontinuous at a or that f has a discontinuity at a.
Q How do you tell if a function is continuous?
A As we saw in the previous tutorial, we can tell whether a function is continuous by looking at its graph. If the graph breaks at some point in the domain, then f has a discontinuity there. If the function is specified algrabcially, sometimes it is easy to tell whether it is continuous by just looking at the formula:
A closed-form function is any function that can be obtained by combining constants, powers of x, exponential functions, radicals, absolute values, trigonometric functions, logarithms (and some other functions you may never see) into a single mathematical formula by means of the usual arithmetic operations and composition of functions. Examples of closed-form functions are:
3x2 - x +1, (x2 - 1)1/26x-2
e(x2 - 1)1/2/x
(log3(4x2 - ex))2/3
They can be as complicated as you like. The following is not a closed form function.
f(x) = -1 if x < -1 x2 + x if -1 x 1 2 - x if 1 < x 2
The reason for this is that f(x) is not specified by a single mathematical expression. What is nice about closed-form functions is the following.
Continuity of Closed Form Functions
Every closed form function is continuous on its domain.
Thus, the limit of a closed-form function at a point on its domain can be obtained by substitution.
Example
f(x) = x2 - 3x4x + 3
is a closed form function, and x = 2 is in its domain. Therefore we can obtain limx→2 f(x) by substitution:
limx2 x2 - 3x4x + 3 = f(2) = - 211 ,
as we saw above.
Q limx0 ex22x-3 Select one does not exist = 0 = -4/3 = -1/3 = 4/3 is undefined = -e/3 none of the above Q lim x3 2x + 3x + 3 =
Q What if f(x) is a closed form function, but the point x = a is not in the domain of the function?
A Then, you either:
1. Use simplification or some other technique to replace f(x) by another closed form function which does have x = a in its domain. This allows you to substitute x = a in the new function to obtain the limit, or
2. Try evaluating the limit numerically or graphically. Note, however, that this may only give you an estimate of the limit, so we suggests trying approach (a) first.
Sometimes, you may need to use both (a) and (b).
Evaluating a Limit Using Simplification
(Similar to Example 2 in Section 3.8 of Applied Calculus, or in Section 11.8 of Finite Mathematics and Applied Calculus )
Let us evaluate
lim x-2 3x2 + x - 10x + 2
Ask yourself the following questions:
1. Is the function f(x) a closed form function?
Answer: yes, since (3x2 + x-10)/(x+2) is a single mathematical formula.
2. Is the value x = a in the domain of f(x)?
Answer: no, since (3(-2)2+(-2)-10)/((-2)+2) is not defined.
Therefore, we consult the above Question/Answer discussion, and simplify the function, if we can.
3x2 + x - 10
x + 2
= (x+2)(3x-5) (x+2)
= 3x-5.
Since we are now left with a closed form function that is defined when x = -2, we can now evaluate the limit by substitution:
3x - 5 lim x-2 3x2 + x-10x + 2 = lim x-2 = 3(-2)-5 = -11.
lim x-1 (x3+1)(x-1)x2 + 3x + 2 Select one does not exist = + infinity = - infinity is undefined = 0 = -2 = -4 = -6 = -8 none of the above
Back in the tutorial for Section 1.2, we looked at the following function:
f(x)= -1 if -4 x < -1 x if -1 x 1 x2-1 if 1 < x 2
This time, we are not showing you the graph right away, and ask you to look at the formula instead. Notice:
1. The function f is not closed form. (It is not defined by a single formula.)
2. Inside each of the separate intervals [-4, -1), (-1, 1) and (1, 2] the function is closed form, and hence continuous. Therefore, the only conceivable points where the function might fail to be continuous are on the boundaries of these intervals where we switch from one formula to another: x = -1 and x = 1.
Q The function f is is not continuous at x = 0. Q The function f is is not continuous at x = -1. Q The function f is is not continuous at x = 1.
Now try the rest of the exercises in Section 3.3 in Applied Calculus or Section 10.3 in Finite Mathematics and Applied Calculus
Top of Page
Last Updated: March, 2007
Copyright © 1999, 2003, 2006, 2007 Stefan Waner
|
# Percentage Questions
FACTS AND FORMULAE FOR PERCENTAGE QUESTIONS
I.Concept of Percentage : By a certain percent , we mean that many hundredths. Thus x percent means x hundredths, written as x%.
To express x% as a fraction : We have , x% = x/100.
Thus, 20% = 20/100 = 1/5;
48% = 48/100 = 12/25, etc.
To express a/b as a percent : We have, $\frac{a}{b}=\left(\frac{a}{b}×100\right)%$ .
Thus, $\frac{1}{4}=\left(\frac{1}{4}×100\right)%=25%$
II. If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is $\left[\frac{R}{\left(100+R\right)}×100\right]%$
If the price of the commodity decreases by R%,then the increase in consumption so as to decrease the expenditure is $\left[\frac{R}{\left(100-R\right)}×100\right]%$
III. Results on Population : Let the population of the town be P now and suppose it increases at the rate of R% per annum, then :
1. Population after n years = $P{\left(1+\frac{R}{100}\right)}^{n}$
2. Population n years ago = $\frac{P}{{\left(1+\frac{R}{100}\right)}^{n}}$
IV. Results on Depreciation : Let the present value of a machine be P. Suppose it depreciates at the rate R% per annum. Then,
1. Value of the machine after n years = $P{\left(1-\frac{R}{100}\right)}^{n}$
2. Value of the machine n years ago = $\frac{P}{{\left(1-\frac{R}{100}\right)}^{n}}$
V. If A is R% more than B, then B is less than A by
$\left[\frac{R}{\left(100+R\right)}×100\right]%$
If A is R% less than B , then B is more than A by
$\left[\frac{R}{\left(100-R\right)}×100\right]%$
Q:
If the cost of production doubles in a period of 3 years, then the corresponding maintenance cost in rupees will be
A) ₹ 75 lakh B) ₹ 12.5 lakh C) ₹ 25 lakh D) ₹ 125 lakh
Answer & Explanation Answer: C) ₹ 25 lakh
Explanation:
Report Error
0 458
Q:
If the production increases to five times of the present cost, then the percentage increase of the cost is
A) 50% B) 400% C) 300% D) 500%
Answer & Explanation Answer: B) 400%
Explanation:
Report Error
2 452
Q:
A number is increased by 30%, then decreased by 30%, then further decreased by 30%. What is the net increase/decrease percent in the number(correct to the nearest integer)?
A) 36% decrease B) 36% increase C) 40% increase D) 40% decrease
Answer & Explanation Answer: A) 36% decrease
Explanation:
Report Error
4 371
Q:
A certain amount invested at a certain rate, compounded annually, grows to an amount in two years, which is a factor of 1.1881 more than to what it would have grown in three years. What is the rate percentage?
A) 8 B) 8.1 C) 9.2 D) 9
Answer & Explanation Answer: D) 9
Explanation:
Report Error
2 356
|
Open in App
Not now
# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.17
• Last Updated : 19 Apr, 2021
### Question 1. ∫dx/√(2x – x2)
Solution:
We have,
Let I = ∫dx/√(2x – x2)
= ∫dx/√(1 – 1 + 2x – x2)
= ∫dx/√{1 – (x2 – 2x + 1)}
= ∫dx/√{12 – (x – 1)2}
Let x – 1 = q …(1)
= ∫dx/√{12 – (q)2}
As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)
So,
= sin-1(q) + C
Now put the value of q from eq(1), we get
= sin-1(x – 1) + C
### Question 2. ∫dx/√(8 + 3x – x2)
Solution:
We have,
Let I = ∫dx/√(8 + 3x – x2)
Here, (8 + 3x – x2) can be written as 8 – (x2 – 3x + 9/4 – 9/4)
= (8 + 9/4) – (x – 3/2)2
= (41/4) – (x – 3/2)2
Let x – 3/2 = q …(1)
As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)
So,
Now put the value of q from eq(1), we get
= sin-1(2x – 3/√41) + C
### Question 3. ∫dx/√(5 – 4x – 2x2)
Solution:
We have,
Let I = ∫dx/√(5 – 4x – 2x2)
= ∫dx/√{2(5/2 – 2x – x2)}
= (1/√2)∫dx/√{5/2 – (x2 – 2x + 1 – 1)}
= (1/√2)∫dx/√{(5/2 + 1) – (x2 – 2x + 1)}
= (1/√2)∫dx/√{7/2 – (x – 1)2}
=
Let x + 1 = q …(1)
As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)
So,
Now put the value of q from eq(1), we get
### Question 4. ∫dx/√(3x2 + 5x + 7)
Solution:
We have,
Let I = ∫dx/√(3x2 + 5x + 7)
= ∫dx/√{3(x2 + 5x/3 + 7/3)}
= (1/√3)∫dx/√{5/2-(x2-2x+1-1)}
### Question 5. ∫dx/√{(x – α)(β – x)}
Solution:
We have,
Let I = ∫dx/√{(x – α)(β – x)}
= ∫dx/√(-x2 +αx + βx – αβ)
= ∫dx/√{-x2 + x(α + β) – αβ}
Let x – (α + β)/2 = q …(1)
As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)
So,
Now put the value of q from eq(1), we get
### Question 6. ∫dx/√(7 – 3x – 2x2)
Solution:
We have,
Let I = ∫dx/√(7 – 3x – 2x2)
= ∫dx/√{2(7/2 – 3x/2 – x2)}
Let x + 3/2 = q …(1)
As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)
So,
Now put the value of q from eq(1), we get
### Question 7. ∫dx/√(16 – 6x – x2)
Solution:
We have,
Let I = ∫dx/√(16 – 6x – x2)
= ∫dx/√{16 – (x2 + 2.3x + 9 – 9)}
= ∫dx/√{25 – (x2 + 2.3x + 9)}
Let x + 3/2 = q …(1)
As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)
So,
Now put the value of q from eq(1), we get
Question 8. ∫dx/√(7 – 6x – x2)
Solution:
We have,
Let I = ∫dx/√(7 – 6x – x2)
= ∫dx/√{7-(x2 + 2.3x + 9 – 9)}
= ∫dx/√{16 – (x2 + 2.3x + 9)}
Let x + 3/2 = q …(1)
As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)
So,
Now put the value of q from eq(1), we get
### Question 9. ∫dx/√(5x2 – 2x)
Solution:
We have,
Let I = ∫dx/√(5x2 – 2x)
= ∫dx/√{5(x2 – 2x/5)}
My Personal Notes arrow_drop_up
Related Articles
|
# Ones, Tens, Hundreds Worksheets
• ###### Train Numbers
Place value is the idea that the value of a digit in a number is determined by its position within that number. These worksheets are will help students understand and practice the concept of place value using the ones, tens, and hundreds places.
The worksheets contain a series of numbers and exercises that require students to identify and manipulate the digits in different place value positions. Being able to identify the ones, tens, and hundreds place values of a number is important because it is a fundamental aspect of the base-ten number system, which is the cornerstone of arithmetic and mathematics.
Understanding place value allows individuals to comprehend the value of a digit based on its position, which is essential for performing calculations correctly. For instance, knowing that in the number 345, the ‘5’ represents five ones, the ‘4’ represents four tens (or forty), and the ‘3’ represents three hundreds (or three hundred), is crucial for grasping how numbers are constructed and interact in mathematical operations.
Here’s a breakdown of what each place value represents:
Ones – This is the rightmost position in a number. It represents the value of individual units or the count of items. For example, in the number 358, the digit “8” is in the ones place, representing 8 individual units.
Tens – The position to the left of the ones place represents tens. It signifies groups of ten. In the number 358, the digit “5” is in the tens place, representing 5 groups of ten, which is 50.
Hundreds – The position to the left of the tens place represents hundreds. It signifies groups of one hundred. In the number 358, the digit “3” is in the hundreds place, representing 3 groups of one hundred, which is 300.
### What Types of Problems Do These Worksheets Have?
Identifying Places – Some questions might simply ask, “What number is in the tens place?” or “How many hundreds are in this number?” These questions are like detective work. You’ve got to find out where the number is hiding!
Adding and Subtracting – You might get questions that ask you to add or subtract numbers. The trick? Always start adding or subtracting from the rightmost side, the ones place, and move to the left. Remember, if you have more than 9 ones or tens, you need to “carry” them over to the next place.
Comparing Numbers – Some challenges might ask which number is bigger or smaller. Is 324 bigger than 235? You’ll start by looking at the hundreds, then the tens, and lastly, the ones to find out.
Filling in the Blanks – These are fun! You might get a number like 2__5 and then a question like, “What number should go in the tens place to make 285?” Time to put on your thinking cap and solve the puzzle!
### Real World Skills
Improving your skills in identifying place values can greatly enhance your mathematical abilities. It forms the basis for addition and subtraction, especially when these operations require regrouping or borrowing. In multiplication and division, a strong grasp of place value assists in understanding how the numbers shift position as they increase or decrease in value. Moreover, place value knowledge is vital in understanding more advanced concepts like decimals and percentages, where the value of digits becomes even more nuanced. In essence, place value is a building block for numerical literacy and is indispensable for mathematical competence in both academic settings and everyday life.
Building a Strong Foundation – Just like when you build a tall tower with blocks, you need to start with a strong base. These basic math skills are the foundation for all the other fun math stuff you’ll learn in the future. And trust me, there’s a lot of cool math out there!
Shopping Smartly – Let’s say you’re buying candies, and one pack has 100 candies, another has 20, and another has 4. If you can understand ones, tens, and hundreds, you’ll quickly know you’re getting 124 candies! That’s a lot of sweetness.
Counting Money – When you start handling money, whether it’s saving up for a toy or counting your weekly allowance, understanding these places will help. If you have 3 one-dollar bills, 4 ten-dollar bills, and 2 hundred-dollar bills, you’ll quickly know how much money you have!
Time Management – When you look at the time, especially in minutes, understanding tens and ones can help you know how many minutes are left in an hour or how many have passed.
In the real world, understanding ones, tens, and hundreds can also help in baking (measuring ingredients), traveling (knowing distances), and even in sports (scoring points). It’s like having a math superpower!
|
## Want to keep learning?
This content is taken from the University of Padova's online course, Advanced Precalculus: Geometry, Trigonometry and Exponentials. Join the course to learn more.
3.22
Skip to 0 minutes and 13 seconds Hello. Welcome to the section “It’s your turn on lengths, areas, and volumes”. OK, we have the following exercise. We have a circle.
Skip to 0 minutes and 28 seconds We know that the length on the circumference of the arc corresponding to an angle equal to pi over 6 is equal to 3 meters. OK. We want to compute the perimeter and the area. OK, of course, to compute the perimeter and the area, we have first of all to compute the radius of this circumference. OK? But what do we know? We know that the arc length on the circumference corresponding to an angle of pi over 6 radians is equal to what? To the radius times pi over 6. Therefore, we know that three is equal to the radius times pi over 6. OK, then we immediately get that the radius is equal to 18 over pi. OK.
Skip to 1 minute and 49 seconds But then, which is the perimeter of the circumference? The perimeter is equal to 2 pi times the radius. That is, 2 pi times 18 over pi, which is 2 times 18, which is 36 meters.
Skip to 2 minutes and 21 seconds And which is the area?
Skip to 2 minutes and 25 seconds It is pi times the square of the radius. Therefore, it’s pi times 18 over pi squared. OK? And 18 squared is 324. And then we have pi over pi squared. Then it remains pi on the bottom. Of course, because it is an area, meters squared– square meters. Hello. Let us consider the second exercise of the section “It’s your turn on lengths, areas, and volumes”. OK, we have a cube. And we know the area of its one face is 90 square meters. OK, and we want to compute the volume and the surface area of the cube. OK, let us start computing the surface area, because what is the surface area?
Skip to 3 minutes and 43 seconds The surface area is equal to 6 times the area of one face. Then it’s equal to 6 times 90, which is 540 square meters.
Skip to 4 minutes and 10 seconds Very easy. And what we can say about the volume of this cube? You know, if you know the length of one side of your cube, then the volume of the cube is the cube of this length. OK, but we know that the area of one face is 90 square meters. Then the length of one side– we call it A– is equal to what? To the square root of 90, which is 3 times the square root of 10 meters. And now that we know the length of one side of our cube, then immediately we can conclude that the volume is equal to what? To A cubed, OK? Which is equal to 3 times the square root of 10 cube– cube meters.
Skip to 5 minutes and 29 seconds That is 27 times 10 times the square root of 10 meters– cube meters. That is, 270 times the square root of 10 cube meters. OK, we can compute the volume also with another reasoning. You see, when you know the area of one face of your cube, then to get the volume of this cube, it is sufficient to multiply this area– 90 square meters– by the length of one side, OK? And we have computed before the length of one side. OK, this was another way to compute the volume. And of course, in following this procedure, you get exactly the same result. Hello. Let us consider now the third exercise of the section “It’s your turn on lengths, areas, and volumes”. Good.
Skip to 6 minutes and 47 seconds We have a right circular cone. And we know the radius, which is equal to 3 centimeters. We know the sides’ area, which is equal to 15 pi square centimeters. And we would like to compute the height of our circular cone. Good. Now, remember that if you have a circular cone of radius r and height h, then the sides’ area is equal to what? To pi r times the square root of r squared plus h squared. Good. Then what do we know? We know that 15 pi has to be equal to what? To pi times the radius, which is three centimeters multiplied by the square of the radius– 9– plus h squared. Good. Now, we can simplify this expression.
Skip to 8 minutes and 24 seconds You know, we can divide by pi on both sides. And we can divide by 3 on both sides. And we get 5 equal to the square root of 9 plus h squared. And now, considering the square of both sides, what do we get? 25 equal to 9 plus h squared. Therefore, we have that h squared is equal to 25 minus 9; that is, 16. And then the height is equal to 4 centimeters. OK, thank you very much for your attention. [IN ITALIAN] Goodbye to everybody
# It's your turn on lengths, areas, and volumes
Do your best in trying to solve the following problems. In any case some of them are solved in the video and all of them are solved in the pdf file below.
