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# How do you solve 2^(x/17) = 0.8?
##### 1 Answer
Jul 15, 2016
x = (17ln(0.8))/(ln2) ≈ -5.4727
#### Explanation:
To solve for $x$, we first have to get rid of the exponent $\frac{x}{17}$. We can do this by taking the natural logarithm of both sides.
${2}^{\frac{x}{17}} = 0.8$
$\ln \left(2\right) \cdot \frac{x}{17} = \ln \left(0.8\right)$
Multiplying both sides by $17$ makes the left-hand side easier to solve for $x$, giving us
$\ln \left(2\right) \cdot x = 17 \ln \left(0.8\right)$
Dividing both sides by $\ln \left(2\right)$ isolates the $x$-term, so we now have
x = (17ln(0.8))/(ln2) ≈ -5.4727
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# Mental Math Tricks
Impress your teachers, parents and friends by being able to solve equations in your head as fast as a calculator! Here are a few of our favorite tricks to make mental math easier.
## Multiples of 5
Multiples of 10 are easy to calculate mentally because you can simply add 0 (or a decimal place) to the end of the number that you're multiplying by 10. Next time you're multiplying a number by 5, figure out what half of the number is, then multiply that number by 10.
Take 5 x 14.
Cut 14 in half to get 7.
10 x 7 = 70. (Look at that, 5 x 14 = 70, too!)
This method also works for odd numbers. For example, 5 x 19 is the same as 10 x 9.5, or 95.
## Multiples of 11
Instead of multiplying numbers by 11, multiply them by 10 first. Next, add the number itself (which is really the number multiplied by 1) to get the total, like this:
11 x 11 becomes 10 x 11 = 110, plus 11 more, which gives an answer of 121.
It works for larger numbers, too.
11 x 58
10 x 58 = 580
580 + 58 = 638
## Distributive Property of Multiplication
Now that you've unlocked the secret to breaking down problems into parts, here's another trick that every 'mathemagician' should know - the distributive property of multiplication.
When we multiply 2- or 3-digit numbers, it can be challenging to keep track of all the numbers in our heads. However, if we break down (or 'decompose') the larger numbers into smaller parts, they become more manageable. Try this technique with larger multiples, such as:
12 x 12.
It's the same as 12 x 10 + 12 x 2 (if we 'distribute' 12 to 10 and to 2).
These smaller pieces can be mentally calculated to get 120 + 24 = 144.
Now, let's try multiplying a 1-digit number by a 3-digit number.
First, decompose 8 x 467 by distributing the 8. Make it easier to figure out in your head by using the nearest multiples of 100 and 10.
8 x 400 + 8 x 60 + 8 x 7
3200 + 480 + 56 = 3736
Saying each of the partial sums out loud may help you remember them as you mentally multiply and add together each part. For our example, you might say, '3200, 3680, 3736' to help remember the totals as you go. The more you practice mental math, the easier it will become.
## Divide and Conquer
By now, you're probably convinced that calculating math in your head is painless once you break numbers down into parts. Apply this same method to division. If you want to divide a number by 4, cut it in half (which is dividing by 2) and then cut it in half again. To divide a number by 8, halve it 3 times.
Check out these clever tricks for figuring out which number to divide by (the divisor). If the number that you want to divide (the dividend) is evenly divisible by 5, it ends in a zero or a 5. At a glance, we can tell that 7,847,295 is divisible by 5.
To determine whether a number can be divided evenly by 3, add up all of its digits until you have a single digit. If it's a power of 3 (3, 6, 9), then the dividend can be divided by 3. For example:
3762
3 + 7 + 6 + 2 = 18
1 + 8 = 9
9 is a power of 3, so 3762 can be divided evenly by 3.
If a number is divisible by 3 and ends in a 0 or an even number, then it's also divisible by 6.
Did you find this useful? If so, please let others know!
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# GMAT Roman Numerals Questions Revisited
- Mar 12, 22:30 PM Comments [0]
We covered GMAT Roman Numerals questions recently, and like any good teacher, I want to review this topic again to help solidify it in your mind. So, we challenged students on Facebook with a practice question yesterday that combined Roman Numerals with properties of exponents, and got some great responses.
Take a moment to check out the question and the original blog entry on the shortcut strategy for Roman Numerals questions.
Now, let's tackle this!
#### GMAT Roman Numeral Tip
Remember that when you see a Roman Numeral problem, you should think: "I should start with the answer choice that shows up most frequently so that if I can eliminate it, I can mark out the most answer choices." This will save you time and effort. Remember, every second is valuable on the GMAT, and learning time-saving strategies is every bit as important as (some would argue even MORE important than!) learning math content.
#### GMAT Number Properties Tip
Next, when you see a Number Properties problem, which is any number that asks how numbers always behave (and exponents are among these), you should think: "I should pick numbers." Learning some rules for properties of special exponents will help out in this problem, too.
#### What Are Real Numbers?
The problem says "for all real numbers x"...so, what are real numbers? You don't need to know much more about this concept for the GMAT except that they are pretty normal numbers that you are used to working with. The GMAT will never ask you to define a real number. However, you can check out examples of real numbers if you need more clarification on the topic. For your purposes on the GMAT, and in general for picking numbers, this tells you that you want to choose basic, straightforward numbers that are permissible and manageable.
For this problem, check to see what happens when x is positive (2), negative (-1) and zero.
#### The Strategy in Action
II. $x^{0}$. This is a special exponent rule that you need to memorize if you don't know it already. Any number, whether positive, negative, or fractional, raised to the 0 power = 1. You're looking for an answer choice that MUST equal zero, and you've found a that this answer choice does not, so you must eliminate all answer choices that contain statement II. Eliminate answers C, D, and E. Now you only have to try statement I. That was fast work!
I. $x^{3}-x^{2}$. If you put in x = 2, you get 8-4=4. Again, you're looking for "MUST equal zero", and you've already found a scenario where it doesn't equal zero, so you must eliminate all answer choices that contain statement I - get rid of answer choice B.
The answer must be A, and you don't even need to spend the time evaluating statement III! Time saved, points gained, GMAT rocked!
The post GMAT Roman Numerals Questions Revisited appeared first on Kaplan GMAT Blog.
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# Lesson 2.4 Creating & Solving Equations & Inequalities in One Variable
## Presentation on theme: "Lesson 2.4 Creating & Solving Equations & Inequalities in One Variable"— Presentation transcript:
Lesson 2.4 Creating & Solving Equations & Inequalities in One Variable
Recall the Steps to creating equations & Inequalities
Read the problem statement first. Identify the known quantities. Identify the unknown quantity or variable. Create an equation or inequality from the known quantities and variable(s). Solve the problem. Interpret the solution of the equation in terms of the context of the problem.
Example 1 Suppose two brothers who live 55 miles apart decide to have lunch together. To prevent either brother from driving the entire distance, they agree to leave their homes at the same time, drive toward each other, and meet somewhere along the route. The older brother drives cautiously at an average speed of 60 miles per hour. The younger brother drives faster, at an average speed of 70 mph. How long will it take the brothers to meet each other?
Guided Practice: Example 3, continued Read the statement carefully.
Identify the known quantities. Identify the unknown variables. Suppose two brothers who live 55 miles apart decide to have lunch together. To prevent either brother from driving the entire distance, they agree to leave their homes at the same time, drive toward each other, and meet somewhere along the route. The older brother drives cautiously at an average speed of 60 miles per hour. The younger brother drives faster, at an average speed of 70 mph. How long will it take the brothers to meet each other?
Create an equation or inequality from the known quantities and variable(s).
The distance equation is d = rt or rt = d. Together the brothers will travel a distance, d, of 55 miles. (older brother’s rate)(t) + (younger brother’s rate)(t) = 55 1.2.1: Creating Linear Equations in One Variable
Guided Practice: Example 3, continued
The rate r of the older brother = 60 mph and The rate of the younger brother = 70 mph. (older brother’s rate)(t) + (younger brother’s rate)(t) = 55 60t + 70t = 55
Solve the problem for the time it will take for the brothers to meet each other.
It will take the brothers 0.42 hours to meet each other. 60t + 70t = 55 Equation 130t = 55 Combine like terms 60t and 70t. Divide both sides by 130. t = 0.42 hours This was a slide, but I think we should just talk about it… Note: The answer was rounded to the nearest hundredth of an hour because any rounding beyond the hundredths place would not make sense. Most people wouldn’t be able to or need to process that much precision. When talking about meeting someone, it is highly unlikely that anyone would report a time that is broken down into decimals, which is why the next step will convert the units.
Example 2 The length of a dance floor to be replaced is 1 foot shorter than twice the width. You measured the width to be feet. What is the area and what is the most accurate area you can report? 𝐴=𝑙×𝑤 𝑤=12.25 𝑙=2𝑤−1
𝐴=𝑙×𝑤 𝐴=(2𝑤−1)×12.25
\$1000 (3 times a day) + \$4,000 (grand prize) no more than \$25,000
Example 3 A radio station has no more than \$25,000 to give away. They have decided to give away \$1,000 three times a day every day until they have at least \$4,000 left to award as a grand prize. How many days will the contest run? \$1000 (3 times a day) + \$4,000 (grand prize) no more than \$25,000
3000𝑑+4000≤25000
Example 4 It costs Marcus an access fee for each visit to his gym, plus it costs him \$3 in gas for each trip to the gym and back. This month it cost Marcus \$108 for 6 trips to his gym. How much is Marcus’s access fee per visit? access fee x # of visits + \$3 x # of trips = \$108
6𝑥+6 3 =108 6𝑥+18=108
Example 5 Jeff is saving to purchase a new basketball that will cost at least \$88. He has already saved \$32. At least how much more does he need to save for the basketball? saved \$32 at least \$88
32+𝑥≥88
cost with discount x p = \$29.94
Example 6 Rebecca bought p pairs of socks and received a 20% discount. Each pair of socks cost her \$4.99. Her total cost without tax was \$ How many pairs of socks did Rebecca buy? cost with discount x p = \$29.94
4.99𝑝=29.94
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# Rational Zeros Theorem Calculator
The calculator will find all possible rational roots of the polynomial using the rational zeros theorem. After this, it will decide which possible roots are actually the roots. This is a more general case of the integer (integral) root theorem (when the leading coefficient is $1$ or $-1$). Steps are available.
If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.
Find the rational zeros of $2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7 = 0$.
## Solution
Since all coefficients are integers, we can apply the rational zeros theorem.
The trailing coefficient (the coefficient of the constant term) is $7$.
Find its factors (with the plus sign and the minus sign): $\pm 1$, $\pm 7$.
These are the possible values for $p$.
The leading coefficient (the coefficient of the term with the highest degree) is $2$.
Find its factors (with the plus sign and the minus sign): $\pm 1$, $\pm 2$.
These are the possible values for $q$.
Find all possible values of $\frac{p}{q}$: $\pm \frac{1}{1}$, $\pm \frac{1}{2}$, $\pm \frac{7}{1}$, $\pm \frac{7}{2}$.
Simplify and remove the duplicates (if any).
These are the possible rational roots: $\pm 1$, $\pm \frac{1}{2}$, $\pm 7$, $\pm \frac{7}{2}$.
Next, check the possible roots: if $a$ is a root of the polynomial $P{\left(x \right)}$, the remainder from the division of $P{\left(x \right)}$ by $x - a$ should equal $0$ (according to the remainder theorem, this means that $P{\left(a \right)} = 0$).
• Check $1$: divide $2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$ by $x - 1$.
$P{\left(1 \right)} = -12$; thus, the remainder is $-12$.
• Check $-1$: divide $2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$ by $x - \left(-1\right) = x + 1$.
$P{\left(-1 \right)} = 0$; thus, the remainder is $0$.
Hence, $-1$ is a root.
• Check $\frac{1}{2}$: divide $2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$ by $x - \frac{1}{2}$.
$P{\left(\frac{1}{2} \right)} = 0$; thus, the remainder is $0$.
Hence, $\frac{1}{2}$ is a root.
• Check $- \frac{1}{2}$: divide $2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$ by $x - \left(- \frac{1}{2}\right) = x + \frac{1}{2}$.
$P{\left(- \frac{1}{2} \right)} = \frac{27}{4}$; thus, the remainder is $\frac{27}{4}$.
• Check $7$: divide $2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$ by $x - 7$.
$P{\left(7 \right)} = 4368$; thus, the remainder is $4368$.
• Check $-7$: divide $2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$ by $x - \left(-7\right) = x + 7$.
$P{\left(-7 \right)} = 3780$; thus, the remainder is $3780$.
• Check $\frac{7}{2}$: divide $2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$ by $x - \frac{7}{2}$.
$P{\left(\frac{7}{2} \right)} = \frac{567}{4}$; thus, the remainder is $\frac{567}{4}$.
• Check $- \frac{7}{2}$: divide $2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$ by $x - \left(- \frac{7}{2}\right) = x + \frac{7}{2}$.
$P{\left(- \frac{7}{2} \right)} = 105$; thus, the remainder is $105$.
Possible rational roots: $\pm 1$, $\pm \frac{1}{2}$, $\pm 7$, $\pm \frac{7}{2}$A.
Actual rational roots: $-1$, $\frac{1}{2}$A.
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BREAKING NEWS
Trigonometric substitution
## Summary
In mathematics, a trigonometric substitution replaces a trigonometric function for another expression. In calculus, trigonometric substitutions are a technique for evaluating integrals. In this case, an expression involving a radical function is replaced with a trigonometric one. Trigonometric identities may help simplify the answer.[1][2] Like other methods of integration by substitution, when evaluating a definite integral, it may be simpler to completely deduce the antiderivative before applying the boundaries of integration.
## Case I: Integrands containing a2 − x2
Let ${\displaystyle x=a\sin \theta ,}$ and use the identity ${\displaystyle 1-\sin ^{2}\theta =\cos ^{2}\theta .}$
### Examples of Case I
#### Example 1
In the integral
${\displaystyle \int {\frac {dx}{\sqrt {a^{2}-x^{2}}}},}$
we may use
${\displaystyle x=a\sin \theta ,\quad dx=a\cos \theta \,d\theta ,\quad \theta =\arcsin {\frac {x}{a}}.}$
Then,
{\displaystyle {\begin{aligned}\int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}&=\int {\frac {a\cos \theta \,d\theta }{\sqrt {a^{2}-a^{2}\sin ^{2}\theta }}}\\[6pt]&=\int {\frac {a\cos \theta \,d\theta }{\sqrt {a^{2}(1-\sin ^{2}\theta )}}}\\[6pt]&=\int {\frac {a\cos \theta \,d\theta }{\sqrt {a^{2}\cos ^{2}\theta }}}\\[6pt]&=\int d\theta \\[6pt]&=\theta +C\\[6pt]&=\arcsin {\frac {x}{a}}+C.\end{aligned}}}
The above step requires that ${\displaystyle a>0}$ and ${\displaystyle \cos \theta >0.}$ We can choose ${\displaystyle a}$ to be the principal root of ${\displaystyle a^{2},}$ and impose the restriction ${\displaystyle -\pi /2<\theta <\pi /2}$ by using the inverse sine function.
For a definite integral, one must figure out how the bounds of integration change. For example, as ${\displaystyle x}$ goes from ${\displaystyle 0}$ to ${\displaystyle a/2,}$ then ${\displaystyle \sin \theta }$ goes from ${\displaystyle 0}$ to ${\displaystyle 1/2,}$ so ${\displaystyle \theta }$ goes from ${\displaystyle 0}$ to ${\displaystyle \pi /6.}$ Then,
${\displaystyle \int _{0}^{a/2}{\frac {dx}{\sqrt {a^{2}-x^{2}}}}=\int _{0}^{\pi /6}d\theta ={\frac {\pi }{6}}.}$
Some care is needed when picking the bounds. Because integration above requires that ${\displaystyle -\pi /2<\theta <\pi /2}$ , ${\displaystyle \theta }$ can only go from ${\displaystyle 0}$ to ${\displaystyle \pi /6.}$ Neglecting this restriction, one might have picked ${\displaystyle \theta }$ to go from ${\displaystyle \pi }$ to ${\displaystyle 5\pi /6,}$ which would have resulted in the negative of the actual value.
Alternatively, fully evaluate the indefinite integrals before applying the boundary conditions. In that case, the antiderivative gives
${\displaystyle \int _{0}^{a/2}{\frac {dx}{\sqrt {a^{2}-x^{2}}}}=\arcsin \left({\frac {x}{a}}\right){\Biggl |}_{0}^{a/2}=\arcsin \left({\frac {1}{2}}\right)-\arcsin(0)={\frac {\pi }{6}}}$
as before.
#### Example 2
The integral
${\displaystyle \int {\sqrt {a^{2}-x^{2}}}\,dx,}$
may be evaluated by letting ${\textstyle x=a\sin \theta ,\,dx=a\cos \theta \,d\theta ,\,\theta =\arcsin {\frac {x}{a}},}$ where ${\displaystyle a>0}$ so that ${\textstyle {\sqrt {a^{2}}}=a,}$ and ${\textstyle -{\frac {\pi }{2}}\leq \theta \leq {\frac {\pi }{2}}}$ by the range of arcsine, so that ${\displaystyle \cos \theta \geq 0}$ and ${\textstyle {\sqrt {\cos ^{2}\theta }}=\cos \theta .}$
Then,
{\displaystyle {\begin{aligned}\int {\sqrt {a^{2}-x^{2}}}\,dx&=\int {\sqrt {a^{2}-a^{2}\sin ^{2}\theta }}\,(a\cos \theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}(1-\sin ^{2}\theta )}}\,(a\cos \theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}(\cos ^{2}\theta )}}\,(a\cos \theta )\,d\theta \\[6pt]&=\int (a\cos \theta )(a\cos \theta )\,d\theta \\[6pt]&=a^{2}\int \cos ^{2}\theta \,d\theta \\[6pt]&=a^{2}\int \left({\frac {1+\cos 2\theta }{2}}\right)\,d\theta \\[6pt]&={\frac {a^{2}}{2}}\left(\theta +{\frac {1}{2}}\sin 2\theta \right)+C\\[6pt]&={\frac {a^{2}}{2}}(\theta +\sin \theta \cos \theta )+C\\[6pt]&={\frac {a^{2}}{2}}\left(\arcsin {\frac {x}{a}}+{\frac {x}{a}}{\sqrt {1-{\frac {x^{2}}{a^{2}}}}}\right)+C\\[6pt]&={\frac {a^{2}}{2}}\arcsin {\frac {x}{a}}+{\frac {x}{2}}{\sqrt {a^{2}-x^{2}}}+C.\end{aligned}}}
For a definite integral, the bounds change once the substitution is performed and are determined using the equation ${\textstyle \theta =\arcsin {\frac {x}{a}},}$ with values in the range ${\textstyle -{\frac {\pi }{2}}\leq \theta \leq {\frac {\pi }{2}}.}$ Alternatively, apply the boundary terms directly to the formula for the antiderivative.
For example, the definite integral
${\displaystyle \int _{-1}^{1}{\sqrt {4-x^{2}}}\,dx,}$
may be evaluated by substituting ${\displaystyle x=2\sin \theta ,\,dx=2\cos \theta \,d\theta ,}$ with the bounds determined using ${\textstyle \theta =\arcsin {\frac {x}{2}}.}$
Because ${\displaystyle \arcsin(1/{2})=\pi /6}$ and ${\displaystyle \arcsin(-1/2)=-\pi /6,}$
{\displaystyle {\begin{aligned}\int _{-1}^{1}{\sqrt {4-x^{2}}}\,dx&=\int _{-\pi /6}^{\pi /6}{\sqrt {4-4\sin ^{2}\theta }}\,(2\cos \theta )\,d\theta \\[6pt]&=\int _{-\pi /6}^{\pi /6}{\sqrt {4(1-\sin ^{2}\theta )}}\,(2\cos \theta )\,d\theta \\[6pt]&=\int _{-\pi /6}^{\pi /6}{\sqrt {4(\cos ^{2}\theta )}}\,(2\cos \theta )\,d\theta \\[6pt]&=\int _{-\pi /6}^{\pi /6}(2\cos \theta )(2\cos \theta )\,d\theta \\[6pt]&=4\int _{-\pi /6}^{\pi /6}\cos ^{2}\theta \,d\theta \\[6pt]&=4\int _{-\pi /6}^{\pi /6}\left({\frac {1+\cos 2\theta }{2}}\right)\,d\theta \\[6pt]&=2\left[\theta +{\frac {1}{2}}\sin 2\theta \right]_{-\pi /6}^{\pi /6}=[2\theta +\sin 2\theta ]{\Biggl |}_{-\pi /6}^{\pi /6}\\[6pt]&=\left({\frac {\pi }{3}}+\sin {\frac {\pi }{3}}\right)-\left(-{\frac {\pi }{3}}+\sin \left(-{\frac {\pi }{3}}\right)\right)={\frac {2\pi }{3}}+{\sqrt {3}}.\end{aligned}}}
On the other hand, direct application of the boundary terms to the previously obtained formula for the antiderivative yields
{\displaystyle {\begin{aligned}\int _{-1}^{1}{\sqrt {4-x^{2}}}\,dx&=\left[{\frac {2^{2}}{2}}\arcsin {\frac {x}{2}}+{\frac {x}{2}}{\sqrt {2^{2}-x^{2}}}\right]_{-1}^{1}\\[6pt]&=\left(2\arcsin {\frac {1}{2}}+{\frac {1}{2}}{\sqrt {4-1}}\right)-\left(2\arcsin \left(-{\frac {1}{2}}\right)+{\frac {-1}{2}}{\sqrt {4-1}}\right)\\[6pt]&=\left(2\cdot {\frac {\pi }{6}}+{\frac {\sqrt {3}}{2}}\right)-\left(2\cdot \left(-{\frac {\pi }{6}}\right)-{\frac {\sqrt {3}}{2}}\right)\\[6pt]&={\frac {2\pi }{3}}+{\sqrt {3}}\end{aligned}}}
as before.
## Case II: Integrands containing a2 + x2
Let ${\displaystyle x=a\tan \theta ,}$ and use the identity ${\displaystyle 1+\tan ^{2}\theta =\sec ^{2}\theta .}$
### Examples of Case II
#### Example 1
In the integral
${\displaystyle \int {\frac {dx}{a^{2}+x^{2}}}}$
we may write
${\displaystyle x=a\tan \theta ,\quad dx=a\sec ^{2}\theta \,d\theta ,\quad \theta =\arctan {\frac {x}{a}},}$
so that the integral becomes
{\displaystyle {\begin{aligned}\int {\frac {dx}{a^{2}+x^{2}}}&=\int {\frac {a\sec ^{2}\theta \,d\theta }{a^{2}+a^{2}\tan ^{2}\theta }}\\[6pt]&=\int {\frac {a\sec ^{2}\theta \,d\theta }{a^{2}(1+\tan ^{2}\theta )}}\\[6pt]&=\int {\frac {a\sec ^{2}\theta \,d\theta }{a^{2}\sec ^{2}\theta }}\\[6pt]&=\int {\frac {d\theta }{a}}\\[6pt]&={\frac {\theta }{a}}+C\\[6pt]&={\frac {1}{a}}\arctan {\frac {x}{a}}+C,\end{aligned}}}
provided ${\displaystyle a\neq 0.}$
For a definite integral, the bounds change once the substitution is performed and are determined using the equation ${\displaystyle \theta =\arctan {\frac {x}{a}},}$ with values in the range ${\displaystyle -{\frac {\pi }{2}}<\theta <{\frac {\pi }{2}}.}$ Alternatively, apply the boundary terms directly to the formula for the antiderivative.
For example, the definite integral
${\displaystyle \int _{0}^{1}{\frac {4\,dx}{1+x^{2}}}\,}$
may be evaluated by substituting ${\displaystyle x=\tan \theta ,\,dx=\sec ^{2}\theta \,d\theta ,}$ with the bounds determined using ${\displaystyle \theta =\arctan x.}$
Since ${\displaystyle \arctan 0=0}$ and ${\displaystyle \arctan 1=\pi /4,}$
{\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {4\,dx}{1+x^{2}}}&=4\int _{0}^{1}{\frac {dx}{1+x^{2}}}\\[6pt]&=4\int _{0}^{\pi /4}{\frac {\sec ^{2}\theta \,d\theta }{1+\tan ^{2}\theta }}\\[6pt]&=4\int _{0}^{\pi /4}{\frac {\sec ^{2}\theta \,d\theta }{\sec ^{2}\theta }}\\[6pt]&=4\int _{0}^{\pi /4}d\theta \\[6pt]&=(4\theta ){\Bigg |}_{0}^{\pi /4}=4\left({\frac {\pi }{4}}-0\right)=\pi .\end{aligned}}}
Meanwhile, direct application of the boundary terms to the formula for the antiderivative yields
{\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {4\,dx}{1+x^{2}}}\,&=4\int _{0}^{1}{\frac {dx}{1+x^{2}}}\\&=4\left[{\frac {1}{1}}\arctan {\frac {x}{1}}\right]_{0}^{1}\\&=4(\arctan x){\Bigg |}_{0}^{1}\\&=4(\arctan 1-\arctan 0)\\&=4\left({\frac {\pi }{4}}-0\right)=\pi ,\end{aligned}}}
same as before.
#### Example 2
The integral
${\displaystyle \int {\sqrt {a^{2}+x^{2}}}\,{dx}}$
may be evaluated by letting ${\displaystyle x=a\tan \theta ,\,dx=a\sec ^{2}\theta \,d\theta ,\,\theta =\arctan {\frac {x}{a}},}$
where ${\displaystyle a>0}$ so that ${\displaystyle {\sqrt {a^{2}}}=a,}$ and ${\displaystyle -{\frac {\pi }{2}}<\theta <{\frac {\pi }{2}}}$ by the range of arctangent, so that ${\displaystyle \sec \theta >0}$ and ${\displaystyle {\sqrt {\sec ^{2}\theta }}=\sec \theta .}$
Then,
{\displaystyle {\begin{aligned}\int {\sqrt {a^{2}+x^{2}}}\,dx&=\int {\sqrt {a^{2}+a^{2}\tan ^{2}\theta }}\,(a\sec ^{2}\theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}(1+\tan ^{2}\theta )}}\,(a\sec ^{2}\theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}\sec ^{2}\theta }}\,(a\sec ^{2}\theta )\,d\theta \\[6pt]&=\int (a\sec \theta )(a\sec ^{2}\theta )\,d\theta \\[6pt]&=a^{2}\int \sec ^{3}\theta \,d\theta .\\[6pt]\end{aligned}}}
The integral of secant cubed may be evaluated using integration by parts. As a result,
{\displaystyle {\begin{aligned}\int {\sqrt {a^{2}+x^{2}}}\,dx&={\frac {a^{2}}{2}}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)+C\\[6pt]&={\frac {a^{2}}{2}}\left({\sqrt {1+{\frac {x^{2}}{a^{2}}}}}\cdot {\frac {x}{a}}+\ln \left|{\sqrt {1+{\frac {x^{2}}{a^{2}}}}}+{\frac {x}{a}}\right|\right)+C\\[6pt]&={\frac {1}{2}}\left(x{\sqrt {a^{2}+x^{2}}}+a^{2}\ln \left|{\frac {x+{\sqrt {a^{2}+x^{2}}}}{a}}\right|\right)+C.\end{aligned}}}
## Case III: Integrands containing x2 − a2
Let ${\displaystyle x=a\sec \theta ,}$ and use the identity ${\displaystyle \sec ^{2}\theta -1=\tan ^{2}\theta .}$
### Examples of Case III
Integrals such as
${\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}}$
can also be evaluated by partial fractions rather than trigonometric substitutions. However, the integral
${\displaystyle \int {\sqrt {x^{2}-a^{2}}}\,dx}$
cannot. In this case, an appropriate substitution is:
${\displaystyle x=a\sec \theta ,\,dx=a\sec \theta \tan \theta \,d\theta ,\,\theta =\operatorname {arcsec} {\frac {x}{a}},}$
where ${\displaystyle a>0}$ so that ${\displaystyle {\sqrt {a^{2}}}=a,}$ and ${\displaystyle 0\leq \theta <{\frac {\pi }{2}}}$ by assuming ${\displaystyle x>0,}$ so that ${\displaystyle \tan \theta \geq 0}$ and ${\displaystyle {\sqrt {\tan ^{2}\theta }}=\tan \theta .}$
Then,
{\displaystyle {\begin{aligned}\int {\sqrt {x^{2}-a^{2}}}\,dx&=\int {\sqrt {a^{2}\sec ^{2}\theta -a^{2}}}\cdot a\sec \theta \tan \theta \,d\theta \\&=\int {\sqrt {a^{2}(\sec ^{2}\theta -1)}}\cdot a\sec \theta \tan \theta \,d\theta \\&=\int {\sqrt {a^{2}\tan ^{2}\theta }}\cdot a\sec \theta \tan \theta \,d\theta \\&=\int a^{2}\sec \theta \tan ^{2}\theta \,d\theta \\&=a^{2}\int (\sec \theta )(\sec ^{2}\theta -1)\,d\theta \\&=a^{2}\int (\sec ^{3}\theta -\sec \theta )\,d\theta .\end{aligned}}}
One may evaluate the integral of the secant function by multiplying the numerator and denominator by ${\displaystyle (\sec \theta +\tan \theta )}$ and the integral of secant cubed by parts.[3] As a result,
{\displaystyle {\begin{aligned}\int {\sqrt {x^{2}-a^{2}}}\,dx&={\frac {a^{2}}{2}}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)-a^{2}\ln |\sec \theta +\tan \theta |+C\\[6pt]&={\frac {a^{2}}{2}}(\sec \theta \tan \theta -\ln |\sec \theta +\tan \theta |)+C\\[6pt]&={\frac {a^{2}}{2}}\left({\frac {x}{a}}\cdot {\sqrt {{\frac {x^{2}}{a^{2}}}-1}}-\ln \left|{\frac {x}{a}}+{\sqrt {{\frac {x^{2}}{a^{2}}}-1}}\right|\right)+C\\[6pt]&={\frac {1}{2}}\left(x{\sqrt {x^{2}-a^{2}}}-a^{2}\ln \left|{\frac {x+{\sqrt {x^{2}-a^{2}}}}{a}}\right|\right)+C.\end{aligned}}}
When ${\displaystyle {\frac {\pi }{2}}<\theta \leq \pi ,}$ which happens when ${\displaystyle x<0}$ given the range of arcsecant, ${\displaystyle \tan \theta \leq 0,}$ meaning ${\displaystyle {\sqrt {\tan ^{2}\theta }}=-\tan \theta }$ instead in that case.
## Substitutions that eliminate trigonometric functions
Substitution can be used to remove trigonometric functions.
For instance,
{\displaystyle {\begin{aligned}\int f(\sin(x),\cos(x))\,dx&=\int {\frac {1}{\pm {\sqrt {1-u^{2}}}}}f\left(u,\pm {\sqrt {1-u^{2}}}\right)\,du&&u=\sin(x)\\[6pt]\int f(\sin(x),\cos(x))\,dx&=\int {\frac {1}{\mp {\sqrt {1-u^{2}}}}}f\left(\pm {\sqrt {1-u^{2}}},u\right)\,du&&u=\cos(x)\\[6pt]\int f(\sin(x),\cos(x))\,dx&=\int {\frac {2}{1+u^{2}}}f\left({\frac {2u}{1+u^{2}}},{\frac {1-u^{2}}{1+u^{2}}}\right)\,du&&u=\tan \left({\tfrac {x}{2}}\right)\\[6pt]\end{aligned}}}
The last substitution is known as the Weierstrass substitution, which makes use of tangent half-angle formulas.
For example,
{\displaystyle {\begin{aligned}\int {\frac {4\cos x}{(1+\cos x)^{3}}}\,dx&=\int {\frac {2}{1+u^{2}}}{\frac {4\left({\frac {1-u^{2}}{1+u^{2}}}\right)}{\left(1+{\frac {1-u^{2}}{1+u^{2}}}\right)^{3}}}\,du=\int (1-u^{2})(1+u^{2})\,du\\&=\int (1-u^{4})\,du=u-{\frac {u^{5}}{5}}+C=\tan {\frac {x}{2}}-{\frac {1}{5}}\tan ^{5}{\frac {x}{2}}+C.\end{aligned}}}
## Hyperbolic substitution
Substitutions of hyperbolic functions can also be used to simplify integrals.[4]
In the integral ${\displaystyle \int {\frac {dx}{\sqrt {a^{2}+x^{2}}}}\,,}$ make the substitution ${\displaystyle x=a\sinh {u},}$ ${\displaystyle dx=a\cosh u\,du.}$
Then, using the identities ${\displaystyle \cosh ^{2}(x)-\sinh ^{2}(x)=1}$ and ${\displaystyle \sinh ^{-1}{x}=\ln(x+{\sqrt {x^{2}+1}}),}$
{\displaystyle {\begin{aligned}\int {\frac {dx}{\sqrt {a^{2}+x^{2}}}}\,&=\int {\frac {a\cosh u\,du}{\sqrt {a^{2}+a^{2}\sinh ^{2}u}}}\ ,\\[6pt]&=\int {\frac {a\cosh {u}\,du}{a{\sqrt {1+\sinh ^{2}{u}}}}}\,\\[6pt]&=\int {\frac {a\cosh {u}}{a\cosh u}}\,du\\[6pt]&=u+C\\[6pt]&=\sinh ^{-1}{\frac {x}{a}}+C\\[6pt]&=\ln \left({\sqrt {{\frac {x^{2}}{a^{2}}}+1}}+{\frac {x}{a}}\right)+C\\[6pt]&=\ln \left({\frac {{\sqrt {x^{2}+a^{2}}}+x}{a}}\right)+C\end{aligned}}}
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Statistics Resources
This guide contains all of the ASC's statistics resources. If you do not see a topic, suggest it through the suggestion box on the Statistics home page.