### Exercise 1.
Let $$\mathcal C$$ be a circle. Assume that the arclength on the circumference corresponding to an angle of $$\pi/6$$ radians at the center is equal to $$3\ m$$. Compute the perimeter and the area of $$\mathcal C$$.
### Exercise 2.
Find the volume and surface area of a cube, if the area of its one face is 90 $$m^2$$.
### Exercise 3.
Find the height of a right circular cone of radius $$3\ cm$$ and side’s area $$15\pi\ cm^2$$.
|
# What Is 15/45 as a Decimal + Solution With Free Steps
The fraction 15/45 as a decimal is equal to 0.333.
The fractions can be converted to their equivalent decimals. Fractions are mostly used to represent rational numbers. The decimal expansion of rational numbers is mostly terminating decimals or non-terminating recurring decimals. The decimal expansion of the given fraction 15/45 represents a non-terminating recurring decimal.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 15/45.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 15
Divisor = 45
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 15 $\div$ 45
This is when we go through the Long Division solution to our problem. The following figure demonstrates the solution for fraction 15/45.
Figure 1
## 15/45 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 15 and 45, we can see how 15 is Smaller than 45, and to solve this division, we require that 15 be Bigger than 45.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 15, which after getting multiplied by 10 becomes 150.
We take this 150 and divide it by 45; this can be done as follows:
150 $\div$ 45 $\approx$ 3
Where:
45 x 3 = 135
This will lead to the generation of a Remainder equal to 150 – 135 = 15. Now this means we have to repeat the process by Converting the 15 into 150 and solving for that:
150 $\div$ 45 $\approx$ 3
Where:
45 x 3 = 135
This, therefore, produces another Remainder which is equal to 150 – 135 = 15. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 150.
150 $\div$ 45 $\approx$ 3
Where:
45 x 3 = 135
Finally, we have a Quotient generated after combining the three pieces of it as 0.333, with a Remainder equal to 15.
Images/mathematical drawings are created with GeoGebra.
|
cg-tower.com
You are watching: What is 60 divided by 4
Here we will show you step-by-step with comprehensive explanation just how to calculation 60 divided by 4 using long division.Before girlfriend continue, keep in mind that in the trouble 60 separated by 4, the numbers are defined as follows:60 = dividend4 = divisorStep 1:Start by setting it up with the divisor 4 ~ above the left side andthe dividend 60 on the right side favor this:
4 ⟌ 6 0
Step 2:
The divisor (4) goes right into the very first digit that the dividend (6), 1 time(s). Therefore, placed 1 top top top:
1 4 ⟌ 6 0
Step 3:
Multiply the divisor by the result in the previous step (4 x 1 = 4) and also write the answer listed below the dividend.
1 4 ⟌ 6 0 4
Step 4:
Subtract the an outcome in the previous step from the first digit the the dividend (6 - 4 = 2) and also write the prize below.
1 4 ⟌ 6 0 - 4 2
Step 5:
Move down the second digit the the dividend (0) like this:
1 4 ⟌ 6 0 - 4 2 0
Step 6:
The divisor (4) goes into the bottom number (20), 5 time(s).
See more: How The Grinch Stole Christmas Theme, Moral Lessons From How The Grinch Stole Christmas
Therefore, placed 5 top top top:
1 5 4 ⟌ 6 0 - 4 2 0
Step 7:
Multiply the divisor by the an outcome in the previous step (4 x 5 = 20) and also write the answer at the bottom:
1 5 4 ⟌ 6 0 - 4 2 0 2 0
Step 8:
Subtract the result in the previous action from the number written over it. (20 - 20 = 0) andwrite the answer at the bottom.
1 5 4 ⟌ 6 0 - 4 2 0 - 2 0 0
You space done, since there room no more digits to move down indigenous the dividend.The prize is the top number and the remainder is the bottom number.Therefore, the answer come 60 separated by 4 calculated using Long department is:15
0 RemainderLong division CalculatorEnter an additional problem for us to explain and solve:/
|
# Algebra II : Graphing Parabolic Inequalities
## Example Questions
### Example Question #1 : Graphing Parabolic Inequalities
Give the solution set of the inequality:
The inequality has no solution
Explanation:
Rewrite in standard form and factor:
The zeroes of the polynomial are therefore , so we test one value in each of three intervals , and to determine which ones are included in the solution set.
:
Test :
False; is not in the solution set.
:
Test
True; is in the solution set
:
Test :
False; is not in the solution set.
Since the inequality symbol is , the boundary points are not included. The solution set is the interval .
### Example Question #2 : Graphing Parabolic Inequalities
Give the set of solutions for this inequality:
This inequality has no solution.
Explanation:
The first step of questions like this is to get the quadratic in its standard form. So we move the over to the left side of the inequality:
This quadratic can easily be factored as. So now we can write this in the form
and look at each of the factors individually. Recall that a negative number times a negative is a positive number. Therefore the boundaries of our solution interval is going to be when both of these factors are negative. is negative whenever , and is negative whenever . Since , one of our boundaries will be . Remember that this will be an open interval since it is less than, not less than or equal to.
Our other boundary will be the other point when the product of the factors becomes positive. Remember that is positive when , so our other boundary is . So the solution interval we arrive at is
### Example Question #3 : Graphing Parabolic Inequalities
Solve for
Explanation:
When asked to solve for x we need to isolate x on one side of the equation.
To do this our first step is to subtract 7 from both sides.
From here, we divide by 4 to solve for x.
### Example Question #4 : Graphing Parabolic Inequalities
Solve for
Explanation:
When asked to solve for y we need to isolate the variable on one side and the constants on the other side.
To do this we first add 9 to both sides.
From here, we divide by -12 to solve for y.
### Example Question #5 : Graphing Parabolic Inequalities
The graphs for the lines and are shown in the figure. The region is defined by which two inequalities?
Explanation:
The region contains only values which are greater than or equal to those on the line , so its values are .
Also, the region contains only values which are less than or equal to those on the line , so its values are .
### Example Question #3 : Graphing Parabolic Inequalities
The graphs of the lines and are shown on the figure. The region is defined by which two inequalities?
Explanation:
The region contains only values which are greater than or equal to those on the line , so its values are .
Similarly, the region contains only values which are less than or equal to those on the line , so its values are .
### Example Question #1 : Graphing Parabolic Inequalities
Which of the following graphs correctly represents the quadratic inequality below (solutions to the inequalities are shaded in blue)?
|
# Dividing Fractions - Divide Fractions and Whole Numbers
As with multiplying fractions, we saw that it is very easy, we just multiply the denominators and multiply the numerators. In the process of division is not much different from multiplication.
To divide two fractions, we convert the division operation into a multiplication operation. but how? There are 3 basic steps in order to perform division.
• First, we convert the division symbol ( or ) to the multiplication symbol ()
• Second: We keep the first number as it is, and flip the second fraction to its reciprocal.
• Third: We perform the multiplication process normally, as we explained in this lesson.
### Note:
Let , , and four non-zero integers.
To divide by we multiply by the reciprocal of . i.e. we multiply by
### Info!
• Note carefully that dividing two fractions has nothing to do with whether the two fractions have the same or different denominators, as in the case of addition and subtraction.
• After dividing two fractions, make sure your answer is reduced to lowest terms.
Let's take some examples of dividing two non-zero fractions.
#### Example1:
Divide the fractions below.
1)
2)
3)
4)
#### Solution:
1) Let's calculate
The reciprocal of is . Thus,
The resulting fraction is in its reduced form since the greatest common factor of 8 and 63 is 1.
2) Let's calculate
The reciprocal of is . Thus,
The resulting fraction is in its reduced form since the greatest common factor of 15 and 88 is 1.
3) Let's calculate
We can rewrite the operation using the division symbol.
The reciprocal of is . Thus,
The resulting fraction is in its reduced form since the greatest common factor of 9 and 35 is 1.
4) Let's calculate
We can rewrite the operation using the division symbol.
The reciprocal of is . Thus,
The resulting fraction is in its reduced form since the greatest common factor of 5 and 6 is 1.
## Dividing Fraction by Whole Number - Dividing Whole Number by Fraction
A whole number can be considered as a fraction so that its denominator is 1, and thus we have to do the division of a fraction by an integer, or vice versa by applying the following property:
### Note:
Let , and three non-zero integers.
Let's take some examples of dividing two non-zero fraction by whole number, and whole number by fraction.
#### Example2:
Divide the fractions below.
1)
2)
3)
4)
#### Solution:
1) To divide a whole number by a fraction, we must find the reciprocal of the fraction and then performe the multiplication of whole number by fraction.
2) By the same way as in the example above:
3) To divide a fraction by a whole number, we must find the reciprocal of the whole number and then performe the multiplication of whole number by fraction.
4) By the same way as in the example above:
|
To find the distance between the two ships, we use trigonometry, specifically the tangent function related to the angles of depression. The angles of depression from the lighthouse to the ships are the same as the angles of elevation from the ships to the top of the lighthouse.
For the Ship at 30° Angle of Depression: Using tan(30°) = 1/√3, the distance to the first ship (d1) is given by 1/√3 = 75/d1. Solving for d1, we get d1 = 75√3.
For the Ship at 45° Angle of Depression: Using tan(45°) = 1, the distance to the second ship (d2) is d2 = 75 m.
The distance between the two ships is d1 − d2 = 75√3 − 75, which simplifies to approximately 44.98 meters. Therefore, the distance between the two ships is about 44.98 meters.
Let’s discuss in detail
## Trigonometric Analysis in Nautical Navigation
Trigonometry, a fundamental branch of mathematics, is extensively used in nautical navigation, particularly for determining distances and angles from a fixed point. The problem at hand involves the use of trigonometry to calculate the distance between two ships as observed from the top of a lighthouse. By analyzing the angles of depression to the ships, we can determine their distances from the lighthouse and subsequently the distance between them. This scenario is a practical demonstration of how trigonometry can be applied in maritime navigation and safety.
### Understanding the Problem: Lighthouse Observation
The problem presents a lighthouse that is 75 meters high. From the top of this lighthouse, the angles of depression to two ships are observed to be 30° and 45°, respectively. The ships are located on the same side of the lighthouse, with one directly behind the other. The objective is to find the distance between these two ships. This setup forms two right-angled triangles, one with each ship, with the lighthouse as the common vertex.
#### The Role of Tangent in Angle of Depression
In trigonometry, the tangent of an angle in a right-angled triangle is the ratio of the length of the opposite side to the length of the adjacent side. The angles of depression from the lighthouse are equivalent to the angles of elevation from the ships to the lighthouse. By applying the tangent function to these angles, we can calculate the distances of the ships from the lighthouse.
Calculating the Distance to the First Ship
For the ship at a 30° angle of depression, we use the tangent function. The equation is tan(30°) = 1/√3 = 75/d1, where d1 is the distance to the first ship. Solving for d1, we find d1 = 75√3 meters. This distance is crucial for determining the relative positioning of the two ships.
##### Determining the Distance to the Second Ship
Next, we apply the tangent function to the 45° angle of depression for the second ship. The equation is tan(45°) = 1 = 75/d2, where d2 is the distance to the second ship. Since
tan(45°) = 1, it follows that d2 is equal to 75 meters.
###### Calculating the Distance Between the Ships
The distance between the two ships is the difference between d1 and d2, which is 75√3 − 75 meters. This calculation simplifies to approximately 44.98 meters. This example illustrates the practical application of trigonometry in maritime navigation, demonstrating how it can provide accurate measurements for safety and operational purposes. Trigonometry proves to be an invaluable tool in nautical scenarios, offering a mathematical approach to solving problems where direct measurement is challenging or impossible.
Discuss this question in detail or visit to Class 10 Maths Chapter 9 for all questions.
Questions of 10th Maths Exercise 9.1 in Detail
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case? The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
|
University Physics Volume 1
# 2.4Products of Vectors
University Physics Volume 12.4 Products of Vectors
## Learning Objectives
By the end of this section, you will be able to:
• Explain the difference between the scalar product and the vector product of two vectors.
• Determine the scalar product of two vectors.
• Determine the vector product of two vectors.
• Describe how the products of vectors are used in physics.
A vector can be multiplied by another vector but may not be divided by another vector. There are two kinds of products of vectors used broadly in physics and engineering. One kind of multiplication is a scalar multiplication of two vectors. Taking a scalar product of two vectors results in a number (a scalar), as its name indicates. Scalar products are used to define work and energy relations. For example, the work that a force (a vector) performs on an object while causing its displacement (a vector) is defined as a scalar product of the force vector with the displacement vector. A quite different kind of multiplication is a vector multiplication of vectors. Taking a vector product of two vectors returns as a result a vector, as its name suggests. Vector products are used to define other derived vector quantities. For example, in describing rotations, a vector quantity called torque is defined as a vector product of an applied force (a vector) and its distance from pivot to force (a vector). It is important to distinguish between these two kinds of vector multiplications because the scalar product is a scalar quantity and a vector product is a vector quantity.
## The Scalar Product of Two Vectors (the Dot Product)
Scalar multiplication of two vectors yields a scalar product.
## Scalar Product (Dot Product)
The scalar product $A→·B→A→·B→$ of two vectors $A→A→$ and $B→B→$ is a number defined by the equation
$A→·B→=ABcosφ,A→·B→=ABcosφ,$
2.27
where $φφ$ is the angle between the vectors (shown in Figure 2.27). The scalar product is also called the dot product because of the dot notation that indicates it.
In the definition of the dot product, the direction of angle $φφ$ does not matter, and $φφ$ can be measured from either of the two vectors to the other because $cosφ=cos(−φ)=cos(2π−φ)cosφ=cos(−φ)=cos(2π−φ)$. The dot product is a negative number when $90°<φ≤180°90°<φ≤180°$ and is a positive number when $0°≤φ<90°0°≤φ<90°$. Moreover, the dot product of two parallel vectors is $A→·B→=ABcos0°=ABA→·B→=ABcos0°=AB$, and the dot product of two antiparallel vectors is $A→·B→=ABcos180°=−ABA→·B→=ABcos180°=−AB$. The scalar product of two orthogonal vectors vanishes: $A→·B→=ABcos90°=0A→·B→=ABcos90°=0$. The scalar product of a vector with itself is the square of its magnitude:
$A→2≡A→·A→=AAcos0°=A2.A→2≡A→·A→=AAcos0°=A2.$
2.28
Figure 2.27 The scalar product of two vectors. (a) The angle between the two vectors. (b) The orthogonal projection $A││A││$ of vector $A→A→$ onto the direction of vector $B→B→$. (c) The orthogonal projection $B││B││$ of vector $B→B→$ onto the direction of vector $A→A→$.
## Example 2.15
### The Scalar Product
For the vectors shown in Figure 2.13, find the scalar product $A→·F→A→·F→$.
### Strategy
From Figure 2.13, the magnitudes of vectors $A→A→$ and $F→F→$ are A = 10.0 and F = 20.0. Angle $θθ$, between them, is the difference: $θ=φ−α=110°−35°=75°θ=φ−α=110°−35°=75°$. Substituting these values into Equation 2.27 gives the scalar product.
### Solution
A straightforward calculation gives us
$A→·F→=AFcosθ=(10.0)(20.0)cos75°=51.76.A→·F→=AFcosθ=(10.0)(20.0)cos75°=51.76.$
For the vectors given in Figure 2.13, find the scalar products $A→·B→A→·B→$ and $F→·C→F→·C→$.
In the Cartesian coordinate system, scalar products of the unit vector of an axis with other unit vectors of axes always vanish because these unit vectors are orthogonal:
$i^·j^=|i^||j^|cos90°=(1)(1)(0)=0,i^·k^=|i^||k^|cos90°=(1)(1)(0)=0,k^·j^=|k^||j^|cos90°=(1)(1)(0)=0.i^·j^=|i^||j^|cos90°=(1)(1)(0)=0,i^·k^=|i^||k^|cos90°=(1)(1)(0)=0,k^·j^=|k^||j^|cos90°=(1)(1)(0)=0.$
2.29
In these equations, we use the fact that the magnitudes of all unit vectors are one: $|i^|=|j^|=|k^|=1|i^|=|j^|=|k^|=1$. For unit vectors of the axes, Equation 2.28 gives the following identities:
$i^·i^=i2=j^·j^=j2=k^·k^=k2=1.i^·i^=i2=j^·j^=j2=k^·k^=k2=1.$
2.30
The scalar product $A→·B→A→·B→$ can also be interpreted as either the product of B with the projection $AǁAǁ$ of vector $A→A→$ onto the direction of vector $B→B→$ (Figure 2.27(b)) or the product of A with the projection $BǁBǁ$ of vector $B→B→$ onto the direction of vector $A→A→$ (Figure 2.27(c)):
$A→·B→=ABcosφ=B(Acosφ)=BAǁ=A(Bcosφ)=ABǁ.A→·B→=ABcosφ=B(Acosφ)=BAǁ=A(Bcosφ)=ABǁ.$
For example, in the rectangular coordinate system in a plane, the scalar x-component of a vector is its dot product with the unit vector $i^i^$, and the scalar y-component of a vector is its dot product with the unit vector $j^j^$:
${A→·i^=|A→||i^|cosθA=AcosθA=AxA→·j^=|A→||j^|cos(90°−θA)=AsinθA=Ay.{A→·i^=|A→||i^|cosθA=AcosθA=AxA→·j^=|A→||j^|cos(90°−θA)=AsinθA=Ay.$
Scalar multiplication of vectors is commutative,
$A→·B→=B→·A→,A→·B→=B→·A→,$
2.31
and obeys the distributive law:
$A→·(B→+C→)=A→·B→+A→·C→.A→·(B→+C→)=A→·B→+A→·C→.$
2.32
We can use the commutative and distributive laws to derive various relations for vectors, such as expressing the dot product of two vectors in terms of their scalar components.