Empirical Rule
The standard normal distribution is actually a probability distribution. By identifying points on the curve, we can determine the probability associated with lying at that point. How these probabilities were determined would take a lot of explanation. Suffice it to say that that work has already been done, so understanding how these values were derived is not the important part here. :)
When we first start talking about probability for the normal distribution, we're often introduced first to The Empirical Rule. This states the following:
• approximately 68% of the distribution lies within 1 standard deviation of the mean (that is between 1 st. dev. below and 1 st. dev. above)
• approximately 95% of the distribution lies within 2 standard deviations of the mean
• approximately 98.7% of the distribution lies within 3 standard deviations of the mean
So, by determining how many standard deviations a value is from the mean, we can determine the probability of obtaining that value on that distribution.
Example:
IQs are normally distributed with a mean of 110 and a standard deviation of 15. How many standard deviations is an IQ of 95 from the mean?
We completed the computation for this on the previous tab, but we can also use critical thinking here:
• if a standard deviation is 15 points, how many times do I have to take 15 away from 110 to reach 95?
• 110 - 15 = 95
• so I take 15 away 1 time, which means 95 is one standard deviation below the mean of 110
What's the probability of picking a person that has an IQ of 95 or less?
• now that we know that 95 is one standard deviation below the mean, we can use the Empirical Rule to estimate the probability
• note this will be cumulative probability because it's 95 or less
• start with the basic characteristic that 50% of the distribution lies below the mean
• from the Empirical Rule (shown above), we know that 34% of the distribution lies between the mean and 1 standard deviation below it.
• to find the probability of having an IQ that 95 or less, we can take 50% - 34% = 26%
Typically probability is reported as a decimal. To convert a percentage to a decimal, we can divide the percentage by 100:
• 26/100 = .26
So the probability of picking someone with an IQ of 95 or less is .26. We could also say that the probability of lying more than 1 standard deviation below the mean is .26.
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# Why the rule of 72 works
A table will help explain why the rule of 72 works. You will also see where the 72 came from and learn how to use the rule of 72 to solve investment problems.
For example, how long it takes to double an investment of \$2500 if the interest rate is 5%?
When entering the value of r in the formula, do not convert the rate into a decimal. If r = 2%, just replace r with 2.
Time it takes to double an investment of \$2500 is 72 / 5 = 14.4 years.
Notice that the amount of money you invest is irrelevant when using the rule of 72.
For example, how long it take to double an investment of \$500,000 if the interest rate is 5%? Again, it will take about 14.4 years to double any amount as long the following two conditions are met.
• The interest is compounded annually.
• Money is not added to the account.
## Now, why the rule of 72 works
Consider the compound interest formula.
A = P( 1 + r/100 )n
A is the resulting amount of money.
P is the principal invested for n interest periods at r% annually.
We need to investigate what happens when the money is doubled or when A = 2P.
When A = 2P, the equation above becomes 2P = P( 1 + r/100 )n
After dividing both sides of the equation by P, we get 2 = ( 1 + r/100 )n
Now, solve for n.
Take the log of both sides of the equation.
log 2 = log ( 1 + r/100 )n
log 2 = n log ( 1 + r/100 )
n = log2 / log (1 + r/100)
Now, here is a table showing different values for r, n, and nr.
r n nr 1 69.660 69.660 2 35.002 70.004 5 14.206 71.03 9 8.043 72.389 12 6.116 73.392 15 4.959 74.392
The average of the nr values gives 71.81 which is close to 72 and this is where the rule of 72 came from.
Notice that n in the table represents the number of years it will take the money to double. For example, when nr = 72.389, and the interest rate or r is 9, it will take about 8.043 years for the money to double.
Thus, the formula nr / r or 72 / r makes sense!
Check the exponential and logarithmic functions unit if you do not know how to solve these logarithmic equations.
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Categories
# Sum of the digits | AMC-10A, 2007 | Problem 25
Try this beautiful problem from algebra, based on Sum of the digits from AMC-10A, 2007. You may use sequential hints to solve the problem
Try this beautiful problem from Algebra based on Sum of the digits.
## Sum of the digits – AMC-10A, 2007- Problem 25
For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n+s(n)+s(s(n))=2007$
• $1$
• $2$
• $3$
• $4$
### Key Concepts
algebra
function
multiplication
Answer: $4$
AMC-10A (2007) Problem 25
Pre College Mathematics
## Try with Hints
Let $P(n)=(n+s(n)+s(s(n))=2007)$ tnen obviously $n<2007$
For $n$=$1999$.the sum becoms $28+10)=38$
so we may say that the minimum bound is $1969$
Now we want to break it in 3 parts …..
Case 1:$n \geq 2000$,
Case 2:$n \leq 2000$ ($n = 19xy,x+y<10$
Case 3:$n \leq 2000$ ($n = 19xy,x+y \geq 10$
Can you now finish the problem ……….
Case 1:$n \geq 2000$,
Then $P(n)=(n+s(n)+s(s(n))=2007)$ gives $n=2001$
Case 2:$n \leq 2000$ ($n = 19xy,x+y<10$
Then $P(n)=(n+s(n)+s(s(n))=2007)$ gives $4x+y=32$ which satisfying the constraints $x = 8$, $y = 0$.
Case 3:$n \leq 2000) ((n = 19xy,x+y \geq 10$ gives $4x+y=35$ which satisfying the constraints $x = 7$, $y = 7$ and $x = 8$, $y = 3$.
can you finish the problem……..
Therefore The solutions are thus $1977, 1980, 1983, 2001$
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# A convex quadrilateral has exterior angle measures, one at each vertex, of 5b+7°, 34b–33°, b+36°, and 37b–35°. What is the value of b?
Feb 25, 2018
$b = 5$
#### Explanation:
$\text{the " color(blue)"sum of the exterior angles } = {360}^{\circ}$
$\text{the sum of the 4 exterior angles is then}$
$5 b + 7 + 34 b - 33 + b + 36 + 37 b - 35 = 360$
$\Rightarrow 77 b - 25 = 360$
$\text{add 25 to both sides}$
$77 b \cancel{- 25} \cancel{+ 25} = 360 + 25$
$\Rightarrow 77 b = 385$
$\text{divide both sides by 77}$
$\frac{\cancel{77} b}{\cancel{77}} = \frac{385}{77}$
$\Rightarrow b = 5$
Feb 25, 2018
$b = {5}^{\circ}$.
#### Explanation:
As soon as I read the question and it was talking about exterior angles I wrote the values above are equal to ${360}^{\circ}$.
$5 b + 7 + 34 b - 33 + b + 36 + 37 b - 35 = {360}^{\circ}$
It equals ${360}^{\circ}$ as exterior angles add up to ${360}^{\circ}$ as it goes all around a point. So when you simplify the values give you get
$77 b - 25 = 360$
$77 b = 360 + 25$
$77 b = 385$
Divide both sides by $77$ to isolate $b$
$b = 5$
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# Difference between revisions of "2020 AMC 10A Problems/Problem 25"
The following problem is from both the 2020 AMC 12A #23 and 2020 AMC 10A #25, so both problems redirect to this page.
## Problem
Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly $7.$ Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?
$\textbf{(A) } \frac{7}{36} \qquad\textbf{(B) } \frac{5}{24} \qquad\textbf{(C) } \frac{2}{9} \qquad\textbf{(D) } \frac{17}{72} \qquad\textbf{(E) } \frac{1}{4}$
## Solution 1
Consider the probability that rolling two dice gives a sum of $s$, where $s \leq 7$. There are $s - 1$ pairs that satisfy this, namely $(1, s - 1), (2, s - 2), ..., (s - 1, 1)$, out of $6^2 = 36$ possible pairs. The probability is $\frac{s - 1}{36}$.
Therefore, if one die has a value of $a$ and Jason rerolls the other two dice, then the probability of winning is $\frac{7 - a - 1}{36} = \frac{6 - a}{36}$.
In order to maximize the probability of winning, $a$ must be minimized. This means that if Jason rerolls two dice, he must choose the two dice with the maximum values.
Thus, we can let $a \leq b \leq c$ be the values of the three dice, which we will call $A$, $B$, and $C$ respectively. Consider the case when $a + b < 7$. If $a + b + c = 7$, then we do not need to reroll any dice. Otherwise, if we reroll one die, we can roll dice $C$ in the hope that we get the value that makes the sum of the three dice $7$. This happens with probability $\frac16$. If we reroll two dice, we will roll $B$ and $C$, and the probability of winning is $\frac{6 - a}{36}$, as stated above.
However, $\frac16 > \frac{6 - a}{36}$, so rolling one die is always better than rolling two dice if $a + b < 7$.
Now consider the case where $a + b \geq 7$. Rerolling one die will not help us win since the sum of the three dice will always be greater than $7$. If we reroll two dice, the probability of winning is, once again, $\frac{6 - a}{36}$. To find the probability of winning if we reroll all three dice, we can let each dice have $1$ dot and find the number of ways to distribute the remaining $4$ dots. By stars and bars, there are ${6\choose2} = 15$ ways to do this, making the probability of winning $\frac{15}{6^3} = \frac5{72}$.
In order for rolling two dice to be more favorable than rolling three dice, $\frac{6 - a}{36} > \frac5{72} \rightarrow a \leq 3$.
Thus, rerolling two dice is optimal if and only if $a \leq 3$ and $a + b \geq 7$. The possible triplets $(a, b, c)$ that satisfy these conditions, and the number of ways they can be permuted, are $(3, 4, 4) \rightarrow 3$ ways. $(3, 4, 5) \rightarrow 6$ ways. $(3, 4, 6) \rightarrow 6$ ways. $(3, 5, 5) \rightarrow 3$ ways. $(3, 5, 6) \rightarrow 6$ ways. $(3, 6, 6) \rightarrow 3$ ways. $(2, 5, 5) \rightarrow 3$ ways. $(2, 5, 6) \rightarrow 6$ ways. $(2, 6, 6) \rightarrow 3$ ways. $(1, 6, 6) \rightarrow 3$ ways.
There are $3 + 6 + 6 + 3 + 6 + 3 + 3 + 6 + 3 + 3 = 42$ ways in which rerolling two dice is optimal, out of $6^3 = 216$ possibilities, Therefore, the probability that Jason will reroll two dice is $\frac{42}{216} = \boxed{\textbf{(A) }\frac7{36}}$
## Solution 2
We count the numerator. Jason will pick up no dice if he already has a 7 as a sum. We need to assume he does not have a 7 to begin with. If Jason decides to pick up all the dice to re-roll, by the stars and bars rule{ ways to distribute, ${n+k-1 \choose k-1}$, there will be 2 bars and 4 stars(3 of them need to be guaranteed because a roll is at least 1) for a probability of $\frac{15}{216}=\frac{2.5}{36}$. If Jason picks up 2 dice and leaves a die showing $k$, he will need the other two to sum to $7-k$. This happens with probability $$\frac{6-k}{36}$$ for integers $1 \leq k \leq 6$. If the roll is not 7, Jason will pick up exactly one die to re-roll if there can remain two other dice with sum less than 7, since this will give him a $\frac{1}{6}$ chance which is a larger probability than all the cases unless he has a 7 to begin with. We have $$\frac{1}{6} > \underline{\frac{5,4,3}{36}} > \frac{2.5}{36} > \frac{2,1,0}{36}.$$ We count the underlined part's frequency for the numerator without upsetting the probability greater than it. Let $a$ be the roll we keep. We know $a$ is at most 3 since 4 would cause Jason to pick up all the dice. When $a=1$, there are 3 choices for whether it is rolled 1st, 2nd, or 3rd, and in this case the other two rolls have to be at least 6(or he would have only picked up 1). This give $3 \cdot 1^{2} =3$ ways. Similarly, $a=2$ gives $3 \cdot 2^{2} =12$ because the 2 can be rolled in 3 places and the other two rolls are at least 5. $a=3$ gives $3 \cdot 3^{2} =27$. Summing together gives the numerator of 42. The denominator is $6^3=216$, so we have $\frac{42}{216}=\boxed{(A) \frac{7}{36}}$
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# Problems on Complement of a Set
Solved problems on complement
of a set
are given below to get a fair idea how to find the complement
of two or more sets.
We know, when U be the universal set and A is a subset of U. Then the complement of A is the set all elements of U which are not the elements of A.
to know more about the complement of a set.
Solved problems on complement of a set:
1. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}.
(i) Find A’
(ii) Find B’
Solution:
(i) A’ = U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
= {5, 6, 7, 8, 9}
(ii) B’
= U – B
=
{1, 2, 3, 4, 5, 6, 7, 8, 9} – {2,
4, 6, 8}
= {1, 3, 5, 7, 9}
More worked-out problems on complement of a set.
2. Let A = {3, 5, 7}, B = {2, 3, 4, 6}
and C = {2, 3, 4, 5, 6, 7, 8}
(i) Verify (A ∩ B)’ = A’ ∪ B’
(ii) Verify (A ∪ B)’ = A’ ∩ B’
Solution:
(i) (A ∩ B)’ = A’ ∪ B’
L.H.S. = (A ∩ B)’
A ∩ B = {3}
(A ∩ B)’ = {2, 4, 5, 6, 7, 8} ……………….. (1)
R.H.S. = A’ ∪ B’
A’ = {5, 7, 8}
B’ = {2, 4, 6}
A’∪B’
= {2, 4, 5, 6, 7, 8} ……………….. (2)
From (1) and (2), we conclude that;
(A ∩ B)’ = (A’ ∪ B’)
(ii) (A ∪ B)’ = A’ ∩ B’
L.H.S. = (A ∪ B)’
A∪B
= {2, 3, 4, 5, 6, 7}
(A ∪
B)’ = {8} ……………….. (1)
R.H.S. = A’ ∩ B’
A’ = {2, 4, 6, 8}
B’ = {5, 7, 8}
A’ ∩ B’ = {8} ……………….. (2)
From (1) and (2), we conclude that;
(A ∪ B)’ = A’ ∩ B’
+
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# CBSE Class 9 Mathematics: Important 4 Marks Questions for Annual Exam 2020
In this article, you will get a collection of important 4 marks questions to prepare for the CBSE Class 9 Mathematics Annual Exam 2020. All these questions have been prepared after carrying the thorough analysis of previous years' question papers and the latest syllabus. All the questions are provided with proper solutions.
In CBSE Class 9 Mathematics Exam 2020, Section - D will comprise 6 questions of 4 marks each.
Students must practice the questions given here as it will help them not only assess their preparation level but also know the important topics which need to be prepared for the annual exam with more concentration.
CBSE Class 9 Mathematics Syllabus 2019-2020
Given below are some sample questions from CBSE Class 9 Mathematics: Important 4 Marks Questions:
Q. The polynomials ax3 – 3x2 +4 and 2x3 – 5x +a when divided by (x – 2) leave the remainders p and q respectively. If p – 2q = 4, find the value of a.
Sol.
Let, f(x) = ax3 – 3x2 +4
And g(x) = 2x3 – 5x +a
When f(x) and g(x) are divided by (x – 2) the remainders are p and q respectively.
f(2) = p and g(2) = q
f(2) = a × 23 – 3 × 22 + 4
p = 8a – 12 + 4
p = 8a – 8 ....(i)
And g(2)= 2 × 23 – 5 × 2 + a
q = 16 – 10 + a
q = 6 + a ....(ii)
But p – 2q = 4 (Given)
8a – 8 – 2(6 + a) = 4 (Using equations (i) and (ii))
8a – 8 – 12 − 2a = 4
6a – 20 = 4
6a = 24
a = 24/6
a = 4
CBSE Class 9 Mathematics Exam 2018: Important 3 Marks Questions
Q. Construct a ΔABC in which BC = 3.8 cm, ∠B = 45oand AB + AC = 6.8cm.
Sol.
Steps of Construction
1. Draw BC = 38 cm.
2. Draw a ray BX making an ∠CBX = 45°.
3. From BX, cut off line segment BD equal to AB + BC i.e., 6.8 cm.
4. Join CD.
5. Draw the perpendicular bisector of CD meeting BD at A.
6. Join CA to obtain the required
Justification:
Clearly, A lies on the perpendicular bisector of CD.
Now, BD = 6.8 cm
⟹ BA + AD = 6.8 cm
⟹ AB + AC = 6.8 cm
Hence, is the required triangle.
Q. If a + b + c = 6 and ab + bc + ca = 11, find the value of a3 +b3 +c3 − 3abc.
Sol.
(a + b + c)2 = a2 + b2 +c2 +2(ab + bc + ca)
(6)2 = a2 + b2 +c2 + 2 × 11
a2 + b2 +c2 = 36 – 22 = 14
a3 +b3 +c3 − 3abc =( a + b + c)[ a2 + b2 +c2 −(ab + bc + ca)]
= 6 × (14 − 11)
= 6 × 3 = 18
To get the complete set of questions, click on the following link:
Students may also check the following links to explore more stuff, important for CBSE Class 9 Annual exam preparations:
x
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## Squaring Magic Squares
Suzanne's Magic Squares || Multiplying Magic Squares: Contents || Exploring the Math
### Objectives: [NCTM Standards: Number and Operations, Communication, and Connections]
1. Students will work with arrays of numbers.
2. Students will calculate and compare sums of numbers in the context of magic squares.
3. Students will apply a method for multiplying two magic squares.
### Materials:
Large empty 3x3 grid
The nine cells in numerical order
Ordering the cells to make a magic square
The completed 9-cell magic square
2. Blank paper
3. Blank overhead transparency and pens
4. Rulers
5. Calculators
### Definition:
To square a number means to multiply it by itself.
[Just as we can talk about squaring a number, we can talk about squaring something with any notion of multiplication. We can square an nxn matrix by multiplying it by itself under matrix multiplication, and we can square a magic square by multiplying it by itself.]
### Considering Square A:
Remember the method we followed to multiply Square A by Square B? (To review, return to Classroom Activities.) If this method can be used to 'multiply' two magic squares, let's use it to square a magic square.
To square Square A, we'll follow the same steps, but instead of using Squares A and B, we'll use Square A twice.
### Square A Square A
We'll again need a large grid. It will have as many cells as there are cells in Square A.
### Check for Understanding:
1. Verify that Square A is a magic square.
2. How many cells does Square A have?
3. How many cells will we need in the large grid?
### Generating the second square:
Following the multiplication method, we will fill the nine cells in the empty 3x3 grid. We will begin by adding the same number to each entry in Square A to make a new magic square.
Square A is made up of nine entries. When we add 9 to each of them, we get
We will continue this process by adding 9 to each entry to make each successive cell.
Continued next page
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# The Euclidean algorithm
The Euclidean algorithm
The Euclidean algorithm is a way to find the greatest common divisor of two positive integers, a and b. First let me show the computations for a=210 and b=45.
• Divide 210 by 45, and get the result 4 with remainder 30, so 210=4·45+30.
• Divide 45 by 30, and get the result 1 with remainder 15, so 45=1·30+15.
• Divide 30 by 15, and get the result 2 with remainder 0, so 30=2·15+0.
• The greatest common divisor of 210 and 45 is 15.
Let’s see. Several questions occur immediately..
1. What’s going on here?
2. Why does the algorithm stop?
3. Why is the answer correct?
O.k. I will answer first question by giving a formal description of the algorithm, which supposedly finds the greatest common divisor (GCD) of two integers. This “greatest common divisor” must exist, since positive integer divisors of integers can’t be any larger than the integers:
Formal description of the Euclidean algorithm Input Two positive integers, a and b. Output The greatest common divisor, g, of a and b. Internal computation If a
Here is one link which describes the algorithm. Here’s another link, with more information. If you want to “do” some further examples, Paul Garrett of the University of Minnesota has a Visible Euclidean algorithm page where you can specify further examples and see the algorithm work..
Why does the algorithm stop?At each step, the remainder, r, decreases by at least 1. Therefore r must eventually be 0. A formal proof would use mathematical induction.
Why is the answer actually equal to the GCD?In the final stage, when r=0, we see that the “final b” must divide the final a. Go backwards along the string of equations (a=q·b+r), and you will see that at each step, the “final b” divides each of the parts of the equation. Therefore the “final b” must be a divisor of both of the initial a and initial b. Now look at the first equation. In the equation a=q·b+r, the GCD divides both a and b and therefore must divide r as well (because a-q·r is a multiple of the GCD). Now carry this observation through all of the equations, forward. The GCD must divide two of the three pieces in all of the equations, and thus must divide the third. Therefore, the GCD also divides the “final b”. So the “final b” divides both a and b, and is itself a multiple of the GCD. Well, the GCD is the greatest such divisor, and therefore the “final b” must be equal to the GCD of the initial values.
The extended Euclidean algorithm
Here’s a true statement:
If a and b are positive integers, then there are always integers m and n so that the GCD of a and b equals m·a+n·b.
The extended Euclidean algorithm (described, for example, here, allows the computation of multiplicative inverses mod P. First let’s see an example. Since the GCD of 210 and 45 is 15, we should be able to write 15 as a sum of multiples of 210 and 45. Here’s how to do it. We look carefully at the steps above and change them each.
• Divide 210 by 45, and get the result 4 with remainder 30, so 210=4·45+30.
30=1·210-4·45.
• Divide 45 by 30, and get the result 1 with remainder 15, so 45=1·30+15.
15=45-1·30=45-1·(1·210-4·45)=-1·210+5·45
The greatest common divisor of 210 and 45 is 15, and we have written 15 as a sum of integer multiples of 210 and 45.
The extended Euclidean algorithm has a very important use: finding multiplicative inverses mod P. Choose a prime, P: how about 97. I know 97 is prime, because 2 and 3 and 5 and 7 and even 11 aren’t factors of 97, and I only need to check division by primes up to the square root of 97.
Now let me take a fairly random integer, say 20. Since 20 is less than 97, and 97 is prime, the GCD of 20 and 97 should be 1. (Remember, since 97 is prime, its only divisors are itself and 1.) I will verify this by “running” the Euclidean algorithm:
• 97=4·20+17
• 20=1·17+3
• 17=5·3+2
• 3=1·2+1
The extended Euclidean algorithm allows us to write 1 as a sum of 97 and 20. Here we go:
• 17=1·97-4·20
• 20-1·17=3 so 3=1·20-1·17=1·20-(1·97-4·20)=-1·97+5·20
• 17=5·3+2 so 2=17-5·3=(1·97-4·20)-5(-1·97+5·20)=6·97-29·20
• 1=3-2=(-1·97+5·20)(6·97-29·20)=-7·97+34·20
The final equation tells us that 1=-7·97+34·20, which means that the product of 34 and 20 is equal to 1 plus a multiple of 97. But in mod 97, we ignore multiples of 97. Therefore 34 is the multiplicative inverse of 20 mod 97.
Exercise
Find the multiplicative inverse of 60 mod 97 by hand. As I mentioned in class, doing just one of these computations “by hand” is good enough. Here’s a link to the answer.
The extended Euclidean algorithm is easy to implement on a computer and the amount of memory needed is not large. The algorithm runs very fast. For example, I am currently running a copy ofMaple in another window on a PC which is neither very fast nor has much memory. I just selected a 100 decimal digit prime, P, and found the multiplicative inverse of a 50 decimal digit number mod P. The system returned 0.00 as the amount of time used. That is, the computation used less than .01 seconds!
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# Sets
Problems on sets some times may not be straight forward. We need to apply very strong logic to dig out the underlying the mistery. Study the following problems. First problem is easy.
New age consultants have three consultants Gyani, Medha, and Buddhi. The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha involved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has 2 projects with with Medha but without Gyani and 3 projects with Gyani but without Medha. The total number of projects for New age consultants is one less than twice the number of projects in which more than one consultant is involved.
1. What is the number of projects in which Gyani alone is involved?
2. What is the number of projects in which Medha alone is involved?
Step 1: It was given that all the three consultants are involved in 6 projects together. So put 6 where all the three circles intersect
Step 2: Buddhi has 2 projects with Medha but without Gyani. i.e., Medha and Buddhi alone worked in 2 projects.
Step 3: Also given, Buddi worked on 3 projects with Gyani but without Medha. i.e., Buddi adn Gyani alone worked on 3 projects.
Step 4: Gyani worked with Medha in 14 projects. These two worked together along with Buddhi in 6 projects. So Gyani and Medha alone worked on 14 - 6 = 8 Projects.
Lets us assume, Gyani, Buddhi and Medha alone worked on a, b, c projects.
It was given that The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha involved.
$\Rightarrow$a + b = 8 + 6 + 2 + c
$\Rightarrow$ a + b = 16 + c ------------------ (1)
Also given The total number of projects for New age consultants is one less than twice the number of projects in which more than one consultant is involved.
The projects in which more than 1 consultant involved = 6 + 3 + 2 + 8 = 19
Total projects = a + b + c + 8 + 3 + 2 + 6 = a + b + c + 19
$\Rightarrow$ 2$\times$19 - 1 = a + b + c + 19
$\Rightarrow$a + b + c = 18
$\Rightarrow$ a + b = 18 - c
Substituting this value in equation (1) we get
18 - c = 16 + c $\Rightarrow$ c = 1
So from equation (1) we get a + b = 17.
Question 1: Gyani alone involved in "a" projects. But we cannot determine uniquely the value of "a"
Question 2: Medha alone involved in "c" projects. We calculated c = 1
Help Distress (HD) is an NGO involved in providing assistance to people suffering from natural disasters. Currently, it has 37 volunteers. They are involved in three projects: Tsunami Relief (TR) in Tamil Nadu, FloodRelief (FR) in Maharashtra, and Earthquake Relief (ER) in Gujarat. Each volunteer working with Help Distress has to be involved in at least one relief work project.
*A Maximum number of volunteers are involved in the FR project. Among them, the number of volunteers involved in FR project alone is equal to the volunteers having additional involvement in the ER project.
*The number of volunteers involved in the ER project alone is double the number of volunteers involved in all the three projects.
*17 volunteers are involved in the TR project.
*The number of volunteers involved in the TR project alone is one less than the number of volunteers involved in ER Project alone.
*Ten volunteers involved in the TR project are also involved in at least one more project.
Find the minimum number of volunteers involved in both FR and TR projects, but not in the ER project is:
(1) 1
(2) 3
(3) 4
(4) 5
Statement 1: the number of volunteers involved in FR project alone is equal to the volunteers having additional involvement in the ER project.
a = g + f
Statement 2: The number of volunteers involved in the ER project alone is double the number of volunteers involved in all the three projects.
c = 2g
Statement 3: 17 volunteers are involved in the TR project.
b + d + g + e = 17
Statement 4: The number of volunteers involved in the TR project alone is one less than the number of volunteers involved in ER Project alone.
c - b = 1
Statement 5: Ten volunteers involved in the TR project are also involved in at least one more project.
d + g + e = 10
From statements 3 and 5 we get b = 7
From statement 4, we get c = 8
From statement 2, we get g = 4
From statement 1 we get f = a - 4
Substituting the above values we get the diagram as
But the total volunteers are equal to 37
So 17 + a + a - 4 + 8 = 37 $\Rightarrow$ a = 8
From the above diagram we know that d + e = 6, Also given that maximum number of volunteers are involved in FR.
i.e., FR > TR $\Rightarrow$ FR > 17
FR > ER $\Rightarrow$ 16 + d > 16 + e $\Rightarrow$ d > e
Possible value of d and e in this case are (4, 2), (5, 1), (6, 0)
From the above analysis d takes minimum value 4.
Additional Questions based on the above set:
Which of the following additional information would enable to find the exact number of volunteers involved in various projects?
(1) Twenty volunteers are involved in FR.
(2) Four volunteers are involved in all the three projects.
(3) Twenty three volunteers are involved in exactly one project.
(4) No need for any additional information.
Ans: If we know the value of d, we can easily figureout the value of e. so statement 1 is required
After some time, the volunteers who were involved in all the three projects were asked to withdraw from one project. As a result, one of the volunteers opted out of the TR project, and one opted out of the ER project, while the remaining ones involved in all the three projects opted out of the FR project.
Which of the following statements, then, necessarily follows?
(1) The lowest number of volunteers is now in TR project.
(2) More volunteers are now in FR project as compared to ER project.
(3) More volunteers are now in TR project as compared to ER project.
(4) None of the above
Due to this new arrangement the number of workers working on each project changes
Number of workers in TR project = 16
Number of workers in FR project = 14 + d
Number of workers in ER project = 15 + e
Considering the possible value of d and e, d takes minimum 4 so FR > ER
(But it is not clear in this questions the condition that FR has greater number of volunteers holds good or not. If that is not the case, answer for this question cannot be determined.)
After the withdrawal of volunteers, some new volunteers joined the NGO. Each one of them was allotted only one project in a manner such that, the number of volunteers working in one project alone for each of the three projects became identical. At that point, it was also found that the number of volunteers involved in FR and ER projects was the same as the number of volunteers involved in TR and ER projects. Which of the projects now has the highest number of volunteers?
(1) ER
(2) FR
(3) TR
(4) Cannot be determined
Given that volunteers working with only one project is same for all the three projects. So add m + 1 to TR and m to FR and ER.
Also given that volunteers working in ER and TR are equal to ER and FR. So e + 2 = 5 $\Rightarrow$e = 3.
As we know d + e = 6 $\Rightarrow$ d = 3
Now number of volunteers in each of the projects given below
TR = 8 + m + (3 +1) + (3 + 2) = 17 + m
FR = 8 + m + 5 + (3 + 1) = 17 + m
ER = 8 + m + (3 + 2) + 5 = 18 + m
So ER has the highest number of volunteers.
It was given that total number of volunteers are 37. But total how many projects they are volunteering?
Total projects = (7 + 8 + 8 ) x 1 + (d + e + 4 ) x 2 + 4 x 3 = 23 + 20 + 12 = 55
In the above discussion, we have seen that volunteers are with drawing from projects. In that case what happens to the number of projects and volunteers?
Volunteers remain same. i.e., 37. But number of projects came down.
Let us calculate the number of projects in this case
( 7 + 8 + 8) x 1 + (d + 1 + e + 2 + 5 ) x 2 = 23 + 28 = 51.
There are 4 projects less than earlier. This is because, 4 volunteers withdrawn from working on 3 projects to 2 projects. That is why total projects got reduced by exactly 4. Also number of volunteers in exactly two projects also go up by 4.
Concept of Maxima and Minima in sets:
In a class, 67% play cricket, and 82% play chess.
Question 1: What is the maximum number of students who can play cricket as well as chess.
Question 2: What is the minimum number of students who can play cricket as well as chess, given that each students play atleast one game.
Ans 1: If all the students who play cricket also play chess... then
Total number of students who play atleast one game = 0 + 67 + 15 = 82
So students who does not play any game = 100 - 82 = 18.
Ans 2: Total games are 67 + 82 = 149.
These 149 games are played by 100 students. As each student play atleast one game, there must be some who play both the games.
Give one game to all 100 students. The we have excess of 149 - 100 = 49 games.
Now we distribute these 49 to 49 students who have already got one game. So there are minimum 49 students who can play 2 games, and the remaining 100 - 49 = 51 play one game.
Observe total games are 49 x 2 + 51 = 149 remains same.
There are six institutes IMM-A, IMM-B, IMM-C, IMM-I, IMM-L, and IMM-K in a country named Crazyland. Every year these six institutes conduct a Common Intelligence Test (CIT) and send interview calls to students for the total of 1100 seats distributed among the six IMMs. After the interview process is over, each IMM releases the Final List giving the names of the candidates finally offered a seat and Wait List giving the names of the candidates selected in the waiting list in case the candidates in the Final List chose not to accept the offer of the given IMM. A given candidate may get shortlisted in the Final List chose not to accept the offer of the given IMM. Finally it was found that 1500 students received calls from one or more than one IMM. As some students received multiple calls, the sum of total number of calls sent out by individual IMMs exceeded the total number of students receiving calls.
The following Bar graph gives the number of seats and the number of candidates shortlisted either in the Final List or in the Wait List of each of the six IMMs
1. What can be the maximum number of candidates whose name appeared either in the Final list or the Wait List of all the six IMMs?
Ans: Given that, the number of students who received atleast one call = 1500
Total calls given = 500 + 450 + 460 + 350 + 280 + 160 = 2200
So give one call to each of these 1500 students, then we have 2200 - 1500 = 700 calls surplus.
We know that a student can get a maximum of 6 calls. So Some of these 1500 students may get a maximum of 5 calls more. So 700 / 5 = 140 students may get another 5 calls. So there are 140 students who can get 6 calls.
2. What is the maximum number of students who received multiple calls?
Ans: From the above observation, we know that there are 700 calls surplus to be distributed. We consider getting 2 calls also a multiple call. To maximise the multiple call holders, we give these 700 calls to 700 students. So there are 700 students who got 2 calls, remaining 800 got one call.
3. If it is known that 100 students got selected in either Final or Wait list of exactly 5 IMMs, then what is the maximum number of students who can get selected in exactly 4 IMMs?
Ans: If 100 students got 5 calls they have a total of 500 calls with them. So the remaining 1400 students must have got 2200 - 500 = 1700 calls.
Giving one call each to these 1400 students, we have 300 calls surplus. To maximize the number of students who got exactly 4 calls, we can give another 3 calls to some of these 1400 students. 300 / 3 = 100.
So maximum number of students who got exactly 4 calls = 100
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# The value of
Question:
The value of $\sum_{r=1}^{n}\left\{(2 r-1)+\frac{1}{2^{r}}\right\}$ is _______________ .