For vector $A→=Axi^+Ayj^+Azk^A→=Axi^+Ayj^+Azk^$ in a rectangular coordinate system, use Equation 2.29 through Equation 2.32 to show that $A→·i^=AxA→·i^=Ax$ $A→·j^=AyA→·j^=Ay$ and $A→·k^=AzA→·k^=Az$.
When the vectors in Equation 2.27 are given in their vector component forms,
$A→=Axi^+Ayj^+Azk^andB→=Bxi^+Byj^+Bzk^,A→=Axi^+Ayj^+Azk^andB→=Bxi^+Byj^+Bzk^,$
we can compute their scalar product as follows:
$A→·B→=(Axi^+Ayj^+Azk^)·(Bxi^+Byj^+Bzk^)=AxBxi^·i^+AxByi^·j^+AxBzi^·k^+AyBxj^·i^+AyByj^·j^+AyBzj^·k^+AzBxk^·i^+AzByk^·j^+AzBzk^·k^.A→·B→=(Axi^+Ayj^+Azk^)·(Bxi^+Byj^+Bzk^)=AxBxi^·i^+AxByi^·j^+AxBzi^·k^+AyBxj^·i^+AyByj^·j^+AyBzj^·k^+AzBxk^·i^+AzByk^·j^+AzBzk^·k^.$
Since scalar products of two different unit vectors of axes give zero, and scalar products of unit vectors with themselves give one (see Equation 2.29 and Equation 2.30), there are only three nonzero terms in this expression. Thus, the scalar product simplifies to
$A→·B→=AxBx+AyBy+AzBz.A→·B→=AxBx+AyBy+AzBz.$
2.33
We can use Equation 2.33 for the scalar product in terms of scalar components of vectors to find the angle between two vectors. When we divide Equation 2.27 by AB, we obtain the equation for $cosφcosφ$, into which we substitute Equation 2.33:
$cosφ=A→·B→AB=AxBx+AyBy+AzBzAB.cosφ=A→·B→AB=AxBx+AyBy+AzBzAB.$
2.34
Angle $φφ$ between vectors $A→A→$ and $B→B→$ is obtained by taking the inverse cosine of the expression in Equation 2.34.
## Example 2.16
### Angle between Two Forces
Three dogs are pulling on a stick in different directions, as shown in Figure 2.28. The first dog pulls with force $F→1=(10.0i^−20.4j^+2.0k^)NF→1=(10.0i^−20.4j^+2.0k^)N$, the second dog pulls with force $F→2=(−15.0i^−6.2k^)NF→2=(−15.0i^−6.2k^)N$, and the third dog pulls with force $F→3=(5.0i^+12.5j^)NF→3=(5.0i^+12.5j^)N$. What is the angle between forces $F→1F→1$ and $F→2F→2$?
Figure 2.28 Three dogs are playing with a stick.
### Strategy
The components of force vector $F→1F→1$ are $F1x=10.0NF1x=10.0N$, $F1y=−20.4NF1y=−20.4N$, and $F1z=2.0NF1z=2.0N$, whereas those of force vector $F→2F→2$ are $F2x=−15.0NF2x=−15.0N$, $F2y=0.0NF2y=0.0N$, and $F2z=−6.2NF2z=−6.2N$. Computing the scalar product of these vectors and their magnitudes, and substituting into Equation 2.34 gives the angle of interest.
### Solution
The magnitudes of forces $F→1F→1$ and $F→2F→2$ are
$F1=F1x2+F1y2+F1z2=10.02+20.42+2.02N=22.8NF1=F1x2+F1y2+F1z2=10.02+20.42+2.02N=22.8N$
and
$F2=F2x2+F2y2+F2z2=15.02+6.22N=16.2N.F2=F2x2+F2y2+F2z2=15.02+6.22N=16.2N.$
Substituting the scalar components into Equation 2.33 yields the scalar product
$F→1·F→2=F1xF2x+F1yF2y+F1zF2z=(10.0N)(−15.0N)+(−20.4N)(0.0N)+(2.0N)(−6.2N)=−162.4N2.F→1·F→2=F1xF2x+F1yF2y+F1zF2z=(10.0N)(−15.0N)+(−20.4N)(0.0N)+(2.0N)(−6.2N)=−162.4N2.$
Finally, substituting everything into Equation 2.34 gives the angle
$cosφ=F→1·F→2F1F2=−162.4N2(22.8N)(16.2N)=−0.439⇒φ=cos−1(−0.439)=116.0°.cosφ=F→1·F→2F1F2=−162.4N2(22.8N)(16.2N)=−0.439⇒φ=cos−1(−0.439)=116.0°.$
### Significance
Notice that when vectors are given in terms of the unit vectors of axes, we can find the angle between them without knowing the specifics about the geographic directions the unit vectors represent. Here, for example, the +x-direction might be to the east and the +y-direction might be to the north. But, the angle between the forces in the problem is the same if the +x-direction is to the west and the +y-direction is to the south.
Find the angle between forces $F→1F→1$ and $F→3F→3$ in Example 2.16.
## Example 2.17
### The Work of a Force
When force $F→F→$ pulls on an object and when it causes its displacement $D→D→$, we say the force performs work. The amount of work the force does is the scalar product $F→·D→F→·D→$. If the stick in Example 2.16 moves momentarily and gets displaced by vector $D→=(−7.9j^−4.2k^)cmD→=(−7.9j^−4.2k^)cm$, how much work is done by the third dog in Example 2.16?
### Strategy
We compute the scalar product of displacement vector $D→D→$ with force vector $F→3=(5.0i^+12.5j^)NF→3=(5.0i^+12.5j^)N$, which is the pull from the third dog. Let’s use $W3W3$ to denote the work done by force $F→3F→3$ on displacement $D→D→$.
### Solution
Calculating the work is a straightforward application of the dot product:
$W3=F→3·D→=F3xDx+F3yDy+F3zDz=(5.0N)(0.0cm)+(12.5N)(−7.9cm)+(0.0N)(−4.2cm)=−98.7N·cm.W3=F→3·D→=F3xDx+F3yDy+F3zDz=(5.0N)(0.0cm)+(12.5N)(−7.9cm)+(0.0N)(−4.2cm)=−98.7N·cm.$
### Significance
The SI unit of work is called the joule $(J)(J)$, where 1 J = 1 $N·mN·m$. The unit $cm·Ncm·N$ can be written as $10−2m·N=10−2J10−2m·N=10−2J$, so the answer can be expressed as $W3=−0.9875J≈−1.0JW3=−0.9875J≈−1.0J$.
How much work is done by the first dog and by the second dog in Example 2.16 on the displacement in Example 2.17?
## The Vector Product of Two Vectors (the Cross Product)
Vector multiplication of two vectors yields a vector product.
## Vector Product (Cross Product)
The vector product of two vectors $A→A→$ and $B→B→$ is denoted by $A→×B→A→×B→$ and is often referred to as a cross product. The vector product is a vector that has its direction perpendicular to both vectors $A→A→$ and $B→B→$. In other words, vector $A→×B→A→×B→$ is perpendicular to the plane that contains vectors $A→A→$ and $B→B→$, as shown in Figure 2.29. The magnitude of the vector product is defined as
$|A→×B→|=ABsinφ,|A→×B→|=ABsinφ,$
2.35
where angle $φφ$, between the two vectors, is measured from vector $A→A→$ (first vector in the product) to vector $B→B→$ (second vector in the product), as indicated in Figure 2.29, and is between $0°0°$ and $180°180°$.
According to Equation 2.35, the vector product vanishes for pairs of vectors that are either parallel $(φ=0°)(φ=0°)$ or antiparallel $(φ=180°)(φ=180°)$ because $sin0°=sin180°=0sin0°=sin180°=0$.
Figure 2.29 The vector product of two vectors is drawn in three-dimensional space. (a) The vector product $A→×B→A→×B→$ is a vector perpendicular to the plane that contains vectors $A→A→$ and $B→B→$. Small squares drawn in perspective mark right angles between $A→A→$ and $C→C→$, and between $B→B→$ and $C→C→$ so that if $A→A→$ and $B→B→$ lie on the floor, vector $C→C→$ points vertically upward to the ceiling. (b) The vector product $B→×A→B→×A→$ is a vector antiparallel to vector $A→×B→A→×B→$.
On the line perpendicular to the plane that contains vectors $A→A→$ and $B→B→$ there are two alternative directions—either up or down, as shown in Figure 2.29—and the direction of the vector product may be either one of them. In the standard right-handed orientation, where the angle between vectors is measured counterclockwise from the first vector, vector $A→×B→A→×B→$ points upward, as seen in Figure 2.29(a). If we reverse the order of multiplication, so that now $B→B→$ comes first in the product, then vector $B→×A→B→×A→$ must point downward, as seen in Figure 2.29(b). This means that vectors $A→×B→A→×B→$ and $B→×A→B→×A→$ are antiparallel to each other and that vector multiplication is not commutative but anticommutative. The anticommutative property means the vector product reverses the sign when the order of multiplication is reversed:
$A→×B→=−B→×A→.A→×B→=−B→×A→.$
2.36
The corkscrew right-hand rule is a common mnemonic used to determine the direction of the vector product. As shown in Figure 2.30, a corkscrew is placed in a direction perpendicular to the plane that contains vectors $A→A→$ and $B→B→$, and its handle is turned in the direction from the first to the second vector in the product. The direction of the cross product is given by the progression of the corkscrew.
Figure 2.30 The corkscrew right-hand rule can be used to determine the direction of the cross product $A→×B→A→×B→$. Place a corkscrew in the direction perpendicular to the plane that contains vectors $A→A→$ and $B→B→$, and turn it in the direction from the first to the second vector in the product. The direction of the cross product is given by the progression of the corkscrew. (a) Upward movement means the cross-product vector points up. (b) Downward movement means the cross-product vector points downward.
## Example 2.18
### The Torque of a Force
The mechanical advantage that a familiar tool called a wrench provides (Figure 2.31) depends on magnitude F of the applied force, on its direction with respect to the wrench handle, and on how far from the nut this force is applied. The distance R from the nut to the point where force vector $F→F→$ is attached is represented by the radial vector $R→R→$. The physical vector quantity that makes the nut turn is called torque (denoted by $τ→)τ→)$, and it is the vector product of the distance between the pivot to force with the force: $τ→=R→×F→τ→=R→×F→$.
To loosen a rusty nut, a 20.00-N force is applied to the wrench handle at angle $φ=40°φ=40°$ and at a distance of 0.25 m from the nut, as shown in Figure 2.31(a). Find the magnitude and direction of the torque applied to the nut. What would the magnitude and direction of the torque be if the force were applied at angle $φ=45°φ=45°$, as shown in Figure 2.31(b)? For what value of angle $φφ$ does the torque have the largest magnitude?
Figure 2.31 A wrench provides grip and mechanical advantage in applying torque to turn a nut. (a) Turn counterclockwise to loosen the nut. (b) Turn clockwise to tighten the nut.
### Strategy
We adopt the frame of reference shown in Figure 2.31, where vectors $R→R→$ and $F→F→$ lie in the xy-plane and the origin is at the position of the nut. The radial direction along vector $R→R→$ (pointing away from the origin) is the reference direction for measuring the angle $φφ$ because $R→R→$ is the first vector in the vector product $τ→=R→×F→τ→=R→×F→$. Vector $τ→τ→$ must lie along the z-axis because this is the axis that is perpendicular to the xy-plane, where both $R→R→$ and $F→F→$ lie. To compute the magnitude $ττ$, we use Equation 2.35. To find the direction of $τ→τ→$, we use the corkscrew right-hand rule (Figure 2.30).
### Solution
For the situation in (a), the corkscrew rule gives the direction of $R→×F→R→×F→$ in the positive direction of the z-axis. Physically, it means the torque vector $τ→τ→$ points out of the page, perpendicular to the wrench handle. We identify F = 20.00 N and R = 0.25 m, and compute the magnitude using Equation 2.35:
$τ=|R→×F→|=RFsinφ=(0.25m)(20.00N)sin40°=3.21N·m.τ=|R→×F→|=RFsinφ=(0.25m)(20.00N)sin40°=3.21N·m.$
For the situation in (b), the corkscrew rule gives the direction of $R→×F→R→×F→$ in the negative direction of the z-axis. Physically, it means the vector $τ→τ→$ points into the page, perpendicular to the wrench handle. The magnitude of this torque is
$τ=|R→×F→|=RFsinφ=(0.25m)(20.00N)sin45°=3.53N·m.τ=|R→×F→|=RFsinφ=(0.25m)(20.00N)sin45°=3.53N·m.$
The torque has the largest value when $sinφ=1sinφ=1$, which happens when $φ=90°φ=90°$. Physically, it means the wrench is most effective—giving us the best mechanical advantage—when we apply the force perpendicular to the wrench handle. For the situation in this example, this best-torque value is $τbest=RF=(0.25m)(20.00N)=5.00N·mτbest=RF=(0.25m)(20.00N)=5.00N·m$.
### Significance
When solving mechanics problems, we often do not need to use the corkscrew rule at all, as we’ll see now in the following equivalent solution. Notice that once we have identified that vector $R→×F→R→×F→$ lies along the z-axis, we can write this vector in terms of the unit vector $k^k^$ of the z-axis:
$R→×F→=RFsinφk^.R→×F→=RFsinφk^.$
In this equation, the number that multiplies $k^k^$ is the scalar z-component of the vector $R→×F→R→×F→$. In the computation of this component, care must be taken that the angle $φφ$ is measured counterclockwise from $R→R→$ (first vector) to $F→F→$ (second vector). Following this principle for the angles, we obtain $RFsin(+40°)=+3.2N·mRFsin(+40°)=+3.2N·m$ for the situation in (a), and we obtain $RFsin(−45°)=−3.5N·mRFsin(−45°)=−3.5N·m$ for the situation in (b). In the latter case, the angle is negative because the graph in Figure 2.31 indicates the angle is measured clockwise; but, the same result is obtained when this angle is measured counterclockwise because $+(360°−45°)=+315°+(360°−45°)=+315°$ and $sin(+315°)=sin(−45°)sin(+315°)=sin(−45°)$. In this way, we obtain the solution without reference to the corkscrew rule. For the situation in (a), the solution is $R→×F→=+3.2N·mk^R→×F→=+3.2N·mk^$; for the situation in (b), the solution is $R→×F→=−3.5N·mk^R→×F→=−3.5N·mk^$.
For the vectors given in Figure 2.13, find the vector products $A→×B→A→×B→$ and $C→×F→C→×F→$.
Similar to the dot product (Equation 2.32), the cross product has the following distributive property:
$A→×(B→+C→)=A→×B→+A→×C→.A→×(B→+C→)=A→×B→+A→×C→.$
2.37
The distributive property is applied frequently when vectors are expressed in their component forms, in terms of unit vectors of Cartesian axes.
When we apply the definition of the cross product, Equation 2.35, to unit vectors $i^i^$, $j^j^$, and $k^k^$ that define the positive x-, y-, and z-directions in space, we find that
$i^×i^=j^×j^=k^×k^=0.i^×i^=j^×j^=k^×k^=0.$
2.38
All other cross products of these three unit vectors must be vectors of unit magnitudes because $i^i^$, $j^j^$, and $k^k^$ are orthogonal. For example, for the pair $i^i^$ and $j^j^$, the magnitude is $|i^×j^|=ijsin90°=(1)(1)(1)=1|i^×j^|=ijsin90°=(1)(1)(1)=1$. The direction of the vector product $i^×j^i^×j^$ must be orthogonal to the xy-plane, which means it must be along the z-axis. The only unit vectors along the z-axis are $−k^−k^$ or $+k^+k^$. By the corkscrew rule, the direction of vector $i^×j^i^×j^$ must be parallel to the positive z-axis. Therefore, the result of the multiplication $i^×j^i^×j^$ is identical to $+k^+k^$. We can repeat similar reasoning for the remaining pairs of unit vectors. The results of these multiplications are
${i^×j^=+k^,j^×k^=+i^,k^×i^=+j^.{i^×j^=+k^,j^×k^=+i^,k^×i^=+j^.$
2.39
Notice that in Equation 2.39, the three unit vectors $i^i^$, $j^j^$, and $k^k^$ appear in the cyclic order shown in a diagram in Figure 2.32(a). The cyclic order means that in the product formula, $i^i^$ follows $k^k^$ and comes before $j^j^$, or $k^k^$ follows $j^j^$ and comes before $i^i^$, or $j^j^$ follows $i^i^$ and comes before $k^k^$. The cross product of two different unit vectors is always a third unit vector. When two unit vectors in the cross product appear in the cyclic order, the result of such a multiplication is the remaining unit vector, as illustrated in Figure 2.32(b). When unit vectors in the cross product appear in a different order, the result is a unit vector that is antiparallel to the remaining unit vector (i.e., the result is with the minus sign, as shown by the examples in Figure 2.32(c) and Figure 2.32(d). In practice, when the task is to find cross products of vectors that are given in vector component form, this rule for the cross-multiplication of unit vectors is very useful.
Figure 2.32 (a) The diagram of the cyclic order of the unit vectors of the axes. (b) The only cross products where the unit vectors appear in the cyclic order. These products have the positive sign. (c, d) Two examples of cross products where the unit vectors do not appear in the cyclic order. These products have the negative sign.
Suppose we want to find the cross product $A→×B→A→×B→$ for vectors $A→=Axi^+Ayj^+Azk^A→=Axi^+Ayj^+Azk^$ and $B→=Bxi^+Byj^+Bzk^B→=Bxi^+Byj^+Bzk^$. We can use the distributive property (Equation 2.37), the anticommutative property (Equation 2.36), and the results in Equation 2.38 and Equation 2.39 for unit vectors to perform the following algebra:
$A→×B→=(Axi^+Ayj^+Azk^)×(Bxi^+Byj^+Bzk^)=Axi^×(Bxi^+Byj^+Bzk^)+Ayj^×(Bxi^+Byj^+Bzk^)+Azk^×(Bxi^+Byj^+Bzk^)=AxBxi^×i^+AxByi^×j^+AxBzi^×k^+AyBxj^×i^+AyByj^×j^+AyBzj^×k^+AzBxk^×i^+AzByk^×j^+AzBzk^×k^=AxBx(0)+AxBy(+k^)+AxBz(−j^)+AyBx(−k^)+AyBy(0)+AyBz(+i^)+AzBx(+j^)+AzBy(−i^)+AzBz(0).A→×B→=(Axi^+Ayj^+Azk^)×(Bxi^+Byj^+Bzk^)=Axi^×(Bxi^+Byj^+Bzk^)+Ayj^×(Bxi^+Byj^+Bzk^)+Azk^×(Bxi^+Byj^+Bzk^)=AxBxi^×i^+AxByi^×j^+AxBzi^×k^+AyBxj^×i^+AyByj^×j^+AyBzj^×k^+AzBxk^×i^+AzByk^×j^+AzBzk^×k^=AxBx(0)+AxBy(+k^)+AxBz(−j^)+AyBx(−k^)+AyBy(0)+AyBz(+i^)+AzBx(+j^)+AzBy(−i^)+AzBz(0).$
When performing algebraic operations involving the cross product, be very careful about keeping the correct order of multiplication because the cross product is anticommutative. The last two steps that we still have to do to complete our task are, first, grouping the terms that contain a common unit vector and, second, factoring. In this way we obtain the following very useful expression for the computation of the cross product:
$C→=A→×B→=(AyBz−AzBy)i^+(AzBx−AxBz)j^+(AxBy−AyBx)k^.C→=A→×B→=(AyBz−AzBy)i^+(AzBx−AxBz)j^+(AxBy−AyBx)k^.$
2.40
In this expression, the scalar components of the cross-product vector are
${Cx=AyBz−AzBy,Cy=AzBx−AxBz,Cz=AxBy−AyBx.{Cx=AyBz−AzBy,Cy=AzBx−AxBz,Cz=AxBy−AyBx.$
2.41
When finding the cross product, in practice, we can use either Equation 2.35 or Equation 2.40, depending on which one of them seems to be less complex computationally. They both lead to the same final result. One way to make sure if the final result is correct is to use them both.
## Example 2.19
### A Particle in a Magnetic Field
When moving in a magnetic field, some particles may experience a magnetic force. Without going into details—a detailed study of magnetic phenomena comes in later chapters—let’s acknowledge that the magnetic field $B→B→$ is a vector, the magnetic force $F→F→$ is a vector, and the velocity $u→u→$ of the particle is a vector. The magnetic force vector is proportional to the vector product of the velocity vector with the magnetic field vector, which we express as $F→=ζu→×B→F→=ζu→×B→$. In this equation, a constant $ζζ$ takes care of the consistency in physical units, so we can omit physical units on vectors $u→u→$ and $B→B→$. In this example, let’s assume the constant $ζζ$ is positive.
A particle moving in space with velocity vector $u→=−5.0i^−2.0j^+3.5k^u→=−5.0i^−2.0j^+3.5k^$ enters a region with a magnetic field and experiences a magnetic force. Find the magnetic force $F→F→$ on this particle at the entry point to the region where the magnetic field vector is (a) $B→=7.2i^−j^−2.4k^B→=7.2i^−j^−2.4k^$ and (b) $B→=4.5k^B→=4.5k^$. In each case, find magnitude F of the magnetic force and angle $θθ$ the force vector $F→F→$ makes with the given magnetic field vector $B→B→$.
### Strategy
First, we want to find the vector product $u→×B→u→×B→$, because then we can determine the magnetic force using $F→=ζu→×B→F→=ζu→×B→$. Magnitude F can be found either by using components, $F=Fx2+Fy2+Fz2F=Fx2+Fy2+Fz2$, or by computing the magnitude $|u→×B→||u→×B→|$ directly using Equation 2.35. In the latter approach, we would have to find the angle between vectors $u→u→$ and $B→B→$. When we have $F→F→$, the general method for finding the direction angle $θθ$ involves the computation of the scalar product $F→·B→F→·B→$ and substitution into Equation 2.34. To compute the vector product we can either use Equation 2.40 or compute the product directly, whichever way is simpler.
### Solution
The components of the velocity vector are $ux=−5.0ux=−5.0$, $uy=−2.0uy=−2.0$, and $uz=3.5uz=3.5$.
(a) The components of the magnetic field vector are $Bx=7.2Bx=7.2$, $By=−1.0By=−1.0$, and $Bz=−2.4Bz=−2.4$. Substituting them into Equation 2.41 gives the scalar components of vector $F→=ζu→×B→F→=ζu→×B→$:
${Fx=ζ(uyBz−uzBy)=ζ[(−2.0)(−2.4)−(3.5)(−1.0)]=8.3ζFy=ζ(uzBx−uxBz)=ζ[(3.5)(7.2)−(−5.0)(−2.4)]=13.2ζFz=ζ(uxBy−uyBx)=ζ[(−5.0)(−1.0)−(−2.0)(7.2)]=19.4ζ.{Fx=ζ(uyBz−uzBy)=ζ[(−2.0)(−2.4)−(3.5)(−1.0)]=8.3ζFy=ζ(uzBx−uxBz)=ζ[(3.5)(7.2)−(−5.0)(−2.4)]=13.2ζFz=ζ(uxBy−uyBx)=ζ[(−5.0)(−1.0)−(−2.0)(7.2)]=19.4ζ.$
Thus, the magnetic force is $F→=ζ(8.3i^+13.2j^+19.4k^)F→=ζ(8.3i^+13.2j^+19.4k^)$ and its magnitude is
$F=Fx2+Fy2+Fz2=ζ(8.3)2+(13.2)2+(19.4)2=24.9ζ.F=Fx2+Fy2+Fz2=ζ(8.3)2+(13.2)2+(19.4)2=24.9ζ.$
To compute angle $θθ$, we may need to find the magnitude of the magnetic field vector,
$B=Bx2+By2+Bz2=(7.2)2+(−1.0)2+(−2.4)2=7.6,B=Bx2+By2+Bz2=(7.2)2+(−1.0)2+(−2.4)2=7.6,$
and the scalar product $F→·B→F→·B→$:
$F→·B→=FxBx+FyBy+FzBz=(8.3ζ)(7.2)+(13.2ζ)(−1.0)+(19.4ζ)(−2.4)=0.F→·B→=FxBx+FyBy+FzBz=(8.3ζ)(7.2)+(13.2ζ)(−1.0)+(19.4ζ)(−2.4)=0.$
Now, substituting into Equation 2.34 gives angle $θθ$:
$cosθ=F→·B→FB=0(24.9ζ)(7.6)=0⇒θ=90°.cosθ=F→·B→FB=0(24.9ζ)(7.6)=0⇒θ=90°.$
Hence, the magnetic force vector is perpendicular to the magnetic field vector. (We could have saved some time if we had computed the scalar product earlier.)
(b) Because vector $B→=4.5k^B→=4.5k^$ has only one component, we can perform the algebra quickly and find the vector product directly:
$F→=ζu→×B→=ζ(−5.0i^−2.0j^+3.5k^)×(4.5k^)=ζ[(−5.0)(4.5)i^×k^+(−2.0)(4.5)j^×k^+(3.5)(4.5)k^×k^]=ζ[−22.5(−j^)−9.0(+i^)+0]=ζ(−9.0i^+22.5j^).F→=ζu→×B→=ζ(−5.0i^−2.0j^+3.5k^)×(4.5k^)=ζ[(−5.0)(4.5)i^×k^+(−2.0)(4.5)j^×k^+(3.5)(4.5)k^×k^]=ζ[−22.5(−j^)−9.0(+i^)+0]=ζ(−9.0i^+22.5j^).$
The magnitude of the magnetic force is
$F=Fx2+Fy2+Fz2=ζ(−9.0)2+(22.5)2+(0.0)2=24.2ζ.F=Fx2+Fy2+Fz2=ζ(−9.0)2+(22.5)2+(0.0)2=24.2ζ.$
Because the scalar product is
$F→·B→=FxBx+FyBy+FzBz=(−9.0ζ)(0)+(22.5ζ)(0)+(0)(4.5)=0,F→·B→=FxBx+FyBy+FzBz=(−9.0ζ)(0)+(22.5ζ)(0)+(0)(4.5)=0,$
the magnetic force vector $F→F→$ is perpendicular to the magnetic field vector $B→B→$.
### Significance
Even without actually computing the scalar product, we can predict that the magnetic force vector must always be perpendicular to the magnetic field vector because of the way this vector is constructed. Namely, the magnetic force vector is the vector product $F→=ζu→×B→F→=ζu→×B→$ and, by the definition of the vector product (see Figure 2.29), vector $F→F→$ must be perpendicular to both vectors $u→u→$ and $B→B→$.
Given two vectors $A→=−i^+j^A→=−i^+j^$ and $B→=3i^−j^B→=3i^−j^$, find (a) $A→×B→A→×B→$, (b) $|A→×B→||A→×B→|$, (c) the angle between $A→A→$ and $B→B→$, and (d) the angle between $A→×B→A→×B→$ and vector $C→=i^+k^C→=i^+k^$.
In conclusion to this section, we want to stress that “dot product” and “cross product” are entirely different mathematical objects that have different meanings. The dot product is a scalar; the cross product is a vector. Later chapters use the terms dot product and scalar product interchangeably. Similarly, the terms cross product and vector product are used interchangeably.
|
# How do you solve using the completing the square method x^2-2x-10=0?
Feb 29, 2016
See solution below.
#### Explanation:
${x}^{2} - 2 x = 10$
$1 \left({x}^{2} - 2 x + m\right) = 10$
$m = {\left(\frac{b}{2}\right)}^{2}$
$m = {\left(- \frac{2}{2}\right)}^{2}$
$m = 1$
$1 \left({x}^{2} - 2 x + 1 - 1\right) = 10$
$1 \left({x}^{2} - 2 x + 1\right) - 1 \left(1\right) = 10$
$1 {\left(x - 1\right)}^{2} = 11$
$\left(x - 1\right) = \pm \sqrt{11}$
$x = \pm \sqrt{11} + 1$
Here are a few things to remember about solving quadratic equations using the completion of square:
1. At step 3, you notice I give you the formula for finding the value of m, which is what will make the expression a perfect square trinomial. "b" is the middle term, with the coefficient x, not ${x}^{2}$!
2. To keep the equation equivalent, you must always add and subtract the value of b inside the parentheses.
3. When you extract the negative value of b from the parentheses, you must multiply it by the number in front of the parentheses (the number that you factored out in step 1)
4. Never forget that when you take the square root of a positive number you must use the $\pm$ sign. Failure to do so will result in only one answer, which is not enough in a quadratic equation (a quadratic equation always has two solutions)
5. At step 1, notice I sent the constant term (10) to the side of the equation with 0. You must always do this: the other side is just for terms a and b, which have coefficients of ${x}^{2}$ and $x$, respectively.
Practice exercises:
1. Solve for x by completion of square.
a) $2 {x}^{2} + 8 x + 15 = 0$
b) $3 {x}^{2} - 4 x + 19 = - 3$
|
You currently have JavaScript disabled on this browser/device. JavaScript must be enabled in order for this website to function properly.
# ZingPath: Volume of Prisms and Pyramids
Searching for
## Volume of Prisms and Pyramids
Learn in a way your textbook can't show you.
Explore the full path to learning Volume of Prisms and Pyramids
Math Foundations
### Learning Made Easy
Observe the changes that occur in the volume of a quadrilateral pyramid when the area of the base, height, and incline change.
### Now You Know
After completing this tutorial, you will be able to complete the following:
• After completing this Activity Object, learners will be able to:
• Recognize that the volume of a quadrilateral pyramid is equal to one-third of the product of its height and the area of the base.
• Recognize that the volume of a quadrilateral pyramid is proportional to both its height and area of the base.
• Recognize that the volume of a quadrilateral pyramid does not change when its incline changes.
### Everything You'll Have Covered
A quadrilateral pyramid is a three-dimensional geometric figure.
A quadrilateral pyramid has a quadrilateral (four-sided) base with four triangular sides meeting at a point (the apex). If the base is a square, it is called a square pyramid.
The volume of a pyramid can be found by using the formula
The volume of a quadrilateral pyramid is equal to one-third of the product of its height (h) and the area of its base (B). To find the volume of a quadrilateral pyramid, first we need to calculate the area of its base (B).
The volume of a pyramid is proportional to both its height and area of its base.
This Activity Object will focus on the changes in a quadrilateral pyramid's volume when other variables are altered. For instance, students will be able to change the area of the base, the height, and the incline of the quadrilateral pyramid, and then observe the results from these changes.
• When only the height is changed
• The volume of a quadrilateral pyramid is proportional to its height. What this means is that if the height is increased the volume increases, and if the height decreases the volume decreases.
• When only the area of the base is changed
• The volume of a pyramid is proportional to the area of its base. What this means is that if the area of its base is increased the volume increases, and if the area of its base decreases the volume decreases.
• When the height AND area of the base are both changed
• Because the volume of a pyramid is proportional to its height and the area of its base, the volume will change when the height and area of the base of the pyramid change.
• When the incline is changed
• The volume of a pyramid does not change when its incline changes. This is because when the incline changes, the area of the base and height do not change.
• The following key vocabulary terms will be used throughout this Activity Object:
• area of the base -the area of the quadrilateral base of the pyramid
• height - the perpendicular distance to the base
• incline - the slant (to the right or left) or slope of a three-dimensional shape
• quadrilateral pyramid - a polyhedron with a four-sided base and four triangular sides meeting at a point (the apex)
• volume of a pyramid
• (V: Volume, B: Area of the base and h: height)
### Tutorial Details
Approximate Time 15 Minutes Pre-requisite Concepts Area of the base, height, incline, quadrilateral pyramid, volume of pyramid Course Math Foundations Type of Tutorial Dynamic Modeling Key Vocabulary area of the base, base, height
|
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
# Center and spread | Lesson
## What are the center and spread of a data set?
A set of quantitative data with varying values may be shown as a list or a data display.
Suppose Pedro asks $7$ friends how many pairs of shoes they own. Pedro can report his findings as a list or a data display such as a dotplot where each dot represents one response.
• List: $1,2,4,4,7,8,8$
• Dotplot:
Center and spread are ways to describe data sets like this.
• Center describes a typical value of a data point. Two measures of center are mean and median.
• Spread describes the variation of the data. Two measures of spread are range and standard deviation.
### What skills are tested?
• Calculating the mean, median, and range from a list of values or a data display
• Comparing the mean, median, range, and standard deviation of data sets. You won't be tested on the formula for standard deviation.
## How is the mean calculated?
The mean is useful for describing the center of data with similar values.
The mean is the average value.
## How is the median calculated?
The median is useful to describe the center of data with
.
The median is the middle value when the data are ordered from least to greatest.
• If the number of values is odd, the median is the middle value.
• If the number of values is even, the median is the average of the two middle values.
## How is the range calculated?
Range measures the total spread of the data.
The range is the difference between the highest and lowest values.
$\text{range}=\text{highest value}-\text{lowest value}$
A larger range indicates a greater spread in the data.
## How is standard deviation used ?
Standard deviation measures the typical spread from the mean.
Standard deviation is the average distance between the mean and a data point.
Larger values of standard deviation indicate greater spread in the data.
TRY: FINDING THE MEDIAN
Station NumberGas Price (per gallon)
$1$$\mathrm{}2.56$
$2$$\mathrm{}2.42$
$3$$\mathrm{}2.65$
$4$$\mathrm{}2.48$
$5$$\mathrm{}2.99$
Tomas visited $5$ gas stations near his house. The price of gas at each station is shown in the table above. What is the median price of gas at the stations Tomas visited?
TRY: CALCULATING THE MEAN
$75,70,45,50,52,68$
A veterinary student is studying newborn giraffes. The list above shows the masses of $6$ newborn giraffes, rounded to the nearest kilogram. What is the mean mass of the newborn giraffes?
Round you answer to the nearest kilogram.
kilograms
TRY: CALCULATING THE RANGE
Tayo got the following scores on her Spanish quizzes: $88,96,94,92,98,58,90$. What is the range of Tayo's quiz scores?
TRY: FINDING THE MEDIAN USING A DATA DISPLAY
The dotplot above shows the volume of juice squeezed from $9$ oranges. What is the median volume of juice squeezed, in fluid ounces?
## Things to remember
Center describes a typical value.
• Mean: Average value
$\text{mean}=\frac{\text{sum of values}}{\text{number of values}}$
• Median: Middle value when data ordered from least to greatest
Spread describes the variation of the data.
$\text{range}=\text{highest value}-\text{lowest value}$
• Standard deviation: Average distance between the mean and a data point
## Want to join the conversation?
• gaergargasfgacvxbcxbxvbdgbdrtgbfgse srgsergerce e ge
|
Paul's Online Notes
Home / Calculus II / Parametric Equations and Polar Coordinates / Parametric Equations and Curves
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.
### Section 3-1 : Parametric Equations and Curves
4. Eliminate the parameter for the following set of parametric equations, sketch the graph of the parametric curve and give any limits that might exist on $$x$$ and $$y$$.
$x = 3\sin \left( t \right)\hspace{0.5in}y = - 4\cos \left( t \right)\hspace{0.25in}\hspace{0.5in}\,\,\,\,\,0 \le t \le 2\pi$
Show All Steps Hide All Steps
Start Solution
First, we’ll eliminate the parameter from this set of parametric equations. For this particular set of parametric equations we will make use of the well-known trig identity,
${\cos ^2}\left( \theta \right) + {\sin ^2}\left( \theta \right) = 1$
We can solve each of the parametric equations for sine and cosine as follows,
$\sin \left( t \right) = \frac{x}{3}\hspace{0.25in}\hspace{0.25in}\cos \left( t \right) = - \frac{y}{4}$
Plugging these into the trig identity gives,
${\left( { - \frac{y}{4}} \right)^2} + {\left( {\frac{x}{3}} \right)^2} = 1\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\frac{{{x^2}}}{9} + \frac{{{y^2}}}{{16}} = 1$
Therefore, the parametric curve will be some or all of the ellipse above.