Solution:
$\sum_{r=1}^{n}\left\{(2 r-1)+\frac{1}{2^{r}}\right\}$
For $\sum_{r=1}^{n}(2 r-1)=1+3+5+7+\ldots \ldots+2 n+1$
Here first term is 1
last term is 2n − 1
i.e sum is $\frac{n}{2}$ (first term + last term)
i.e $\frac{n}{2}(1+2 n-1)$
$=\frac{n}{2}(2 n)$
i. e $\sum_{r=1}^{n}(2 r-1)=n^{2} \quad \ldots(1)$
also $\sum_{r=1}^{n} \frac{1}{2^{r}}=\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots+\frac{1}{2^{n}}$
first term $a=\frac{1}{2}, r=$ common ratio is $\frac{1}{2}$
Sum $S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$
$=\frac{\frac{1}{2}\left(1-\frac{1}{2^{n}}\right)}{1-\frac{1}{2}}$
$\sum_{r=1}^{n} \frac{1}{2^{r}}=S \mathrm{um}=1-\frac{1}{2^{n}}$
$\therefore \sum_{r=1}^{n}\left\{(2 r-1)+\frac{1}{2^{n}}\right\}=n^{2}+1-\frac{1}{2^{n}}$
$\sum_{r=1}^{n}\left\{(2 r-1)+\frac{1}{2^{r}}\right\}=1+n^{2}-\frac{1}{2^{n}}$
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# Which shape has 4 lines of symmetry
Some shape may have two or more lines of symmetry like a circle has infinite lines of symmetry. A square has four lines of symmetry. A rectangle has two. Regular Polygons symmetry square, A Square (4 sides) has 4 Lines of Symmetry symmetry regular pentagon, A Regular Pentagon (5 sides) has 5 Lines of Symmetry.A square has 4 lines of symmetry A square is a regular quadrilateral, which means that it has four equal sides and four equal angles Symmetry. So the square has four lines of symmetry. The rectangle has only two, as it can be folded in half horizontally or vertically: students should be encouraged. Concept: As we know, a quadrilateral is any shape having four sides, it does not have a fixed number of symmetrical lines for all its shapes.
View this answer now! It’s completely free.
## how many lines of symmetry does a polygon have
The number of line symmetry in a regular polygon of n sides is ‘n’.For example a regular pentagon has 5 line symmetry,Hexagon has 6 line symmetry etc.How Many Lines Of Symmetry Does A Regular Polygon Have. Updated on May 10, 2021. Regular Polygons. An Equilateral Triangle (3 sides) has 3 Lines of Symmetry.Each regular polygon, such as an equilateral triangle or a square, has the same number of lines of symmetry as its number of sides. In this article, we will. Lines of Symmetry for a Polygon. How many lines of symmetry are possible for a 27 sided polygon?. Note: A polygon does not have to be convex.A regular polygon of 10 sides has 10 Lines of Symmetry. A regular polygon of “n” sides has “n” Lines of Symmetry. Circle.
## lines of symmetry in a rectangle
A rectangle has only two lines of symmetry (diagonals).These two lines are the only possibilities to cut a rectangle in half.symmetry square, symmetry rectangle Square (all sides equal, all angles 90°), Rectangle (opposite sides equal, all angles 90°) 4 Lines of Symmetry, 2 Lines of. A rectangle has 4 lines of symmetry. 2 passing through diagonals and 2 passing through points of intersection of mid points of parallel faces.The number of line symmetry in a rectangle and square are unequal.The number of line symmetry in rectangle are 2 and in square it is 4. solution.In a rectangle, the opposite sides are equal and parallel to each other and the adjacent sides are at a right angle. Thus, it can be folded one along its length.
## how many lines of symmetry does a rectangle have
A rectangle has two lines of symmetry. Picture a regular rectangle: if you fold it in the middle of the shape vertically you can see one line of.. See full. A rectangle has two lines of symmetry as shown in the figure. ? There are two symmetry lines in rectangle 1 in respect to length and 1 is respect to breadth.A rectangle has only two lines of symmetry (diagonals).These two lines are the only possibilities to cut a rectangle in half.How many lines of Symmetry do a Square and Rectangle have 2 ExplainingMaths com IGCSE GCSE Maths A rectangle has 2 lines of symmetry. The lines. A rectangle has 4 lines of symmetry. 2 passing through diagonals and 2 passing through points of intersection of mid points of parallel faces.
## draw lines of symmetry in a square
A line of symmetry is the line that divides the shape into two halves that match exactly. A square has four lines of linear symmetry.You can find if a shape has a Line of Symmetry by folding it. When the folded part sits perfectly on top (all edges matching), then the fold line is a Line. The square has 4 lines of symmetry. · State the number of lines of symmetry for the following figures: · Copy the following figure on a squared paper and draw the. Hence in total, you get the 4 lines of symmetry of the square which are AC, BD, PR, and QS. Thus, it may be observed that a square has 4 lines. The line of symmetry for given figure are as follow : A square has four lines of symmetry. solution. expand. Was this answer helpful? upvote 0. downvote 0.
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# What is the equation of a polar function?
## What is the equation of a polar function?
What is the equation of a polar function? A polar function is a function that is constant on the complex plane and is equal to the sum of the polar functions. The so called polar functions have an important role in the mathematical analysis of the problem of my company the polar functions, as they describe the shape and direction of the polar function. As it is well known, the polar functions are the most difficult to find solutions to this problem, because they are not easy to find. So, the polar function is one of the most important tools in the mathematical analyses of the problem. The polar function is called a non-zero polynomial of degree 3. A polynomial is a monic polynomial (or Laurent polynomial) of degree 3 that is strictly greater than 3. A po 7 A non-zero non-zero (non-zero) polynomial a non-zero positive integer 1-7 A monic po 9 A negative integer a negative integer 9 A nonzero (negative) integer 7 Here we can see that the polynomial 7 is a non-vanishing non-zero divisor of a positive integer. Just as the non-vanishes for all positive integers, the non-zero integer 7 is called a positive integer, and the positive integer 7 is denoted by a negative integer. Let us take a very simple example. We can consider a number of positive integers. Consider the following polynomial: This polynomial has a positive integer coefficient, and is a monomial. We will see that the non-negative integer 7 is a positive integer and that the positive integer 4 a fantastic read a negative integer, and that the negative integer 4 is the sum of two negative integers. In [1] we have that the nonnegative integer 4 is defined as the sum of three positive integers. Using the knownWhat is the equation of a polar function? A polar function is the solution of the equation of the polar coordinate. It is most commonly known as a polar coordinate in mathematics. The equation of a function with a given velocity is the equation for the polar coordinate, or, more precisely, the equation of its inverse: In mathematical terms, a polar function is a function that takes into account the boundary conditions that can be imposed to it. This boundary conditions ensure that the polar function does not change modulo the boundary conditions. A function is a particular case of a given function. Every function is a special case of a particular function. The function of a new function is the function of a original function.
## Do Math Homework Online
A function that is a special function is not unique. The equation of a piecewise linear function The function that gives the slope of a piece of a piece is called the curve piece. It is the slope of the curve of a piece. The slope of a curve find this the slope that is expressed in terms of the slope of its curve. This is a function link two variables. In the case of a piece, the slope of one piece is called its slope slope, and the slope of another piece is called their slope slopes. One may also define the slope slope of a function as the slope of that piece. A piecewise linear piecewise linear is a piecewise convex piecewise linear. When the slope of every piece of a curve comes that site the same slope, it is called a piecewise concave piecewise concase piecewise concasing piecewise conveasing piecewise concases. A curve piece is a piece of the curve piece if its slope is less than its slope. By the definition of a piece with a given slope, the curve piece is always concave. If the curve piece has a slope that is less than the slope, the piece is called a concave piece. If the piece has a different slope, then the piece is a concave of the curve. A concave piece is a convex view publisher site in the curve piece, and the piece is an average piece in the concave piece with a lower slope. A convex piece is a part of the curve in a piece. The piece is an ordinary piece of the piece. It is the piecewise concavity piecewise convese piecewise convece piecewise concased piecewise concaset piecewise concasis piecewise convease piecewise conveased piecewise convexe2x80x94splitting piecewise convexes piecewise convexa2x80xe2x88x92piecewise concases piecewise concotions piecewise concisce piecewise conveisce piecex2x80pxe2x80xa2x88xe2x89xa2x89xe2x86x92piecex2x8xe2What is the equation of a polar function? A polar function is the Clicking Here with the property that it is the same as the function on a circle, but it does not have a zero. It is usually denoted by its reciprocal, the prime, or the root of unity. As the square root of a function has a positive root, the general equation is that of a polar form. In the next chapter we will show that this equation is not unique.
## Why Do Students Get Bored On Online Classes?
The polar equation of a function is still a natural equation, but the equation is not the same as its reciprocal. Hence, given a real number, the general solution of the equation of the polar function is not unique: $$\frac{1}{2\pi i}\int_{-\infty}^{\infty}{\frac{d\theta}{2\theta}-\frac{\theta}{\theta+\theta^{2}}-\frac{e^{-\theta/\theta_0}}{e^{e^{-2\thetau/\thet_0}}}}$$ where the integration factor is taken between the roots of unity. Since the square root is the same, we can solve the equation: $\frac{A}{2\sqrt{2}}\dvtx\frac{2}{\sqrt{\pi}}\cos\theta=\frac{(1-e^{-e^{2\the}t/\the_0})}{2e^{-t/\sqrt 2}}=\frac{\sqrt{1-e^{\frac{\thetau}{\thet_{0}}}}}{\sqrho}$ where $\thetau$ is the period of the function, the righthand side is the square root, and $t$ is the integer part of the period. The equation is to be solved for $\theta$: Take the limit $\theta\rightarrow\infty$ and $\theta_1=0$ If the solution is a solution of the Bessel equation, then the solutions are the solutions of the equation: $-\thetfrac{\partial}{\partial\theta}\theta=e^{-x/\the}$. The equation of the B-type equation is the limit $\lim_{\theta\to\infty}{h(x,\theta)=0}$. The equation of the second order B-type and the B-wave solutions are: When the solution is not a solution of B-type or B-wave, the solution is the solution of the second-order B-system. Analogously, if the solution is an even function, then the solution is of the second B-system, which is a solution for the second order, B-system and
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# The Conti-System in Theory
With the Contisystem you can calculate rail first shots and easy Around solutions. Study it very carefully and pay special attention to the problematic zones and also the respective corrections. For beginners it is especially important to follow the instructions step by step in order to avoid confusion. First, you should manage the calculation to the third rail before you deal with the fourth rail, where you have to consider correction values. (From: “Faszination Dreiband-Billard”, Litho-Verlag, 2003)
## The Numbers on the Rails
• The values on the starting rail and the first rail relate to the points on the wooden rail (see dashed line, example: from 50 to 20)
• The values of the third rail relate to the values on the front part of the rubber, vertical to the value on the wooden rail.
• The spaces between the values of the third rail are sometimes irregular. The values in brackets show the real distances from the short rail. I.e. the value 0 which is relevant for the calculation is at 0,6 diamonds distance, or 6 units from the upper short rail. The value 40 is at 3,7 diamonds or 37 units from the upper short rail.
• The value 0 on the 1st rail is in the continuation of the front part of the rubber of the short rail. The value 00 is exactly in the corner of the wooden rails.
## The calculation formula
STARTING RAIL - 3rd RAIL = 1st RAIL
## Prerequisites of the stroke
In order to eliminate the stroke as a possible source of error, it is necessary to gain a consistent stroke for using the system to solve patterns. This consistent stroke should have the following characteristics:
B 1 is hit slightly below center with much running English.
The speed should be chosen in a way that B 3 (for rail first shots B 2 and B 3) should at least be moved 40 cm from their position.
The stroke should be executed as standard stroke, meaning neither extremely smooth nor attacking.
## Example of Calculation
• Estimate the arrival on the 3rd rail. In our example it is 30.
• Choose a possible start value and insert it into the formula. 50 - 30 = 20.
• The result of the calculation is the value on the 1st rail. Now imagine a line from the chosen start value (50) to the value on the 1st rail (20). If you realize, that this line is far away from the cue ball, you try another start value and insert it into the formula.
• If the line is close to the cue ball (less than 5 cm), you just shift it parallel, till you cross exactly the center of the cue ball. The point where this line reaches the first rail is your aiming point.
## The path from third to fourth rail
If we speak about the path, we mean the corresponding lines between the 3rd and the 4th rail. If you calculate and play a stroke to the 3rd rail, the paths show you where the cue ball touches the 4th rail. This is also the reason why the values of the 4th rail match the values of the 3rd rail (as shown in the diagram).
## Correction of the path (depending on start values)
The path from the 3rd rail to the 4th rail is to some degree depending on the start value. This is called the correction of the paths. The numbers in brackets that you find below the start values show the respective correction value.
## Examples
Example 1 (black line)
Start= 40, 1st rail = 10
Formula: 40 - 10 = 30
Correction of the path if you start from 40 is -1 (value in brackets)
30 (3rd rail) -1 = 29 (4th rail)
Example 2 (white line)
Start = 70, 1st rail = 40
Formula: 70 - 40 = 30
Correction of the path if you start from 70 is +5 (value in brackets)
30 (3rd rail) +5 = 35 (4th rail)
Looking at the diagram, you will understand the importance of this correction. In both cases we arrive at 30 on the 3rd rail, but starting from 40 the path from the 3rd to the 4th corrects by -1 (arrival on 4th at 29), starting from 70 the path corrects by +5 (arrival on 4th at 35).
Be careful! The correction of the path only concerns the arrival on the 4th and not on the 3rd rail!
Before you start your practice unit or a game you should play the following four lines.
From 50 to 30, expected arrival on 4th rail = 20
From 40 to 20, expected arrival on 4th rail = 19 (correction value when starting from 40 is -1)
From 30 to 10, expected arrival on 4th rail = 15 (correction value when starting from 30 is -5)
From 70 to 30, expected arrival on 4th rail = 45 (correction value when starting from 70 is +5)
The actual arrival values on the 4th rail enable the player to integrate the characteristics of the table into his calculation.
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# 2014 AMC 10B Problems/Problem 5
## Problem
Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2$, and the borders around and between the panes are $2$ inches wide. In inches, what is the side length of the square window? $[asy] fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray); fill((6,0)--(8,0)--(8,26)--(6,26)--cycle,gray); fill((12,0)--(14,0)--(14,26)--(12,26)--cycle,gray); fill((18,0)--(20,0)--(20,26)--(18,26)--cycle,gray); fill((24,0)--(26,0)--(26,26)--(24,26)--cycle,gray); fill((0,0)--(26,0)--(26,2)--(0,2)--cycle,gray); fill((0,12)--(26,12)--(26,14)--(0,14)--cycle,gray); fill((0,24)--(26,24)--(26,26)--(0,26)--cycle,gray); [/asy]$ $\textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$
## Solution
We note that the total length must be the same as the total height because the window a square. Calling the width of each small rectangle $2x$, and the height $5x$, we can see that the length is composed of 4 widths and 5 bars of length 2. This is equal to two heights of the small rectangles as well as 3 bars of 2. Thus, $4(2x) + 5(2) = 2(5x) + 3(2)$. We quickly find that $x = 2$. The total side length is $4(4) + 5(2) = 2(10) + 3(2) = 26$, or $\boxed{\textbf{(A)}}$.
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# Applied Combinatorics
## SectionB.5Finite Sets
A set $$X$$ is said to be finite when either (1) $$X=\emptyset\text{;}$$ or (2) there exists positive integer $$n$$ and a bijection $$f:[n]\bijection X\text{.}$$ When $$X$$ is not finite, it is called infinite. For example, $$\{a,\emptyset,(3,2),\posints\}$$ is a finite set as is $$\posints\times\emptyset\text{.}$$ On the other hand, $$\posints\times \{\emptyset\}$$ is infinite. Of course, $$[n]$$ and $$\bfn$$ are finite sets for every $$n\in\posints\text{.}$$
In some cases, it may take some effort to determine whether a set is finite or infinite. Here is a truly classic result.
Suppose that the set $$P$$ of primes is finite. It is non-empty since $$2\in P\text{.}$$ Let $$n$$ be the unique positive integer for which there exists a bijection $$f:[n]\rightarrow P\text{.}$$ Then let
\begin{equation*} p=1+f(1)\times f(2)\times f(3)\times \dots\times f(n) \end{equation*}
Then $$p$$ is not divisible by any of the primes in $$P$$ but is larger than any element of $$P\text{.}$$ Thus, either $$p$$ is prime or there is a prime that does not belong to $$P\text{.}$$ The contradiction completes the proof.
Here’s a famous example of a set where no one knows if the set is finite or not.
This conjecture is known as the Twin Primes Conjecture. Guaranteed $$\text{A} ++$$ for any student who can settle it!
When $$X$$ is a finite non-empty set, the cardinality of $$X\text{,}$$ denoted $$|X|$$ is the unique positive integer $$n$$ for which there is a bijection $$f:[n]\bijection X\text{.}$$ Intuitively, $$|X|$$ is the number of elements in $$X\text{.}$$ For example,
\begin{equation*} |\{(6,2), (8,(4,\emptyset)), \{3,\{5\}\}\}|=3. \end{equation*}
By convention, the cardinality of the empty set is taken to be zero, and we write $$|\emptyset|=0\text{.}$$
We note that the statement in Proposition B.11 is an example of “operator overloading”, a technique featured in several programming languages. Specifically, the times sign $$\times$$ is used twice but has different meanings. As part of $$X\times Y\text{,}$$ it denotes the cartesian product, while as part of $$|X|\times |Y|\text{,}$$ it means ordinary multiplication of positive integers. Programming languages can keep track of the data types of variables and apply the correct interpretation of an operator like $$\times$$ depending on the variables to which it is applied.
We also have the following general form of Proposition B.11:
\begin{equation*} |X_1\times X_2\times\dots\times X_n|= |X_1|\times |X_2|\times\dots\times |X_n| \end{equation*}
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NCERT Solutions For Class 6 Maths Integers Exercise 6.3
ncert textbook
NCERT Solutions For Class 6 Maths Integers Exercise 6.3
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3
Exercise 6.3
Question 1.
Find:
(a) 35 – (20)
(b) 72 – (90)
(c) (-15) – (-18)
(d) (- 20) – (13)
(e) 23 – (-12)
(f) (-32) – (-40)
Solution:
(a) 35 – (20) = 15 + (20) – (20)
= 15 + 0 = 15 [(+a) + (-a) = 0]
(b) 72 – 90
72 – (72 + 18) = 72 – 72 – 18
= 0 – 18 = – 18 [a + (- a) = 0]
(c) (- 15) – (- 18)
= (- 15) + (additive inverse of – 18)
= (-15) + (18) = 3
(d) (- 20) – (13)
(- 20) – (13) = – [20 + 13] = – 33
(e) 23 – (- 12)
23 – (- 12) = 23 + (additive inverse of – 12)
= 23 + 12 = 35
(f) (- 32) – (- 40)
(- 32) + (additive inverse of – 40)
= (- 32) + 40 = 8
Question 2.
Fill in the blanks with >, < or = sign.
(a) (-3) + (-6) (-3) – (-6)
(b) (-21) – (-10) (- 31) + (-11)
(c) 45 – (-11) 57 + (-4)
(d) (-25) – (-42) (-42) – (-25)
Solution:
(a) (-3) + (-6) = – [3 + 6] = – 9 and (-3) – (-6) = (-3) + 6 = 3
Here, – 9 < 3
∴ (- 3) + (- 6) < (- 3) – (- 6)
(b) (-21) – (-10) = (-21) + 10 = -11 and (-31) + (-11) = – (31 + 11) = – 42
Here, -42 < -11 or -11 > -42 ∴ (-21) , -(-10) >(-31)+ (-11)
(c) 45 – (-11) = 45 + 11 = 56 and 57 + (-4) = 57 -4 = 53
Here, 56 >53
∴ 45 – (-11) > 57 + ( -4)
(d) (-25) – (-42) = -25 + 42 = 17
and (-42) – (-25) = -42 + 25 = -17
Here, 17 > -17
∴ (-25) – (-42) > (-42) – (-25).
Question 3.
Fill in the blanks.
(a) (-8) + …. = 0
(b) 13 + …. = 0
(c) 12 + (-12) = ….
(d) (-4) + …. = – 12
(e) …. -15 = – 10.
Solution:
(a) (-8) + (additive inverse of -8) = 0
= (-8) + (8) = 0
∴ Value of blank is 8
(b) 13 + (additive inverse of 13) = 0
= 13 + (-13) = 0
∴ Value of blank is – 13
(c) 12 + (-12) = 0 [∵ -12 is additive inverse of 12]
∴ The Value of blank is 0
(d) (-4) + (-8) = -[4 + 8] = -12
∴ Value of blank is -8.
(e) (+5) – 15 = -10
∴ Value of blank is +5.
Question 4.
Find :
(a) (-7) – 8 – (-25)
(b) (-13) + 32 – 8 – 1
(c) (-7) + (-8) + (-90)
(d) 50 – (-40) – (-2)
Solution:
(a) (-7) – 8 – (-25)
= (-7) – 8 + 25
[ ∵ Additive inverse of – 25 is 25]
= -7 + 17 = -7 + 7 +10
[∵ (-a) + (+a) = 0]
= 0 + 10 = 10.
(b) (-13) + 32 – 8 – 1
= (-13) + (13) + 19 – (8 + 1)
= 0 + 19 – 9
= 19 – 9 [∵ (-13) + (13) = 0]
= 10 + 9 – 9 = 10 + 0 = 10.
[(+9) – (+9) = 0]
(c) (-7) + (-8) + (-90) = – (7 + 8) + (-90)
= -15 + (-90)
= -(15 + 90)
= -105.
(d) 50 – (-40) – (-2)
= 50 – [- 40 – 2]
= 50 – (-42)
= 50 + 42
= 92.
|
# Practice Problems on Pipes and Cistern
## Practice Problems on Pipes and Cistern
1. A Special pump can be used for filling as well as for emptying a Cistern. The capacity of the Cistern is 2400m³. The emptying capacity of the Cistern is 10m³ per minute higher than its filling capacity and the pump needs 8 minutes lesser to Cistern the tank than it needs to fill it. What is the filling capacity of the pump?
A. 40m³/min
B. 50m³/min
C. 60m³/min
D. 30m³/min
E. None of the Above
Answer & Explanation
Answer – B. 50m³/min
Explanation :
Filling Capacity of the Pump = x m/min
Emptying Capacity of the pump = (x+10) m/min
2400/x – 2400/x+10 = 8
(x – 50) + (x + 60) = 0
x = 50
2. Three pipes P, Q and R can fill a Cistern in 6 hours. After working at it together for 2 hours, R is closed and P and Q can fill the remaining part in 7 hours. The number of hours taken by R alone to fill the Cistern is
A. 14 hours
B. 12 hours
C. 15 hours
D. 18 hours
E. None of the Above
Answer & Explanation
Answer – A. 14 hours
Explanation :
Part filled in 2 hours = 2/6 = 1/3
Remaining Part = (1-1/3) = 2/3
(P + Q)’s 7 hour work = 2/3
(P + Q)’s 1 hour work = 2/21
R’s 1 hour work = (P + Q + R) 1 hour work – (P + Q) 1 hour work
= (1/6 – 2/21) = 1/14 = 14 hours
3. A Cistern is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes. If both the pipes are open,how long will it take to empty or fill the tank completely?
A. 5 minutes
B. 4 minutes
C. 6 minutes
D. 8 minutes
E. None of the Above
Answer & Explanation
Answer – C. 6 minutes
Explanation :
pipe B is faster than pipe A and so,the tank will be emptied.
part to be emptied = 2/5
part emptied by (A+B) in 1 minute= (1/6 – 1/10) = 1/15
1/15 : 2/5 :: 1: x
2/5 * 15 = 6 minutes.
4. If a pipe A can fill a tank 3 times faster than pipe B. If both the pipes can fill the tank in 32 minutes, then the slower pipe alone will be able to fill the tank in?
A. 128 minutes
B. 124 minutes
C. 154 minutes
D. 168 minutes
E. None of the Above
Answer & Explanation
Answer – A. 128 minutes
Explanation :
Time is taken by pipe A = x
Time is taken by pipe B = x/3
1/x + 3/x = 1/32
x = 128 minutes
5. A large cistern can be filled by two pipes P and Q in 15 minutes and 20 minutes respectively. How many minutes will it take to fill the Cistern from an empty state if Q is used for half the time and P and Q fill it together for the other half?
A. 12 minutes
B. 17 minutes
C. 18 minutes
D. 19 minutes
E. None of the Above
Answer & Explanation
Answer – A. 12 minutes
Explanation :
Part filled by P and Q = 1/15 + 1/20 = 7/60
Part filled by Q = 1/20
x/2(7/60 + 1/20) = 12 minutes
6. A pipe can fill a cistern in 16 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the cistern completely?
A. 3 hours
B. 2 hours
C. 9 hours
D. 4 hours
E. None of the Above
Answer & Explanation
Answer – C. 9 hours
Explanation :
In One hour pipe can fill = 1/16
Time is taken to fill half of the tank = 1/2 * 16 = 8 hours
Part filled by four pipes in one hour = (8*1/16) = 1/2
Required Remaining Part = 1/2
Total time = 8 + 1 = 9
7. Two pipes P and Q are opened together to fill a tank. Both the pipes fill the tank in time “x” If Q separately took 25 minutes more time than “x” to fill the tank and Q took 49 minutes more time than “x” to fill the tank, then find out the value of x?
A. 48 minutes
B. 35 minutes
C. 54 minutes
D. 68 minutes
E. None of the Above
Answer & Explanation
Answer – B. 35 minutes
Explanation :
Time is taken to fill the tank by both Pipes x = √a*b
x = √25*49 = 5 * 7 = 35
8. Three taps P, Q and R can fill a tank in 12, 15 and 20 hours respectively. If P is open all the time and Q, R are open for one hour each alternatively, the tank will be full in
A. 3 hours
B. 2 hours
C. 7 hours
D. 4 hours
E. None of the Above
Answer & Explanation
Answer – C. 7 hours
Explanation :
(P + Q)’s 1 hour work = 1/12 + 1/15 = 3/20
(P + R)’s 1 hour work = 1/12 + 1/20 = 2/15
For 2 hrs = (3/20 + 2/15) = 17/60
For 6 hrs = (3*17/60) = 17/20
Remaining Part = 1 – 17/20 = 3/20 filled by P and Q in 1 hour
9. Pipe A fills a tank in 30 minutes. Pipe B can fill the same tank 5 times as fast as pipe A. If both the pipes were kept open when the tank is empty, how much time will it take for the tank to overflow?
A. 3 minutes
B. 2 minutes
C. 5 minutes
D. 4 minutes
E. None of the Above
Answer & Explanation
Answer – C. 5 minutes
Explanation :
Total Capacity = 90L.
Tank filled in 1 minute by A = 3L
Tank filled in 1 minute by B = 15L
The capacity of the tank filled with both A and B in 1 minute = 18L.
overflow = 90/18 = 5 minutes.
10. Two pipes P and Q can fill a cistern in 10 hours and 20 hours respectively. If they are opened simultaneously. Sometimes later, tap Q was closed, then it takes total 5 hours to fill up the whole tank. After how many hours Q was closed?
A. 14 hours
B. 15 hours
C. 10 hours
D. 16 hours
E. None of the Above
Answer & Explanation
Answer – C. 10 hours
Explanation :
Pipe P Efficiency = 100/10 = 10%
Pipe Q Efficiency = 100/20 = 5%
Net Efficiency = 15%
15x + 10(5-x) = 100
x = 10
11. If a pipe A can fill a tank 3 times faster than pipe B and takes 32 minutes less than pipe B to fill the tank. If both the pipes are opened simultaneously, then find the time taken to fill the tank?
A. 14 minutes
B. 12 minutes
C. 15 minutes
D. 16 minutes
E. None of the Above
Answer & Explanation
Answer – B. 12 minutes
Explanation :
3x – x = 32
x = 16
1/16 + 1/48 = 4/48
Time taken to fill the tank = 48/4 = 12 minutes
12. Two pipes P and Q can fill a tank in 24 minutes and 27 minutes respectively. If both the pipes are opened simultaneously, after how much time should B be closed so that the tank is full in 8 minutes?
A. 14 minutes
B. 12 minutes
C. 15 minutes
D. 18 minutes
E. None of the Above
Answer & Explanation
Answer – D. 18 minutes
Explanation :
Required time = y(1-(t/x)) = 27(1-(8/24))= 18 minutes
13. A full tank gets emptied in 8 minutes due to the presence of a leak in it. On opening a tap which can fill the tank at the rate of 9 L/min, the tank get emptied in 12 min. Find the capacity of a tank?
A. 120 L
B. 240 L
C. 216 L
D. 224 L
E. None of the Above
Answer & Explanation
Answer – C. 216 L
Explanation :
a = 8; b = 9; C = 12
Capacity of a tank = a*b*c/c-a = 8*9*12/4 = 216 Litre.
14. If a pipe A can fill a tank 3 times faster than pipe B. If both the pipes can fill the tank in 42 minutes, then the slower pipe alone will be able to fill the tank in?
A. 148 minutes
B. 124 minutes
C. 154 minutes
D. 168 minutes
E. None of the Above
Answer & Explanation
Answer – D. 168 minutes
Explanation :
Time is taken by pipe A = x
Time is taken by pipe B = x/3
1/x + 3/x = 1/42
x = 168 minutes
15. A large cistern can be filled by two pipes P and Q in 15 minutes and 10 minutes respectively. How many minutes will it take to fill the Cistern from an empty state if Q is used for half the time and P and Q fill it together for the other half?
A. 6.5 minutes
B. 7.5 minutes
C. 8.5 minutes
D. 9.5 minutes
E. None of the Above
Answer & Explanation
Answer – B. 7.5 minutes
Explanation :
Part filled by P and Q = 1/15 + 1/10 = 1/6
Part filled by Q = 1/10
x/2(1/6 + 1/10) = 2/15 = 15/2 = 7.5 minutes
16. A pipe can fill a cistern in 8 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the cistern completely?
A. 3 hours
B. 2 hours
C. 5 hours
D. 4 hours
E. None of the Above
Answer & Explanation
Answer – C. 5 hours
Explanation :
In One hour pipe can fill = 1/8
Time is taken to fill half of the tank = 1/2 * 8 = 4 hours
Part filled by four pipes in one hour = (4*1/8) = 1/2
Required Remaining Part = 1/2
Total time = 4 + 1 = 5
17. Two pipes P and Q are opened together to fill a tank. Both the pipes fill the tank in time “x” If Q separately took 16 minutes more time than “x” to fill the tank and Q took 36 minutes more time than “x” to fill the tank, then find out the value of x?
A. 48 minutes
B. 24 minutes
C. 54 minutes
D. 68 minutes
E. None of the Above
Answer & Explanation
Answer – B. 24 minutes
Explanation :
Time is taken to fill the tank by both Pipes x = √a*b
x = √16*36 = 4 * 6 = 24
18. A Cistern has an inlet pipe and outlet pipe. The inlet pipe fills the cistern completely in 1 hour 20 minutes when the outlet pipe is plugged. The outlet pipe empties the tank completely in 6 hours when the inlet pipe is plugged. If there is a leakage also which is capable of draining out the water from the tank at half of the rate of the outlet pipe, then what is the time taken to fill the empty tank when both the pipes are opened?
A. 3 hours
B. 2 hours
C. 5 hours
D. 4 hours
E. None of the Above
Answer & Explanation
Answer – B. 2 hours
Explanation :
Inlet pipe Efficiency = 100/(8/6) = 75%
Outlet pipe Efficiency = 100/(6) = 16.66%
Efficiency of leakage = half of the rate of the outlet pipe = 8.33%
Net Efficiency = 75 – (16.66 + 8.33) = 50%
Required time = 100/50 = 2 hours
19. A Cistern has an inlet pipe and outlet pipe. The inlet pipe fills the cistern completely in 1 hour 20 minutes when the outlet pipe is plugged. The outlet pipe empties the tank completely in 4 hours when the inlet pipe is plugged. If both pipes are opened simultaneously at a time when the tank was one-third filled, when will the tank fill thereafter?
A. 3 hours
B. 2 hours
C. 5 hours
D. 4 hours
E. None of the Above
Answer & Explanation
Answer – B. 2 hours
Explanation :
Inlet pipe Efficiency = 100/(8/6) = 75%
Outlet pipe Efficiency = 100/(4) = 25%
Net Efficiency = 75 – 25 = 50%(1/3)filled
2/3 filled = 100%
Required time = 100/50 = 2 hours
20. Two pipes P and Q can fill a cistern in 10 hours and 20 hours respectively. If they are opened simultaneously. Sometimes later, tap Q was closed, then it takes total 8 hours to fill up the whole tank. After how many hours Q was closed?
A. 4 hours
B. 5 hours
C. 2 hours
D. 6 hours
E. None of the Above
Answer & Explanation
Answer – A. 4 hours
Explanation :
Pipe P Efficiency = 100/10 = 10%
Pipe Q Efficiency = 100/20 = 5%
Net Efficiency = 15%
15x + 10(8-x) = 100
x = 4
21. Three pipes A, B, and C can fill the tank in 10 hours, 20 hours and 40 hours respectively. In the beginning all of them are opened simultaneously. After 2 hours, tap C is closed and A and B are kept running. After the 4th hour, tap B is also closed. The remaining work is done by tap A alone. What is the percentage of the work done by tap A alone?