We have to be careful when eliminating the parameter from a set of parametric equations. The graph of the resulting equation in only $$x$$ and $$y$$ may or may not be the graph of the parametric curve. Often, although not always, the parametric curve will only be a portion of the curve from the equation in terms of only $$x$$ and $$y$$. Another situation that can happen is that the parametric curve will retrace some or all of the curve from the equation in terms of only $$x$$ and $$y$$ more than once.
The next few steps will help us to determine just how much of the ellipse we have and if it retraces the ellipse, or a portion of the ellipse, more than once.
Before we proceed with the rest of the problem let’s fist note that there is really no set order for doing the steps. They can often be done in different orders and in some cases may actually be easier to do in different orders. The order we’ll be following here is used simply because it is the order that I’m used to working them in. If you find a different order would be best for you then that is the order you should use.
Show Step 2
At this point we can get a good idea on what the limits on $$x$$ and $$y$$ are going to be so let’s do that. Note that often we won’t get the actual limits on $$x$$ and $$y$$ in this step. All we are really finding here is the largest possible range of limits for $$x$$ and $$y$$. Having these can sometimes be useful for later steps and so we’ll get them here.
We can use our knowledge of sine and cosine to get the following inequalities. Note as discussed above however that these may not be the limits on $$x$$ and $$y$$ we are after.
$\begin{array}{ccc} - 1 \le \sin \left( t \right) \le 1 & \hspace{1.0in} & - 1 \le \cos \left( t \right) \le 1\\ - 3 \le 3\sin \left( t \right) \le 3 & \hspace{1.0in} & 4 \ge - 4\cos \left( t \right) \ge - 4\\ - 3 \le x \le 3 & \hspace{1.0in} & - 4 \le y \le 4\end{array}$
Note that to find these limits in general we just start with the appropriate trig function and then build up the equation for $$x$$ and $$y$$ by first multiplying the trig function by any coefficient, if present, and then adding/subtracting any numbers that might be present (not needed in this case). This, in turn, gives us the largest possible set of limits for $$x$$ and $$y$$. Just remember to be careful when multiplying an inequality by a negative number. Don’t forget to flip the direction of the inequalities when doing this.
Now, at this point we need to be a little careful. What we’ve actually found here are the largest possible inequalities for the limits on $$x$$ and $$y$$. This set of inequalities for the limits on $$x$$ and $$y$$ assume that the parametric curve will be completely traced out at least once for the range of $$t$$’s we were given in the problem statement. It is always possible that the curve will not trace out a full trace in the given range of $$t$$’s. In a later step we’ll determine if the parametric curve does trace out a full trace and hence determine the actual limits on $$x$$ and $$y$$.
Before we move onto the next step there are a couple of issues we should quickly discuss.
First, remember that when we talk about the parametric curve tracing out once we are not necessarily talking about the ellipse itself being fully traced out. The parametric curve will be at most the full ellipse and we haven’t determined just yet how much of the ellipse the parametric curve will trace out. So, one trace of the parametric curve refers to the largest portion of the ellipse that the parametric curve can possibly trace out given no restrictions on $$t$$.
Second, if we can’t completely determine the actual limits on $$x$$ and $$y$$ at this point why did we do them here? In part we did them here because we can and the answer to this step often does end up being the limits on $$x$$ and $$y$$. Also, there are times where knowing the largest possible limits on $$x$$ and/or $$y$$ will be convenient for some of the later steps.
Finally, we can sometimes get these limits from the sketch of the parametric curve. However, there are some parametric equations that we can’t easily get the sketch without doing this step. We’ll eventually do some problems like that.
Show Step 3
Before we sketch the graph of the parametric curve recall that all parametric curves have a direction of motion, i.e. the direction indicating increasing values of the parameter, $$t$$ in this case.
There are several ways to get the direction of motion for the curve. One is to plug in values of $$t$$ into the parametric equations to get some points that we can use to identify the direction of motion.
Here is a table of values for this set of parametric equations. In this case we were also given a range of $$t$$’s and we need to restrict the $$t$$’s in our table to that range.
$$t$$ $$x$$ $$y$$
0 0 -4
$$\frac{\pi }{2}$$ 3 0
$$\pi$$ 0 4
$$\frac{{3\pi }}{2}$$ -3 0
$$2\pi$$ 0 -4
Now, this table seems to suggest that the parametric equation will follow the ellipse in a counter clockwise rotation. It also seems to suggest that the ellipse will be traced out exactly once.
However, tables of values for parametric equations involving sine and/or cosine equations can be deceptive.
Because sine and cosine oscillate it is possible to choose “bad” values of $$t$$ that suggest a single trace when in fact the curve is tracing out faster than we realize and it is in fact tracing out more than once. We’ll need to do some extra analysis to verify if the ellipse traces out once or more than once.
Also, just because the table suggests a particular direction doesn’t actually mean it is going in that direction. It could be moving in the opposite direction at a speed that just happens to match the points you got in the table. Go back to the notes and check out Example 5. Plug in the points we used in our table above and you’ll get a set of points that suggest the curve is tracing out clockwise when in fact it is tracing out counter clockwise!
Note that because this is such a “bad” way of getting the direction of motion we put it in its own step so we could discuss it in detail. The actual method we’ll be using is in the next step and we’ll not be doing table work again unless it is absolutely required for some other part of the problem.
Show Step 4
As suggested in the previous step the table of values is not a good way to get direction of motion for parametric curves involving trig function so let’s go through a much better way of determining the direction of motion. This method takes a little time to think things through but it will always get the correct direction if you take the time.
First, let’s think about what happens if we start at $$t = 0$$ and increase $$t$$ to $$t = \pi$$.
As we cover this range of $$t$$’s we know that cosine starts at 1, decreasing through zero and finally stops at -1. So, that means that $$y$$ will start at $$y = - 4$$ (i.e. where cosine is 1), go through the $$x$$-axis (i.e. where cosine is zero) and finally stop at $$y = 4$$ (i.e. where cosine is -1). Now, this doesn’t give us a direction of motion as all it really tells us that $$y$$ increases and it could do this following the right side of the ellipse (i.e. counter clockwise) or it could do this following the left side of the ellipse (i.e. clockwise).
So, let’s see what the behavior of sine in this range tells us. Starting at $$t = 0$$ we know that sine will be zero and so $$x$$ will also be zero. As $$t$$ increases to $$t = \frac{\pi }{2}$$ we know that sine increases from zero to one and so $$x$$ will increase from zero to three. Finally, as we further increase $$t$$ to $$t = \pi$$ sine will decrease from one back to zero and so $$x$$ will also decrease from three to zero.
So, taking the $$x$$ and $$y$$ analysis above together we can see that at $$t = 0$$ the curve will start at the point $$\left( {0, - 4} \right)$$. As we increase $$t$$ to $$t = \frac{\pi }{2}$$ the curve will have to follow the ellipse with increasing $$x$$ and $$y$$ until it hits the point $$\left( {3,0} \right)$$. The only way we can reach this second point and have the correct increasing behavior for both $$x$$ and $$y$$ is to move in a counter clockwise direction along the right half of the ellipse.
If we further increase $$t$$ from $$t = \frac{\pi }{2}$$ to $$t = \pi$$ we can see that $$y$$ must continue to increase but $$x$$ now decreases until we get to the point $$\left( {0,4} \right)$$ and again the only way we can reach this third point and have the required increasing/decreasing information for $$y$$/$$x$$ respectively is to be moving in a counter clockwise direction along the right half.
We can do a similar analysis increasing $$t$$ from $$t = \pi$$ to $$t = 2\pi$$ to see that we must still move in a counter clockwise direction that takes us through the point $$\left( { - 3,0} \right)$$ and then finally ending at the point $$\left( {0, - 4} \right)$$.
So, from this analysis we can see that the curve must be tracing out in a counter clockwise direction.
This analysis seems complicated and maybe not so easy to do the first few times you see it. However, once you do it a couple of times you’ll see that it’s not quite as bad as it initially seems to be. Also, it really is the only way to guarantee that you’ve got the correct direction of motion for the curve when dealing with parametric equations involving sine and/or cosine.
If you had trouble visualizing how sine and cosine changed as we increased $$t$$ you might want to do a quick sketch of the graphs of sine and cosine and you’ll see right away that we were correct in our analysis of their behavior as we increased $$t$$.
Show Step 5
Okay, in the last step notice that we also showed that the curve will trace the ellipse out exactly once in the given range of $$t$$’s. However, let’s assume that we hadn’t done the direction analysis yet and see if we can determine this without the direction analysis.
This is actually pretty simple to do, or at least simpler than the direction analysis. All it requires is that you know where sine and cosine are zero, 1 and -1. If you recall your unit circle it’s always easy to know where sine and cosine have these values. We’ll also be able to verify the ranges of $$x$$ and $$y$$ found in Step 2 were in fact the actual ranges for $$x$$ and $$y$$.
Let’s start with the “initial” point on the curve, i.e. the point at the left end of our range of $$t$$’s, $$t = 0$$ in this case. Where you start this analysis is really dependent upon the set of parametric equations, the parametric curve and/or if there is a range of $$t$$’s given. Good starting points are the “initial” point, one of the end points of the curve itself (if the curve does have endpoints) or $$t = 0$$. Sometimes one option will be better than the others and other times it won’t matter.
In this case two of the options are the same point so it seems like a good point to use.
So, at $$t = 0$$ we are at the point $$\left( {0, - 4} \right)$$. We know that the parametric curve is some or all of the ellipse we found in the first step. So, at this point let’s assume it is the full ellipse and ask ourselves the following question. When do we get back to this point? Or, in other words, what is the next value of $$t$$ after $$t = 0$$ (since that is the point we choose to start off with) are we back at the point $$\left( {0, - 4} \right)$$?
Before doing this let’s quickly note that if the parametric curve doesn’t get back to this point we’ll determine that in the following analysis and that will be useful in helping us to determine how much of the ellipse will get traced out by the parametric curve.
Okay let’s back to the analysis. In order to be at the point $$\left( {0, - 4} \right)$$ we know we must have $$\sin \left( t \right) = 0$$ (only way to get $$x = 0$$!) and we must have $$\cos \left( t \right) = 1$$ (only way to get $$y = - 4$$!). For $$t > 0$$ we know that $$\sin \left( t \right) = 0$$ at $$t = \pi ,2\pi ,3\pi , \ldots$$ and likewise we know that $$\cos \left( t \right) = 1$$ at $$t = 2\pi ,4\pi ,6\pi , \ldots$$. The first value of $$t$$ that is in both lists is $$t = 2\pi$$ and so this is the next value of $$t$$ that will put us at that point.
This tells us several things. First, we found that the parametric equation will get back to the initial point and so it is possible for the parametric equation to trace out the full ellipse.
Secondly, we got back to the point $$\left( {0, - 4} \right)$$ at the very last $$t$$ from the range of $$t$$’s we were given in the problem statement and so the parametric curve will trace out the ellipse exactly once for the given range of $$t$$’s.
Finally, from this analysis we found the parametric curve traced out the full ellipse in the range of $$t$$’s given in the problem statement and so we know now that the limits of $$x$$ and $$y$$ we found in Step 2 are in fact the actual limits on $$x$$ and $$y$$ for this curve.
As a final comment from this step let’s note that this analysis in this step was a little easier than normal because the argument of the trig functions was just a $$t$$ as opposed to say 2$$t$$ or $$\frac{1}{3}t$$ which does make the analysis a tiny bit more complicated. We’ll see how to deal with these kinds of arguments in the next couple of problems.
Show Step 6
Finally, here is a sketch of the parametric curve for this set of parametric equations.
For this sketch we included the points from our table because we had them but we won’t always include them as we are often only interested in the sketch itself and the direction of motion.
Also, because the problem asked for it here are the formal limits on $$x$$ and $$y$$ for this parametric curve.
$- 3 \le x \le 3\hspace{0.25in}\hspace{0.25in} - 4 \le y \le 4$
As a final set of thoughts for this problem you really should go back and make sure you understand the processes we went through in Step 4 and Step 5. Those are often the best way of getting at the information we found in those steps. The processes can seem a little mysterious at first but once you’ve done a couple you’ll find it isn’t as bad as they might have first appeared.
Also, for the rest of the problems in this section we’ll build a table of $$t$$ values only if it is absolutely necessary for the problem. In other words, the process we used in Step 4 and 5 will be the processes we’ll be using to get direction of motion for the parametric curve and to determine if the curve is traced out more than once or not.
You should also take a look at problems 5 and 6 in this section and contrast the number of traces of the curve with this problem. The only difference in the set of parametric equations in problems 4, 5 and 6 is the argument of the trig functions. After going through these three problems can you reach any conclusions on how the argument of the trig functions will affect the parametric curves for this type of parametric equations?
|
# 4th Grade Math Multiplying 2-digit by 2-digit Numbers Using the Area Model / Box Method.
## Presentation on theme: "4th Grade Math Multiplying 2-digit by 2-digit Numbers Using the Area Model / Box Method."— Presentation transcript:
4th Grade Math Multiplying 2-digit by 2-digit Numbers Using the Area Model / Box Method
Slide 1 Here's the Problem Let's go with 23 x 45
Slide 3 Expanded Form 23 becomes 20 + 3 45 becomes 40 + 5
page 4 Drawing Time! Draw a window with 4 panels. 20 and 3 go on the top 40 and 5 go on the side
Okay, Let's Multiply! ❖ The top left square... ❖ Is under the 20... ❖ And beside the 40... ❖ So multiply 20 x 40 ❖ 800 in the 1st square
Square 2 ❖ The top right square... ❖ Is under the 3... ❖ And beside the 40 ❖ Multiply 3 x 40 ❖ 120
Down on the Corner ❖ Bottom left square... ❖ Under the 20... ❖ Beside the 5 ❖ 5 x 20 ❖ 100
The Last Panel ❖ The bottom right square... ❖ Is under the 3... ❖ And beside the 5 ❖ 3 x 5 ❖ 15
How much are those numbers in the Window? ❖ These 4 numbers are the partial products ❖ Add them together to get the final answer ❖ Thanks to the commutative property the order that you add them doesn't matter ❖ In this example, we are adding across first, then adding our two answers
Add it up! ❖ Adding the top row ❖ 800 + 120 ❖ 920
Bottoms Up! ❖ Adding the bottom Row ❖ 100 + 15 ❖ 115
Summing it Up ❖ Add the two sums together ❖ 920 + 115 ❖ 1,035
Almost the last page This is it! 23 x 45 = 1,035
This is the last page I hope this helps The end (Told you it was the last page)
Download ppt "4th Grade Math Multiplying 2-digit by 2-digit Numbers Using the Area Model / Box Method."
Similar presentations
|
Education.com
# Triangles and the Pythagorean Theorem Study Guide
(not rated)
## Introduction to Triangles and the Pythagorean Theorem
### Lesson Summary
In this lesson, you will learn the special names for the sides of a right triangle. You will also learn how to use the Pythagorean theorem to find missing parts of a right triangle and to determine whether three segments will make a right triangle.
The right triangle is very important in geometry because it can be used in so many different ways. The Pythagorean theorem is just one of the special relationships that can be used to help solve problems and find missing information. Right triangles can be used to find solutions to problems involving figures that are not even polygons.
## Parts of a Right Triangle
In a right triangle, the sides that meet to form the right angle are called the legs. The side opposite the right angle is called the hypotenuse. The hypotenuse is always the longest of the three sides. It is important that you can correctly identify the sides of a right triangle, regardless of what position the triangle is in.
## Review of Squares and Square Roots
Before you study the Pythagorean theorem, let's first review squares and square roots. Just like addition and subtraction are inverses, so are squares and square roots. In other words, they "undo" each other. To square a number, you multiply it by itself. For example, 52 means two factors of five, or five times five, which is 25. Written algebraically, it looks like this: 52 = 5 × 5 = 25. A common mistake is to say that you multiply by two, since two is the exponent (the small raised number). But the exponent tells you how many times to use the base (bottom number) as a factor.
Twenty-five is a perfect square. It can be written as the product of two equal factors. It would be helpful for you to learn the first 16 perfect squares. When completed, the following chart will be a useful reference. It is not necessarily important that you memorize the chart, but you need to understand how the numbers are generated. Even the most basic calculators can help you determine squares and square roots of larger numbers.
## The Pythagorean Theorem
The Pythagorean theorem is one of the most famous theorems in mathematics. The Greek mathematician Pythagoras (circa 585–500 b.c.) is given credit for originating it. Evidence shows that it was used by the Egyptians and Babylonians for hundreds of years before Pythagoras.
The Pythagorean theorem can be used to solve many real-life problems. Any unknown length can be found if you can make it a part of a right triangle. You need to know only two of the sides of a right triangle to find the third unknown side. A common mistake is always adding the squares of the two known lengths. You add the squares of the legs only when you are looking for the hypotenuse. If you know the hypotenuse and one of the legs, then you subtract the square of the leg from the square of the hypotenuse. Another common mistake is forgetting to take the square root as your final step. You just need to remember that you are solving not for the square of the side, but for the length of the side.
150 Characters allowed
### Related Questions
#### Q:
See More Questions
### Today on Education.com
#### WORKBOOKS
May Workbooks are Here!
#### ACTIVITIES
Get Outside! 10 Playful Activities
#### PARENTING
7 Parenting Tips to Take the Pressure Off
Welcome!
|
# Common Core: High School - Algebra : Polynomial Identities and Numerical Relationships: CCSS.Math.Content.HSA-APR.C.4
## Example Questions
### Example Question #21 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4
Use FOIL for the following expression.
Explanation:
The first step is to rewrite the problem as follows.
Now we multiply the first parts of the first and second expression together.
Now we multiply the first term of the first expression with the second term of the second expression.
Now we multiply the second term of the first expression with the first term of the second expression.
Now we multiply the last terms of each expression together.
Now we add all these results together, and we get.
### Example Question #21 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4
Use FOIL for the following expression.
Explanation:
The first step is to rewrite the problem as follows.
Now we multiply the first parts of the first and second expression together.
Now we multiply the first term of the first expression with the second term of the second expression.
Now we multiply the second term of the first expression with the first term of the second expression.
Now we multiply the last terms of each expression together.
Now we add all these results together, and we get.
### Example Question #191 : Arithmetic With Polynomials & Rational Expressions
Use FOIL for the following expression.
Explanation:
The first step is to rewrite the problem as follows.
Now we multiply the first parts of the first and second expression together.
Now we multiply the first term of the first expression with the second term of the second expression.
Now we multiply the second term of the first expression with the first term of the second expression.
Now we multiply the last terms of each expression together.
Now we add all these results together, and we get.
### Example Question #23 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4
Use FOIL for the following expression.
Explanation:
The first step is to rewrite the problem as follows.
Now we multiply the first parts of the first and second expression together.
Now we multiply the first term of the first expression with the second term of the second expression.
Now we multiply the second term of the first expression with the first term of the second expression.
Now we multiply the last terms of each expression together.
Now we add all these results together, and we get.
### Example Question #24 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4
Use FOIL for the following expression.
Explanation:
The first step is to rewrite the problem as follows.
Now we multiply the first parts of the first and second expression together.
Now we multiply the first term of the first expression with the second term of the second expression.
Now we multiply the second term of the first expression with the first term of the second expression.
Now we multiply the last terms of each expression together.
Now we add all these results together, and we get.
### Example Question #25 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4
Use FOIL for the following expression.
Explanation:
The first step is to rewrite the problem as follows.
Now we multiply the first parts of the first and second expression together.
Now we multiply the first term of the first expression with the second term of the second expression.
Now we multiply the second term of the first expression with the first term of the second expression.
Now we multiply the last terms of each expression together.
Now we add all these results together, and we get.
### Example Question #26 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4
Use FOIL for the following expression.
Explanation:
The first step is to rewrite the problem as follows.
Now we multiply the first parts of the first and second expression together.
Now we multiply the first term of the first expression with the second term of the second expression.
Now we multiply the second term of the first expression with the first term of the second expression.
Now we multiply the last terms of each expression together.
Now we add all these results together, and we get.
### Example Question #28 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4
Use FOIL for the following expression.
Explanation:
The first step is to rewrite the problem as follows.
Now we multiply the first parts of the first and second expression together.
Now we multiply the first term of the first expression with the second term of the second expression.
Now we multiply the second term of the first expression with the first term of the second expression.
Now we multiply the last terms of each expression together.
Now we add all these results together, and we get.
### Example Question #29 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4
Use FOIL for the following expression.
Explanation:
The first step is to rewrite the problem as follows.
Now we multiply the first parts of the first and second expression together.
Now we multiply the first term of the first expression with the second term of the second expression.
Now we multiply the second term of the first expression with the first term of the second expression.
Now we multiply the last terms of each expression together.
Now we add all these results together, and we get.
### Example Question #27 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4
Use FOIL for the following expression.
Explanation:
The first step is to rewrite the problem as follows.
Now we multiply the first parts of the first and second expression together.
Now we multiply the first term of the first expression with the second term of the second expression.
Now we multiply the second term of the first expression with the first term of the second expression.
Now we multiply the last terms of each expression together.
Now we add all these results together, and we get.
|
# GRE Math : Other Number Line
## Example Questions
### Example Question #1 : Number Line
The range of the earnings for architecture graduates is , and the range of the salaries for engineering graduates is .
Which of the following statements individually provide(s) sufficient additional information to determine the range of the salaries of all graduates between the two professions?
A: The median salary for the engineers is greater than that of the architects.
B: The average (arithmetic mean) of the engineers is greater than that of the architects.
C: The lowest salary of the engineers is less than the lowest of the architects.
B only
A and C only
C only
A only
A, B, and C
C only
Explanation:
The provision of the bottom-end of the engineering range is the only additional information that provides us a fixed endpoint from which we can build off of by supplementing with the ranges provided in the question, to give us the full range between both engineering and achitecture graduates. See the diagram provided to understand how this can be done.
Even if the mean and medians were provided, these additional values give us no information on the endpoints of the salaries, and the question only asks for the range.
### Example Question #2 : Number Line
What's the distance between and on a number line?
Explanation:
Let's draw a number line.
Since a number line is straight and contains the numbers consecutively, we just subtract from to get
### Example Question #3 : Number Line
Which of the following answer best fits in the picture below?
Explanation:
Open circles mean the values are excluded from the set.
The number line shows the set is between and exclusive.
The only value in that set would be
### Example Question #4 : Number Line
If , then where on the number line lies ?
Explanation:
Because a number line contains both positive and negative integers, we need to consider both possibilities.
is and that value is the same as . Therefore we eliminate the choice because will always be greater than those values raised to the power.
Next is . We elminate both the positive and negative range of . If we look at the difference between and , it's over .
Then, we should guess that will definitely be greater than so therefore answer is .
Remember, a negative value raised to an even power will always have a positive value.
### Example Question #5 : Number Line
If perimeter of equilateral triangle is , what is the height of the triangle?
Explanation:
Since perimeter of equilateral triangle is and we have three equal sides, we just divide that vaue by to get . To find height, we can set-up a proportion.
The height is opposite the angle . Side opposite is and the side of equilateral triangle which is opposite is .
Cross multiply.
Divide both sides by
Let's simplify by factoring out to get a final answer of
Tired of practice problems?
Try live online GRE prep today.
|
# Cos 30 value
cos 30 value
## What is the value of cos 30°?
The value of \cos 30^\circ is a fundamental trigonometric value that is frequently used in mathematics and various applications. To determine this value, one can rely on either trigonometric identities or geometric methods.
### Solution By Steps:
1. Using the Unit Circle:
• The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane. The cosine of an angle is the x-coordinate of the point where the terminal side of the angle intersects the unit circle.
• For 30^\circ (which is equivalent to \frac{\pi}{6} radians), the coordinates of the corresponding point on the unit circle are \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right).
• Hence, \cos 30^\circ = \frac{\sqrt{3}}{2}.
2. Using Special Triangles:
• Consider an equilateral triangle with each angle measuring 60^\circ and each side of length 2.
• If the triangle is bisected, it forms two 30^\circ-60^\circ-90^\circ right triangles.
• In such a right triangle, the hypotenuse is 2, the side opposite the 30^\circ angle (the shorter leg) is 1, and the side adjacent to the 30^\circ angle (the longer leg) is \sqrt{3}.
• The cosine of an angle in a right triangle is given by the ratio of the length of the adjacent side to the hypotenuse:
\cos 30^\circ = \frac{\text{length of adjacent side}}{\text{length of hypotenuse}} = \frac{\sqrt{3}}{2}
### Mathematical Expression:
For accurate representation, the value of \cos 30^\circ is:
\cos 30^\circ = \frac{\sqrt{3}}{2}
|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Linear Inequalities in Two Variables
## Graph inequalities like 2x + 3 < y on the coordinate plane
Estimated16 minsto complete
%
Progress
Practice Linear Inequalities in Two Variables
Progress
Estimated16 minsto complete
%
Linear Inequalities in Two Variables
Did you know that in European hockey leagues, a player gets 2 points for a goal and 1 point for an assist? Suppose a player's contract stipulates that he receives a bonus if he gets more than 100 points. What linear inequality could you write to represent this situation? How would you graph this inequality? In this Concept, you'll learn to graph linear inequalities in two variables so that you can properly analyze scenarios such as this one.
### Guidance
When a linear equation is graphed in a coordinate plane, the line splits the plane into two pieces. Each piece is called a half plane . The diagram below shows how the half planes are formed when graphing a linear equation.
A linear inequality in two variables can also be graphed. Instead of graphing only the boundary line $(y=mx+b)$ , you must also include all the other ordered pairs that could be solutions to the inequality. This is called the solution set and is shown by shading, or coloring, the half plane that includes the appropriate solutions.
When graphing inequalities in two variables, you must remember when the value is included ( $\le$ or $\ge$ ) or not included ( $<$ or $>$ ). To represent these inequalities on a coordinate plane, instead of shaded or unshaded circles, we use solid and dashed lines. We can tell which half of the plane the solution is by looking at the inequality sign.
• $>$ The solution is the half plane above the line.
• $\ge$ The solution is the half plane above the line and also all the points on the line.
• $<$ The solution is the half plane below the line.
• $\le$ The solution is the half plane below the line and also all the points on the line.
The solution of $y > mx+b$ is the half plane above the line. The dashed line shows that the points on the line are not part of the solution.
The solution of $y \ge mx+b$ is the half plane above the line and all the points on the line.
The solution of $y < mx+b$ is the half plane below the line.
The solution of $y \le mx+b$ is the half plane below the line and all the points on the line.
#### Example A
Graph the inequality $y \ge 2x-3$ .
Solution:
This inequality is in slope-intercept form. Begin by graphing the line. Then determine the half plane to color.
• The inequality is $\ge$ , so the line is solid.
• According to the inequality, you should shade the half plane above the boundary line.
In general, the process used to graph a linear inequality in two variables is:
Step 1: Graph the equation using the most appropriate method.
• Slope-intercept form uses the $y-$ intercept and slope to find the line.
• Standard form uses the intercepts to graph the line.
• Point-slope uses a point and the slope to graph the line.
Step 2: If the equal sign is not included, draw a dashed line. Draw a solid line if the equal sign is included.
Step 3: Shade the half plane above the line if the inequality is “greater than.” Shade the half plane under the line if the inequality is “less than.”
#### Example B
Julian has a job as an appliance salesman. He earns a commission of $60 for each washing machine he sells and$130 for each refrigerator he sells. How many washing machines and refrigerators must Julian sell to make $1,000 or more in commission? Solution: Determine the appropriate variables for the unknown quantities. Let $x =$ number of washing machines Julian sells and let $y =$ number of refrigerators Julian sells. Now translate the situation into an inequality. $60x + 130y \ge 1,000$ . Graph the standard form inequality using its intercepts. When $x=0, y=7.692$ . When $y=0, x=16.667$ . The line will be solid. We want the ordered pairs that are solutions to Julian making more than$1,000, so we shade the half plane above the boundary line.
Graphing Horizontal and Vertical Linear Inequalities
Linear inequalities in one variable can also be graphed in the coordinate plane. They take the form of horizontal and vertical lines; however, the process is identical to graphing oblique , or slanted, lines.
#### Example C
Graph the inequality $x>4$ on : 1) a number line and 2) the coordinate plane.
Solution:
Remember what the solution to $x>4$ looks like on a number line.
The solution to this inequality is the set of all real numbers $x$ that are larger than four but not including four.
On a coordinate plane, the line $x=4$ is a vertical line four units to the right of the origin. The inequality does not equal four, so the vertical line is dashed. This shows the reader that the ordered pairs on the vertical line $x=4$ are not solutions to the inequality.
The inequality is looking for all $x-$ coordinates larger than four. We then color the half plane to the right, symbolizing $x>4$ .
Graphing absolute value inequalities can also be done in the coordinate plane. To graph the inequality $|x|\ge 2$ , we can recall a previous Concept and rewrite the absolute value inequality.
$x \le -2$ or $x \ge 2$
Then graph each inequality on a coordinate plane.
In other words, the solution is all the coordinate points for which the value of $x$ is smaller than or equal to –2 and greater than or equal to 2. The solution is represented by the plane to the left of the vertical line $x=-2$ and the plane to the right of line $x=2$ .
Both vertical lines are solid because points on the line are included in the solution.
### Guided Practice
A pound of coffee blend is made by mixing two types of coffee beans. One type costs $9.00 per pound and another type costs$7.00 per pound. Find all the possible mixtures of weights of the two different coffee beans for which the blend costs $8.50 per pound or less. Solution: Begin by determining the appropriate letters to represent the varying quantities. Let $x =$ weight of$9.00 per pound coffee beans in pounds and let $y =$ weight of $7.00 per pound coffee beans in pounds. Translate the information into an inequality. $9x+7y \le 8.50$ . Because the inequality is in standard form, it will be easier to graph using its intercepts. When $x=0, y=1.21$ . When $y=0, x=0.944$ . Graph the inequality. The line will be solid. We shade below the line. We graphed only the first quadrant of the coordinate plane because neither bag should have a negative weight. The blue-shaded region tells you all the possibilities of the two bean mixtures that will give a total less than or equal to$8.50.
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Graphing Inequalities (8:03)
1. Define half plane.
2. In which cases would the boundary line be represented by a dashed line when graphing an inequality?
3. In which cases would the boundary line be represented by a solid line when graphing an inequality?
4. What is a method to help you determine which half plane to color when graphing an inequality?
In 5–26, graph each inequality in a coordinate plane.
1. $x < 20$
2. $y \ge -5$
3. $y \le 6$
4. $|x| > 10$
5. $|y| \le 7$
6. $y \le 4x+3$
7. $y > -\frac{x}{2}-6$
8. $y\le -\frac{1}{2} x+5$
9. $3x-4y \ge 12$
10. $x+7y < 5$
11. $y < -4x+4$
12. $y > \frac{7}{2}x+3$
13. $6x+5y>1$
14. $6x-5y\le 15$
15. $2x-y<5$
16. $y+5 \le -4x+10$
17. $x-\frac{1}{2}y \ge 5$
18. $y+4 \le -\frac{x}{3}+5$
19. $5x-2y > 4$
20. $30x+5y < 100$
21. $y \ge -x$
22. $6x-y < 4$
23. Lili can make yarn ankle bracelets and wrist bracelets. She has 600 yards of yarn available. It takes 6 yards to make one wrist bracelet and 8 yards to make one ankle bracelet. Find all the possible combinations of ankle bracelets and wrist bracelets she can make without going over her available yarn.
24. An ounce of gold costs $670 and an ounce of silver costs$13. Find all possible weights of silver and gold that make an alloy (combination of metals) that costs less than $600 per ounce. 25. A phone company charges 50 cents per minute during the daytime and 10 cents per minute at night. How many daytime minutes and nighttime minutes would you have to use to pay more than$20.00 over a 24-hour period?
26. Jesu has $30 to spend on food for a class barbeque. Hot dogs cost$0.75 each (including the bun) and burgers cost $1.25 (including bun and salad). Plot a graph that shows all the combinations of hot dogs and burgers he could buy for the barbecue, spending less than$30.00.
27. At the local grocery store, strawberries cost $3.00 per pound and bananas cost$1.00 per pound. If I have $10 to spend between strawberries and bananas, draw a graph to show what combinations of each I can buy and spend at most$10.
### Vocabulary Language: English Spanish
boundary line
boundary line
The boundary line is an equation $(y=mx+b)$ that creates two half planes.
half plane
half plane
When a linear equation is graphed in a coordinate plane, the line splits the plane into two pieces. Each piece is called a half plane.
solution set
solution set
All ordered pairs that could be solutions to an inequality, which are shown graphically by shading, or coloring, the half plane that includes the appropriate solutions.
Cartesian Plane
Cartesian Plane
The Cartesian plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin.
Linear Inequality
Linear Inequality
Linear inequalities are inequalities that can be written in one of the following four forms: $ax + b > c, ax + b < c, ax + b \ge c$, or $ax + b \le c$.
Slope-Intercept Form
Slope-Intercept Form
The slope-intercept form of a line is $y = mx + b,$ where $m$ is the slope and $b$ is the $y-$intercept.
### Explore More
Sign in to explore more, including practice questions and solutions for Linear Inequalities in Two Variables.
Please wait...
Please wait...
|
1. Identity
sinA / (1+cosA) = tan 0.5A
2. Hello, Punch!
We need a few identities: .$\displaystyle \begin{array}{cccccccccc}\sin^2\!\frac{A}{2} &=& \dfrac{1-\cos A}{2} && \Rightarrow && \sin\frac{A}{2} &=& \sqrt{\dfrac{1-\cos A}{2}} \\ \\[-3mm] \cos^2\!\frac{A}{2} &=& \dfrac{1+\cos A}{2} && \Rightarrow && \cos\frac{A}{2} &=& \dfrac{1+\cos A}{2}\end{array}$
Prove: .$\displaystyle \frac{\sin A}{1+\cos A}\: =\: \tan \tfrac{A}{2}$
The right side is: .$\displaystyle \tan\frac{A}{2} \;=\;\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} \;=\;\frac{\sqrt{\dfrac{1-\cos A}{2}}}{\sqrt{\dfrac{1+\cos A}{2}}} \;=\;\sqrt{\frac{1-\cos A}{1+\cos A}}$
Multiply by $\displaystyle \frac{1+\cos A}{1 + \cos A}:\quad\sqrt{\frac{1-\cos A}{1+\cos A}\cdot\frac{1+\cos A}{1+\cos A}} \;=\;\sqrt{\frac{1-\cos^2\!A}{(1+\cos A)^2}} \;=\;\sqrt{\frac{\sin^2\!A}{(1+\cos A)^2}}$
Therefore: .$\displaystyle \tan\frac{A}{2} \;=\;\frac{\sin A}{1 + \cos A}$
3. Originally Posted by Soroban
Hello, Punch!