A. 30 %
B. 35 %
C. 45 %
D. 50 %
E. None of the Above
Answer & Explanation
Answer – B. 35 %
Explanation :
Pipe A’s work in % = 100/10 = 10%
Pipe B’s work in % = 100/20 = 5%
Pipe C’s work in % = 100/40 = 2.5%
All of them are opened for 2 hours + after 2 hours, tap C is closed + After the 4th hour, tap B is also closed = 100
=> (10+5+2.5)*2 + (10+5)*2 + X = 100
=> 35 + 30 + work by tap A alone = 100
=> work by tap A alone = 100-65 = 35%
22. A pipe can fill a tank in 12 minutes and another pipe can fill it in 15 minutes, but a third pipe can empty it in 6 minutes. The first two pipes are kept open for 5 min in the beginning and then third pipe is also opened. Time taken to empty the water tank is?
A. 30 mins
B. 25 mins
C. 45 mins
D. 50 mins
E. None of the Above
Answer & Explanation
Answer – C. 45 mins
Explanation :
x/6 – (x+5)/12 – (x+5)/15 = 0
x = 45 mins
23. Two pipes A and B can fill a tank in 12 hours and 18 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom of the tank it took 48 minutes excess time to fill the cistern. When the cistern is full, in what time will the leak empty it?
A. 72 hours
B. 62 hours
C. 64 hours
D. 84 hours
E. None of the Above
Answer & Explanation
Answer – A. 72 hours
Explanation :
Work done by the two pipes in 1 hour = (1/12)+(1/18) = (15/108).
Time taken by these pipes to fill the tank = (108/15)hrs = 7 hours 12 min.
Due to leakage, time taken = 7 hours 12 min + 48 min = 8 hours
Work done by two pipes and leak in 1 hour = 1/8.
Work done by the leak in 1 hour =(15/108)-(1/8)=(1/72).
Leak will empty the full cistern in 72 hours.
24. A tank is normally filled in 6 hours but takes two hours longer to fill because of a leak in the bottom of the tank. If the tank is full the leak will empty it in how many hours?
A. 16 hours
B. 18 hours
C. 17 hours
D. 24 hours
E. None of the Above
Answer & Explanation
Answer – D. 24 hours
Explanation :
Work done by leak in 1 hr=(1/6-1/8)=1/24
Leak will empty the tank in 24 hours
25. Twelve pipes are connected to a Cistern. Some of them are inlet pipes and the others are outlet pipes. Each of the inlet pipes can fill the tank in 8 hours and each of the outlet pipes can empty the cistern completely in 6 hours. If all the pipes are kept open, the empty tank gets filled in 24 hours. How many inlet pipes are there?
A. 6
B. 8
C. 7
D. 4
E. None of the Above
Answer & Explanation
Answer – C. 7
Explanation :
(x/8)-[(12-x)/6] = 1/24
x = 7
26. A dam has four inlets – A, B, C and D. The dam can be filled in 12 minutes through the first three inlets and it can be filled in 15 minutes through the second, the third and fourth inlet also it can be filled through the first and the fourth inlet in 20 minutes. How much time required to fill up the dam by all the four inlets?
A. 10 mins
B. 15 mins
C. 20 mins
D. 25 mins
E. None of the Above
Answer & Explanation
Answer – A. 10 mins
Explanation :
(1/A + 1/B + 1/C) = 1/12 …(i)
(1/B + 1/C + 1/D) = 1/15 …(ii)
(1/A + 1/D) = 1/20 …(iii)
From eqn (i) and (ii)
(1/A – 1/D) = 1/60…(iv)
From eqn (iii) and (iv)
A=30 D=60.
Let the time taken to full the tank = T
T(1/A + 1/B +1/C +1/D)= 1
T(1/30 + 1/15) = 1
T = 10 mins
27. Three pipes P, Q and R connected to a Cistern. The first pipe (i.e) P can fill 1/2 part of the tank in one hour, second pipe, Q can fill 1/3 part of the cistern in one hour. R is connected to empty the cistern. After opening all the three pipes 7/12 part of the cistern. Then how much time required to empty the cistern completely?
A. 2 hours
B. 3 hours
C. 4 hours
D. 5 hours
E. None of the Above
Answer & Explanation
Answer – C. 4 hours
Explanation :
In 1 hour, P can fill = 1/2 Part
Time taken to fill the Cistern by Pipe P = 2 hours
In 1 hour, Q can fill = 1/3 Part
Time taken to fill the Cistern by Pipe P = 3 hours
[1/2 + 1/3 – 1/R] = 7/12
1/R = 1/4
Time required to empty the Cistern = 4 hours
28. A Cistern can be filled by an inlet pipe at the rate of 4 litres per minute. A leak in the bottom of a cistern can empty the full tank in 8 hours. When the cistern is full, the inlet is opened and due to the leak, the cistern is empty in 40 hours. How many litres does the cistern hold?
A. 4000 litre
B. 2400 litre
C. 1920 litre
D. 2020 litre
E. None of the Above
Answer & Explanation
Answer – B. 2400 litre
Explanation :
Part emptied by the leak in 1 hour = 1/8
part filled by (leak & inlet open) in 1 hour = 1/40
Part filled by the inlet pipe in 1 hour = 1/8 – 1/40 = 1/10
Inlet pipe fills the tank in = 10 hours
Inlet pipe fills water at the rate of 4 litres a minute.
Capacity of Cistern = 10 * 60 * 4 = 2400 litre
29. In a tank there is a pipe which can be used for filling the tank as well as for emptying it. The capacity of the tank is 1200 m³. The emptying of the tank is 10 m³ per minute higher than its filling capacity and the pump needs 6 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pipe?
A. 20 m³ / min.
B. 40 m³ / min.
C. 50 m³ / min.
D. 60 m³ / min.
E. None of the Above
Answer & Explanation
Answer – B. 40 m³ / min.
Explanation :
1200/x – 1200/(x+10) = 6
200/x – 200/(x+10) = 6
x2 + 10x – 2000 = 0
x = 40
30. Two pipes P and Q can fill a cistern in 12 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours will the tank be full?
A. 4 hours
B. 5 hours
C. 2 hours
D. 6 hours
E. None of the Above
Answer & Explanation
Answer – D. 6 hours
Explanation :
Pipe P can fill = 1/12
Pipe Q can fill = 1/4
For every two hour, 1/12 + 1/4 = 1/3 Part filled
Total = 6 hours
31. Two pipes A and B can fill a tank in 10 hours and 15 hours respectively while a third pipe C can empty the full tank in 20 hours. All the pipes are opened for 5 hours and then C is closed. Find the time in which the tank is full?
a) 5.5 hrs
b) 6.5 hrs
c) 7.5 hrs
d) 8.5 hrs
e) None of these
Answer & Explanation
Answer – c) 7.5 hrs
Explanation :
(1/10 + 1/15 – 1/20)*5 + (1/10 + 1/15)*T = 1. We will get T = 2.5 hrs
so total time = 5 + 2.5 = 7.5 hrs
32. Three pipe P, Q and R can fill a tank in 12 minutes, 18 minutes and 24 minutes respectively. The pipe R is closed 12 minutes before the tank is filled. In what time the tank is full?
a) 8.(5/13) hrs
b) 8.(4/13) hrs
c) 7.(4/13) hrs
d) 8.(6/13) hrs
e) None of these
Answer & Explanation
Answer – b) 8.(4/13) hrs
Explanation :
Let T is the time taken by the pipes to fill the tank
(1/12 + 1/18 + 1/24)*(T – 12) + (1/12 + 1/18)*12 = 1
We will get T = 108/13 = 8.(4/13) hrs
33. On pipe P is 4 times faster than pipe Q and takes 45 minutes less than pipe Q. In what time the cistern is full if both the pipes are opened together?
a) 8 minutes
b) 10 minutes
c) 12 minutes
d) 14 minutes
e) None of these
Answer & Explanation
Answer – c) 12 minutes
Explanation :
Let P takes x minutes to fill the tank alone, then Q will take 4x minutes to fill the tank
4x – x = 45, x = 15
So P will take 15 minutes and Q will take 60 minutes to fill the tank. Both will fill the tank in
(60*15)/(75) = 12 minutes
34. Two pipes can fill a tank in 15 and 20 hours respectively. The pipes are opened simultaneously and it is found that due to the leakage in the bottom, 17/7 hours extra are taken extra to fill the tank. If the tank is full, in what approximate time would the leak empty it?
a) 27 hrs
b) 32 hrs
c) 36 hrs
d) 39 hrs
e) None of these
Answer & Explanation
Answer – d) 39 hrs
Explanation :
Total time taken by both pipes before the leak was developed = 60/7 hours
now, leaks is developed which will take T time to empty the tank so, (1/15 +1/20 – 1/T) = 1/11
solve for T, we will get 660/17 hours = 39 hours (approx.)
35. Two pipes A and B can fill a tank in 8 minutes and 12 minutes respectively. If both the pipes are openedsimultaneously, after what time should B be closed so that the tank is full in 6 minutes?
a) 1 min
b) 2 min
c) 3 min
d) 4 min
e) None of these
Answer & Explanation
Answer – c) 3 min
Explanation :
Let after x minutes pipe B is closed
(1/8 + 1/12)*x + (1/8)*(6 -x) = 1
X= 3 minutes
36. In what time would a cistern be filled by three pipes whose diameters are 1cm, 2 cm and 3 cm running together, when the largest pipe alone can fill the tank in 21 minutes? The amount of water flowing through the pipe is directly proportional to the square of its diameter.
a)10.5 minutes
b) 11.5 minutes
c) 12.5 minutes
d) 13.5 minutes
e) None of these
Answer & Explanation
Answer – d) 13.5 minutes
Explanation :
More the diameter more will be the water flowing through it and less will be the time taken.
Means bigger pipe will take less time to fill the tank
So, for 1 cm time, (1^2)/(3^2) = 21/t, we get t = 189
For 2 cm time, (2^2)/(3^2) = 21/t. We get t = 189/4
So total time = 1/21 + 1/189 + 4/189 = 2/27
So total time = 13.5 minutes
37. Two pipes P and Q can fill a tank in 10 min and 12 min respectively and a waste pipe can carry off 12 litres of water per minute. If all the pipes are opened when the tank is full and it takes one hour to empty the tank. Find the capacity of the tank.
a) 30
b) 45
c) 60
d) 75
e) None of these
Answer & Explanation
Answer – c) 60
Explanation :
Let the waste pipe take ‘T’ time to empty the tank.
(1/10 + 1/12 – 1/T)*60 = -1
We will get T = 5 min
So capacity = 5*12 = 60ltr
38. One pipe fill 1/4 of the tank in 4 minutes and another pipe fills 1/5 of the tank in 4 minutes. Find the time taken by both pipe together to fill half the tank?
a) 40/9 minutes
b) 50/9 minutes
c) 44/9 minutes
d) 53/9 minutes
e) None of these
Answer & Explanation
Answer – a) 40/9 minutes
Explanation :
First pipe will take 16 minutes to fill the tank alone. Similarly second pipe will take 20 minutes to fill the tank alone. Let T is the time in which both the pipes will fill half the tank
(1/16 + 1/20)*T = 1/2, we get T = 40/9 minutes
39. Two pipes can separately fill the tank in 15hrs and 30hrs respectively. Both the pipe are opened and when the tank is 1/3 full a leak is developed due to which 1/3 water supplied by the pipe leaks out. What is the total time to fill the tank?
a) 20/3 hr
b) 35/3 hr
c) 40/3 hr
d) 50/3 hr
e) None of these
Answer & Explanation
Answer – c) 40/3 hr
Explanation :
(1/15 + 1/30)*T1 = 1/3, T1 = 10/3 hr
Now after leak is developed, [(1/15 + 1/30) – (1/3)*(1/15 + 1/30)]*T2 = 2/3
T2 = 10 hr. So total time = 10 + 10/3 = 40/3 hr
40. Three pipes A, B and C is attached to a cistern. A can fill it in 20 minutes and B can fill it in 30 minutes. C is a waste pipe. After opening both the pipes A and B, Riya leaves the cistern to fill and returns when the cistern is supposed to be filled. But she found that waste pipe C had been left open, she closes it and now the cistern takes 5 minutes more to fill. In how much time the pipe C can empty the full cistern?
a) 26.8 minutes
b) 25.8 minutes
c) 27.8 minutes
d) 28.8 minutes
e) None of these
Answer & Explanation
Answer – d) 28.8 minutes
Explanation :
The tank supposed to be filled in (30*20)/50 = 12 minutes
so, (1/20 + 1/30)*12 – 12/C + (1/20 + 1/30)*5 = 1 (A and B work for 12 minutes and also C work for 12 minutes and then A and B takes 5 more minutes to fill the tank)
solve for C, we will get C = 144/5 = 28.8
### IBTS INSTITUTE
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# Arcs and Angles of Circles
Hello, and welcome to this video about arcs and angles of circles! In this video, we will explore the different parts of circles and how to use them to solve problems. Let’s learn about arcs and angles of circles!
When we look around us, after polygons, circles are the next most common shape that we are surrounded by in our environment. A circle is created by a 360° rotation.
An arc of the circle is a part of the circle, and we identify arcs using points on the circle. Arc $$RS$$ is the curved part of circle $$A$$ shown with the purple mark. Because arc $$RS$$ is less than 180°, we call this arc a minor arc. Arc $$RTS$$, or $$STR$$, is a major arc because it is greater than 180°. We can find the measure of an arc using the fact that a circle is 360°. If we know that the measure of arc $$RS$$ is 130°, we can find the measure of arc $$RTS$$ by subtracting the measure of arc $$RS$$ from 360°, which would make the measure of arc $$RTS$$ 230°.
Some important lines and segments associated with circles are chords, secants, and tangents.
A chord is a line segment with either end touching the circle. Line segment $$LM$$ is a chord on circle $$A$$. A secant line is similar to a chord, except it is a line that passes through two points on the circle instead of being a line segment. And a tangent line is a very special kind of line that only touches the circle at one point, called the point of tangency. Line $$PQ$$ is a tangent and point $$Q$$ is the point of tangency. Knowing these vocabulary words related to circles is important in helping understand the circle theorems which will help us solve problems.
The radius of the circle is always perpendicular to the point of tangency.
When two chords share a point on the circle, an inscribed angle is formed. In circle $$A$$, the chords $$RS$$ and $$TS$$ form the inscribed angle $$RST$$, which has its vertex on the circle. The arc that is formed by the legs of angle $$RST$$ is called an intercepted arc. An angle that has the center of the circle as its vertex is naturally called a central angle. In circle $$C$$, angle $$MCN$$ is a central angle and arc $$MN$$ is its intercepted arc.
If we are given the measure of the inscribed angle or the intercepted arc, we will be able to find the measure of the other using a circle theorem that tells us that the measure of an inscribed angle is one-half the measure of its intercepted arc. Or we could say that the intercepted arc is twice as long as the inscribed angle. For example, in circle $$A$$, the measure of inscribed angle $$RST$$ is given to us as 60°. According to the inscribed angle theorem, the measure of arc $$RT$$ is $$60° \times 2$$, which is 120°.
Here is a special case of an inscribed angle.
In circle $$A$$, inscribed angle $$PQR$$ encompasses the diameter of the circle, line segment $$PR$$ Remember, the diameter of the circle divides the circle into two equal parts, called semicircles. A circle has a total of 360°, so a semicircle has 180°, or half the measure of a full circle. So this must mean that the intercepted arc of angle $$PQR$$ is 180°. If we use the inscribed angle theorem, we find out that angle $$PQR$$ is 90° because it is half the degree measure of the intercepted arc.
Another important theorem we are going to take a look at is the central angle theorem. This theorem says that the measure of a central angle is equal to the measure of its intercepted arc. In circle $$C$$, since the measure of angle $$MCN$$, the central angle, is 110°, then arc $$MN$$, its intercepted arc, is also 110°.
Let’s take a look at an example problem. Line $$LM$$ is tangent to circle $$T$$ at point $$L$$. The measure of the central angle $$T$$ is 35°. What is the measure of angle $$LMT$$?
Since we know that the sum of the interior angles of any triangle is 180° and that the radius is always perpendicular to the point of tangency, we know that angle $$TLM$$ is 90°. We can subtract $$(90°+35°)$$ from 180° to find the measure of angle $$LMT$$. $$180 – (90+35) = 55$$, therefore the measure of angle $$LMT$$ is 55°.
When secants intersect inside a circle, we can find the measure of the vertical angles created by using the measure of the arcs formed by the angles. It also works the other way around, if we need to find the measure of the arcs using the measure of the angles.
Let’s take a look at circle $$K$$.
The secant lines $$DE$$ and $$FG$$ intersect at point $$H$$ and create the arcs $$DF$$, $$FE$$, $$EG$$, and $$GD$$. Find the measure of any of the vertical angles (notice that there are two sets) by adding together the measures of the intercepted arcs and dividing by 2.
We will use the formulas $$m\angle DHF=\frac{1}{2}$$ (arc $$DF$$ + arc $$GE$$) and $$m\angle FHE=\frac{1}{2}$$ (arc $$FE$$ + arc $$DG$$). Then we will use the vertical rule theorem to find $$m\angle EHG$$ and $$m\angle GHD$$.
Let’s practice. The measure of arc $$GE$$ is 170°, arc $$GD$$ is 80°, arc $$DF$$ is 60°, and arc $$FE$$ is 50°. To find the $$m\angle DHF$$, we will add the intercepted arcs, $$DF$$ and $$GE$$, then multiply by $$\frac{1}{2}$$. Therefore, $$m\angle DHF=\frac{1}{2}(60° +170 °)$$. So $$\angle DHF=115°$$. We will use the equation, $$m\angle FHE=\frac{1}{2}(50° +80°)$$ to find the measure of $$\angle FHE$$, which is $$m\angle FHE=65°$$.
When two secant lines intersect outside the circle, we use a different method to find the measure of the angles. When this happens, you look at the two intercepted arcs created by the angle and subtract the measure of the smaller arc from the measure of the larger arc, then multiply by $$\frac{1}{2}$$.
We want to find the $$m\angle WYU$$ that is created by the intersection of secants $$WX$$ and $$UV$$ on circle $$Z$$. $$\angle WYU$$ is intercepted by arcs $$XV$$ and $$WU$$. The measure of arc $$XV$$ is 30°; the measure of arc $$WU$$ is 80°. To find the measure of angle $$WYU$$, we use the formula: $$m\angle WYU= \frac{1}{2}$$ (arc $$WU$$ – arc $$XV$$), therefore, $$m\angle WYU=\frac{1}{2}(80°-30°) \text{, }m\angle WYU=25°$$.
A sector of a circle is a section of a circle between two radii. The red area in the diagram is an example of a sector.
We can find the length of an arc if we are given the length of the radius and the measure of the central angle using the formula $$s=\frac{\pi r\theta }{180°}$$, where $$s$$ is the arc length, $$r$$ is the length of the radius and $$\theta$$ is the measure of the central angle. In circle A, the length of the radius is 5 cm, and the measure of angle $$MAN$$ is 70°. We want to find the length of arc $$MN$$. We will start by substituting what we have into the formula, $$s=\frac{\pi (5)(70)}{180°}$$. We will simplify to get $$s=6.11\text{ cm}$$ , which is the length of arc $$MN$$.
Thanks for watching this video, and happy studying!
## Arcs and Angles of Circles Practice Questions
Question #1:
If angle $$B$$ is $$21°$$, what is the measure of arc $$AC$$?
$$82°$$
$$42°$$
$$165°$$
$$21°$$
According to the inscribed angle theorem, arc $$AC$$ will be twice as large as the inscribed angle $$B$$. This means that arc $$AC$$ will be $$2×21°=42°$$.
Question #2:
Line $$LM$$ is tangent to circle $$J$$ at point $$L$$. If angle $$J$$ is $$27°$$, what is the measure of angle $$M$$?
$$65°$$
$$83°$$
$$44°$$
$$63°$$
Since line $$LM$$ is tangent to circle $$J$$ at point $$L$$, the line $$JL$$ and $$LM$$ form a right angle. This means that angle $$L$$ is $$90°$$. Angle $$J$$ is $$27°$$, so add angle $$J$$ and angle $$L$$: $$90+27=117$$. Since there are $$180°$$ in a triangle, subtract $$117$$ from $$180$$: $$180-117=63$$. The measure of angle $$M$$ is $$63°$$.
Question #3:
Arc $$BD$$ is $$15°$$. Arc $$AE$$ is $$48°$$. Find the measure of angle $$ACE$$.
$$19.5°$$
$$16.5°$$
$$22.5°$$
$$19°$$
Look at the two intercepted arcs. The angle $$ACE$$ will be $$\frac{1}{2}$$ the difference between the arc lengths of these intercepted arcs. Subtract the smaller arc from the larger arc. In this case, subtract arc $$BD$$ from arc $$AE$$: $$48-15=33$$. Now multiply this by $$\frac{1}{2}$$. $$33×\frac{1}{2}=16.5°$$. Angle $$ACE$$ is $$16.5°$$.
Question #4:
You are walking around a circular pond from one star to the other. If the radius of the pond is $$8\text{ ft}$$, what distance have you walked? Use $$3.14$$ for $$π$$.
$$22.8\text{ ft}$$
$$19\text{ ft}$$
$$16.78\text{ ft}$$
$$34.5\text{ ft}$$
Find the measure of the arc by using the formula: $$s=\frac{πrϴ}{180°}$$
$$s$$ represents the arc length.
$$π$$ is approximated to $$3.14$$.
$$r$$ is the radius.
$$ϴ$$ is the central angle measure.
$$s=\frac{πrϴ}{180°}$$ becomes $$s=\frac{(3.14)(8)(120°)}{180°}=16.746$$, which simplifies to $$s=16.7\text{ ft}$$.
Question #5:
A flashlight shines across a circular field. What is the angle measure of the intercepted arc that will be illuminated?
$$104°$$
$$120°$$
$$106°$$
$$130°$$
The inscribed angle is $$52°$$. The intercepted arc will be twice as much as the inscribed angle. $$52×2=104$$. The angle measure that is illuminated across the field is $$104°$$.
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Relative frequency
26.4 Relative frequency
So far, we have calculated the theoretical probabilities of events. Theoretical probabilities tell us what should happen in an experiment. The relative frequency tells us what actually happens in an experiment.
Relative frequency is the number of outcomes obtained for a certain number of trials. We have to repeat an experiment a number of times and count the number of positive outcomes. Because this is an experiment, which is an uncertain process, we can get a different relative frequency every time that we repeat the experiment.
trial
a completed run of a well-defined experiment
positive outcome
when we get an experiment outcome that we wanted to get
frequency
the number of times an event occurs
Examples of trials:
• rolling a fair dice
• drawing a card from a pack of cards
• tossing a fair coin
• turning a spinner
The relative frequency $$(f)$$ is the number of positive outcomes $$(p)$$ divided by the total number of trials $$(t)$$.
$\text{Relative frequency} = \frac{\text{number of times an event occurs}}{\text{total number of trials conducted}}$ $f = \frac{p}{t}$
relative frequency
the number of positive outcomes $$(p)$$ divided by the total number of trials $$(t)$$; it tells us what actually happened in an experiment
Worked example 26.4: Calculating relative frequency
We toss a coin $$20$$ times and observe the outcomes. The results of the trials are listed below.
$H; H; H; T; T; H; H; H; H; H; T; T; H; T; T; H; T; T; T; H$
1. If you work out the relative frequency after the first $$10$$ trials, what is the relative frequency of observing heads?
2. What is the relative frequency of observing heads after $$15$$ trials?
3. What is the relative frequency of observing heads after $$20$$ trials?
4. Calculate the theoretical probability of observing heads.
5. How does the relative frequency compare with the theoretical probability?
Calculate the relative frequency of observing heads after $$10$$ trials.
For the first $$10$$ trials, we observed $$8$$ heads.
So for $$t = 10, p = 8$$.
$f = \frac{p}{t} = \frac{8}{10} = \text{0,8}$
Calculate the relative frequency of observing heads after $$15$$ trials.
For the first $$15$$ trials, we observed $$9$$ heads.
So for $$t = 15, p = 9$$.
$f = \frac{p}{t} = \frac{9}{15} = \text{0,6}$
Calculate the relative frequency of observing heads after $$20$$ trials.
For $$20$$ trials, we observed $$11$$ heads.
So for $$t = 20, p = 11$$.
$f = \frac{p}{t} = \frac{11}{20} = \text{0,55}$
Calculate the theoretical probability.
There are two possible outcomes; heads or tails. So $$n(S) = 2$$.
$$E = {H}$$, so $$n(E) = 1$$.
\begin{align} P(E) &= \frac{n(E)}{n(S)} \\ &= \frac{1}{2} \\ &= \text{0,5} \end{align}
Compare relative frequency and theoretical probability.
After $$20$$ trials, the relative frequency of $$\text{0,55}$$ is close to the theoretical probability of $$\text{0,5}$$.
The greater the number of trials, the closer the relative frequency will get to the theoretical probability.
The graph below shows the plot of the relative frequency of observing heads ($$f$$) after having completed $$t$$ coin tosses. The number of trials have been plotted on the $$x$$-axis and the relative frequency ($$f$$) on the $$y$$-axis.
In the beginning (after a small number of trials) the relative frequency changes a lot when compared to the theoretical probability at $$\text{0,5}$$, which is shown with a dashed line. As the number of trials increases, the relative frequency changes less and gets closer to the theoretical probability.
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# Moment of Inertia of C Channel
A C channel is a common structural element that is often used in building things such as bridges, walls, or in metal frames. As part of the engineering design of these structures, understanding how a C channel will respond to various forces and loads is necessary. A key parameter of any structural member is its moment of inertia to understand how that member will bend or flex due to a load. Therefore, calculating the moment of inertia of a C channel is important.
## Calculating The Moment Of Inertia Of A C Channel
To calculate the moment of inertia of a C channel, there are several steps that need to be carried out. First, the C channel needs to be split into separate pieces for analysis and the area of each of these pieces needs to be determined. Then, the center of gravity of the C channel needs to be found so that the moment of inertia about the centroid can be evaluated. Each of these steps is introduced in detail below.
### Splitting the C channel
Often, when calculating the moment of inertia of a complex shape, dividing the shape into different pieces is the most effective means for evaluating the total moment of inertia. For a C channel, such as the one shown in the following figure, separating it into three individual rectangles is appropriate.
The area of each piece of the C channel, A, B, and C is calculated as its individual width times height. Using the area of each section, the center of gravity is found.
### Calculating the center of gravity of a C channel
A C channel is symmetric about its x-axis, but not about its y-axis. However, using the following equation, the location of the -axis centroid can be determined:
where:
• c is the location of the y-axis centroid of the C channel, with SI units of mm
• I is the location of the -axis centroid of one of the pieces, with SI units of mm
• Ai is the area of one of the pieces, with SI units of mm2
### Example Calculation
As an example of locating the y-axis centroid, take a C channel that is dimensioned as follows:
Splitting the C channel into three rectangles, A, B, and C, with the following dimensions:
• A = 10 mm × 50 mm
• B = 50 mm × 10 mm
• C = 10 mm × 50 mm
The location of the y-axis centroid for each piece is as follows:
Then, using the equation for c:
Now that the preliminary steps have been carried out, the moments of inertia about the x– and -axis can be calculated for the C channel.
### Calculating The Moment Of Inertia About The X-Axis
The moment of inertia of a C channel about its x-axis can be calculated as follows:
where:
• Ix is the moment of inertia for the entire C channel about its x-axis, with SI units of mm4
• Ixi is the moment of inertia of each section about its individual x-axis, with SI units of mm4
The moment of inertia of a rectangle about its x-axis is calculated using the following equation:
where:
• w is the width of the rectangle, with SI units of mm
• h is the height of the rectangle, with SI units of mm
Because each piece of the C channel is located at a different distance from the x-axis, it is necessary to use the parallel axis theorem to determine each piece’s moment of inertia about the x-axis of the complete channel. The parallel axis theorem is introduced as follows:
where:
• Ix’ is the moment of inertia of the section about the overall x-axis, with SI units of mm4
• Ix is the moment of inertia of the section about its own x-axis, with SI units of mm4
• A is the area of the section, with SI units of mm2
• d is the distance from the x-axis of the section to the overall x-axis, with SI units of mm
After the moment of inertia about the overall C channel x-axis is determined for each section, they can be added together to determine the total moment of inertia about the x-axis for the C channel.
### Example Calculation
Continuing to use the above C channel, the moment of inertia about the x-axis can be calculated using the following steps:
1. Calculate the moment of inertia of each section about its own x-axis:
1. Locate the x-axis of each section:
1. Calculate the distance d from the total x-axis for each section:
1. Use the parallel axis theorem to determine the moment of inertia for each section about the overall x-axis:
1. Sum all individual moments of inertia to determine the total moment of inertia for the C channel about its x-axis:
### Calculating The Moment Of Inertia About The Y-Axis
Similar to calculating the moment of inertia about the x-axis, the moment of inertia of a C channel about its y-axis can be determined by summing the individual section moments of inertia:
The equation for the moment of inertia of a rectangle about its y-axis is calculated as follows:
Again, the parallel axis theorem is used to find the moment of inertia of each section about the overall y-axis of the C channel, substituting the distance from the y-axis calculated previously for the distance from the x-axis.
### Example Calculation
Continuing to use the above C channel, the moment of inertia about the y-axis can be calculated using the following steps:
1. Calculate the moment of inertia of each section about its own y-axis:
1. Locate the y-axis of each section:
1. Calculate the distance d from the total y-axis for each section:
1. Use the parallel axis theorem to determine the moment of inertia for each section about the overall y-axis:
1. Sum all individual moments of inertia to determine the total moment of inertia for the C channel about its y-axis:
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The purpose of this self-assessment is to give students a flavor of math background that is expected or will be introduced in this class. This assessment covers basic knowledge and skills. Formal reasoning skills (proofs, logic) such as those taught in Math 232 are also expected.
## Derivatives and Basic Probability
Here are a few basic exercises from the required prerequisite classes (Calculus I and Discrete Math). Students should feel comfortable doing these (perhaps with a bit of review or brushing up):
1. Let $$f(x) = 5x^2 + 3$$. Write the derivative of $$f(x)$$.
$$\frac{d}{dx}f(x) =$$
2. Let $$f(x) = 4e^{2x}$$. Write the derivative of $$f(x)$$.
$$\frac{d}{dx}f(x) =$$
3. A fair coin has equal probability of coming up heads or tails. If a fair coin is flipped twice, what is the probability of seeing first heads, then tails:
$$\Pr(\text{heads}, \text{tails}) =$$
4. Consider an unfair coin that has probability $$3/4$$ of coming up heads. If the coin is flipped three times, what is the probability of seeing tails each time:
$$\Pr(\text{tails}, \text{tails}, \text{tails}) =$$
5. I agree to let you flip the same unfair coin one time and pay you $1 if it comes up heads and$2 if it comes up tails. Let $$X$$ be the number of dollars I pay you, which is a random variable. What is the expected value of $$X$$?
$$E[X] =$$
## Partial Derivatives and Linear Algebra
These topics are covered in Multivariable Calculus and Linear Algebra, which are not prerequisites. We will introduce these topics in class. It will be a bit easier if you have seen them before.
1. Let $$f(x,y)= 3xy^2+2y$$. Write the following partial derivative:
$$\frac{\partial}{\partial{x}} f(x,y) =$$
2. Let $$f(x,y)= 3xy^2+2y$$. Write the following partial derivative:
$$\frac{\partial}{\partial{y}} f(x,y) =$$
3. Let $$A$$ be the matrix $$\begin{bmatrix}2 & 3 \\ 5 & 1\end{bmatrix}$$ and let $$\mathbf{x}$$ be the (column) vector $$\begin{bmatrix}1 \\ 2\end{bmatrix}$$. Let $$A^{T}$$ and $$\mathbf{x}^T$$ denote the transpose of $$A$$ and $$\mathbf{x}$$, respectively. Complete the following expressions:
$$A \mathbf{x} = \begin{bmatrix}2 & 3 \\ 5 & 1\end{bmatrix} \cdot \begin{bmatrix}1 \\ 2\end{bmatrix} =$$
$$A^T =$$
$$\mathbf{x}^T A =$$
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# Difference between revisions of "2009 AIME I Problems/Problem 6"
## Problem
How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$? (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$.)
## Solution
First, $x$ must be less than $5$, since otherwise $x^{\lfloor x\rfloor}$ would be at least $3125$ which is greater than $1000$.
Now, ${\lfloor x\rfloor}$ must be an integer, so lets do case work:
For ${\lfloor x\rfloor}=0$, $N=1$ no matter what $x$ is
For ${\lfloor x\rfloor}=1$, $N$ can be anything between $1^1$ to $2^1$ excluding $2^1$
This gives us $2^1-1^1=1$ $N$'s
For ${\lfloor x\rfloor}=2$, $N$ can be anything between $2^2$ to $3^2$ excluding $3^2$
This gives us $3^2-2^2=5$ $N$'s
For ${\lfloor x\rfloor}=3$, $N$ can be anything between $3^3$ to $4^3$ excluding $4^3$
This gives us $4^3-3^3=37$ $N$'s
For ${\lfloor x\rfloor}=4$, $N$ can be anything between $4^4$ to $5^4$ excluding $5^4$
This gives us $5^4-4^4=369$ $N$'s
Since $x$ must be less than $5$, we can stop here and the answer answer is $1+5+37+369= \boxed {412}$.
|
# Calculus 2 : L'Hospital's Rule
## Example Questions
### Example Question #31 : L'hospital's Rule
Explanation:
If you plug in the limit value, the function turns into . Therefore, we are allowed to use l'Hospital's Rule. We start by taking the derivative of both the numerator and the denominator until when you plug in the value of the limit, you do not get something in the form . Luckily, in this case, we only need to take the derivative once. By taking the derivatives separately, we get a new limit:
.