We need a few identities: .$\displaystyle \begin{array}{cccccccccc}\sin^2\!\frac{A}{2} &=& \dfrac{1-\cos A}{2} && \Rightarrow && \sin\frac{A}{2} &=& \sqrt{\dfrac{1-\cos A}{2}} \\ \\[-3mm]$$\displaystyle \cos^2\!\frac{A}{2} &=& \dfrac{1+\cos A}{2} && \Rightarrow && \cos\frac{A}{2} &=& \dfrac{1+\cos A}{2}\end{array}$
The right side is: .$\displaystyle \tan\frac{A}{2} \;=\;\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} \;=\;\frac{\sqrt{\dfrac{1-\cos A}{2}}}{\sqrt{\dfrac{1+\cos A}{2}}} \;=\;\sqrt{\frac{1-\cos A}{1+\cos A}}$
Multiply by $\displaystyle \frac{1+\cos A}{1 + \cos A}:\quad\sqrt{\frac{1-\cos A}{1+\cos A}\cdot\frac{1+\cos A}{1+\cos A}} \;=\;\sqrt{\frac{1-\cos^2\!A}{(1+\cos A)^2}} \;=\;\sqrt{\frac{\sin^2\!A}{(1+\cos A)^2}}$
Therefore: .$\displaystyle \tan\frac{A}{2} \;=\;\frac{\sin A}{1 + \cos A}$
Hello! Thanks a lot thank you very much!
4. lol Soroban, guess you can explain it better than me.
http://www.mathhelpforum.com/math-he...-identity.html
5. Just a question, do we need the memorize all the formulas, or the formulas are given in the formula sheet during the examination?
6. Originally Posted by Punch
Just a question, do we need the memorize all the formulas, or the formulas are given in the formula sheet during the examination?
It up to the teacher. He or she may let you use a formula sheet or you may have to memorize them. You could also learn to derive them.
|
# Quadratic Equations Set – 2
1. I. 4x2 + 27x + 18 = 0,
II. 2y2 – 7y + 3 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Answer & Explanation
Option B
Solution:
4x2 + 27x + 18 = 0
4x2 + 24x + 3x + 18 = 0
So x = -3/4, -6
2y2 – 7y + 3 = 0
2y2 – 6y – y + 3 = 0
So y = 1/2, 3
Put all values on number line and analyze the relationship
-6… -3/4… 1/2… 3
2. I. 3x2 – 2x – 8 = 0,
II. 6y2 – 17y + 10 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Answer & Explanation
Option D
Solution:
3x2 – 2x – 8 = 0
3x2 – 6x + 4x – 8 = 0
So x = -4/3, 2
6y2 – 17y + 10 = 0
6y2 – 12y – 5y + 10 = 0
So y = 5/6, 2
Put all values on number line and analyze the relationship
-4/3… 2… 5/6
3. I. 32 + 11x + 6 = 0,
II. 5y2 + 16y + 3 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Answer & Explanation
Option E
Solution:
32 + 11x + 6 = 0
32 + 9x + 2x + 6 = 0
So x = -3, -2/3
5y2 + 16y + 3 = 0
5y2 + 15y + y + 3 = 0
So y = -1/5, -3
Put all values on number line and analyze the relationship
-3 …. -2/3 ….-1/5
Since the common value (-3) is not in between other 2 values, there is no relationship between x and y.
4. I. 4x2 – 11x + 6 = 0,
II. 6y2 – 29y + 28 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Answer & Explanation
Option E
Solution:
4x2 – 11x + 6 = 0
4x2 – 8x – 3x + 6 = 0
So x = 3/4, 2
6y2 – 29y + 28 = 0
6y2 – 8y – 21y + 28 = 0
So y = 4/3, 7/2
Put all values on number line and analyze the relationship
3/4 …. 4/3….. 2…. 7/2
5. I. 3x2 – 25x + 52 = 0,
II. 3y2 – 8y – 16 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Answer & Explanation
Option C
Solution:
3x2 – 25x + 52 = 0
3x2 – 12x – 13x + 52 = 0
So x = 4, 13/3
3y2 – 8y – 16 = 0
3y2 – 12y + 4y – 16 = 0
So y = 4, -4/3
Put all values on number line and analyze the relationship
-4/3 …. 4…. 13/3
6. I. 8x2 + 10x + 3 = 0,
II. 3y2 + 70y + 40 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Answer & Explanation
Option A
Solution:
8x2 + 10x + 3 = 0
8x2 + 4x + 6x + 3 = 0
So x = -3/4, -1/2
3y2 + 70y + 40 = 0
3y2 + 30y + 40y + 40 = 0
So y = -10, -4/3
Put all values on number line and analyze the relationship
-10 … -4/3 …. -3/4 …. -1/2
7. I. 50x2 ¬¬– 95x + 42 = 0,
II. 50y2 – 65y + 21 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relationship cannot be determined
Answer & Explanation
Option C
Solution:
50x2 ¬¬– 95x + 42 = 0
50x2 ¬¬– 60x – 35x + 42 = 0
So x = 7/10, 6/5
50y2 – 65y + 21 = 0
50y2 – 65y + 21 = 0
So y = 3/5, 7/10
Put all values on number line and analyze the relationship
3/5…. 7/10…. 6/5
8. I. 5x2 – 13x + 6 = 0,
II. 3y2 – 22y – 35 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Answer & Explanation
Option B
Solution:
5x2 – 13x + 6 = 0
5x2 – 10x – 3x + 6 = 0
So x = 3/5, 2
3y2 – 22y – 35 = 0
3y2 – 15y – 7y – 35 = 0
So y = 7/3, 5
Put all values on number line and analyze the relationship
3/5…. 2…. 7/3… 5
9. I. 3x2 – 4x – 15 = 0,
II. 5y2 – 11y – 18 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Answer & Explanation
Option E
Solution:
3x2 – 4x – 15 = 0
3x2 – 9x + 5x – 15 = 0
So x = -5/3, 3
5y2 – 11y – 18 = 0
5y2 – 15y + 6y – 18 = 0
So y = -6/5, 3
Put all values on number line and analyze the relationship
-5/3….. -6/5…. 3
Since the common value (3) is not in between other 2 values, there is no relationship between x and y.
10. I. 2x2 + 5x – 12 = 0,
II. 2y2 – 19y + 35 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relationship cannot be determined
Answer & Explanation
Option B
Solution:
2x2 + 5x ¬– 12 = 0
2x2 + 8x ¬– 3x – 12 = 0
So x = -4 , 3/2
2y2 – 19y + 35 = 0
2y2 – 14y – 5y + 35 = 0
So y = 5/2, 7
Put all values on number line and analyze the relationship
-4……. 3/2…… 5/2…
|
Thanks to theidioms.com
# Learn Linear Algebra for Data Science (Mini-Course)
## Learn Linear Algebra for Data Science (Mini-Course)
### Dot Products and Matrix Multiplication
While working with matrices, there are two major forms of multiplicative operations: dot products and matrix multiplication. A dot product takes the product of two matrices and outputs a single scalar value. On the other hand, matrix multiplication takes the product of two matrices and outputs a single matrix. In this lesson, we will be discussing these two operations and how they work.
#### Dot Product in matrices
Matrix dot products (also known as the inner product) can only be taken when working with two matrices of the same dimension. When taking the dot product of two matrices, we multiply each element from the first matrix by its corresponding element in the second matrix and add up the results.
If we take two matrices and such that = , and , then the dot product is given as,
#### Matrix Multiplication
Two matrices can be multiplied together only when the number of columns of the first matrix is equal to the number of rows in the second matrix.
For matrix multiplication, we take the dot product of each row of the first matrix with each column of the second matrix that results in a matrix of dimensions of the row of the first matrix and the column of the second matrix. For example, for two matrices and , if has a dimension , and has a dimension , matrix multiplication is possible and the resulting matrix is of dimension .
For an easier understanding, let us suppose matrices and to be of dimensions each. Taking matrix and matrix , the matrix multiplication of is given as,
Now, let us find the value of ,
Hence, we also conclude that the matrix multiplication is not commutative, i.e., .
Matrix multiplication has a wide range of applications in Linear Algebra as well as Data Science. In this lesson, we discussed some of the major multiplicative operations performed on matrices. Such operations are usually applied to matrices that represent image data in the field of data science. In the upcoming chapter, we will learn about some matrix transformation methods and the concept of determinants.
|
Parallel & Perpendicular Lines
# Parallel & Perpendicular Lines
## Parallel & Perpendicular Lines
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Parallel & Perpendicular Lines Identifying and graphing March, 2011 Ms. Adler BEGIN
2. Review: Parallel Lines Parallel lines are two lines that run in the same direction; the two lines lie in the same plane and NEVER intersect. What exactly makes them parallel?
3. REVIEW: SLOPE Two lines are parallel because they have equal slopes. What is slope?
4. REVIEW: SLOPE Slope is rise over run. It is the change in y divided by the change in x.
5. review Using pencil and paper, answer this review question from last week: What is the slope between the points (0,1) and (2,7)? 3 1/4 5 1/5 4
6. INCORRECT. Remember that slope is Rise over run. TRY THIS PROBLEM AGAIN. HINT
7. HINT If you are still having trouble, remember that we learned to use this formula: BACK TO REVIEW QUESTION
8. CORRECT! Great job! You know that slope is equal to (y2-Y1)/(x2-x1)
9. Perpendicular Lines parallel perpendicular This is new material, so pay close attention! Perpendicular lines are the opposite of parallel lines. While parallel lines never cross, perpendicular lines cross and create a right angle at their intersection.
10. Take a look at the perpendicular lines shown here. What makes them different from the parallel lines we studied earlier? Pay close attention to the differences in the slope. You can see this in the slope-intercept form. PERPENDICULAR LINES
11. Reciprocals A reciprocal is the “upside down” version of a number. It is the number that you multiply by to get 1. For example, let’s find the reciprocal of 3: 3 can be re-written as 3/1. Now flip it. It’s reciprocal would be 1/3! 3/1 × 1/3 = 1
12. Reciprocals, cont’d A reciprocal is the number you multiply by to get 1. Another way to think of this is the number you get when you divide 1 by the original number. Ex) 1=3r (r is for reciprocal) 1/3=r Ex) 1=1/4×r 4=r
13. mini quiz! What is the reciprocal of 7? -7 1/7 -1/7 14 1/14
14. INCORRECT. To find the reciprocal, divide 1 by the original number. TRY THIS PROBLEM AGAIN
15. CORRECT! Great job! You know that 1/7 is the reciprocal of 7 because 1/7×7=1
16. Slopes Recall that parallel lines have the same slope. Perpendicular lines have slopes which are NEGATIVE reciprocals of one another. Ex) parallel slopes: 2 and 2 perpendicular slopes: 2 and -1/2
17. mini quiz! Which of the following pairs of slopes could be those of a pair of perpendicular lines? 1/4 and 4 -3 and -3 2 and 1/2 -3/2 and -3/2 5 and -1/5
18. INCORRECT. Hint: Recall that perpendicular slopes are negative opposites of one another. try this problem again.
19. INCORRECT. Hint: Remember that perpendicular slopes are reciprocals. Try this problem again!
20. CORRECT! Good job! You remembered that parallel lines have identical slopes, but perpendicular lines have slopes that are negative reciprocals of one another!
21. Review of equation forms STANDARD EQUATION Ax+By+C=0 SLOPE-INTERCEPT FORM y=mx+b POINT-SLOPE FORM y-y1=m(x-x1) Example of each form. All of these examples are equal: -4x+y-2=0 y=4x+2 y-2=4(x-0) With pencil and paper, practice manipulating these equations and discovering how they relate before moving on to the next slide.
22. mini quiz! Find the equation of the line which has slope m=-2 and includes the point (3,1/2). Put the final answer in STANDARD equation form. 2x+y-13/2=0 y=-2x+13/2 y=-2x-13/2 2x+y+13/2=0
23. INCORRECT. Hint: STANDARD equation form is ax+by+c=o. Slope-intercept form is y=mx+b. TRY THIS PROBLEM AGAIN.
24. INCORRECT. Hint: When starting this problem, try putting it into point-slope form first. Y-y1=m(x-x1) TRY THIS PROBLEM AGAIN.
25. CORRECT. GREAT WORK! YOU KNEW TO PLUG THE SLOPE AND X AND Y-VALUES INTO POINT-SLOPE FORM AND MANIPULATE INTO STANDARD EQUATION FORM, AX+BY+C=0.
26. How to Graph We have learned in class how to graph equations, but let’s review and apply it to our current lesson. First, manipulate the equation into point slope form.
27. How to Graph When you have the graph in slope intercept form (y=mx+b), plot the y-intercept. From that point on the y-axis, apply the slope to determine the next point. Once you have the two points, you can use the slope to find the next point, and draw the line.
28. Graphing an Actual Problem For instance, the line y=2x+2 has a y-intercept value of 2, which means that the line crosses the y-axis at point (0,2). Then, since the slope is 2, and slope is equal to the change in x over the change in y (rise over run), and 2 is equal to 2/1, we can “rise” 2 units and “run” 1 unit, which gives us the point (1,4). Now that we have two points, we can draw a straight line through them, giving us our final graph.
29. Think It Over What have you learned so far? This presentation has addressed the following: Slope Parallel lines Reciprocals Perpendicular lines Equation forms Graphing
30. Evaluation It is important that you apply the information that you have learned to more complex problems. The quiz that follows will be not be taken for a grade. However, I want you to work out every problem on looseleaf paper and turn your work in to the sub at the end of class.
31. QUIZ Make sure that you have pencil and paper. Put your name and date on the sheet you turn in. You may not talk to your neighbors. BEGIN QUIZ
32. QUIZ Questions: 1 2 3 4 5 6 7 8 9 10 11 12 FINISHED
33. QUIZ QUESTION 1 What is the negative reciprocal of 3/2? 2/3 -2/3 3/2 3 -3/2
34. INCORRECT. To find the reciprocal, divide 1 by the original number. In this problem, make sure you’re finding the negative reciprocal. TRY THIS PROBLEM AGAIN
35. CORRECT! Great job! you correctly identified the negative reciprocal of the original number! BACK TO QUIZ
36. QUIZ QUESTION 2 WHAT IS THE SLOPE OF THE LINE THAT RUNS THROUGH THE POINTS (-1,6) and (3,3)? -4/3 2/9 -1/4 -3/4 9/2
37. INCORRECT. Remember that slope is Rise over run. try this problem again.
38. CORRECT! Great job! You know how to calculate slope: (y2-Y1)/(x2-x1) BACK TO QUIZ
39. QUIZ QUESTION 3 Identify two lines that are parallel: 4x+y+7=0 y=4x-2 y=3x+7 (1/6)x+(1/2)y=0 y=-3x+1 6x+2y-1=0 5x+y-3=0 y=(-1/5)x+4
40. INCORRECT. Remember that parallel lines have equal slopes, and perpendicular lines have negative reciprocal slopes. (You most likely missed this question because of arithmetic errors, so double check your work.) try this problem again.
41. CORRECT! Great job! you know the qualities of slopes of parallel lines and correctly identified the pair of parallel lines! BACK TO QUIZ
42. QUIZ QUESTION 4 Identify the equation in STANDARD equation form of a line with slope -2 that passes through the point (-1,4). 2x+y-2=0 y=-2x+2 2x+y+6=0 -2x+y+2=0
43. INCORRECT. Make sure that the line is in STANDARD equation form. It is most effective to start with point-slope form and then manipulate. HINT try this problem again.
44. HINT STANDARD equation form: Ax+by+c=0 Point-slope form: Y-y1=m(x-x1) try this problem again.
45. CORRECT! Great job! You used the correct equation form and properly manipulated the information! BACK TO QUIZ
46. QUIZ QUESTION 5 Identify the two slopes which would create perpendicular lines when graphed. 4 and -1/4 2 and 2 -1 and 1 6 and 1/3 1/7 and 7
47. INCORRECT. Recall that perpendicular lines have slopes which are negative reciprocals of one another. try this problem again.
48. CORRECT! Great job! You know what a reciprocal is and applied your knowledge to identify the perpendicular slopes! BACK TO QUIZ
49. QUIZ QUESTION 6 Identify the slope of the line. 0 4 1/4 2 1/2
50. INCORRECT. Remember that slope is Rise over run. try this problem again.
|
# Equations of Concentric Circles
We will learn how to form the equation of concentric circles.
Two circles or more than that are said to be concentric if they have the same centre but different radii.
Let, x$$^{2}$$ + y$$^{2}$$ + 2gx + 2fy + c = 0 be a given circle having centre at (- g, - f) and radius = $$\mathrm{\sqrt{g^{2} + f^{2} - c}}$$.
Therefore, the equation of a circle concentric with the given circle x$$^{2}$$ + y$$^{2}$$ + 2gx + 2fy + c = 0 is
x$$^{2}$$ + y$$^{2}$$ + 2gx + 2fy + c' = 0
Both the circle have the same centre (- g, - f) but their radii are not equal (since, c ≠ c')
Similarly, the equation of a circle with centre at (h, k) and radius equal to r, is (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = r$$^{2}$$.
Therefore, the equation of a circle concentric with the circle (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = r$$^{2}$$ is (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = r$$_{1}$$$$^{2}$$, (r$$_{1}$$ ≠ r)
Assigning different values to r$$_{1}$$ we shall have a family of circles each of which is concentric with the circle (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = r$$^{2}$$.
Solved example to find the equation of a concentric circle:
Find the equation of the circle which is concentric with the circle 2x$$^{2}$$ + 2y$$^{2}$$ + 3x - 4y + 5 = 0 and whose radius is 2√5 units.
Solution:
2x$$^{2}$$ + 2y$$^{2}$$ + 3x - 4y + 5 = 0
⇒ x$$^{2}$$ + y$$^{2}$$ + 3/2x - 2y + $$\frac{5}{2}$$ = 0 ………………..(i)
Clearly, the equation of a circle concentric with the circle (i) is
x$$^{2}$$ + y$$^{2}$$ + $$\frac{3}{2}$$x - 2y + c = 0 ……………………..(ii)
Now, the radius of the circle (ii) = $$\sqrt{(\frac{3}{2})^{2} + (-2)^{2} - c}$$
By question, $$\sqrt{\frac{9}{4} + 4 - c}$$ = 2√5
⇒ $$\frac{25}{4}$$ - c = 20
⇒ c = $$\frac{25}{4}$$ - 20
c = -$$\frac{55}{4}$$
Therefore, the equation of the required circle is
x$$^{2}$$ + y$$^{2}$$ + $$\frac{3}{2}$$x - 2y - $$\frac{55}{4}$$ = 0
⇒ 4x$$^{2}$$ + 4y$$^{2}$$ + 6x - 8y - 55 = 0.
`
The Circle
|
New
New
Year 3
# Scale divisions derived from multiplication facts by 10
I can use unitising and multiplying and dividing by 10 to scale division facts.
New
New
Year 3
# Scale divisions derived from multiplication facts by 10
I can use unitising and multiplying and dividing by 10 to scale division facts.
Share activities with pupils
Share function coming soon...
## Lesson details
### Key learning points
1. Division is the inverse of multiplication
2. If you know that 3 groups of 4 are equal to 12 then there are 3 groups of 4 in 12
3. If you know there are 3 groups of 4 in 12 then there are 3 groups of 4 tens in 12 tens
### Common misconception
Lack of fluency in 2s, 4s and 8s times tables and not understanding the divisibility rules or the problems.
To clearly model use of gattegno chart or place value chart to highlight scaling up and down by 10. Highlight key words of worded problems to understand the problem and the operation required for the calculation.
### Keywords
• Scale / scaling - Scaling is when a given quantity is made ___ times the size. In this lesson, scaling will involve making values 10 times the size.
• Inverse - Inverse means the opposite in effect. The reverse of.
### Licence
This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).
## Starter quiz
### 6 Questions
Q1.
Select the numbers which are multiples of both 4 and 8
2
4
13
Q2.
8 × 9 can help you to solve which of these multiplication equations?
Correct answer: 8 × 90 =
Correct answer: 80 × 9 =
6 × 5 =
5 × 7 =
Q3.
Match the expressions on the left which can help you to solve the expressions on the right.
Correct Answer:8 × 4,8 × 40
8 × 40
Correct Answer:4 × 6,60 × 4
60 × 4
Correct Answer:2 × 8,20 × 8
20 × 8
Correct Answer:6 × 9,6 × 90
6 × 90
Q4.
If 4 × 4 = 16 then 40 × 4 =
Q5.
Complete: If 8 × 8 = 64 then 80 × 8 =
Q6.
Andeep has collected 5 packets of 8 guinea cards. Izzy has collected 10 times as many as Andeep. How many guinea cards does Izzy have?
## Exit quiz
### 6 Questions
Q1.
Which equations does this array represent?
Correct answer: 3 × 4 =
Correct answer: 4 × 3 =
3 × 5 =
3 + 4 =
3 + 3 + 3 =
Q2.
Which division equation does this array represent?
20 ÷ 3 =
Correct answer: 20 ÷ 4 =
Correct answer: 20 ÷ 5 =
20 + 5 =
Q3.
Fill in the gap in this diagram showing related facts. 80 ÷ 4 = ___
2
|
<meta http-equiv="refresh" content="1; url=/nojavascript/">
# Points in the Coordinate Plane
%
Progress
Practice Points in the Coordinate Plane
Progress
%
Points in the Coordinate Plane
Have you ever wanted to find a sunken ship? Take a look at this dilemma.
Gina and Cameron are very excited because they are going to go on a dive to see a sunken ship. The dive is quite shallow which is unusual because most dives are found at depths that are too deep for two junior divers. However, this one is at 40 feet, so the two divers can go to see it.
They have the following map to chart their course. Cameron wants to figure out exactly how far the boat will be from the sunken ship. Each square represents 160 cubic feet of water.
First, he makes a note of the coordinates. Then he can use the map to calculate the distance.
We use coordinate grids like this one all the time. Use the information in this lesson to help Cameron figure out the coordinates of his boat and the sunken ship. Then you will be able to estimate the distance between them.
### Guidance
Previously we worked with integers. We used both horizontal (left-to-right) and vertical (up-and-down) number lines. Imagine putting a horizontal and a vertical number line together. In doing this, you could create a coordinate plane .
In a coordinate plane like the one shown, the horizontal number line is called the $x-$ axis . The vertical number line is called the $y-$ axis . The point at which both of these axes meet is called the origin .
We can use coordinate planes to represent points, two-dimensional geometric figures, or even real-world locations. If you think about a map, you will realize that you see a coordinate plane on a map. Then you use coordinates to find different locations. Let’s look at how we can use a coordinate plane.
How do we name points on a coordinate plane?
Each point on a coordinate plane can be named by an ordered pair of numbers, in the form $(x, y)$ .
• The first number in an ordered pair identifies the $x-$ coordinate . That coordinate describes the point's position in relation to the $x-$ axis.
• The second number in an ordered pair identifies the $y-$ coordinate . That coordinate describes the point's position in relation to the $y-$ axis.
You can remember that the $x-$ coordinate is listed before the $y-$ coordinate in an ordered pair $(x, y)$ , because $x$ comes before $y$ in the alphabet.
#### Example A, B, C
Triangle $ABC$ has vertices $A (-2, -5), \ B(0, 3)$ , and $C(6, -3)$ . Graph triangle $ABC$ on a coordinate plane. Label the coordinates of its vertices.
Solution: Here are the steps to graphing the triangle.
• To plot vertex $A$ at (-2, -5), start at the origin. Move 2 units to the left and then 6 units down. Plot and label point $A$ .
• To plot vertex $B$ at (0, 3), start at the origin. The $x-$ coordinate is zero, so do not move to the left or right. From the origin, simply move 3 units up. Plot and label point $B$ .
• To plot vertex $C$ at (6, -3), start at the origin. Move 6 units to the right and then 3 units down. Plot and label point $C$ .
Connect the vertices with line segments to show the sides of the triangle, as shown.
Here is the original problem once again. Use this information to help Cameron with the coordinates and the distance.
Gina and Cameron are very excited because they are going to go on a dive to see a sunken ship. The dive is quite shallow which is unusual because most dives are found at depths that are too deep for two junior divers. However, this one is at 40 feet, so the two divers can go to see it.
They have the following map to chart their course. Cameron wants to figure out exactly how far the boat will be from the sunken ship. Each square represents 160 cubic feet of water.
First, he makes a note of the coordinates. Then he can use the map to calculate the distance.
First, here are the coordinates of each item on the map.
The dive boat is marked at (4, 8).
The sunken ship is marked at (-3, 7).
Notice the arrows. Once they get to the sunken ship, Gina and Cameron will swim up 1 unit and over 6 units.
$1 + 6 = 7$
If each unit = 160 cubic feet, then we can multiply $160 \times 7$
Gina and Cameron will swim through 1120 cubic feet of water from the sunken ship to the boat.
### Vocabulary
Coordinate Plane
The coordinate plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin. The coordinate plane is also called a Cartesian Plane.
$x -$ axis
The $x-$ axis is the horizontal axis in the coordinate plane, commonly representing the value of the input or independent variable.
$y -$ axis
The $y$ -axis is the vertical number line of the Cartesian plane.
Origin
The origin is the point of intersection of the $x$ and $y$ axes on the Cartesian plane. The coordinates of the origin are (0, 0).
$x-$ coordinate
The $x-$ coordinate is the first term in a coordinate pair, commonly representing the value of the input or independent variable.
$y-$ coordinate
The $y-$ coordinate is the second term in a coordinate pair, commonly representing the value of the output or dependent variable.
### Guided Practice
Here is one for you to try on your own.
This coordinate grid shows locations in Jimmy's city. Name the ordered pair that represents the location of the city park.
Here are the steps to figuring out the coordinates of the city park.
• Place your finger at the origin.
• Next, move your finger to the right along the $x-$ axis until your finger is lined up above the point representing the city park. You will need to move 2 units to the left to do that. Moving to the left along the $x-$ axis means that you are moving in a negative direction. Your finger will point to a negative integer, -2, so that is the $x-$ coordinate.
• Now, move your finger down from the $x-$ axis until your finger reaches the point for the city park. You will need to move 6 units down to do that. Moving down parallel to the $y-$ axis means that you are moving in a negative direction. Your finger will be aligned with the negative integer, -6, on the $y-$ axis, so, that is the $y-$ coordinate.
The arrows below show how you should have moved your finger to find the coordinates.
So, the ordered pair (-2, -6) names the location of the city park.
### Explore More
Directions : Use what you have learned to complete this practice section.
1. Name the ordered pair that represents each of these points on the coordinate plane.
Directions: Below is a map of an amusement park. Name the ordered pair that represents the location of each of these rides.
2. roller coaster
3. Ferris wheel
4. carousel
5. log flume
Directions: Name the ordered pairs that represent the vertices of triangle $FGH$ .
6. $F$
7. $G$
8. $H$
Directions: Name the ordered pairs that represent the vertices of pentagon $ABCDE$ .
9. $A$
10. $B$
11. $C$
12. $D$
13. $E$
14. On the grid below, plot point $V$ at (-6, 4).
15. On the grid below, plot point a triangle with vertices $R (4, -1), \ S (4, -4)$ , and $T (-3, -4)$ .
### Vocabulary Language: English
$x-$axis
$x-$axis
The $x-$axis is the horizontal axis in the coordinate plane, commonly representing the value of the input or independent variable.
$x-$coordinate
$x-$coordinate
The $x-$coordinate is the first term in a coordinate pair, commonly representing the value of the input or independent variable.
$y-$axis
$y-$axis
The $y-$axis is the vertical number line of the Cartesian plane.
Abscissa
Abscissa
The abscissa is the $x-$coordinate of the ordered pair that represents a plotted point on a Cartesian plane. For the point (3, 7), 3 is the abscissa.
Cartesian Plane
Cartesian Plane
The Cartesian plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin.
Coordinate Plane
Coordinate Plane
The coordinate plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin. The coordinate plane is also called a Cartesian Plane.
Ordinate
Ordinate
The ordinate is the $y$-coordinate of the ordered pair that represents a plotted point on a Cartesian plane. For the point (3, 7), 7 is the ordinate.
Origin
Origin
The origin is the point of intersection of the $x$ and $y$ axes on the Cartesian plane. The coordinates of the origin are (0, 0).
|
## RWM102 Study Guide
### 6a. Determine the number of solutions of a given system of linear equations
• How do you verify if a point is a solution to a system of equations?
A solution to a system of equations is just like the solution to a single linear equation, except that the point must satisfy both equations in order to be considered the solution to the system of equations.
For example, the system of equations:
$x+2y=13$
$3x-y=-11$
Let's check if the point (-1,7) is a solution. To check, first we will substitute the point into the first equation.
$-1+2(7)=-1+14=13$
So far, the point works, but we must make sure it works in the other equation as well:
$3(-1)-7=-3-7=-10$
Since this does not satisfy both equations, (-1,7) is not a solution to this system.
To review, see Checking Solutions for Systems of Linear Equations
### 6b. Classify systems of linear equations according to the number of solutions
• How do you know the number of solutions of a system of linear equations?
Systems of linear equations can have 0, 1, or infinite solutions. Most commonly, two lines intersect at only one point, meaning the system has 1 solution. If the two lines are parallel, then they never intersect, and therefore the system has no solution. Finally, if the system has two equations that are actually representative of the same line, then all the points on each line are also a solution to the other equation, meaning there are infinitely many solutions.
To review, see Using Graphs to Solve Linear Equations.
### 6c. Solve systems of linear equations using graphing, substitution, or elimination
• How do you solve a system of linear equations with graphing?
• How do you solve a system of linear equations with substitution?
• How do you solve a system of linear equations with elimination?
Systems of linear equations can be solved through 3 methods, each with advantages and disadvantages.
First, systems of linear equations can be solved by graphing. To use graphing, you only need to graph each line on the same coordinate plane, and then find the point where the lines cross. That point is the solution to the system.
For example, consider the following system of equations:
$y=3x-1$
$y=-x+3$
We can graph both lines and look for the point where they intersect.
Since the lines intersect at (1,2), that is the solution to the system. The graphing method works well when the solution is a lattice point, with whole number values, but is not as effective if the answers are fractions or decimals.
Another method is substitution. Substitution is an algebraic method, rather than the geometric method of graphing. In substitution, we solve one equation for either $x$ or $y$, and then substitute that value into the other equation to find the value of one variable. Once we have found that value, we can substitute it to find the value of the other variable.
For example, let us once again consider our example:
$y=3x-1$
$y=-x+3$
Since $y=-x+3$ in the second equation, we can replace the $y$ in the first equation with that value:
$-x+3=3x-1$
Now we can solve for $x$. $4=4x$, therefore $x=1$. We can now use that value to find the value of $y$: $y=3(1)-1=2$. Therefore the solution is (1,2)
Finally, we can solve a system of equations by elimination. This method is best for systems where one variable can't be isolated that easily. In this method, we multiply one or both equations so that when we add them together, one of the variables cancels. This will allow us to solve for one variable, and then as we did with substitution, we can use that value to find the other remaining value.
For example consider the following system of equations:
$3x+2y=7$
$-x-4y=6$
First, we will multiply the top equation by 2, so that when we add the equations, the $y$ terms will cancel:
$6x+4y=14$
$-x-4y=6$
Now we add the two equations together and solve for $x$:
$5x=20$, $x=4$
Now that we know $x=4$, we can substitute into one of the original equations to find $y$:
$3(4)+2y=7$
Now we can solve for $y$:
$2y=-5$, $y=-52$
Therefore the solution to this system of linear equations is (4, -52)
### 6d. Locate on a coordinate plane all solutions of a given system of inequalities
• How do you graph the solutions to a system of linear inequalities?
When solving a system of inequalities, graph the solution to each inequality, and shade the side with the solutions. When you have done both, look for the area where the shading overlaps. That is the area with the solutions that work for both inequalities, and are therefore the solutions to the system of inequalities.
For example:
To review, see Using Graphs to Solve Linear Equations
### 6e. Create systems of equations and use them to solve world problems
• How do you use systems of equations to solve real-world problems?
As we have seen, systems of equations are helpful in solving real-world problems. When given a real-world problem, we can create a system of equations to find the solution. For example, consider the following problem:
Juan is considering two cell phone plans. The first company charges $120 for the phone and$30 per month for the calling plan that Juan wants. The second company charges $40 for the same phone but charges$45 per month for the calling plan that Juan wants. After how many months would the total cost of the two plans be the same?
First, we need to create two linear equations to represent the problem:
First company: $120+30x=y$
Second company: $40+45x=y$
Since 120 and 40 are the fixed costs, they are the constants, and the monthly cost is the coefficient of $x$, since each month you have to pay that amount.
Now we can solve this system of equations in any of the ways we have already learned, such as elimination, substitution, or graphing.
In this case, we can use substitution to get $x$:
$120+30x=40+45x$
We can solve this to find $80=15x$, $x=\dfrac{80}{15}=5\dfrac{1}{3}$ months.
#### 6f. Use systems of inequalities to model word problems and interpret their solutions in the context of the problem
• How can you use systems of inequalities to solve word problems?
Sometimes, a system of equations isn't appropriate for our problem. Instead, a system of inequalities should be used. For example, consider the following problem:
Jake does not want to spend more than $50 on bags of fertilizer and peat moss for his garden. Fertilizer costs$2 a bag and peat moss costs \$5 a bag. Jake's van can hold at most 20 bags.
First, we must create our inequalities. We will let Fertilizer $= x$, and Peat Moss $= y$:
$x+y \leq 20$
$2x+5y \geq 50$
In addition to these two inequalities that we can create from the problem, remember that $x \geq 0$ and $y \geq 0$, since Jake obviously can't have a negative number of bags of something.
We can now graph the solution to this system and then interpret the answers:
As you can see in the solution above, the area with the diagonal lines is the solution to our system of equations. Every point in that area is a solution. For example, (5,5) is a solution, meaning Jake could buy 5 bags of fertilizer and 5 bags of peat moss.
### Unit 6 Vocabulary
This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course.
• system of equations
• intersect
• parallel
• graphing
• lattice point
• substitution
• elimination
• system of inequalities
|
Comment
Share
Q)
# The mean weight of $500$ male students in a certain college in $151$ pounds and the standard deviation is $15$ pounds. Assuming the weights are normally distributed, find how many students weigh between $120$ and $155$ pounds
Comment
A)
Toolbox:
• Standard normal distribution:
• In a standard normal distribution $\mu=0,\sigma ^2=1$
• The random variable $X$ can be converted to the standard normal variable $Z$ by the transformation
• $Z=\large\frac{X-\mu}{\sigma}$
Step 1:
Let $X$ be the random variable denoting the weight of a male student in the college.
$X\sim N(150,15^2)$
Step 2:
To find the probability that a student weighs between 120 and 155 pounds.
To find $P(120 < X < 155)$
Step 3:
Let $Z$ be the standard normal variable
$Z=\large\frac{X-\mu}{\sigma}$
$\;\;=\large\frac{X-151}{15}$
When $X=120$
$Z=\large\frac{120-151}{15}$
$\;\;=\large\frac{-31}{15}$
$\;\;=-2.07$
When $X=155$
$Z=\large\frac{155-151}{15}$
$\;\;=\large\frac{-4}{15}$
$\;\;=0.27$
Step 4:
$P(120 < X < 155)=P(-2.07 < Z < 0.27)$
$\qquad\qquad\qquad\quad\;\;=P(-2.07 < Z < 0)+P(0 < Z < 0.27)$
$\qquad\qquad\qquad\quad\;\;=P(0 < Z < 2.07)+P(0 < Z < 0.27)$ by symmetry
$\qquad\qquad\qquad\quad\;\;=0.4808+0.1064$
$\qquad\qquad\qquad\quad\;\;=0.5872$
Step 5:
The number of male students in the college=500
$\therefore$ the number of students expected to weigh between 120 and 155 pounds =$500\times 0.5872$
$\Rightarrow 244$(approx)
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.