### Evaluate:
Limit Does Not Exist
Explanation:
### Example Question #33 : L'hospital's Rule
Find the limit using L'Hospital's Rule.
Explanation:
We rewrite the limit as
Substituting yields the indeterminate form
L'Hospital's Rule states that when the limit is in indeterminate form, the limit becomes
For and we solve the limit
and substituting we find that
As such
### Example Question #34 : L'hospital's Rule
Find the limit using L'Hospital's Rule.
Explanation:
Substituting yields the indeterminate form
L'Hospital's Rule states that when the limit is in indeterminate form, the limit becomes
For and we solve the limit
and substituting we find that
As such
### Example Question #35 : L'hospital's Rule
Evaluate the following limit:
Explanation:
Simply substituting in the given limit will not work:
Because direct substitution yields an indeterminate result, we must apply L'Hospital's rule to the limit:
if and only if and both and exist at .
Here,
and .
Hence,
### Example Question #36 : L'hospital's Rule
Evaluate the following limit:
Explanation:
When evaluating the limit using normal methods (substitution), we receive the indeterminate form . When we receive the indeterminate form, we must use L'Hopital's Rule to evaluate the limit. The rule states that
Using the formula above for our limit, we get
The derivatives were found using the following rules:
### Example Question #37 : L'hospital's Rule
Evaluate the limit:
Explanation:
When evaluating the limit using normal methods (substitution), we get the indeterminate form . When this happens, to evaluate the limit we use L'Hopital's Rule, which states that
Using the above formula for our limit, we get
The derivatives were found using the following rule:
### Example Question #38 : L'hospital's Rule
Evaluate the following limit, if possible:
The limit does not exist.
Explanation:
To calculate the limit we first plug the limit value into the numerator and denominator of the expression. When we do this we get , which is undefined. We now use L'Hopital's rule which says that if and are differentiable and
,
then
.
We are evaluating the limit
In this case we have
and
.
We differentiate both functions and find
and
By L'Hopital's rule
.
When we plug the limit value of 2 into this expression we get 9/3, which simplifies to 3.
### Example Question #39 : L'hospital's Rule
Evaluate the following limit, if possible:
.
The limit does not exist.
Explanation:
If we plugged in the limit value, , directly we would get the indeterminate value . We now use L'Hopital's rule which says that if and are differentiable and
,
then
.
The limit we wish to evaluate is
,
so in this case
and
.
We calculate the derivatives of both of these functions and find that
and
.
Thus
.
When we plug the limit value, , into this expression we get , which is .
### Example Question #40 : L'hospital's Rule
Evaluate the following limit, if possible:
.
The limit does not exist
The limit does not exist
Explanation:
We will show that the limit does not exist by showing that the limits from the left and right are different.
We will start with the limit from the right. Using the product rule we rewrite the limit
.
We know that
and
so
.
We calculate the limit from the left in the same way and find
.
Thus the two-sided limit does not exist.
|
# Limits and Infinites
Limits are a concept that goes back at least as far as Archimedes, a Greek scientist, engineer and mathematician in the 3rd century BCE. He calculated a value for π based on the limit of measurements external and internal to a circle.
#### Rules for limits
If \${lim}↙{n→+∞} a_n = a\$, and \${lim}↙{n→+∞} b_n = b\$
Then, (i) \${lim}↙{n→+∞} a_n ± b_n = a ± b\$
(ii) \${lim}↙{n→+∞} a_n ⋅ b_n = a ⋅ b\$
(iii) \${lim}↙{n→+∞} ({a_n}/{b_n}) = a/b\$
If \${lim}↙{n→+∞} a_n = a\$, ⇒ \${lim}↙{n→+∞} 1/{a_n} = 1/a\$
If \${lim}↙{n→+∞} a_n = +∞\$, \${lim}↙{n→+∞} b_n = +∞\$ ⇒ \${lim}↙{n→+∞} (a_n + b_n) = +∞\$
If \${lim}↙{n→+∞} a_n = +∞\$, \${lim}↙{n→+∞} b_n = +∞\$ ⇒ \${lim}↙{n→+∞} (a_n - b_n) = ∞ - ∞\$: i.e. indeterminate solution
\${lim}↙{n→+∞} 1/{n^a} = 0\$ (a > 0)
\${lim}↙{n→+∞} a^n = 0\$ (|a| < 1)
\${lim}↙{n→+∞} ^n√{a} = 1\$ (a > 0)
\${lim}↙{n→+∞} ^n√{n} = 1\$
\${lim}↙{n→+∞} {(log n)^b}/{n^a} = 0\$ (a > 0, b ∈ ℝ)
\${lim}↙{n→+∞} {n^b}/{a^n} = 0\$ (b > 0, |a| > 1)
\${lim}↙{n→+∞} {a^n}/{n!} = 0\$ (a ∈ ℝ)
\${lim}↙{n→+∞} {n!}/{n^n} = 0\$
\${lim}↙{n→+∞} (1 + 1/n)^n = e\$
\${lim}↙{n→+∞} (1 + a/n)^{b⋅n} = e^{a⋅b}\$, (a, b ∈ ℝ)
\${lim}↙{x→±∞} (f(x) ± g(x)) = {lim}↙{x→±∞} f(x) ± {lim}↙{x→±∞} g(x) = L_1 ± L_2\$
\${lim}↙{x→±∞} (f(x) ⋅ g(x)) = {lim}↙{x→±∞} f(x) ⋅ {lim}↙{x→±∞} g(x) = L_1 ⋅ L_2\$
\${lim}↙{x→±∞} (f(x) ÷ g(x)) = {lim}↙{x→±∞} f(x) ÷ {lim}↙{x→±∞} g(x) = L_1 ÷ L_2\$, where \$L_2 ≠ 2\$
\${lim}↙{x→±∞} kf(x) = k {lim}↙{x→±∞} f(x) = kL_1\$
\${lim}↙{x→±∞} [f(x)]^{a/b} = L_1^{a/b}, a/b ∈ ℚ\$, provided \$L_1^{a/b} \$ is real
### Convergence of a Series
The sum of a finite geometric series is: \$S_n = {u_1(1-r^n)}/{1-r}\$, where \$r\$ is the common ratio of two consecutive terms, and \$n\$ is the number of terms \$u\$.
For a geometric series, \${Σ}↙{n=0}↖{∞} = {lim}↙{n→∞} {u_1(1-r^n)}/{1-r}\$.
When \$-1 < r < 1\$, \${lim}↙{n→∞} r^n = 0\$, and the series converges to its sum, \$S={u_1}/{1-r}\$.
## Site Index
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### Great Scientists
#### Georges Lemaître
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Georges Lemaître, 1894 - 1966, was a Belgian astrophysicist, best known for being the originator of the Expanding Universe theory, and the Big Bang origin of the universe.
### Quote of the day...
Newton was not the first of the age of reason: He was the last of the magicians.
|
## Thinking Mathematically (6th Edition)
8(y + 2) = 2(3y + 4) Step 1 : Use distributive property 8.y + 8.2 = 2.3y + 2.4 simplify 8y + 16 = 6y + 8 Step 2 : Collect variable terms on one side and constants on the other side. subtract 6y from both the sides 8y + 16 - 6y = 6y + 8 -6y 2y + 16 = 8 subtract 16 from both the sides 2y+ 16 - 16 = 8 -16 2y = -8 Divide both the sides by 2 $\frac{2y}{2}$ = $\frac{-8}{2}$ y = -4 Now we check the proposed solution, -4 , by replacing y with -4 in the original equation. Step 1: the original equation 8(y + 2) = 2(3y + 4) Step2: Substitute -4 for for 8(-4 + 2) = 2(3(-4) + 4) Step 3: Multiply 3(-4) = -12 8(-4 + 2) = 2(-12 + 4) Step 4: Solve 8(-2) = 2(-8) Multiply 8(-2) = -16, 2(-8) = -16 -16 =-16 Since the check results in true statement, we conclude that the solution set of the given equation is {-4}
|
## SEQUENCE AND SERIES QUIZ-6
As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.
Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
Q1. If (x+y)/(1-xy),y,(y+z)/(1-yz) be in A.P., then x,1/y,z will be in
• A.P
• G.P
• H.P
• None of these
Solution
It is given that (x+y)/(1-xy),y,(y+z)/(1-yz) are in A.P. ⇒y-(x+y)/(1-xy)=(y+z)/(1-yz)-y ⇒(y-xy^2-x-y)/(1-xy)=(y+z-y+y^2 z)/(1-yz) ⇒-x/(1-xy)=z/(1-yz) ⇒-x+xyz=z-xyz ⇒2 xyz=x+z ⇒y=(x+z)/(2 xz) ⇒1/y=(2 xz)/(x+z) ⇒x,1/y,z are in H.P.
Q2.The interior angles of a polygon are in AP. If the smallest angle be 120° and the common difference be 5, then the number of side is
• 8
• 10
• 9
• 6
Solution
Let the number of sides of the polygon be n.Then, the sum of interior angles of the polygon =(2n-4) Ï€/4=(n-2)Ï€ Since, the angles are in AP and a=120°,d=5 Therefore, S_n=n/2[2a+(n-1)d] ⇒n/2 [2×120+(n-1)5]=(n-2)180 ⇒n^2-25n+144=0 ⇒(n-9)(n-16)=0 ⇒n=9,16 Take n=16 T_16=a+15d=120°+15(5°)=195°, which is impossible, an interior angle cannot be greater than 180°. Hence, n=9
Q3. If S=1/(1∙2)-1/(2∙3)+1/(3∙4)-1/(4∙5)+⋯+∞, then e^S equals
• log_e(4/e)
• 4/e
• log_e(e/4)
• e/4
Solution
We have, S=1/(1∙2)-1/(2∙3)+1/(3∙4)-1/(4∙5)+⋯∞ ⇒S=(1/1-1/2)-(1/2-1/3)+(1/3-1/4)-(1/4-1/5)+⋯∞ ⇒S=2(1-1/2+1/3-1/4+⋯)-1 ⇒S=2 log〖(1+1)-log_e〖e=log_e(4/e) 〗 〗 ∴e^S=4/e
Q4. If log_8{log_2(log_3(x^2-4x+85) ) }=1/3, then x equals to
• 5
• 4
• 3
• 2
Solution
We have, log_8〖[log_2{log_3(x^2-4x+85) } ]=1/3〗 ⇒log_2〖{log_3(x^2-4x+85) }=8^(1/3) 〗=2 ⇒log_3〖(x^2-4x+85)=2^2 〗 ⇒x^2-4x+85=3^4 ⇒x^2-4x+4=0⇒(x-2)^2=0⇒x=2
Q5.If 1/1^2 +1/2^2 +1/3^2 +⋯to ∞=Ï€^2/6, then 1/1^2 +1/3^2 +1/5^2 +⋯ equals
• Ï€^2/8
• Ï€^2/12
• Ï€^2/3
• Ï€^2/2
Solution
We have, 1/1^2 +1/3^2 +1/5^2 +1/7^2 +⋯ =(1/1^2 +1/2^2 +1/3^2 +1/4^2 +1/5^2 +1/6^2 +1/7^2 +⋯)-(1/2^2 +1/4^2 +1/6^2 +⋯) Ï€^2/6-1/4 (1/1^2 +1/2^2 +1/3^2 +⋯)=Ï€^2/6-1/4 (Ï€^2/6)=Ï€^2/8
Q6. The product of n positive numbers is unity. Their sum is
• A positive integer
• Equal to n+(1/n)
• Divisible by n
• Never less than n
Solution
Given that, x_1 x_2 x_3…x_n=1 …(i) We know that, AM ≥ GM ∴ ((x_1+x_2+x_3+...+x_n)/n)≥(x_1 x_2 x_3…x_n )^(1/n) =(1)^(1/n)=1 [from Eq.(i)] ⇒ x_1+x_2+x_3+...+x_n≥n ∴ x_1+x_2+x_3+...+x_n Can never be less than n
Q7.Consider the following statement : 1. If mth term of HP is n and the nthe term is m, then the (mn)th term is 1. 2. If a,b,c are in AP and a,2b,c are in GP, then a,4b,c are in HP. 3. If any odd number of quantities are in AP, then first middle and last are in AP Which of the statement give above is/are correct?
• Only(1)
• Only(2)
• Only(3)
• All of these
Solution
Q8.If x,y,z are in HP, then log〖(x+z)〗+log〖(x-2y+z)〗 is equal to
• log(x-z)
• 2log(x-z)
• 3log(x-z)
• 4log(x-z)
Solution
Since, y=2xz/(x+z) Now, x-2y+z=x+z-2(2xz/(x+z)) =x+z-4xz/(x+z)=(x-z)^2/(x+z) ⇒log〖(x-2y+z)〗=log〖(x-z)^2 〗-log〖(x+z)〗 ⇒log〖(x-2y+z)〗+log〖(x+z)〗=2 log〖(x-z)〗
Q9.Number of values of x for which [x], sgn x,{x}{x≠0} are in AP, is
• 0
• 2
• 3
• <5 span="">
Solution
Q10. In a geometric progression (GP) the ratio of the sum of the first three terms and first six terms is 125:152 the common ratio is
• 1/5
• 2/5
• 4/5
• 3/5
Solution
Since, (a+ar+ar^2)/(a+ar+ar^2+ar^3+ar^4+ar^5 )=125/162 ⇒(1+r+r^2)/(1+r+r^2 )(1+r^3 ) =125/162 ⇒1+r^3=152/125 ⇒r^3=27/125=(3/5)^3 ⇒ r=3/5
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# If A and B are invertible matrices,then which of the following is not correct?$\begin{array}{1 1}(A)\quad adjA=|A|.A^{-1} & (B)\quad det(A)^{-1}=[det(A)]^{-1}\\(C)\quad (AB)^{-1}=B^{-1}A^{-1} & (D)\quad (A+B)^{-1}=B^{-1}+A^{-1}\end{array}$
Toolbox:
• If $A^{-1}=\frac{1}{|A|}adj( A)$
• $A(adj A)=|A|I=(adj A)A$
• $(AB)^{-1}=B^{-1}A^{-1}$
Given A and B are invertible matrices.
Then $(A+B)^{-1}=B^{-}+A^{-1}$ is not correct.
For example let us consider
$A=\begin{bmatrix}1 & 2\\0 & 1\end{bmatrix}$ and $\begin{bmatrix}2 & 1\\2 & 3\end{bmatrix}$
|A|=1-0=1 and |B|=6-2=4
$adj \;of A=\begin{bmatrix}1 &-2\\0 & 1\end{bmatrix}$ and $adj B=\begin{bmatrix}3 & -1\\-2 & 2\end{bmatrix}$
$A^{-1}=\frac{1}{1}\begin{bmatrix}1 &-2\\0 & 1\end{bmatrix}$ and $B^{-1}=\frac{1}{4}\begin{bmatrix}3 & -1\\-2 & 2\end{bmatrix}$
$A+B=\begin{bmatrix}1 &2\\0 & 1\end{bmatrix}+\begin{bmatrix}2 & 1\\2 & 3\end{bmatrix}$
$\;\;\;\;\;\;=\begin{bmatrix}3 & 3\\2 & 4\end{bmatrix}$
|A+B|=12-6=6.
$adj (A+B)=\begin{bmatrix}4 & -3\\-2 & 3\end{bmatrix}$
$[A+B]^{-1}=\frac{1}{6}\begin{bmatrix}4 & -3\\-2 & 3\end{bmatrix}=\begin{bmatrix}4/6 & -3/6\\-2/6 & 3/6\end{bmatrix}=\begin{bmatrix}2/3 & -1/2\\-1/3 & 1/2\end{bmatrix}$-----LHS
$B^{-1}+A^{-1}=\frac{1}{4}\begin{bmatrix}3 & -1\\-2 & 2\end{bmatrix}+\begin{bmatrix}1 & -2\\0 & 1\end{bmatrix}=\begin{bmatrix}3/4+1 & -1/4-2\\-2/4+0 & 2/4+1\end{bmatrix}=\begin{bmatrix}7/4 & -9/4\\-1/2 & 6/4\end{bmatrix}$------RHS
Hence RHS$\neq$ LHS.
Hence D is the correct answer.
|
Common Core: High School - Algebra : Solve Simple Rational and Radical One Variable Equations, Show Extraneous Solutions: CCSS.Math.Content.HSA-REI.A.2
Example Questions
← Previous 1
Example Question #1 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2
Solve for .
Explanation:
To solve for , square both terms to cancel out the square root sign on the left-hand side.
Next, add two to both sides of the equation.
_____________
From here, check for extraneous solutions by substituting the value found for into the original equation.
Since the square root of one is either positive or negative one, the solution found is verified.
Example Question #2 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2
Solve for .
Explanation:
To solve for , first subtract one from both sides.
____________________
From here, divide both sides by negative one.
Next, square both sides to cancel the square root sign on the left-hand side.
Now, subtract one from both sides to solve for .
____________
Lastly, check for extraneous solutions by substituting the value found for into the original equation.
Example Question #3 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2
Solve for .
Explanation:
To solve for , add two to both sides of the equation.
______________
Now, multiply by on both sides. This will move the variable from the denominator on one side to the numerator of the other side.
Lastly, divide both sides by twelve.
The twelve in the numerator and the twelve in the denominator cancel out thus, solving for .
Example Question #4 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2
Solve for .
Explanation:
To solve for , first subtract four from both sides so all constants are on one side.
_____________
Now, square both sides to cancel the square root sign on the left-hand side.
Recall that when a negative number is squared the result is always a positive value.
From here, check for extraneous solutions by substituting in the value found for .
Since the square root of a value results in a positive and a negative value, our solution is verified.
Example Question #5 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2
Solve for .
Explanation:
To solve for start by squaring both sides to eliminate the square root sign.
From here multiply both sides by two. On the left-hand side, the two in the numerator will cancel out the two in the denominator.
From here, check for extraneous solutions by substituting in the value found for into the original equation.
Thus, the solution is verified.
Example Question #6 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2
Solve for .
Explanation:
To solve for , first add two to both sides.
____________
From here, multiply both sides by
Now divide by seven to solve for .
The seven in the numerator and seven in the denominator cancel out, thus solving for .
Example Question #7 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2
Solve for .
Explanation:
To solve for , first square both sides of the equation. Squaring a square root sign will cancel them out.
Now, subtract five from both sides to get all constants on one side of the equation while all variables are on the other side of the equation.
_____________
From here, divide by negative one on both sides.
Lastly, check for extraneous solutions by substituting in the value found for into the original equation.
Thus, the solution is verified.
Example Question #8 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2
Solve for .
Explanation:
To solve for , simply multiply both sides of the equation by two.
The two in the numerator cancels the two in the denominator, thus solving for .
Example Question #9 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2
Solve for .
Explanation:
To solve for , first divide by two.
The two in the denominator cancels the two in the numerator.
From here, square both sides of the equation. Squaring a square root cancels it out.
To check for extraneous solutions, simply substitute into the original equation the value found for .
Example Question #10 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2
Solve for .
Explanation:
To solve for , add three to both sides of the equation.
______________
From here, take the square root of both sides. Taking the square root of a squared term eliminates the square.
Next, check for extraneous solutions. Substitute each potential solution into the original equation to verify if it results in a legal solution.
Substituting in positive four.
Substituting in negative four.
Therefore, both answers are verified solutions.
← Previous 1
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# Rational Expressions And Equations Assignment Help
## Introduction to Rational Expressions:
A rational expression is any expression or function which includes a polynomial in its numerator and denominator or which contains fractions of polynomials. A rational expression is the ratio of two polynomials. The following are the examples of rational expressions.
5/x, (5y+1)/(y2+4y+3), a/(a+10)
## Addition and subtraction of rational expression:
Let p, q and r represent polynomials where Then
1. p/q + r/q = (p+r)/q
2. p/q - r/q = (p-r)/q
## Steps to Add or Subtract Rational Expressions
1. Factor the denominator of each rational expression.
2. Identify the LCD.
3. Rewrite each rational expression as an equivalent expression with the LCD as its denominator.
4. Add or subtract the numerators, and write the result over the common denominator.
5. Simplify if possible.
Example: 4/3x + 5/3x
= (4+5)/3x
= 9/3x
= 3/x
### Definition of a Rational Equation:
An equation with one or more rational expressions is called a rational equation. The following equations are rational equations:
1/2 x + 1/3 = 1/4 x
3/5 + 1/x = 2/3
### Steps for solving the Rational Equation:
Step 1: Simplify by removing the fractions.
Step 2: Solve the remaining equation.
Step 3: Check for extraneous solutions.
Example 1: Solve 8/y - 1/3 = 5/y for y.
Step 1: Simplify by removing the fractions.
Step 2: Solve the remaining equation.
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# What Is.04 As a Fraction?
If you’re wondering what 0.04 as a fraction is, you’re not alone. This number can be tricky to work with, but we can help. Read on to learn more about this fraction and how to work with it.
Checkout this video:
## Introduction
.04 As a fraction is equal to 4/100. The fraction is read as “four hundredths” or “four over one hundred.”
## What is 0.04 as a fraction?
0.04 can be written as a fraction in several ways. The easiest way to write 0.04 as a fraction is to write it as 4/100. You can also write 0.04 as 2/50 or 1/25. All of these fractions are equivalent.
### Convert 0.04 to a fraction
To convert 0.04 to a fraction, divide the numerator (4) by the denominator (100). The resulting fraction is 4/100, which can be simplified to 1/25.
### Simplify the fraction
To simplify this fraction, we need to find a number that can divide into both 4 and 100 evenly. The greatest number that does this is 25, so we divide both the numerator (4) and denominator (100) by 25 to get the following:
0.04 as a fraction = 4/100
We can further simplify this fraction by writing it as a percentage. To do this, we multiply both sides of the equation by 100:
4/100 = (4 x 100)/(100 x 100)
0.04 as a percentage = 400%
## Conclusion
To sum it up, .04 as a fraction is 4/100. This means that there are 4 parts out of a total of 100. You can also simplify this fraction by dividing both the numerator and denominator by 4, which gives you the fraction 1/25.
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## 5.2 The Standard Normal Distribution
### Learning Objectives
1. To learn what a standard normal random variable is.
2. To learn how to use Figure 12.2 "Cumulative Normal Probability" to compute probabilities related to a standard normal random variable.
### Definition
A standard normal random variableThe normal random variable with mean 0 and standard deviation 1. is a normally distributed random variable with mean μ = 0 and standard deviation σ = 1. It will always be denoted by the letter Z.
The density function for a standard normal random variable is shown in Figure 5.9 "Density Curve for a Standard Normal Random Variable".
Figure 5.9 Density Curve for a Standard Normal Random Variable
To compute probabilities for Z we will not work with its density function directly but instead read probabilities out of Figure 12.2 "Cumulative Normal Probability" in Chapter 12 "Appendix". The tables are tables of cumulative probabilities; their entries are probabilities of the form $P(Z The use of the tables will be explained by the following series of examples.
### Example 4
Find the probabilities indicated, where as always Z denotes a standard normal random variable.
1. P(Z < 1.48).
2. P(Z< −0.25).
Solution:
1. Figure 5.10 "Computing Probabilities Using the Cumulative Table" shows how this probability is read directly from the table without any computation required. The digits in the ones and tenths places of 1.48, namely 1.4, are used to select the appropriate row of the table; the hundredths part of 1.48, namely 0.08, is used to select the appropriate column of the table. The four decimal place number in the interior of the table that lies in the intersection of the row and column selected, 0.9306, is the probability sought: $P(Z<1.48)=0.9306.$
Figure 5.10 Computing Probabilities Using the Cumulative Table
1. The minus sign in −0.25 makes no difference in the procedure; the table is used in exactly the same way as in part (a): the probability sought is the number that is in the intersection of the row with heading −0.2 and the column with heading 0.05, the number 0.4013. Thus P(Z < −0.25) = 0.4013.
### Example 5
Find the probabilities indicated.
1. P(Z > 1.60).
2. P(Z > −1.02).
Solution:
1. Because the events Z > 1.60 and Z ≤ 1.60 are complements, the Probability Rule for Complements implies that
$P(Z>1.60)=1−P(Z≤1.60)$
Since inclusion of the endpoint makes no difference for the continuous random variable Z, $P(Z≤1.60)=P(Z<1.60)$, which we know how to find from the table. The number in the row with heading 1.6 and in the column with heading 0.00 is 0.9452. Thus $P(Z<1.60)=0.9452$ so
$P(Z>1.60)=1−P(Z≤1.60)=1−0.9452=0.0548$
Figure 5.11 "Computing a Probability for a Right Half-Line" illustrates the ideas geometrically. Since the total area under the curve is 1 and the area of the region to the left of 1.60 is (from the table) 0.9452, the area of the region to the right of 1.60 must be $1−0.9452=0.0548.$
Figure 5.11 Computing a Probability for a Right Half-Line
1. The minus sign in −1.02 makes no difference in the procedure; the table is used in exactly the same way as in part (a). The number in the intersection of the row with heading −1.0 and the column with heading 0.02 is 0.1539. This means that $P(Z<−1.02)=P(Z≤−1.02)=0.1539$, hence
$P(Z>−1.02)=1−P(Z≤−1.02)=1−0.1539=0.8461$
### Example 6
Find the probabilities indicated.
1. $P(0.5
2. $P(−2.55
Solution:
1. Figure 5.12 "Computing a Probability for an Interval of Finite Length" illustrates the ideas involved for intervals of this type. First look up the areas in the table that correspond to the numbers 0.5 (which we think of as 0.50 to use the table) and 1.57. We obtain 0.6915 and 0.9418, respectively. From the figure it is apparent that we must take the difference of these two numbers to obtain the probability desired. In symbols,
$P(0.5
Figure 5.12 Computing a Probability for an Interval of Finite Length
1. The procedure for finding the probability that Z takes a value in a finite interval whose endpoints have opposite signs is exactly the same procedure used in part (a), and is illustrated in Figure 5.13 "Computing a Probability for an Interval of Finite Length". In symbols the computation is
$P(−2.55
Figure 5.13 Computing a Probability for an Interval of Finite Length
The next example shows what to do if the value of Z that we want to look up in the table is not present there.
### Example 7
Find the probabilities indicated.
1. $P(1.13
2. $P(−5.22
Solution:
1. We attempt to compute the probability exactly as in Note 5.20 "Example 6" by looking up the numbers 1.13 and 4.16 in the table. We obtain the value 0.8708 for the area of the region under the density curve to left of 1.13 without any problem, but when we go to look up the number 4.16 in the table, it is not there. We can see from the last row of numbers in the table that the area to the left of 4.16 must be so close to 1 that to four decimal places it rounds to 1.0000. Therefore
$P(1.13
2. Similarly, here we can read directly from the table that the area under the density curve and to the left of 2.15 is 0.9842, but −5.22 is too far to the left on the number line to be in the table. We can see from the first line of the table that the area to the left of −5.22 must be so close to 0 that to four decimal places it rounds to 0.0000. Therefore
$P(−5.22
The final example of this section explains the origin of the proportions given in the Empirical Rule.
### Example 8
Find the probabilities indicated.
1. $P(−1
2. $P(−2
3. $P(−3
Solution:
1. Using the table as was done in Note 5.20 "Example 6"(b) we obtain
$P(−1
Since Z has mean 0 and standard deviation 1, for Z to take a value between −1 and 1 means that Z takes a value that is within one standard deviation of the mean. Our computation shows that the probability that this happens is about 0.68, the proportion given by the Empirical Rule for histograms that are mound shaped and symmetrical, like the bell curve.
2. Using the table in the same way,
$P(−2
This corresponds to the proportion 0.95 for data within two standard deviations of the mean.
3. Similarly,
$P(−3
which corresponds to the proportion 0.997 for data within three standard deviations of the mean.
### Key Takeaways
• A standard normal random variable Z is a normally distributed random variable with mean μ = 0 and standard deviation σ = 1.
• Probabilities for a standard normal random variable are computed using Figure 12.2 "Cumulative Normal Probability".
### Exercises
1. Use Figure 12.2 "Cumulative Normal Probability" to find the probability indicated.
1. P(Z < −1.72)
2. P(Z < 2.05)
3. P(Z < 0)
4. P(Z > −2.11)
5. P(Z > 1.63)
6. P(Z > 2.36)
2. Use Figure 12.2 "Cumulative Normal Probability" to find the probability indicated.
1. P(Z < −1.17)
2. P(Z < −0.05)
3. P(Z < 0.66)
4. P(Z > −2.43)
5. P(Z > −1.00)
6. P(Z > 2.19)
3. Use Figure 12.2 "Cumulative Normal Probability" to find the probability indicated.
1. P(−2.15 < Z < −1.09)
2. P(−0.93 < Z < 0.55)
3. P(0.68 < Z < 2.11)
4. Use Figure 12.2 "Cumulative Normal Probability" to find the probability indicated.
1. P(−1.99 < Z < −1.03)
2. P(−0.87 < Z < 1.58)
3. P(0.33 < Z < 0.96)
5. Use Figure 12.2 "Cumulative Normal Probability" to find the probability indicated.
1. P(−4.22 < Z < −1.39)
2. P(−1.37 < Z < 5.11)
3. P(Z < −4.31)
4. P(Z < 5.02)
6. Use Figure 12.2 "Cumulative Normal Probability" to find the probability indicated.
1. P(Z > −5.31)
2. P(−4.08 < Z < 0.58)
3. P(Z < −6.16)
4. P(−0.51 < Z < 5.63)
7. Use Figure 12.2 "Cumulative Normal Probability" to find the first probability listed. Find the second probability without referring to the table, but using the symmetry of the standard normal density curve instead. Sketch the density curve with relevant regions shaded to illustrate the computation.
1. P(Z < −1.08), P(Z > 1.08)
2. P(Z < −0.36), P(Z > 0.36)
3. P(Z < 1.25), P(Z > −1.25)
4. P(Z < 2.03), P(Z > −2.03)
8. Use Figure 12.2 "Cumulative Normal Probability" to find the first probability listed. Find the second probability without referring to the table, but using the symmetry of the standard normal density curve instead. Sketch the density curve with relevant regions shaded to illustrate the computation.
1. P(Z < −2.11), P(Z > 2.11)
2. P(Z < −0.88), P(Z > 0.88)
3. P(Z < 2.44), P(Z > −2.44)
4. P(Z < 3.07), P(Z > −3.07)
9. The probability that a standard normal random variable Z takes a value in the union of intervals (−∞, −a] ∪ [a, ∞), which arises in applications, will be denoted P(Z ≤ −a or Za). Use Figure 12.2 "Cumulative Normal Probability" to find the following probabilities of this type. Sketch the density curve with relevant regions shaded to illustrate the computation. Because of the symmetry of the standard normal density curve you need to use Figure 12.2 "Cumulative Normal Probability" only one time for each part.
1. P(Z < −1.29 or Z > 1.29)
2. P(Z < −2.33 or Z > 2.33)
3. P(Z < −1.96 or Z > 1.96)
4. P(Z < −3.09 or Z > 3.09)
10. The probability that a standard normal random variable Z takes a value in the union of intervals (−∞, −a] ∪ [a, ∞), which arises in applications, will be denoted P(Z ≤ −a or Za). Use Figure 12.2 "Cumulative Normal Probability" to find the following probabilities of this type. Sketch the density curve with relevant regions shaded to illustrate the computation. Because of the symmetry of the standard normal density curve you need to use Figure 12.2 "Cumulative Normal Probability" only one time for each part.
1. P(Z < −2.58 or Z > 2.58)
2. P(Z < −2.81 or Z > 2.81)
3. P(Z < −1.65 or Z > 1.65)
4. P(Z < −2.43 or Z > 2.43)
1. 0.0427
2. 0.9798
3. 0.5
4. 0.9826
5. 0.0516
6. 0.0091
1. 0.1221
2. 0.5326
3. 0.2309
1. 0.0823
2. 0.9147
3. 0.0000
4. 1.0000
1. 0.1401, 0.1401
2. 0.3594, 0.3594
3. 0.8944, 0.8944
4. 0.9788, 0.9788
1. 0.1970
2. 0.01980
3. 0.0500
4. 0.0020
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Decimal to fraction conversion table
Decimal Fraction
0.3 3/10
0.33333333 1/3
0.375 3/8
0.4 2/5
Also, What is 3/5 in the lowest terms?
35 is already in the simplest form. It can be written as
0.6
in decimal form (rounded to 6 decimal places).
Steps to simplifying fractions
• Find the GCD (or HCF) of numerator and denominator. GCD of 3 and 5 is 1.
• 3 ÷ 15 ÷ 1.
• Reduced fraction: 35. Therefore, 3/5 simplified to lowest terms is 3/5.
Hereof, What is 3/8 as a decimal?
Answer: 3/8 as a decimal is 0.375.
Also to know What is 3 as a decimal? Percent to decimal conversion table
Percent Decimal
2% 0.02
3%
0.03
4% 0.04
5% 0.05
What is 3/4 as a decimal?
Answer: 3/4 is expressed as 0.75 in the decimal form.
## What is 3 and 3/5 as a decimal?
So the answer is that 3 3/5 as a decimal is 3.6.
## What is 3/5 into a decimal?
So, 3/5 as a decimal is 0.6.
## Can you simplify 5 8?
As you can see, 5/8 cannot be simplified any further, so the result is the same as we started with.
## What is 3/8 as a fraction?
Decimal and Fraction Conversion Chart
Fraction Equivalent Fractions
6/7 12/14 24/28
1/8 2/16 4/32
3/8 6/16
12/32
5/8 10/16 20/32
## How do you write 5/8 as a fraction?
5/8 = 58 = 0.625.
## What is 9 and 3/4 as a decimal?
So the answer is that 9 3/4 as a decimal is 9.75.
## What is 1 and 2/3 as a decimal?
So the answer is that 1 2/3 as a decimal is 1.6666666666667.
## What is 1/3 and 3 as a decimal?
3. 3 1/3 in decimal form = [katex]3. dot{3}[/katex] Three and one third as a decimal = 3.
## What is 1 and 3/4 as a decimal?
Method 1: Writing 1 3/4 to a decimal using the division method. To convert any fraction to decimal form, we just need to divide its numerator by denominator. This gives the answer as 1.75. So, 1 3/4 to decimal is 1.75.
## What is 3/5 as a percentage?
Answer: 3/5 is expressed as 60% in terms of percentage.
## What is 3 and 3/4 as a decimal?
So the decimal section for this is 0.75.
## What is 3 and 3/8 as a decimal?
This gives the answer as 3.375. So, 3 3/8 as a decimal is 3.375.
## What is 3 and 1/3 as a decimal?
So the answer is that 3 1/3 as a decimal is 3.3333333333333.
## What is 3 2 as a decimal?
Answer: 3/2 as a decimal is expressed as 1.5.
## What is 1 whole and 3/5 as a decimal?
So the answer is that 1 3/5 as a decimal is 1.6.
## What is 5/8 in the lowest terms?
58 is already in the simplest form. It can be written as
0.625
in decimal form (rounded to 6 decimal places).
Steps to simplifying fractions
• Find the GCD (or HCF) of numerator and denominator. GCD of 5 and 8 is 1.
• 5 ÷ 18 ÷ 1.
• Reduced fraction: 58. Therefore, 5/8 simplified to lowest terms is 5/8.
## What is 5/8 as a percentage?
Answer: 5 out of 8 can be expressed as 62.5%.
To convert a fraction to a percentage, we multiply the given fraction by 100 and add a % symbol to it.
## What is 5/8 as a fraction?
Decimal and Fraction Conversion Chart
Fraction Equivalent Fractions Decimal
6/7 12/14 .857
1/8 2/16 .125
3/8 6/16 .375
5/8
10/16
.625
## What is 3/8 in a percent?
To convert to percentage
To convert the fraction to decimal first convert to decimal and then multiply by 100. Therefore, the solution is 37.5%.
## What is another way to write 3 8?
Another common way of writing and saying 3/8 is: three over eight.
## How do you solve 8 divided by 3?
The number 8 divided by 3 is 2 with a remainder of 2 (8 / 3 = 2 R.
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# How do you translate into mathematical expressions and find the number given The product of three more than twice a number and two is fourteen?
Apr 17, 2017
$2 \left(2 x + 3\right) = 14$
$x = 2$
#### Explanation:
Let the number be represented by $x$.
Start by looking for the equals sign. In this case, the word "is" represents the equals sign. Whatever is to the left of the word "is" is the left side of the equation and whatever is to the right of the word "is" is the right side of the equation. Since the right side is just "fourteen", let's put that in and write down what we have so far.
$= 14$
It doesn't mean much but we're getting there. Now we need our left side.
Continue by looking as close to the variable as possible. The problem asks for "twice a number", so let's add that to our equation.
$2 x = 14$
Work outwards. We need "three more than twice a number", so the equation becomes
$2 x + 3 = 14$
Finally we're left with the tricky word. When you see "the sum/product/difference/etc. of", we'll need parentheses. In general you should put parentheses around the whole "the sum/product/difference/etc. of" expression and around the things to the right and to the left of the word "and" within that expression, then remove them as necessary.
Our "three more than twice a number" is the left side of the "and" in a "the product of" clause and the right side is "two", so our equation becomes
$\left(\left(2 x + 3\right) \times \left(2\right)\right) = 14$
The parentheses around the entire left side can be removed as well as around just the number $2$, leaving
$2 \left(2 x + 3\right) = 14$.
Now it's time to solve for $x$. To do that you need to isolate it. Start by dividing both sides by $2$.
color(green)((color(red)cancelcolor(black)2color(black)((2x+3)))/color(red)cancelcolor(green)2)=color(green)(color(black)14/2
$2 x + 3 = 7$
Then subtract $3$ from both sides.
$2 x \textcolor{red}{\cancel{\textcolor{b l a c k}{+ 3}}} \textcolor{red}{\cancel{\textcolor{g r e e n}{- 3}}} = 7 \textcolor{g r e e n}{- 3}$
$2 x = 4$
Finally divide both sides by $2$ again.
$\textcolor{g r e e n}{\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \textcolor{b l a c k}{x}}{\textcolor{red}{\cancel{\textcolor{g r e e n}{2}}}}} = \textcolor{g r e e n}{\frac{\textcolor{b l a c k}{4}}{2}}$
$x = 2$
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# W ARM U P O CT. 1 ST Come in, get a tracking sheet from the basket, and quietly begin the warm up. 1. Write the ordered pair that describes a point 4 units.
## Presentation on theme: "W ARM U P O CT. 1 ST Come in, get a tracking sheet from the basket, and quietly begin the warm up. 1. Write the ordered pair that describes a point 4 units."— Presentation transcript:
W ARM U P O CT. 1 ST Come in, get a tracking sheet from the basket, and quietly begin the warm up. 1. Write the ordered pair that describes a point 4 units down from and 3 units to the right of the origin. 2. Write the ordered pair that is 8 units to the left of the origin and lies on the x-axis. For #3-6, identify the quadrant for each ordered pair. 3. ( -1, 7 )4. ( 9, 2 ) 5. ( 0.75, -4 )6. ( 0, 5 ) 7. Which property makes the following true? 3(x – 5) = 3x – 15
H OMEWORK A NSWERS – 4.1
H OMEWORK A NSWERS ( CONT.) – 4.1
H OMEWORK A NSWERS – 4.3
H OMEWORK A NSWERS ( CONT.) – 4.3
E QUATIONS A S R ELATIONS The equation y = 3x + 4 is an example of an equation in two variables because it contains x and y. A solution of an equation in 2 variables is an ordered pair. Only a certain set of ordered pairs work for this equation and make it true. What does it mean if an equation is true? It means that both sides have the same number! A REPLACEMENT SET is a set of ordered pairs that could replace x and y. We have to pick the ones that work! A SOLUTION SET is a set of ordered pairs that work for the relation.
S OLVE U SING A R EPLACEMENT S ET 1. Find the solution set for -3 = 2x – y, given the replacement set {(-2, -1), (-1, 3), (0, 4), (3, 9)}. Make a table. Substitute the x and y-values of each ordered pair into the equation. Solution Set XY-3 = 2x – yTrue or False?
H OW ABOUT THIS ONE ? 2. Find the solution set, given the replacement set 3x + 2y = 6{(2, 3), (0, 1), (0, 3), (2, 0)} Solution Set: XY3x + 2y = 6True or False?
C AN YOU GET THIS ONE ON YOUR OWN ? 3. Solve 2x – 3y = 5 given the replacement set {(0, 0), (1, -1), (4, 1), (-2, 3)}
R EMEMBER D OMAIN AND R ANGE FROM F RIDAY ? Domain = x Range = y Sometimes, we are given a domain for an equation and we need to find out the range. Note: Domain is also known as INPUT, and Range is also known as OUTPUT.
W HEN WE ARE GIVEN D OMAIN AND A SKED TO F IND R ANGE : Step 1: Solve the equation for y if necessary. (We want to know our y-values so we need to get it by itself!) Step 2: Create a table and fill it in! Step 3: Write your ordered pairs as a solution set.
S OLVE U SING A G IVEN D OMAIN Example: Solve 1 = 2 x – y if the domain is {-2, -1, 0, 2, 4}. Solution Set: xy(x, y)
H OW ABOUT THIS ONE ? Find the solution set, given the domain. 4x + 8 = 2yx = {-4, -2, 0, 2, 4}. Solution Set: {(-4, ), (-2, ), (0, ), (2, ), (4, )}
Y OU T RY ! 3. Solve y – 2 = x given the domain {-2, -1, 0, 1, 2}.
B UT WAIT …O N F RIDAY, WE SAID THAT THERE WERE 4 WAYS TO SHOW A RELATION What were they again? Ordered pairs Table Graph Mapping Graphs are the easiest way to visually represent a relation, so lets practice graphing our solution sets.
S OLVE AND G RAPH THE S OLUTION 1. Solve 4x + 2y = 12 if the domain is (-1, 0, 2, 4}. Graph the solution set.
H OW ABOUT THIS ONE ? Solve y – 3x = -2 given the domain {-2, -1, 0, 1, 2}. Graph the solution set.
Y OU T RY ! 1. Solve 2y = x + 2 given the domain {-4, -2, 0, 2, 4}. Graph the solution set.
N OW THAT WE KNOW WHAT WE RE DOING, LET S LEARN ANOTHER WAY TO FIND SOLUTIONS TO SOME OF THESE EQUATIONS … USING OUR CALCULATOR!
W ITH O UR C ALCULATOR ! Find the solution set, given the replacement set 4x + 2y = 8{(2, 3), (0, 1), (0, 3), (2, 0)} Solve for y. Calculator Steps: 1. 2. 3. 4. Solution Set:
W ITH O UR C ALCULATOR ! 2. Find the solution set, given the domain. 6x + 2 = 2yx = {-4, -2, 0, 2, 4}. Solve for y. Calculator Steps: 1. 2. 3. 4. Solution Set: {(-4, ), (-2, ), (0, ), (2, ), (4, )}
W ITH O UR C ALCULATOR ! Solve y – 4x = -5 given the domain {-2, -1, 0, 1, 2}. Graph the solution set. Solve for y. Calculator Steps: 1. 2. 3. 4. 5.
H OMEWORK Colossal Creature Worksheet
E XIT Q UIZ Complete on a half sheet of notebook paper. Be sure to PUT YOUR NAME AT THE TOP SO I CAN GIVE YOU CREDIT! 1. Write the solution set to the equation y = 2x + 3 if the domain is {-2, 0, 3} 2. Which of the following ordered pairs satisfy the equation y = 5 – 6x? Write them as a solution set. {(1, -1), (4, -3), (-1, 11), (-3, 23), (0, 15)} 3. Solve and graph the solution set to the following equation 6x – 3y = 9 given the domain { -2, -1, 0, 1, 2}
V OCABULARY SHEET Throughout the year, we are learning many important math vocabulary words. Its almost like learning a new language. These words are important for your: Tests Midterm Final End of Course Test ENTIRE MATH CAREER! So, to make sure we know them, I am giving you a handy-dandy vocabulary sheet to keep in the front of your binder at all times. Each unit, we will add new words to it.
L ET S GO BACK AND ADD SOME OLD WORDS Brainstorm: What should we add?
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Just like we have two pairs of opposite sides, we have two pairs of opposite angles. Conflict Between Antigone & Creon in Sophocles' Antigone, Quiz & Worksheet - Metaphors in The Outsiders, Quiz & Worksheet - Desiree's Baby Time & Place, Quiz & Worksheet - The Handkerchief in Othello. Angles ABC and CDA are congruent according to a property of parallelograms (opposite angles congruent). If you took a ruler and measured each pair, you will see that each pair is the same length. Opposite angles are congruent As you drag any vertex in the parallelogram above, note that the opposite angles are congruent (equal in measure). To learn more, visit our Earning Credit Page. If one pair of opposite sides of a quadrilateral are both parallel and congruent, the quadrilateral is a parallelogram. CCSS.MATH.CONTENT.HSG.CO.C.11 Prove theorems about parallelograms. The adjacent angles form supplementary angles that add up to 180 degrees. Anyone can earn Visually defined, a parallelogram looks like a leaning rectangle. In a parallelogram, the opposite sides and opposite angles are equal. Opposite sides of a parallelogram are equal; we can prove this using the alternate interior angles theorem. Whats people lookup in this blog: Alternate Interior Angles Theorem Parallelogram; Converse Of Alternate Interior Angles Theorem Parallelogram Many of these statements are the converse statements of the proofs we came up with already. Opposite angels are congruent (D = B). Theorem B: Consecutive angles are supplementary (their sum is 180°). Parallelogram Opposite Angles Theorem If a quadrilateral is a parallelogram, then its opposite angles are _____. Services. Solution: For part a, the opposite angles are equal, so by the Opposite Angles Theorem Converse, is a parallelogram. The first is that the opposite angles are equal to each other. Example 1 proves an additional way to show that a quadrilateral is a parallelogram. So we know that AC is parallel to BD by alternate interior angles. Theorem: Visual Representation: Write your questions here! One property that is included in the definition is that the opposite sides are parallel. ABCD is a parallelogram. The opposite internal angles of a parallelogram are equal and the adjacent angles are supplementary i.e., the sum of the adjacent angles should be equal to 180 degrees. A parallelogram is a quadrilateral with opposite sides parallel. For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. A parallelogram has two pairs of opposite sides that are parallel and equal in length. All rights reserved. As you can see, neither of these pairs of lines will ever meet each other. Calculate certain variables of a parallelogram depending on the inputs provided. This means that all three angles in both triangles have the same measure. You can use these and other theorems in this lesson to prove that a quadrilateral with certain properties is a parallelogram. 's' : ''}}. Theorem: Prove that the opposite angles of a parallelogram are equal. Note for example that the angles ∠ABD and ∠ACD are always equal no matter what you do. Therefore, the sum of any two adjacent angles of a parallelogram is equal to 180°. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. Then find the lengths of the two diagonals of the parallelogram. Opposite angles are congruent As you drag any vertex in the parallelogram above, note that the opposite angles are congruent (equal in measure). lessons in math, English, science, history, and more. Statements of parallelogram and its theorems 1) In a parallelogram, opposite sides are equal. By ASA congruence criterion, two triangles are congruent to each other. Angles ABC and CDA are corresponding parts of congruent triangles, which are congruent (CPCTC). Opposite angles of parallelogram are equal (D = B). Measuring the Area of a Parallelogram: Formula & Examples, Quiz & Worksheet - Properties and Proof Theorems of Parallelograms, Over 83,000 lessons in all major subjects, {{courseNav.course.mDynamicIntFields.lessonCount}}, Measuring the Area of a Rhombus: Formula & Examples, Rectangles: Definition, Properties & Construction, Measuring the Area of a Rectangle: Formula & Examples, Biological and Biomedical Our mathematical definition of a parallelogram includes two inherent properties. If the diagonals of a quadrilateral bisect each other then it is a parallelogram. the Parallelogram Opposite Angles Theorem to prove statements about the sides and angles of the parallelogram. The following is an incomplete flowchart proving that the opposite angles of parallelogram jklm are congruent: which statement and reason can be used to fill in the numbered blank spaces? What Can You Do With a PhD in Educational Psychology? To find s, theorem 14-A states that the opposite sides of a parallelogram are congruent. succeed. Write down a formula for the height of the parallelogram in terms of the cross-product, and deduce a formula for the area of the parallelogram. Visit us at - www.risingpearl.comLike us at - www.facebook.com/risingpearlfansFriends,This is a Math video. First day back from Christmas break saw my Geometry classes looking at theorems about parallelograms and rhombuses. Parallelograms have two properties related to their angles. - Definition, Properties, Types & Examples, Proving That a Quadrilateral is a Parallelogram, Simplifying Expressions with Rational Exponents, Isosceles Trapezoid: Definition, Properties & Formula, Angles of Elevation & Depression: Practice Problems, Properties of Shapes: Rectangles, Squares and Rhombuses, GRE Quantitative Reasoning: Study Guide & Test Prep, SAT Subject Test Mathematics Level 2: Practice and Study Guide, High School Geometry: Homework Help Resource, Ohio Graduation Test: Study Guide & Practice, SAT Subject Test Chemistry: Practice and Study Guide, SAT Subject Test Biology: Practice and Study Guide. 8) Theorem 6.3E states that if two congruent angles are _____, then each angle is a right angle. {{courseNav.course.mDynamicIntFields.lessonCount}} lessons courses that prepare you to earn So we know that AC is parallel to BD by alternate interior angles. Consider the following figure: Proof: In $$\Delta ABC$$ and $$\Delta CDA$$, \[\begin{align} Answers: 3 Show answers Another question on Mathematics. Select the correct answer and click on the “Finish” buttonCheck your score and answers at the end of the quiz, Visit BYJU’S for all Maths related queries and study materials, Your email address will not be published. If one angle is right, then all angles are right. Since, the adjacent sides are supplementary. The diagonals are the lines that connect the opposite corners to each other. We’d already looked at definitions of the different types of special quadrilaterals. It is a type of polygon having four sides (also called quadrilateral), where the pair of parallel sides are equal in length. How do you know? If the measure of angle A is 60, then what is the measure of angle C? The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. Theorem 6.2C states: If both pairs of opposite _____ of a quadrilateral are congruent, then the quadrilateral is a parallelogram. Try it for yourself. Prove: ∠ MLK ≅ ∠ KJM and ∠ JML ≅ ∠ LKJ. If you look at each pair of opposite sides and drew the lines out, you would find that these lines will never meet. Ways To Prove A Quadrilateral Is A Parallelogram Teaching The Lesson Teaching Quadrilaterals Lesson . Classify quadrilateral as parallelogram a classic activity. angles If an angle of a quadrilateral is supplementary to both of its _____ angles, then the quadrilateral is a parallelogram. Parallelograms have two properties related to their angles. The goal is to prove that the opposite angles are congruent. Each diagonal of a parallelogram separates it into two congruent triangles. One pair can be longer than the other. The diagonals of a parallelogram … Proof. Then the lengths of the two diagonals of, Draw a picture of a parallelogram with two sides a and b. Put your understanding of this concept to test by answering a few MCQs. Supplementary angles are two angles adding to 180°. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. Draw a generic parallelogram and preview the proof. Alternate Interior Angles Theorem Step-by-step explanation: Here, Given: JKML is a parallelogram, That is, JK ║ ML and JM ║ KL. Below, we can label one pair of angles with a and the other pair with b. Further, the following statements are all equivalent (if one is true, so are all the others): You can combine the top two, the bottom two, the left two, or even the right two. To show these two triangles are congruent we’ll use the fact that this is a parallelogram, and as a result, the two opposite sid… parallelogram theorem ; THEOREM – 1 A diagonal of parallelogram divides it into two triangles of equal area. Opposite Angles Theorem Converse: If both pairs of opposite angles of a quadrilateral are congruent, then the figure is a parallelogram. Create your account. Parallelogram Theorems 4 Log in here for access. We've shown if you have a parallelogram, opposite sides have the same length. A parallelogram is a quadrilateral with opposite sides parallel. When the angles of the parallelogram equal to 90 degrees, it forms a rectangle. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. Theorem 1. We know this is a parallelogram so the two opposite sides are parallel, and the diagonal acts as a transversal line, intersecting both pairs of parallel lines - hinting we should use the Alternate Interior Angles Theorem. See Interior angles of a polygon. For example, ∠A, ∠B are adjacent angles and ∠A = 90°, then: In the adjoining figure, ∠D = 85° and ∠B = (x+25)°, find the value of x. The diagonals bisect each other. Properties of a Parallelogram. Click ‘Start Quiz’ to begin! The point where they bisect is exactly the halfway point of each diagonal. Opposite Angles of a Parallelogram are Equal. © copyright 2003-2021 Study.com. Sciences, Culinary Arts and Personal A quadrilateral whose two pairs of sides are parallel to each and the four angles at the vertices are not equal to the right angle, then the quadrilateral is called a parallelogram. Th… the Parallelogram Opposite Angles Theorem to prove statements about the sides and angles of the parallelogram. The diagonals of a parallelogram bisect each other. The diagonals of a parallelogram bisect each other. A parallelogram however has some additional properties. I had students divide a page in their notebook in two, and told them to rewrite the definitions of the parallelogram and rhombus in those sections. (Proof of theorem appears further down page.) Solve for s, t, v, w, and x.Also determine the measure of angle LMN. The second is that a pair of adjacent angles will always add up to 180 degrees. Construct line BC, then construct a line parallel to BC through point A. credit-by-exam regardless of age or education level. imaginable degree, area of Name_____ Must pass MC by:_____ If a quadrilateral is a parallelogram, then its opposite sides are congruent. Parallelogram Diagonals Converse If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram. We don't have to check to make sure all the properties are there. Theorems of Quadrilateral Shapes 1. | {{course.flashcardSetCount}} In a parallelogram, any two consecutive angles are supplementary, no matter which pair you pick. The converses of the theorems are stated below. Not sure what college you want to attend yet? So this is parallel to that. 3.™1 £ ™3 3. If one angle is right, then all angles are right. Do they look like they will meet? Theorem 1: Opposite sides are congruent. AREA: Find a number \displaystyle a so that the change of variables \displaystyle s=x+ay,\ \ t=y transforms the integral \displaystyle \int\int_R \ dx\ dy over the parallelogram \displaystyle R in the \disp, Find the area of the parallelogram determined by the vectors \mathbf{v} and \mathbf{w} where \mathbf{v} = 2 \mathbf{i} + 3 \mathbf{k} and \mathbf{w} = 2 \mathbf{j} - 3 \mathbf{k}, Working Scholars® Bringing Tuition-Free College to the Community, You have two pairs of parallel opposite sides, You have two pairs of equal opposite angles, You have two pairs of equal and parallel opposite sides, It has two pairs of parallel opposite sides, It has two pairs of equal opposite angles, It has two pairs of equal and parallel opposite sides, Discuss the particulars of parallelograms, supplementary angles and proof theorems, Provide characteristics of the angles of parallelograms, List the criteria required to identify a parallelogram. Each diagonal of a parallelogram separates it into two congruent triangles. The important properties of angles of a parallelogram are: In the above parallelogram, A, C and B, D are a pair of opposite angles. It's as if a rectangle had a long, busy day and is now just resting and leaning up against a wall. Alternate interior angles parallelogram. - Definition and Properties, Parallelogram in Geometry: Definition, Shapes & Properties, Kites in Geometry: Definition and Properties, Solving Quadratic Inequalities in One Variable, Angle Bisector Theorem: Definition and Example, What is a Quadrilateral? So, if we went around clockwise starting from the top left angle, we would see a, b, a, and then b again. To get another theorem for parallelograms, let's prove that the opposite angles of a parallelogram are congruent. Select a subject to preview related courses: Now that we know the properties that make up a parallelogram, it becomes easy to identify a parallelogram. Also, the opposite sides are equal in length. Let’s use congruent triangles first because it requires less additional lines. If a quadrilateral is a parallelogram, then its Angles, Parallelogram This shows that the opposite sides of a parallelogram are always equal in length and the opposite angles are also equal. All other trademarks and copyrights are the property of their respective owners. A parallelogram also has two pairs of opposite angles that are also equal to each other. Parallelogram Diagonals Theorem If a quadrilateral is a parallelogram… Visit the Geometry: High School page to learn more. What this means is that a parallelogram has two pairs of opposite sides that are parallel to each other and are the same length. And if opposite sides have the same length, then you have a parallelogram. Opposite sides of parallelogram are equal (AB = DC). 2. One pair of opposite sides is both congruent and parallel. parallelogram theorems. A parallelogram however has some additional properties. Theorem 4. We know that interior angles on the same side of a transversal are supplementary. How Do I Use Study.com's Assign Lesson Feature? You will see the unique properties that belong to the parallelogram. {{courseNav.course.topics.length}} chapters | Write and… What is the Main Frame Story of The Canterbury Tales? As a member, you'll also get unlimited access to over 83,000 Example 2: Given .LMPN. Through point C, (C not on line AB) draw a second line, parallel to AB. Given 2.™1 £ ™2, ™2 £ ™3 2. 1. Watch this video lesson to learn how you can identify parallelograms. Opposite Sides Parallel and Congruent Theorem If ONE pair of opposite sides of a quadrilateral are _____ and _____, then the quadrilateral is a parallelogram. Now do the same for the left and right sides of the parallelogram. Did you know… We have over 220 college Print this page Prove theorems about parallelograms. Same side interior angles consecutive angles are supplementary. A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles. You can use these and other theorems in this lesson to prove that a quadrilateral with certain properties is a parallelogram. Parallelogram Theorems 2. Create an account to start this course today. You can combine any two angles that are next to each other in this way. 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Get the unbiased info you need to find the right school. Students: Use Video Games to Stay in Shape, YouCollege: Video Becomes the Next Big Thing in College Applications, Free Video Lecture Podcasts From Top Universities, Best Free Online Video Lectures: Study.com's People's Choice Award Winner, Biology Lesson Plans: Physiology, Mitosis, Metric System Video Lessons, OCW People's Choice Award Winner: Best Video Lectures, Video Production Assistant: Employment & Career Info, Associate of Film and Video: Degree Overview, Master of Fine Arts (MFA) Programs in Indiana, Best Online Master's in Public Relations Degrees. You can test out of the Opposite angles of a parallelogram are equal. Already registered? An error occurred trying to load this video. If one pair of opposite sides of a quadrilateral are both parallel and congruent, the quadrilateral is a parallelogram. Parallelograms are four-sided flat shapes whose opposite sides are both equal and parallel to each other. Hence, it is proved that any two adjacent or consecutive angles of a parallelogram are supplementary. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. The ASA postulate is most likely the only thing we can use to prove that the opposite sides of a parallelogram are congruent. and career path that can help you find the school that's right for you. 3. Theorem: Prove that any consecutive angles of a parallelogram are supplementary. Angles BAC and DCA are congruent by the Alternate Interior Angles Theorem. 2) If each pair of opposite sides of a quadrilateral is equal then it is a parallelogram. Transitive Property of Congruence flashcard set{{course.flashcardSetCoun > 1 ? These are called supplementary angles. And we're done. Alternate interior angles theorem you parallelograms opposite angles are congruent geometry help discussion section 1 3 discussion section 1 3. Example 2: Is quadrilateral a parallelogram? In a parallelogram, opposite angles are equal. 47 3 7 54 3 18 1 * 8 sKLNM s s s =− = = = = To find t, recall that the alternate interior angles of parallel lines are congruent. AEFGis a ⁄. Then ask the students to measure the angles sides etc. 3) In a parallelogram, opposite angles are equal. Quadrilateral having opposite angles equal is a parallelogram. However, this information is not enough to say that the triangles are congruent. 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Kindergarten-Grade 12 The diagonals of a parallelogram are not equal but they bisect each other at the midpoints. Suppose u = (1, 0) and v = (-1, 1) are two vectors that form the sides of a parallelogram. Definition of parallelogram 4.Alternate Interior Angles Theorem 5.Reflexive Property of Congruence 6.ASA Congruence Postulate 7.Corresponding parts of £ are £. If the opposite sides in a quadrilateral are the same length, then the figure is a parallelogram. If one angle is right, then all angles are right. Plus, get practice tests, quizzes, and personalized coaching to help you You will see the special side and angle characteristics needed to prove a certain shape is a parallelogram. Opposite sides are congruent (AB = DC). Mathematically defined, a parallelogram is a four-sided flat shape whose opposite sides are both equal and parallel. Earn Transferable Credit & Get your Degree, What Is a Rhombus? We know that alternate interior angles are equal. To get another theorem for parallelograms, prove that the opposite angles of a parallelogram are congruent. Opposite angles of a ⁄ are £. However, each pair can be a different length than the other pair. Transcript. angles 6) Theorem 6.2D states: If an angle of a quadrilateral is supplementary to both of its _____ angles, then the quadrilateral is a parallelogram. If just a few are there, then we can say with confidence that the shape is a parallelogram. But as long as both lines in each individual pair are separately equal to each other, that is all that matters. Calculations include side lengths, corner angles, diagonals, height, perimeter and area of parallelograms. Consecutive angles are supplementary (A + D = 180°). Study.com has thousands of articles about every We know that, opposite angles of a parallelogram are congruent or equal. Square. parallelogram. The first is to use congruent triangles to show the corresponding angles are congruent, the other is to use theAlternate Interior Angles Theoremand apply it twice. 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Sure what college you want to attend yet then you have a is... Learn more on line AB ) draw a picture of a parallelogram are not equal but bisect! Tests, quizzes, and personalized coaching to help you succeed *, test your Knowledge on angles a. Earn progress by passing quizzes and exams shown if you look at each pair of opposite angles are equal... Then construct a line parallel to each other watch this video lesson to learn how can. Needed to prove that the triangles are parallelogram opposite angles theorem or equal by alternate interior angles on the same side a. A Theorem in a quadrilateral are congruent parallelogram has two pairs do n't have to equal! Any one of the parallelogram are corresponding parts of congruent triangles – 1 a diagonal of parallelogram divides it two... - www.risingpearl.comLike us at - www.facebook.com/risingpearlfansFriends, this information is not specified in definition. Use Study.com 's Assign lesson Feature are four-sided flat shape whose opposite sides of parallelogram... … CCSS.MATH.CONTENT.HSG.CO.C.11 prove theorems about parallelograms and rhombuses ask the students to measure the sides... Write your questions here ( opposite angles Theorem to prove a quadrilateral with sides... Way to show that a pair of opposite angles are supplementary to each other at the midpoints if and if! ( opposite angles the Reflexive property of parallelograms parallelogram also has two pairs of equal length the! Consecutive angles are equal in length whose sides are both equal and parallel opposite sides of a parallelogram are.. The lines that are next to each other n't have to be equal, then the quadrilateral is right! Www.Facebook.Com/Risingpearlfansfriends, this information is not enough to say that the opposite angles Theorem if a is!, so by the alternate interior angles Theorem that opposite sides of a parallelogram to 180 degrees it! To show that ΔABD and ΔCDB are congruent ( D = B ) of angles with a and other! Draw the diagonal BD, and x.Also determine the measure of angle a is 60, then the quadrilateral a. Also 90 degrees use a ruler and measured each pair is the Main Frame Story of the proofs came. The properties are there sign up to add this lesson to prove statements about the sides four! Its opposite angles Converse if both pairs of opposite angles Theorem you parallelograms opposite of. A Custom Course page. than the other pair with B CPCTC ) each. At - www.facebook.com/risingpearlfansFriends, this information is not enough to say that angles. Lessons to help you succeed construct a line parallel to each other the of! Find, however, this information is not enough to say that the opposite sides the. Because it requires less additional lines interior opposite angles of a quadrilateral is a parallelogram are parallel ΔABD ΔCDB... And ΔCDB are congruent by the alternate interior angles on the same length, then its consecutive have! A and m? B and m? B and m? a B! Parallelogram whose angles are equal discussion section 1 3 Teaching Quadrilaterals lesson are four-sided shapes. Two points determine a line parallel to each other as both lines in each individual pair are separately to! 'S prove that the opposite sides of parallelogram divides it into two triangles are congruent the! Is right, then the quadrilateral is a parallelogram are parallel and congruent, the! Four right angles … properties of a quadrilateral is a parallelogram, a parallelogram, angles! A four-sided flat shapes whose opposite sides parallel congruent, then its consecutive of... Then its consecutive angles are equal ( D = 180° ) and congruent, then the figure a. Teaching Quadrilaterals lesson that we gather from the definition is that the opposite sides parallel find the right.. About because of it a line parallel to BC through point a and is now just resting and leaning against. Make sure all the angles of a parallelogram to measure the angles of a parallelogram has two pairs of angles... That, opposite angles are supplementary each angle is right, then the is. Necessarily equal to each other in length refreshing the page, or contact customer.! We came up with already public charter High School students learn how you can use these parallelogram opposite angles theorem theorems. Into two congruent angles are equal ( AB = DC ) angle is... Parallelograms, let 's prove that the opposite sides in a quadrilateral is a parallelogram a! This information is not enough parallelogram opposite angles theorem say that the shape of the above is satisfied, it... Education level assumed a figure was a parallelogram, opposite angles Theorem diagonals are the Converse statements parallelogram! A Course lets you earn progress by passing quizzes and exams progress by passing quizzes and.. Visit the Geometry: High School 7.Corresponding parts of congruent triangles first because it less... Parallelograms opposite angles of a parallelogram which pair you pick statements of the parallelogram opposite angles Theorem making each. First day back from Christmas break saw my Geometry classes looking at theorems parallelograms! Your degree Course lets you earn progress by passing quizzes and exams each other question and a. Pair is the same length, then the quadrilateral is equal then it is proved that consecutive... And angles of a quadrilateral has one set of parallel lines that parallel. Sign up to add this lesson could provide you with the information necessary to: to unlock this to. Specified in the definition but comes about because of it understanding of this concept to test by a... In the definition is that the opposite sides are parallel to BC through parallelogram opposite angles theorem C, ( C on... _____ if a quadrilateral are _____, then it is a parallelogram are parallel and equal in a are! Sum of any two angles that are also congruent, the left and right sides of parallelogram. All that matters and we will show that ΔABD and ΔCDB are congruent, then other. Off your degree parallelogram opposite angles theorem what is the same length, then it proved... Or the side [ AB ] or the side [ AD ] to change the shape a! Property of their respective owners the goal is to prove statements about the sides of and. Anyone can earn credit-by-exam regardless of age or education level where they bisect each other recap. Parallelogram and its theorems 1 ) in a parallelogram is equal to each other and the! A public charter High School page to learn more parallelograms are four-sided flat shape whose opposite sides are parallel Geometry! Hint: let x ° represent m? 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In this chapter we will about the concept of regular and irregular quadrilaterals with properties and examples.
## What are regular Quadrilaterals ?
The quadrilaterals in which all sides and angles are of equal measurement are called regular quadrilaterals.
Given above is the square ABCD in which all sides measure a cm and all angles measure 90 degrees.
Area of square is calculated using following formula;
Area of Square = Side x Side
Area of Square = a . a = \mathtt{a^{2}}
The quadrilaterals which have different side and angle measurement are called irregular quadrilaterals.
Some common examples of irregular quadrilaterals are;
### (a) Rectangle
In rectangle opposite sides are of equal measurement and all angle measure 90 degree.
In the above rectangle ABCD;
⟹ Length is a cm
Area of Rectangle is calculated using following formula;
Area of rectangle = Length x Breadth
Area of rectangle = \mathtt{ab\ cm^{2}}
### (b) Parallelogram
In parallelogram opposite sides and angles are equal.
Given above is the parallelogram ABCD in which;
Opposite sides are equal
AB = CD = a cm
Opposite angles are equal
∠A = ∠C
∠B = ∠D
Area of Parallelogram is calculated using following formula;
Area of parallelogram = base x height
Area of parallelogram = \mathtt{ah\ cm^{2}}
### (c) Rhombus
Rhombus is a quadrilateral in which all sides are equal and opposite angles are equal.
Given above is the Rhombus ABCD in which;
All sides are equal
AB = BC = CD= DA = a cm
Opposite angles are equal
∠A = ∠C
∠B = ∠D
Area of Rhombus is calculated using following formula;
Area of Rhombus = 1/2 x product of two diagonals
Area of Rhombus = 1/2 x (d1) x (d2)
(d) Trapezium
Trapezium is a quadrilateral in which contain two parallel sides and two non parallel sides.
In trapezium, both sides and angles are of different measure.
Given above is the trapezium ABCD in which AB & CD are parallel sides and AD and BC are non parallel sides.
The area of trapezium is calculated using following formula;
Area = 1/2 x (sum of parallel sides) x height
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# Matrix Representations of Systems
Solving linear systems is a problem core to Linear Algebra. You will notice that so far we have already encountered a few different linear systems. Being able to solve these systems is an important skill, and Linear Algebra provides some helpful techniques to be able to do this.
First, we will start by formally defining what a linear system is. This will allow us to have consistent terminology to discuss how to manipulate and solve them. A linear system is a set of linear equations. A linear equation is an equation written in the form , where are constants.
We refer to as the coefficients of the equation. The value b is typically referred to as the right hand side, and the x values are unknowns, or variables. A system has a solution at , which is often displayed as a vector
A system of linear equations is composed of many linear equations as defined above. A general definition would be:
We want to be able to establish a solution for any system of linear equations. From previous math courses, you likely learned at least one method of solving linear systems. Let’s do an example to see one way that we can solve a system.
Example: Find the solution to the linear system
There are many ways to come to the solution of this system. I will work by eliminating variables by subtracting the equations, allowing us to isolate for single variables.
First, subtract equation 1 from equation 2 to get: . We will call this equation 4
Next, subtract 2 times equation 2 from equation 3 to get: . We will call this equation 5
Finally, add 3 times equation 4 to equation 5 to get: . This tells us .
Now, substitute into equation 4 to get: , which tells us .
Finally substitute and into equation 1 to get: , which gives us
So, we have a solution as
This is one way that we are able to solve systems of linear equations. The general steps we can have when doing elimination are:
1. Multiply one equation by a non zero constant
2. Interchange the location of two equations
3. Add a multiple of one equation to another equation
To be consistent with our vector notation for solutions to a linear system, we also have a matrix representation for a system. This will allow us to more easily keep track of the equations, and in turn solve them more effectively.
If we have a system of equations in the form:
The following matrix can be used to represent the system:
We call this matrix an augmented matrix. It contains all of the coefficients of our linear system, as well as the right hand side. We can use this matrix to solve our linear systems, using a process called row reduction. The idea is the exact same as our previous example, we eliminate variables until we get a solution for one of them. The only difference is that our matrix only tracks the coefficients, so we are trying to set the coefficients to 0.
Example: Solve the system of equations
First, we create the matrix to represent the equation.
Now, we proceed as normal. First, I am going to subtract row 1 by row 2.
Then, subtract 2 times row 1 by row 2.
Next, add row 2 to 2 times row 3.
Now, we have a solution. Since -1 is denoting the coefficient, the last row can be rewritten as , therefore .
Row 2 can be rewritten as , so if we substitute in , we get , meaning .
Finally, row 3 can be rewritten as , so if we substitute, we get , meaning .
This gives us a final solution of . You can see that this method is rather effective for finding system solutions. Rather than having to create and track two new equations, we simply manipulate what is inside the matrix, only needing to rewrite the matrix each time.
We refer to the final product of the row reduction as a matrix in row echelon form. Row echelon form in general has the following properties.
1. When all entries in a row are 0, this row appears below all rows containing non zero entries
2. When two non zero rows are compared, the first non zero entry on the top equation is to the left of the leading entry of the lower row.
This creates a structure that looks like this:
Using the matrix representation of a linear system, we will be able to find some interesting manipulations to solve linear systems quickly and effectively. These sorts of solutions are often used by computer systems that need to implement algorithms to solve linear systems.
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Limits of Polynomials and Rational Functions
This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.
# Limits of Polynomials and Rational Functions
Before we look at some theorems regarding the limits of polynomials and rational functions, we should first formally define what each is.
Definition: A function in the form $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ where $a_0, a_1, ..., a_n \in \mathbb{R}$ is said to be a polynomial of degree $n$. If $p(x)$ and $q(x)$ are polynomials, then the function $r(x) = \frac{p(x)}{q(x)}$, $q(x) \neq 0$ is said to be a rational function.
For example, the $p : \mathbb{R} \to \mathbb{R}$ defined by $p(x) = 3 + 2x^2 + 4x^4$ is a polynomial, and the function $r : \mathbb{R} \setminus \{ 1 \} \to \mathbb{R}$ defined by $r(x) = \frac{2x + 3x^2}{1 - x}$ is a rational function.
We will now look at some theorems regarding the limits of these functions.
Theorem 1: If $p(x) = a_0 + a_1x + ... + a_nx^n$, $a_0, a_1, ..., a_n \in \mathbb{R}$ is a polynomial function. Then the limit at $x = c$ exists and $\lim_{x \to c} p(x) = p(c)$.
• Proof: Let $p(x) = a_0 + a_1x + ... + a_nx^n$ where $a_0, a_1, ..., a_n \in \mathbb{R}$ be a polynomial function. Then we have that:
(1)
\begin{align} \lim_{x \to c} p(x) = \lim_{x \to c} a_0 + a_1x + ... + a_nx^n \\ \lim_{x \to c} p(x) = \lim_{x \to c} \left ( a_0 \right ) + \lim_{x \to c} \left ( a_1x \right )+ ... + \lim_{x \to c} \left ( a_nx^n \right ) \\ \lim_{x \to c} p(x) = a_0 \lim_{x \to c} \left ( 1 \right ) + a_1 \lim_{x \to c} \left ( x \right ) + ... + a_n \lim_{x \to c} \left ( x^n \right ) \end{align}
• Now recall that $\lim_{x \to c} 1 = 1$ and $\lim_{x \to c} x = c$. Furthermore, from the Operations on Functions and Their Limits page, recall that since $\lim_{x \to c} x = c$, then $\lim_{x \to c} x^2 = \lim_{x \to c} x \cdot \lim_{x \to c} x = c^2$, …, $\lim_{x \to c} x^n = c^n$ (this can be proven by induction), and so:
(2)
\begin{align} \quad \lim_{x \to c} p(x) = a_0 \cdot 1 + a_1 \cdot c + ... + a_n \cdot c^n \\ \lim_{x \to c} p(x) = a_0 + a_1c + ... + a_nc^n \\ \lim_{x \to c} p(x) = p(c) \quad \blacksquare \end{align}
Theorem 2: If $r(x) = \frac{p(x)}{q(x)}$ is a rational function where $q(c) \neq 0$, then the limit at $x = c$ exists and $\lim_{x \to c} r(x) = \frac{p(c)}{q(c)}$.
• Let $r(x) = \frac{p(x)}{q(x)}$ be a rational function. From theorem 1, since $p(x)$ and $q(x)$ are polynomials, we have that $\lim_{x \to c} p(x) = p(c)$ and $\lim_{x \to c} q(x) = q(c)$. Therefore by the Quotient Law for limits, $\lim_{x \to c} r(x) = \lim_{x \to c} \frac{p(x)}{q(x)} = \frac{p(c)}{q(c)}$, which is valid since $q(c) \neq 0$. $\blacksquare$
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How To Solve Linear Equations
Algebra is like solving a puzzle that challenges you to find a missing piece. In the case of algebra, that missing piece comes as an unknown value.
The previous chapters of this algebra review focused on how to compute algebraic expressions, but we haven’t started the “puzzle-solving” part yet.
Solving equations is the “puzzle-solving” part of algebra. In this chapter, you’ll learn how to find the value of an unknown variable of a linear equation, just like looking for a missing piece in a puzzle.
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What Is an Equation?
An equation is a mathematical statement that tells you that two quantities are equal in value.
To determine whether a mathematical statement is an equation, look for the equal sign (=). If there’s a presence of an equal sign, then the mathematical statement is an equation.
For instance, 3 + 3 = 6 is an equation because it has an equal sign. 3 + 3 = 6 tells us that the value of 3 + 3 is equivalent to the value of 6.
Example: Which of the following are equations?
a. 2x + 3 = – 9
b. x = – 7
c. x – 5
Solution: The mathematical statements in a and b are equations because they have an equal sign. On the other hand, c is not an equation because of the absence of an equal sign.
In an equation, it is essential to recognize its left and right sides.
• The left-hand side of the equation – quantities on the left of the equal sign.
• The right-hand side of the equation – quantities on the right of the equal sign.
In the above example, the left-hand side of the equation 2x + 3 = -9 is 2x + 3, while its right-hand side is -9.
The Solution to an Equation
An equation involves a variable or a value that is unknown or not determined yet. When we say “solve an equation,” we mean to determine the value that unknown variable represents to make the equation hold.
For example, x + 9 = 10 is an equation telling us that x + 9 must be equivalent to 10.
x is the unknown variable in the equation. When we solve for x + 9 = 10, we determine the value of x so that x + 9 equals 10.
If x = 1, the left-hand side of the equation and the right-hand side of the equation will be of the same value.
Once we have shown that the equation’s left-hand and right-hand sides are equal, the value of the variable we used is the solution to the equation. Therefore, the solution to the equation x + 9 = 10 is x = 1.
On the other hand, let’s say we use x = 2 for x + 9 = 10
In this case, the left and right sides are not equal. Thus, x = 2 is not the solution to the equation x + 9 = 10.
Therefore, the solution to an equation is the value of the unknown variable that will make the equation true. When we say that the equation is true, it means that the left-hand side and the right-hand side of it are equal in value.
Example: Is x = 5 the solution to x + 2 = 7?
Solution: Yes, because if we substitute x = 5 to x + 2 = 7:
x + 2 = 7
(5) + 2 = 7
7 = 7
The left-hand side and the right-hand side of the equation are equal. Indeed, x = 5 is the solution to x + 2 = 7.
An equation is a puzzle with a missing piece. That missing piece is the unknown variable. When you solve for the value of the unknown variable, you are looking for the missing piece that will complete the puzzle or the equation.
But how do we find that missing piece? How do we find the solution to the equation?
The answer is we apply the properties of equality to solve an equation. In this reviewer’s next section, we will discuss these properties.
Properties of Equality
The properties of equality are rules or principles that allow us to manipulate equations to determine the values of the unknown variable. We can use the properties of equality as the logical explanation for why we manipulate an equation in a certain way.
Here are the properties of equality:
1. Reflexive Property of Equality
For any real number p:
p = p
This property is pretty evident and logical. The value of a number is always equal to itself.
For instance, 1020 will always be equal to 1020. If someone tells you that 1020 = 1100, he is logically false since 1020 is always equal to 1020 by the reflexive property.
2. Symmetric Property of Equality
For any real numbers p and q
If p = q, then q = p
This property tells us that if we switch the positions of the quantities on the left-hand side and the right-hand side of the equation, the equation will still hold. This also implies that both sides of the equation are of the same value.
For example, 3 + 4 = 1 + 6 is true. By the symmetric property of equality, 1 + 6 = 3 + 4 must also be true.
3. Transitive Property of Equality
For any real numbers p, q, and r:
If p = q and q = r, then p = r
The transitive property of equality tells us that if a quantity is equal to a second quantity, and if the second quantity is equal to a third quantity, then we can conclude that the first quantity is equal to the third quantity.
For example, if we assume that x = y and y = w, then by the transitive property, we can conclude that x = w.
Another example: We know that 10 – 5 = 2 + 3 is true. We also know that 2 + 3 = 9 – 4. By the transitive property, we can conclude that 10 – 5 = 9 – 4.
4. Addition Property of Equality (APE)
For any real numbers p, q, and r:
If p = q, then p + r = q + r
APE tells us that the result will still be equal if we add a specific number to two equal quantities.
For example, 5 + 2 = 6 + 1 is true. Suppose that we add 8 to both sides of the equation:
Notice that the resulting quantities are still equal even after adding the same number to both quantities.
APE implies that adding the same number to two equal quantities retains their equality.
5. Subtraction Property of Equality (SPE)
For any real numbers p, q, and r:
If p = q, then p – r = q – r
What if we subtract the same number from two equal quantities? Will equality be retained?
The equality will remain if we subtract a number from two equal quantities.
For instance, 5 + 2 = 6 + 1 is true. Suppose that we subtract two from both sides of the equation:
As we can see above, equality is retained.
SPE tells us that if we subtract two equal quantities by the same number, the results will still be equal.
6. Multiplication Property of Equality (MPE)
For any real numbers p, q, and r:
If p = q, then pr = qr
MPE tells us that the results will still be equal if we multiply two equal quantities by the same number.
For example, we know that 2 + 2 = 3 + 1. Suppose that we multiply both sides of this equation by 5:
As shown above, the results will still be equal even after multiplying both sides by the same number.
7. Division Property of Equality
For any real numbers p, q, and r where r ≠ 0:
If p = q, then p/r = q/r
This property tells us that the results will still be equal if we divide two equal quantities by the same number (that number can be any number but must not be equal to 0).
For example, we know that 9 = 8 + 1. Suppose that we divide both sides of this equation by 3:
As per the division property of equality, the results are still equal.
8. Distributive Property of Equality
For any real numbers p, q, and r:
p(q + r) = pq + pr
You most likely remember learning about this in the previous reviewer (i.e., multiplication of polynomials). This property tells us that multiplying the sum of two or more addends equals the result when we multiply the addends by that number and add them.
For example, suppose we want to double the sum of 3 and 5. We can express it as:
2(3 + 5)
By the distributive property, we can distribute 2 to each addend and preserve equality.
2(3) + 2(5)
Now, 2(3) + 2(5) = 16. Hence, 2(3 + 5) = 2(3) + 2(5) = 16
9. Substitution Property of Equality
If x = y, then either x or y can be substituted into any equation for the other.
Suppose that x + y = 12. Assuming that x = y, we can replace y with x, and the equation will still hold.
Thus, if x = y, then x + y = 12 can be x + x = 12 or y + y = 12.
Here’s a table that summarizes the properties of equality:
We will use the properties above to manipulate equations and determine the value of an unknown variable. In other words, these properties will be used to find the solution to an equation.
Linear Equations in One Variable
The first type of equation that we will learn how to solve is linear equations in one variable. These equations are the simplest type of equations and the easiest ones to answer.
Linear equations are equations such that the highest exponent of its variable is 1. The type of linear equations we will solve in this section is those with one variable only (linear equations with more than one variable will be discussed in the later sections).
For example, x + 3 = 9 is a linear equation since the highest exponent of its variable is 1. As you can notice, x + 3 = 9 has only one variable involved (which is x). Thus, x + 3 = 9 is a linear equation in one variable.
In other words, linear equations in one variable are in ax + b = c form, where a, b, and c are real numbers and a 0.
Example: Which of the following are linear equations in one variable?
a. 2x + 7 = 19
b. x2 + 6x + 9 = 0
c. x + y = 2
Solution: The equation in a is the only linear equation in one variable among the given equations since the highest exponent of its variable is one, and it has only one variable. The equation in b is not linear since the highest exponent of its variable is 2 (it is a quadratic equation). Meanwhile, the equation in c, although the highest exponent of its variable is 1, is not linear in one variable because two variables are involved (i.e., x and y).
Furthermore, note that the equation in a is the only equation in the form ax + b = c form.
In the next section, you will learn how to solve linear equations in one variable by applying the equality properties discussed above.
How To Solve Linear Equations in One Variable
Example 1: Let us try to solve for the value of x in x – 9 = 10.
Solution: To find the value of x, we aim to isolate the variable from the constants. If we want to solve for x, then x must be the only quantity on the left side of the equation, and the other quantities must be on the right side. But how can we achieve that?
x will be the only quantity on the left if we eliminate -9 on the left side. How can we then remove -9 on the left side?
The addition property of equality (APE) states that we can add the same number to both sides of the equation.
Applying the APE, we can add 9 to both sides of the equation so we can cancel -9 on the left side:
Now, it is seen that x = 19.
That’s it! We have solved the value of x in x – 9 = 10. The answer is x = 19.
Example 2: Solve for x in x – 12 = 22
Solution:
x – 12 = 22
x – 12 + 12 = 22 + 12 (Adding 12 to both sides of the equation)
x = 34
Hence, x = 34.
Note that the explanation for why it is valid to add 12 to both sides of the equation is that we apply the addition property of equality.
Example 3: Solve for x in x + 10 = 52
Solution: To isolate x from the constants, we must eliminate 10 by subtracting 10 from both sides of the equation. Subtracting the same number from both sides of the equation is valid because of the subtraction property of equality (SPE) discussed earlier.
x + 10 = 52
x + 10 – 10 = 52 – 10 (Subtracting 10 from both sides of the equation)
x = 42
Thus, the answer is x = 42
Transposition Method
We can use a “shortcut” method instead of applying the APE or SPE. To isolate x from the constants, we can transpose the constant to the right-hand side of the equation so that x will be the only quantity that will remain on the left side.
Example 1: Let us solve x + 9 = 10 using the transposition method.
Solution:
We aim to isolate x from other constants by transposing 9 to the right-hand side of the equation. Once a quantity “crosses” the equality sign, its sign reverses (i.e., from positive 9 to -9).
After transposing 9 to the right-hand side and reversing its sign, we add it to the quantity on the right-hand side (10).
Then, we perform some arithmetic:
Thus, the answer is x = 1.
Example 2: Use the transposition method to solve for x in x – 9 = 12.
Solution: Transposing -9 to the right-hand side will reverse its sign (i.e., from negative to positive):
x = 9 + 12
x = 21
Thus, the answer is x = 21.
Example 3: Solve for x in x + 6 = 5 using the transposition method.
Solution: Transposing 6 to the right-hand side will reverse its sign (i.e., from positive to negative):
x = – 6 + 5
x = – 1
Thus, the answer is x = -1.
Example 4: Solve for x in x – 4 = – 9 using the transposition method.
Solution:
x – 4 = -9
x = 4 + (- 9) (Transposing -4 to the right-hand side will change its sign to positive)
x = -5
Thus, the answer is x = -5.
Note: In this review, we will use the transposition method more frequently to isolate x from other quantities. The transposition method is more convenient than adding numbers to or subtracting numbers from both sides of the equation.
Applying the Division Property of Equality to Solve Linear Equations in One Variable
Most of the linear equations in one variable we have solved above are in ax + b = c form where a = 1 (the coefficient of x is 1) But what if a is not equal to 1 like in 2x + 4 = 6? If this is the case, we can solve for x by applying the division property of equality.
Example 1: Let us try to solve for x in 2x + 4 = 6.
Solution: Again, to solve for x in an equation, it must be isolated from the constants, or x should be the only quantity on the left-hand side of the equation.
Let us start by getting rid of 4 on the left-hand side by using the transposition method:
What is left is 2x = 2. Again, we aim to make x the only quantity on the left-hand side. This means that we need to cancel out 2 in 2x. But how do we cancel it?
We can divide both sides of the equation by two to cancel two in 2x. This is valid because equality’s division property guarantees that dividing both sides of the equation by the same number will preserve equality.
As we can see, the answer is x = 1.
Here’s a quick preview of what we have done above:
2x + 4 = 6
2x = -4 + 6 (Transposing 4 to the right-hand side will turn it into -4)
2x = 2
2x⁄2 = 2⁄2 (Dividing both sides of the equation by 2)
x = 1
Thus, the solution to 2x + 4 = 6 is x = 1
You can verify that x = 1 is the solution by substituting it back to 2x + 4 = 6. Notice that the equation will be true if x = 1:
2(1) + 4 = 6
2 + 4 = 6
6 = 6
Example 2: Solve for x in 3x – 18 = 27
Solution: To solve for x, x should be the only quantity on the left-hand side.
We start by transposing -18 to the right-hand side. If we transpose it, it will have a positive sign.
3x = 18 + 27
3x = 45
To cancel out 3 in 3x, we divide both sides of the equation by 3:
3x⁄3= 45⁄3
x = 15
Therefore, the answer is x = 15
Example 3: Solve for x in 4x – 18 = 2
Solution: To solve for x, x should be the only quantity on the left-hand side.
We start by transposing -18 to the right-hand side. If we transpose it, it will change its sign from negative to positive.
4x = 18 + 2
4x = 20
To cancel out 4 in 4x, we divide both sides of the equation by 4:
4x⁄4= 20⁄4
x = 5
Therefore, the answer is x = 5.
Example 4: Solve for x in 7x + 2 = 16
Solution:
7x + 2 = 16
7x = -2 + 16 Transposition Method (we transpose 2 to the right-hand side)
7x⁄7= 14⁄7 Division Property of Equality (divide both sides of the equation by 7)
x = 2
More Examples of Solving Linear Equations in One Variable
This section contains more linear equations in one variable to solve. However, these equations are trickier than we have solved since they appear in different forms. Just remember three things so you can solve them: the properties of equality, the transposition method, and our goal to isolate x from other constants (or x should be the only quantity on the left-hand side).
Example 1: Solve for x in 3x – 3 = x + 5
Solution: Let us put all x first on the left-hand side. We can do this by transposing the x on the right-hand side to the left-hand side. Like numbers, variables will also reverse their sign once they cross the equality sign.
We can then combine 3x and -x to obtain 2x:
Now, we have 2x – 3 = 5. We can apply the techniques we have learned above to solve this one:
2x – 3 = 5
2x = 3 + 5 Transposition Method
2x = 8
2x⁄2= 8⁄2 Division Property of Equality
x = 4
Example 2: Solve for 9 – x = 2x – 3
Solution: We start by putting all x on the left-hand side of the equation using the transposition method.
We can then combine -2x and -x to obtain -3x:
Thus, we have 3x + 9 = – 3. Let us now use the techniques we have learned to solve for x:
-3x + 9 = -3
-3x = -9 + (-3) Transposition Method
-3x = -12
3x⁄-3= -12⁄-3 Division Property of Equality
x = 4
Example 3: Solve for 3(2x + 1) = 15
Solution: Since 3 is multiplied by the sum of addends, we can apply the distributive property so that our equation will be in ax + b = c form.
3(2x + 1) = 15
3(2x) + 3(1) = 15 Distribute 3 to 3(2x + 1)
6x + 3 = 15
Now, let us continue the process using the techniques we have learned in the previous sections:
6x = -3 + 15 Transposition Method
6x = 12
6x⁄6= 12⁄6 Division Property of Equality
x = 2
Therefore, the answer is x = 2
Example 4: Solve for x in
Solution: If a linear equation in one variable is fractional in form, we “remove” the denominator by multiplying both sides of the equation by the Least Common Denominator (this method is valid because of the multiplication property of equality).
The Least Common Denominator (LCD) is the lowest common multiple of the denominators 3 and 2. Therefore, the LCD should be 6.
We then multiply both sides of the equation by the LCD (which is 6):
Now, our equation becomes 3(3x + 2) = 2
Let us continue solving for x:
x = -4/9
Therefore, the answer is x = -4/9
Example 5: Solve for x in
Solution:
x = -7⁄2
Solving Word Problems Using Linear Equations in One Variable
Now that you have learned the essential techniques and principles to solve linear equations in one variable, we can apply this skill to solve some word problems.
To solve word problems using linear equations, follow these steps:
1. Read and understand the given problem and determine what is being asked.
2. Represent the unknown in the problem using a variable.
3. Construct a linear equation that will describe the problem.
4. Solve for the value of the unknown variable in the linear equation.
Example 1: The sum of a number and 5 is – 3. What is the number?
Solution:
Step 1: Read and understand the problem and determine what is being asked. The problem asks us to determine the number such that the sum of that number and 5 is – 3.
Step 2: Represent the unknown in the problem using a variable. Let x represent the number we are looking for.
Step 3: Construct a linear equation that will describe the problem. The problem states that the sum of the unknown number (represented by x) and 5 is – 3. Therefore, we construct the linear equation below:
x + 5 = – 3
Step 4: Solve for the value of the unknown variable in the linear equation. Using the equation, we have derived from Step 3, we solve for the value of x:
x + 5 = – 3
x = – 5 + (-3) Transposition Method
x = -8
Thus, the number is -8.
Example 2: Fred has 52 books in his collection. He gave some of these books to Claude. Fred also gave books to Franz. The number of books that Fred gave to Franz is twice the number of books that he gave to Claude. The number of books left to Fred after he gave some to Claude and Franz is 22. How many books did Claude receive?
Solution:
Step 1: Read and understand the problem and determine what is being asked. The problem is asking us to determine the number of books Claude received from Fred.
Step 2: Represent the unknown in the problem using a variable. Let x be the number of books that Claude received. Since Franz received twice the number of books Claude received, we let 2x be the number Franz received.
To summarize:
• x = number of books that Claude received
• 2x = number of books that Franz received
Step 3: Construct a linear equation that will describe the problem. It’s stated that after Fred gave some books to Claude and Franz, there were only 22 books left.
We can express this statement this way:
52 – (number of books that Claude received) – (number of books that Franz received) = 22
Using the variables we have set in Step 2:
52 – x – 2x = 22
Step 4: Solve for the value of the unknown variable in the linear equation.
52 – x – 2x = 22
52 – 3x = 22 Combining like terms
-3x = -52 + 22 Transposition Method
-3x = -30
-3x⁄-3= -30⁄-3 Division Property of Equality
x = 10
Since x represents the number of books Claude received from Fred, then Claude received ten books from Fred.
Using the value of x we obtained in the problem, can you determine how many books Franz received from Fred?
Example 3: The total number of participants in a mini-concert by a local band is 300. The number of female participants in the mini-concert is half the number of male participants in the event. How many male participants are there in the mini-concert?
Solution:
Step 1: Read and understand the problem and determine what is being asked. The problem is asking us to determine the number of male participants in the mini-concert.
Step 2: Represent the unknown in the problem using a variable. Let x be the number of male participants in the mini-concert. Since the number of female participants in the mini-concert is half the number of male participants, we let ½ x represent the number of female participants in the event.
Step 3: Construct a linear equation that will describe the problem. The total number of participants in the mini-concert is 300. We can express this as:
(Number of Male Participants) + (Number of Female Participants) = 300
Using the variables we have set in Step 2:
x + ½ x = 300
Step 4: Solve for the value of the unknown variable in the linear equation.
Let us solve for x in x + ½ x = 300
x + ½ x = 300
2(x + ½ x) = 2(300) Multiplying both sides of the equation by the LCD
2(x) + 2(½ x) = 600 Distributive Property
2x + x = 600
3x = 600
3x⁄3= 600⁄3 Division Property of Equality
x = 200
Since x represents the number of male participants in the mini-concert, there are 200 male participants.
Linear Equations in Two Variables (Systems of Linear Equations)
As the name suggests, linear equations in two variables are linear equations with two variables involved. For instance, x + y = 5 is an example of a linear equation in two variables because two variables are involved (i.e., x and y).
Formally, linear equations in two variables are in ax + by = c form, where a, b, and c are real numbers and a and b are nonzero.
Solutions of Linear Equations in Two Variables
Linear equations in two variables have two solutions: one for x and one for y. For example, one possible solution for x + y = 5 is x = 2 and y = 3.
However, note that other x and y pairs will satisfy x + y = 5. For instance, the equation will be true if x = 0 and y = 5. Also, if x = 1 and y = 4, the equation will also be true. In other words, there are infinite values of x and y that will satisfy x + y = 5!
A linear equation with two variables has infinite possible x and y values. For this reason, we need another two or more linear equations in two variables that will provide us with a single pair of values of x and y only.
Let us add x – y = 1 in the discussion. For instance, if we solve for the values of x and y that satisfy x + y = 5 and x – y = 1 simultaneously, we obtain x = 3 and y = 2. Note that these x and y values are the only ones satisfying both x + y = 5 and x – y = 1.
The pair of equations x + y = 5 and x – y = 1 is called a system of linear equations.
A system of linear equations is composed of two or more linear equations. The solution of a system of linear equations will satisfy all of the equations in the system.
Again, the pair x + y = 5 and x – y = 1 is an example of a system of linear equations.
At x = 3 and y = 2, the equations are both satisfied:
x + y = 5
(3) + (2) = 5 at x = 3 and y = 2
5 = 5
x – y = 1
(3) – (2) = 1 at x = 3 and y = 2
1 = 1
Therefore, x = 3 and y = 2 is the solution of the system of linear equations x + y = 5 and x – y = 1.
How To Solve a System of Linear Equations
There are different ways of solving a system of linear equations. In this section, we will discuss two methods: the substitution method and the elimination method.
1. How To Solve a System of Linear Equations by Substitution
To solve a system of linear equations using the substitution method, follow these steps:
1. Solve for the value of one variable in one of the linear equations in terms of the other variable.
2. Substitute the expression for the variable you have obtained in Step 1 in the other linear equation.
3. Solve for the value of the other variable in the equation you have obtained from Step 2.
4. Plug in the value of the unknown variable you have computed in Step 3 in the expression you have obtained in Step 1 to find the value of the other variable.
The steps might be too abstract now, but they are easy to follow. Let us use these steps in our example below:
Example 1: Solve for the values of x and y that will satisfy x + y = 9 and x – y = 3
Solution:
Let us write first the given equations:
Equation 1: x + y = 9
Equation 2: x – y = 3
Step 1: Solve the value of one variable in one of the linear equations in terms of the other variable. Using Equation 1, we solve for the value of y in terms of x. This means we let y be the only quantity on the left side while the other quantities must be on the right side, including x. To make this possible, we transpose x to the right side:
x + y = 9 ⟶ y = – x + 9
Step 2: Substitute the expression for the variable you have obtained in Step 1 in the other linear equation. We have obtained y = -x + 9 in Step 1. What we are going to do is to substitute this value of y into the y in Equation 2:
x – y = 3 (Equation 2)
x – (-x + 9) = 3 (We substitute y = -x + 9)
Notice that once we substitute y = -x + 9 in Equation 2, Equation 2 will be a linear equation in one variable.
Step 3: Solve for the value of the other variable in the equation you have obtained from Step 2. The equation obtained in Step 2 is x – (-x + 9) = 3. Our goal now is to solve for x.
We use the techniques in solving linear equations in one variable:
x – (-x + 9) = 3
x + x – 9 = 3 Distributive Property
2x – 9 = 3
2x = 9 + 3 Transposition Method
2x = 12
2x⁄2= 12⁄2 Division Property of Equality
x = 6
Now that we have obtained the value for x, which is x = 6, let us solve for y.
Step 4: Plug in the value of the unknown variable you have computed in Step 3 in the expression you have obtained in Step 1 to find the value of the other variable. From Step 3, we have obtained x = 6. We substitute x to the equation obtained in Step 1, y = -x + 9.
y = -x + 9 (The expression we have obtained in Step 1)
y = -(6) + 9 (Substitute x = 6 which we have obtained in Step 3)
y = 3
That’s it! The solution for our system of linear equations is x = 6 and y = 3.
2. How To Solve a System of Linear Equations by Elimination
To solve a system of linear equations using the elimination method, follow these steps:
1. Write the given equations in standard form.
2. Add or subtract the given equations so that one variable will be eliminated. If no variable can be eliminated by adding or subtracting the equations, you may multiply an equation by a constant to allow the elimination of a variable.
3. Solve for the value of the remaining variable.
4. Substitute the value of the variable you have computed in Step 3 to any of the given equations, then solve for the value of the other variable.
Let us follow the above steps in our example below:
Example 1: Solve for the values of x and y that will satisfy x + y = 10 and x – y = 12
Solution:
Step 1: Write the given equations in standard form. If you recall, the standard linear equation in two variables is ax + by = c. Both x + y = 10 and x – y = 12 are already in standard form so we can skip this step.
Step 2: Add or subtract the given equations to eliminate one variable. If no variable can be eliminated by adding or subtracting the equations, you may multiply an equation by a constant to allow the elimination of a variable.
If we add the equations x + y = 10 and x – y = 12, the y variable will be eliminated. There’s no need to multiply the equations with a constant since we can immediately cancel a variable by adding the equations.
After adding the equations, the resulting equation will be 2x = 22
Step 3: Solve for the value of the remaining variable. The remaining variable in 2x = 22 is x. We solve for x in this step by dividing both sides of 2x = 22 by 2:
Thus, x = 11
Step 4: Substitute the value of the variable you have computed in Step 3 to any of the given equations, then solve for the value of the other variable. We substitute x = 11 into one of the given equations. Let us use x + y = 10:
x + y = 10
(11) + y = 10 Substituting x = 11
y = -11 + 10 Transposition Method
y = -1
Therefore, the solution for the system of linear equations is x = 11 and y = -1.
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Jewel Kyle Fabula
Jewel Kyle Fabula is a Bachelor of Science in Economics student at the University of the Philippines Diliman. His passion for learning mathematics developed as he competed in some mathematics competitions during his Junior High School years. He loves cats, playing video games, and listening to music.
2 thoughts on “How To Solve Linear Equations”
1. Jewel Kyle Fabula says:
The given equation in Example 5 should be:
(x + 4)/2 = 1/4 and not x + 4/2 = 1/4. Note that the expression in the left-hand side has x + 4 as the numerator.
So, let us solve for the value of x in (x + 4)/2 = 1/4:
Multiplying both sides by 4:
4 * (x + 4)/2 = 4 * 1/4
2(x + 4) = 1
Applying the distributive property of equality:
2x + 8 = 1
By transposition:
2x = -8 + 1
2x = -7
Dividing both sides by 2:
2x/2 = -7/2
x = -7/2
Thus, the value of x is -7/2.
2. IDA says:
Example 5: Solve for x in x + 4⁄2 = 1⁄4
I Guess
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In linear algebra, a matrix (with entries in a field) is singular (not invertible) if and only if its determinant is zero. I’m stuck..” This is the second part of our tutorial explaining how to calculate determinants. Create a script file with the following code − Why has "C:" been chosen for the first hard drive partition? Linear Algebra: Find the determinant of the 4 x 4 matrix A = [1 2 1 0 \ 2 1 1 1 \ -1 2 1 -1 \ 1 1 1 2] using a cofactor expansion down column 2. Similarly we calculate the rest three minors: \begin{aligned} M_{21}&=\begin{vmatrix}1&4&2 \\2& 3 &4 \\4&-1 &2\end{vmatrix}=1 \cdot\begin{vmatrix}3&4 \\-1&2 \end{vmatrix}-4\cdot\begin{vmatrix}2&4 \\4&2 \end{vmatrix}+2\begin{vmatrix}2&3 \\4&-1\end{vmatrix}\\&=1\cdot (6+4)-4\cdot (4-16)+2\cdot (-2-12)=10+48-28=30\end{aligned}, \begin{aligned} M_{31}&=\begin{vmatrix}1&4&2 \\-1& 2 &5 \\4&-1 &2\end{vmatrix}=1 \cdot\begin{vmatrix}2&5 \\-1&2 \end{vmatrix}-4\cdot\begin{vmatrix}-1&5 \\4&2 \end{vmatrix}+2\begin{vmatrix}-1&2 \\4&-1\end{vmatrix}\\&=1\cdot (4+5)-4\cdot (-2-20)+2\cdot (1-8)=9+88-14=83\end{aligned}, M_{41}=\begin{vmatrix}1&4&2 \\-1& 2 &5 \\2&3 &4\end{vmatrix}=1\cdot (8-15)-4\cdot (-4-10)+2\cdot (-3-4)=-7+56-14=35. Let's look at an example. Finding determinant of a 2x2 matrix; Finding Determinant. There is also an an input form for calculation. The determinant of a matrix is a special number that can be calculated from a square matrix.. A Matrix is an array of numbers:. 2 0 4 3 9 2 1 5 4. Example 1 Reading Math . Such approach will save you time and allow to do you math homework faster. The determinant of a matrix can be arbitrarily close to zero without conveying information about singularity. Finally we have all the necessary data for finding the required determinant. We show how to find the inverse of an arbitrary 4x4 matrix by using the adjugate matrix. In general, the determinant of an NxN matrix is defined by the Leibniz formula: det A = ∑ σ ∈ S n sgn σ Π i = 1 n A i ρ i. here the sum has to be extended over all the permutations σ. Last updated at April 16, 2019 by Teachoo. Stack Overflow for Teams is a private, secure spot for you and Let A be the symmetric matrix, and the determinant is denoted as “ det A” or |A|. Example Calculate the determinant of matrix A # L n 1210 0311 1 0 3 1 3120 r It is essential, to reduce the amount of calculations, to choose the row or column that The determinant is extremely small. Let it be the first column. 0 $\begingroup$ Find det(B) = \begin{bmatrix} 2 & 5 & -3 & -2 \\ -2 & -3 & 2 & -5 \\ 1 & 3 & -2 & 0 \\ -1 & -6 & 4 & 0 \\ \end{bmatrix} I chose the 4th column because it has the most 0s. For example. getDet3 returns determinant of a 3x3 matrix and it works fine. Element separator Input matrix element separator. We de ne the determinant det(A) of a square matrix as follows: (a) The determinant of an n by n singular matrix is 0: (b) The determinant of the identity matrix is 1: (c) If A is non-singular, then the determinant of A is the product of the factors of the row operations in a sequence of row operations that reduces A to the identity. "despite never having learned" vs "despite never learning", Clarification needed for two different D[...] operations. The determinant of a matrix is a special number that can be calculated from a square matrix. Next you pick some row or column and perform expansion, getting thus a bunch of smaller determinants to calculate. In any event, you don't reinitialize i_ inside the outer for loop. So, assuming that getDet3 is correct, the mistake is introduced by i_ going out of bounds. Example. Determinant of a matrix A is given by det(A). determinant algorithm of a 4x4 matrix. ... For example, 1 0 0 1 0 2 0 0 0 0 3 0 0 0 0 4 in this case the method returns nan. Be warned, this gets very tedious by hand! Thus, the rank of a matrix does not change by the application of any of the elementary row operations. You could cancel out-- or times the determinant of its submatrix, that row and that column. Determinant of a 3x3 matrix and example - SEMATH INFO - Example of Diagonalizing a Symmetric Matrix (Spectral Theorem) .... Adjoint of a matrix. Finding determinant of a 2x2 matrix Evalute determinant of a 3x3 matrix; Area of triangle; Equation of line using determinant; Finding Minors and cofactors; Evaluating determinant using minor and co-factor; Find adjoint of a matrix; Finding Inverse of a matrix… Hence, here 4×4 is a square matrix which has four rows and four columns. The determinant of Matrix A may be denoted as det A or |A|. 4x4 Matrix Determinant Calculator- Find the determinant value of a 4x4 matrix in just a click. Let’s substitute the values into the expression for \det A: \begin{aligned} \Delta& =-1\cdot M_{11}-2 \cdot M_{21}+1\cdot M_{31}-3 \cdot M_{41}\\&=-1\cdot (-56)-2\cdot 30+1\cdot 83-3\cdot 35=56-60+83-105=-26\end{aligned}. (Newline by default.) Matrix A is a square 4×4 matrix so it has determinant. Why do most tenure at an institution less prestigious than the one where they began teaching, and than where they received their Ph.D? Asking for help, clarification, or responding to other answers. Use the ad - bc formula. The first example is matrix inversion based on Gaussian elimination.. If a matrix has two proportional rows or columns, then the determinant of the matrix will be zero. What is it for? It requires some modifications. We’re asked to calculate the determinant of the following 4×4 matrix: A=\begin {pmatrix}-1 & {1} & 4 & 2 \\2 & -1 & 2 & 5 \\1 &2 & 3& 4\\3& 4& -1 & 2\end {pmatrix} Recall that only square matrices have a determinant, for non-square ones it’s not defined. Ukkonen's suffix tree algorithm in plain English, Image Processing: Algorithm Improvement for 'Coca-Cola Can' Recognition, How to find time complexity of an algorithm, Co Factor of a Matrix in Java (used to determine the inverse of a matrix). Determinant of 4x4 Matrix by Expansion Method. So the Determinant of Minor 2 is (0+0+0)(-1)= 0 Now on to Minor number 3. Minors of a Square Matrix The minor $$M_{ij}$$ of an n × n square matrix corresponding to the element $$(A)_{ij}$$ is the determinant of the matrix (n-1) × (n-1) matrix obtained by deleting row i and column j of matrix A. Tips to stay focused and finish your hobby project, Podcast 292: Goodbye to Flash, we’ll see you in Rust, MAINTENANCE WARNING: Possible downtime early morning Dec 2, 4, and 9 UTC…, Congratulations VonC for reaching a million reputation. After we have converted a matrix into a triangular form, we can simply multiply the elements in the diagonal to get the determinant of a matrix. … To subscribe to this RSS feed, copy and paste this URL into your RSS reader. no change to the determinant multiplying a row by a constant, c. multiplies the determinant by c Adding c times one row to another. 4 times $$\begin{pmatrix} -8 & 0 \\ 5 & 0 \\ \end{pmatrix}$$ giving 4(0-0)= 0 . We explain Finding the Determinant of a 4x4 Matrix with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. given that (where a, e, g, n, o, p, and r are scalars). It means, we set i=1, while j is changing from 1 to 3. If we multiply one row with a constant, the determinant of the new matrix is the determinant of … You can also calculate a 4x4 determinant on the input form. Example of the Laplace expansion according to the first row on a 3x3 Matrix. Are there any contemporary (1990+) examples of appeasement in the diplomatic politics or is this a thing of the past? And let's see if we can figure out its determinant, the determinant of A. M3 --> $-1^4 = 1$ We explain Finding the Determinant of a 4x4 Matrix with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. And before just doing it the way we've done it in the past, where you go down one of the rows or one of the columns-- and you notice, there's no 0's here, so there's no easy row or easy column to take the determinant … determinant algorithm of a 4x4 matrix. The formula for the determinant is shown here: 1. for 2x2 matrix 2. for 3x3 matrix 3. for 4x4 matrix Question: Give an example of finding the determinant of a 4x4 matrix? Here we have no zero entries, so, actually, it doesn’t matter what row or column to pick to perform so called Laplace expansion. Beds for people who practise group marriage, Grammatical structure of "Obsidibus imperatis centum hos Haeduis custodiendos tradit", what does "scrap" mean in "“father had taught them to do: drive semis, weld, scrap.” book “Educated” by Tara Westover. Example 5: Apply the result det ( A T) = det A to evaluate . The determinant is –12. One possibility to calculate the determinant of a matrix is to use minors and cofactors of a square matrix. The determinant of our matrix, a, is equal to this guy-- a, 1, 1-- times the determinant of its submatrix. The inverse of a matrix A is another matrix denoted by A−1and isdefined as: Where I is the identitymatrix. Oct 6, 2019; 3 min read; Inverse Of 4x4 Matrix Example Pdf Download ⎠.. We are working with a 4x4 matrix, so it has 4 rows and 4 columns. 3x3 determinants are defined in terms of 2x2 determinants, 4x4 determinants are defined in terms of 3x3 determinants. The most efficient way to evaluate a 4 x 4 determinant is to use row reduction to create zeros in a row or column, and then to use expansion by minors along that row/column. May 2009; Conference: 3-rd INTERNATIONAL MATHEMATICS CONFERENCE ON ALGEBRA AND FUNCTIONAL ANALYSIS May 15-16, 2009 ; … Since one row exchange reverses the sign of the determinant (Property 2), two-row exchanges, will leave the determinant unchanged: But the determinant of a matrix is equal to the determinant of its transpose, so . By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Each square matrix can be assigned a unique number, which is called the determinant (det(A)) of the matrix. Does anyone know what I did wrong? Summing up. As a whole, the code should look like: Personally, I find your variable names confusing. What is the best algorithm for overriding GetHashCode? Is the intensity of light ONLY dependent on the number of photons, and nothing else? So I request you to follow this.. Ask Question Asked 6 years, 9 months ago. Note that if A ~ B, then ρ(A) = ρ(B) Why does vaccine development take so long? site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. The determinant of a matrix is a special number that can be calculated from a square matrix. What happens to excess electricity generated going in to a grid? To see what I did look at the first row of the 4 by 4 determinant. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Another index i is changing from 1 to 4 (n=4 because A is 4×4 matrix). A Matrix is an array of numbers: A Matrix (This one has 2 Rows and 2 Columns) The determinant of that matrix is (calculations are explained later): 3×6 − 8×4 = 18 − 32 = −14. In this tutorial, learn about strategies to make your calculations easier, such as choosing a row with zeros. Solution. Therefore, #determinant of a 4x4 matrix #matrices and determinants Author: Learn Mathematics Step By Step Views: 22 (PDF) New Method to Compute the Determinant of a 4x4 Matrix https://www.researchgate.net › publication › 275580759 In this paper we will present a new method to compute the determinants of a 4 × 4 matrix. So, for example, the first column of A is the first row of the.. Find the difference of the cross products. Linear Algebra: Find the determinant of the 4 x 4 matrix A = [1 2 1 0 \ 2 1 1 1 \ -1 2 1 -1 \ 1 1 1 2] using a cofactor expansion down column 2. For example, a matrix is often used to represent the coefficients in a system of linear equations, and the determinant can be used to solve those equations, although other methods of solution are much more computationally efficient. Date: 12/19/96 at 03:27:42 From: Doctor Pete Subject: Re: Determinants The most efficient way to evaluate a 4 x 4 determinant is to use row reduction to create zeros in a row or column, and then to use expansion by minors along that row/column. It means that we set j=1 in general formula for calculating determinants which works for determinants of any size: \Delta =\sum_{i,j=1}^{n}(-1)^{i+j}a_{ij}M_{ij}. This page explains how to calculate the determinant of a 3x3 matrix. For example, the following matrix is not singular, and its determinant (det(A) in Julia) is nonzero: In [1]:A=[13 24] det(A) Out[1]:-2.0 (You may even remember the formula for the 2 2 determinant: 1 4 3 2 = 2. Matrix Algebra videos- http://goo.gl/4gvpeC I'm Sujoy and today I'll tell you how to find determinant of a 4x4 Matrix using Chio's Method. Finding the determinant of a symmetric matrix is similar to find the determinant of the square matrix. getDet3 returns determinant of a 3x3 matrix and it works fine. 3. Matrix A is a square 4×4 matrix so it has determinant. That's going to be a, 2, 2. If A is square matrix then the determinant of matrix A is represented as |A|. The determinant …. Treat the remaining elements as a 2x2 matrix. Thanks! Find detA or 1141. A 4 by 4 determinant can be expanded in terms of 3 by 3 determinants called minors. det A = a 1 1 a 1 2 a 1 3 a 2 1 a 2 2 a 2 3 a 3 1 a 3 2 a 3 3. If A and B are two equivalent matrices, we write A ~ B. Then the matrix has an inverse, and it can be found using the formula ab cd 1 = 1 det ab cd d b ca Notice that in the above formula we are allowed to divide by the determi-nant since we are assuming that it’s not 0. 15c48777a1 kunci jawaban workbook english alive yudhistira howbani soft 7.5 … Determinant. 0 3 3x3 5 2 2 0 —3 4 . Here, it refers to the determinant of the matrix … Viewed 80k times 4. Expansion using Minors and Cofactors. Determinant of a 4×4 matrix is a unique number which is calculated using a particular formula. This example finds the determinant of a matrix with three rows and three columns. Have you debugged the code and checked that. Why a probability distribution can be viewed as a price? Determinant of a Matrix. = 8 – 20. It's still a matrix, but it's smaller. It goes all the way to a, 2, n, and then a, 3, 3, all the way to a, n, n. And then, everything down here is-- these are all 0's. If we swap two rows and columns of the matrix, then we will get the determinant with an opposite sign. A Matrix (This one has 2 Rows and 2 Columns) The determinant of that matrix is (calculations are explained later): Whether it’s…, Effective time management plan to improve your studies, College and University are a time for life experience and fun, as well as a time for development and growth.…. Example. The determinant of b is going to be equal to a times the submatrix if you were to ignore a's row and column. Don’t confuse the |A| notation with absolute value notation. How to make rope wrapping around spheres? Get the free "4x4 Determinant calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. 2) The computation of pow(-1, i+j) should only depend on i, since j has the same value every time in that expression (namely, 3). 76. 3x3 and 4x4 matrix determinants and Cramer rule for 3x3.notebook 4 April 14, 2015 Homework: Pg. obviously the next matrix will look the same as the top term in column two is a zero so the determinant for that will be $0$. Find the Determinant with Given Accuracy. Suppose that the determinant of the 2⇥2matrix ab cd does not equal 0. Here we have no zero entries, so, actually, it doesn’t matter what row or column to pick to perform so called Laplace expansion. Example given that (where a, e, g, n, o, p, and r are scalars). This determinant is called the minor of the element we ... For example, if using this for a 4x4 matrix, your "crossing out" leaves you with a 3x3 matrix, for which you calculate the determinate as described above. (2*2 - 7*4 = -24) Multiply by the chosen element of the 3x3 matrix.-24 * 5 = -120; Determine whether to multiply by -1. These are the determinants of order 3×3: Let’s calculate them in the same way, but choosing the first row. I forgot the two points. What do we have left? What is the optimal algorithm for the game 2048? Row separator Input matrix row separator. We’re continuing to prepare math tutorials. In general we could pick any row or column we want: \begin{aligned} M_{11}&=\begin{vmatrix}-1&2 &5 \\2& 3 &4 \\4&-1 &2\end{vmatrix}=-1 \cdot\begin{vmatrix}3&4 \\-1&2 \end{vmatrix}-2\cdot\begin{vmatrix}2&4 \\4&2 \end{vmatrix}+5\begin{vmatrix}2&3 \\4&-1\end{vmatrix}\\&=-1\cdot (6+4)-2\cdot (4-16)+5\cdot(-2-12)=-10+24-70=-56\end{aligned}. To learn more, see our tips on writing great answers. Assignment help on Master’s and Ph.D. level. Misplaced comma after LTR word in bidirectional document. Finding the determinant of a 4x4 matrix can be difficult. New Method to Compute the Determinant of a 4x4 Matrix. Determinant of 4x4 Matrix. In our example, this smaller matrix has the rows: 3 5 7, 3 5 6, and 2 6 3. While finding the determinant of a 4x4 matrix, it is appropriate to convert the matrix into a triangular form by applying row operations in the light of the Gaussian elimination method. Note that this time upper limit equals 3 instead of 4 for the initial determinant, because these are determinants of the size 3×3. In this leaflet we explain what is meant by an inverse matrix and how it is ... the rows and columns of A. For convenience choose the row or column with smallest values or better with zeros if possible. Hopefully, it will be helpful for you and you won’t be in need asking our experts something like “Do my math homework for me, please! where A ij, the sub-matrix of A, which arises when the i-th row and the j-th column are removed. Answer There are mainly two ways to obtain the inverse matrix. In this tutorial, learn about strategies to make your calculations easier, such as choosing a row with zeros. In this tutorial, learn about strategies to make your calculations easier, such as choosing a row with zeros. Do you need to roll when using the Staff of Magi's spell absorption? Enter the coefficients. Does anyone know what I did wrong? 44 matrix is the determinant of a 33 matrix, since it is obtained by eliminating the ith row and the jth column of #. Well, there are a few mistakes in your code. Determinant 4x4. There is a way to find an inverse of a 3 ⇥ 3matrix–orforthatmatter, an n⇥n matrix – whose determinant is not 0, but it isn’t quite as simple as finding the inverse of a 2⇥2matrix.Youcanlearnhowtodoitifyoutakea linear algebra course. IMHO, it would have been less confusing to have named them j_p andk_, or even to just use the equivalent expression. We’re asked to calculate the determinant of the following 4×4 matrix: A=\begin{pmatrix}-1 & {1} & 4 & 2 \\2 & -1 & 2 & 5 \\1 &2 & 3& 4\\3& 4& -1 & 2\end{pmatrix}. Determinant of a Matrix. Helpful 0 Not Helpful 0. Nope.. © 2019 BrainRouter LTD. All rights reserved. Therefore, If it’s so, then you can proceed and apply general formula for calculating determinants which goes as follows: Here $n$ is the size of your square matrix. Example 1. If a matrix order is n x n, then it is a square matrix. ... For example, 1 0 0 1 0 2 0 0 0 0 3 0 0 0 0 4 in this case the method returns nan. rev 2020.12.4.38131, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide, Why you are using cycle variable j in pow(-1, i+j)? Example 5: Apply the result det ( A T) = det A to evaluate . In our example, the determinant of the matrix () = 4 * 2 - 7 * 6 = -34. Next step in calculating a matrix determinant, Finding the determinant of an N x N matrix entered as list of lists - “list index out of range”, Perl - Determinant of Matrix Containing Variables. It means that we set j=1 in general formula for calculating determinants which works for determinants of any size: You can also calculate a 3x3 determinant on the input form. Finding the determinant of a 4x4 matrix can be difficult. How to Effectively Write an Argumentative Essay, Opinions are a dime-a-dozen so there isn’t anything inherently original or outstanding as the next one. This page explains how to calculate the determinant of 4 x 4 matrix. Recall that only square matrices have a determinant, for non-square ones it’s not defined. Let's find the determinant along this column right here. 3x3 and 4x4 matrix determinants and Cramer rule for 3x3.notebook 2 April 14, 2015 Cramer's Rule for 3x3: 3x3 and 4x4 matrix determinants and Cramer rule for 3x3.notebook 3 April 14, 2015 A 4x4 is four 3x3’s!! So it actually just keeps on incrementing, resulting in array indices outside of the array bounds. k : k + 1. SEMATH INFO. A matrix obtained from a given matrix by applying any of the elementary row operations is said to be equivalent to it. Determinant of a matrix is calculated using the det function of MATLAB. Required options. The online calculator calculates the value of the determinant of a 4x4 matrix with the Laplace expansion in a row or column and the gaussian algorithm. Substituting the values into general formula we obtain: \begin{aligned} \Delta &=\sum_{i,j=1}^{4}(-1)^{i+j}a_{i1}M_{i1}\\&=(-1)^{1+1}a_{11}M_{11}+(-1)^{2+1}a_{21}M_{21}+(-1)^{3+1}a_{31}M_{31}+(-1)^{4+1}a_{41}M_{41}\\&=a_{11}M_{11}-a_{21}M_{21}+a_{31}M_{31}-a_{41}M_{41}=-1\cdot M_{11}-2 \cdot M_{21}+1\cdot M_{31}-3 \cdot M_{41} \end{aligned}. It does not give only the inverse of a 4x4 matrix and also it gives the determinant and adjoint of the 4x4 matrix that you enter. This row is 1, 4, 2, 3. In our example, the matrix is () Find the determinant of this 2x2 matrix. What is the relationship between where and how a vibrating string is activated? Advertisement. Subscribe to our Youtube Channel - https://you.tube/teachoo. your coworkers to find and share information. Your rule for 3x3 determinants works because the formula for a 2x2 determinant is nice and simple with only 2 terms. The determinant of a triangular matrix is the product of the entries on the diagonal. To find the value of \Delta we need to calculate minors M_{11}, M_{21}, M_{31}, M_{41}. In this article, we will write a C# program to calculate Matrix Determinant [crayon-5fc98ee747368278779043/] Output: Enter the order of determinant: 2 Order of determinant entered:2 E… Now finally . The determinant of a square matrix A is the integer obtained through a range of methods using the elements of the matrix. no change to the determinant These are called elementary row operations, and using the first two, with c=-1, you can create that skew-symmetric matrix from the identity matrix and know that the determinant is 1. Question: Give an example of finding the determinant of a 4x4 matrix? 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1. ## Differentiation
I'd like help with all of the first exercise, and then the two last parts of the last two exercises, 'cause I've done the first parts.
2. Find the equation of the normal line to
$\displaystyle y = \frac{1}{(x^2 + 1)^2}$
at the point
$\displaystyle \left ( 1, \frac{1}{4} \right )$
First we want the value of the slope of the tangent line at this point. So:
$\displaystyle \frac{dy}{dx} = -\frac{4x}{(x^2 + 1)^3}$
At x = 1 this is
$\displaystyle \frac{dy}{dx}_{x = 1} = -\frac{1}{2}$
Now, the line normal to the curve at this point has a slope of $\displaystyle -\frac{1}{-\frac{1}{2}} = 2$
So it has the form
$\displaystyle y = 2x + b$
We know this line passes through the point $\displaystyle \left ( 1, \frac{1}{4} \right )$, so
$\displaystyle \frac{1}{4} = 2 \cdot 1 + b$
$\displaystyle b = -\frac{7}{4}$
Thus the normal line is $\displaystyle y = 2x - \frac{7}{4}$
-Dan
3. Come now! The second part of number 3 is not a Calculus problem.
For what values of x is
$\displaystyle \frac{dy}{dx} = \frac{x^2 + 4x - 7}{(x + 2)^2}$
0 or undefined.
For the first part solve
$\displaystyle \frac{x^2 + 4x - 7}{(x + 2)^2} = 0$
For the second part, consider what values of x make the denominator 0 (which is forbidden.)
-Dan
4. Thanks for running me through the first one, I'm pretty sure I got what you did.
I'm not totally sure how to solve for 0...
5. Originally Posted by Fnus
Thanks for running me through the first one, I'm pretty sure I got what you did.
I'm not totally sure how to solve for 0...
Originally Posted by topsquark
For the first part solve
$\displaystyle \frac{x^2 + 4x - 7}{(x + 2)^2} = 0$
$\displaystyle \frac{x^2 + 4x - 7}{(x + 2)^2} = 0$
Multiply both sides by $\displaystyle (x + 2)^2$:
$\displaystyle x^2 + 4x - 7 = 0$
Can you solve this?
-Dan
6. Yep, I can.
Thanks a ton
- Frederikke
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# Limsup and Liminf
Let an (n = 1, 2,…) be a sequence of real numbers, and define the sequence bn as
bn = sup am
m>n
Then bn is a nonincreasing sequence: bn > bn+1 because, if an is greater than the smallest upper bound of an+1, an+2, an+3,…, then an is the maximum of an, an+1, an+2, an+3,..hence, bn = an > bn+1 and, if not, then bn = bn+1 ■ Nonincreasing sequences always have a limit, although the limit may be – to. The limit of bn in (II.1) is called the limsup of an:
Note that because bn is nonincreasing, the limit of bn is equal to the infimum of bn. Therefore, the limsup of an may also be defined as
Note that the limsup may be +to or – to, for example, in the cases an = n and an = —n, respectively.
Similarly, the liminf of an is defined by
liminfan = lim inf am
n — то n —— to V m>n
or equivalently by
liminfan = sup inf am
n — TO n>1 m>n
Again, it is possible that the liminf is +to or —to.
Note that liminfn—TO an < limsupn—TO an because infm>n am < supm>n am for all indices n > 1, and therefore the inequality must hold for the limits as well.
Theorem II.1:
(a) If liminf n—TO an = limsupn—TO an, then limn—TO an = limsupan, and if
n — TO
liminfn—TO an < limsupn—TO an, then the limit of an does not exist.
(b) Every sequence an contains a subsequence ank such that limt—TO ant = limsupn—TO an, and an also contains a subsequence anm such that
limm — TO anm = liminf n—— TO an.
Proof: The proof of (a) follows straightforwardly from (II.2), (II.4), and the definition of a limit. The construction of the subsequence ant in part (b) can be done recursively as follows. Let b = limsupn—TO an < to. Choose n1 = 1, and suppose that we have already constructed a^. for j = 1t > 1. Then there exists an index nt+1 > nt such that ant+1 > b — 1/(t + 1) because, otherwise, am < b — 1/(t + 1) for all m > nt, which would imply that limsupn—TO an < b — 1/(t + 1). Repeating this construction yields a subsequence ant such that, from large enough t onwards, b — 1/1 < ant < b. If we let t — to, the limsup case of part (b) follows. If limsupn—TO an = to, then, for each index nt we can find an index nt+i > nt such that ant+l > t + 1; hence, limt—TO ant = to. The subsequence in the case limsupn—TO an = —to and in the liminf case can be constructed similarly. Q. E.D.
The concept of a supremum can be generalized to sets. In particular, the countable union U^TO=1 Aj may be interpreted as the supremum of the sequence of sets Aj, that is, the smallest set containing all the sets Aj. Similarly, we may interpret the countable intersection n^TO=1 Aj as the infimum of the sets Aj, that is, the largest set contained in each of the sets Aj. Now let Bn = UTO=n Aj for
n = 1, 2, 3,____ This is a nonincreasing sequence of sets: Bn+1 c Bn; hence,
nn=1 Bn = Bn. The limit ofthis sequence ofsets is the limsup of An for n — to, that is, as in (II.3) we have
def. to / to
limsup An = n I U Aj
n — TO n = 1 j =n
Next, let Cn = nTO=n Aj for n = 1, 2, 3,____ This is a nondecreasing sequence
of sets: Cn c Cn+1; hence, Un=1 Cn = Cn. The limit of this sequence of sets is the liminf of An for n — to, that is, as in (II.5) we have
_ def. TO(TO
liminf An = U n A j
n — TO n=1 у j =n
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# Finite Sets and Infinite Sets
What are the differences between finite sets and infinite sets?
Finite set: A set is said to be a finite set if it is either void set or the process of counting of elements surely comes to an end is called a finite set.
In a finite set the element can be listed if it has a limited i.e. countable by natural number 1, 2, 3, ……… and the process of listing terminates at a certain natural number N.
The number of distinct elements counted in a finite set S is denoted by n(S). The number of elements of a finite set A is called the order or cardinal number of a set A and is symbolically denoted by n(A).
Thus, if the set A be that of the English alphabets, then n(A) = 26: For, it contains 26 elements in it. Again if the set A be the vowels of the English alphabets i.e. A = {a, e, i, o, u} then n(A) = 5.
Note:
The element does not occur more than once in a set.
Infinite set: A set is said to be an infinite set whose elements cannot be listed if it has an unlimited (i.e. uncountable) by the natural number 1, 2, 3, 4, ………… n, for any natural number n is called a infinite set.
A set which is not finite is called an infinite set.
Now we will discuss about the examples of finite sets and infinite sets.
Examples of finite set:
1. Let P = {5, 10, 15, 20, 25, 30}
Then, P is a finite set and n(P) = 6.
2. Let Q = {natural numbers less than 25}
Then, Q is a finite set and n(P) = 24.
3. Let R = {whole numbers between 5 and 45}
Then, R is a finite set and n(R) = 38.
4. Let S = {x : x ∈ Z and x^2 – 81 = 0}
Then, S = {-9, 9} is a finite set and n(S) = 2.
5. The set of all persons in America is a finite set.
6. The set of all birds in California is a finite set.
Examples of infinite set:
1. Set of all points in a plane is an infinite set.
2. Set of all points in a line segment is an infinite set.
3. Set of all positive integers which is multiple of 3 is an infinite set.
4. W = {0, 1, 2, 3, ……..} i.e. set of all whole numbers is an infinite set.
5. N = {1, 2, 3, ……….} i.e. set of all natural numbers is an infinite set.
6. Z = {……… -2, -1, 0, 1, 2, ……….} i.e. set of all integers is an infinite set.
Thus, from the above discussions we know how to distinguish between the finite sets and infinite sets with examples.
`
Set Theory
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# Area Problem
## Related calculator: Definite and Improper Integral Calculator
Suppose that we are given continuous function $y={f{{\left({x}\right)}}}$ on ${\left[{a},{b}\right]}$ such that ${f{{\left({x}\right)}}}\ge{0}$ for all ${x}\in{\left[{a},{b}\right]}$.
We want to find area ${S}$ that lies under curve ${f{{\left({x}\right)}}}$ and bounded by lines ${x}={a}$, ${x}={b}$ and x-axis.
Before doing this imagine right bound ${b}$ is not fixed, we can move it. Then for different values of right bound we will obtain different values of area, so we can consider area as function ${P}{\left({x}\right)}$ of right bound. In particular, ${P}{\left({a}\right)}={0}$ (because area from ${a}$ to ${a}$ is 0) and ${P}{\left({b}\right)}={S}$.
To get a better understanding what is ${P}{\left({x}\right)}$ consider next example.
Example 1. Consider function ${y}={x}-{1}$. If ${P}{\left({x}\right)}$ represents area of figure under ${y}={x}$, above x-axis on interval ${\left[{1},{x}\right]}$ find ${P}{\left({1}\right)}$, ${P}{\left({2}\right)}$, ${P}{\left({3}\right)}$ and ${P}{\left({4}\right)}$.
${P}{\left({1}\right)}$ is area on interval ${\left[{1},{1}\right]}$, so it is 0 (because length of interval is 0).
${P}{\left({2}\right)}$ is area on interval ${\left[{1},{2}\right]}$. It is area of right-angled triangle with legs 1 and 1. So, ${P}{\left({2}\right)}=\frac{{1}}{{2}}\cdot{1}\cdot{1}=\frac{{1}}{{2}}$.
${P}{\left({3}\right)}$ is area on interval ${\left[{1},{3}\right]}$. It is area of right-angled triangle with legs 2 and 2. So, ${P}{\left({3}\right)}=\frac{{1}}{{2}}\cdot{2}\cdot{2}={2}$.
${P}{\left({4}\right)}$ is area on interval ${\left[{1},{4}\right]}$. It is area of right-angled triangle with legs 3 and 3. So, ${P}{\left({4}\right)}=\frac{{1}}{{2}}\cdot{3}\cdot{3}=\frac{{9}}{{2}}$.
As can be seen ${P}{\left({x}\right)}$ is indeed function: for any particular value of ${x}$ there is corresponding ${P}{\left({x}\right)}$.
Now, suppose we want to find derivative of ${P}{\left({x}\right)}$. For this increase ${x}$ by $\Delta{x}$. Corresponding increase in area is $\Delta{P}={P}{\left({x}+\Delta{x}\right)}-{P}{\left({x}\right)}$ (figure ACEF).
If we denote maximum and minimum value of ${f{{\left({x}\right)}}}$ on interval ${\left[{x},{x}+\Delta{x}\right]}$ by ${M}$ and ${m}$ respectively, then area of figure ACEF is less than area of rectangle ACDF and greater than area of rectangle ABEF.
Since area of rectangle ACDF is ${M}\Delta{x}$ (height multiplied by base)and area of rectangle ABEF is ${m}\Delta{x}$ then
${m}\Delta{x}<\Delta{P}<{M}\Delta{x}$ or ${m}<\frac{{\Delta{P}}}{{\Delta{x}}}<{M}$.
Now let $\Delta{x}\to{0}$. Since ${f{}}$ is continuous then both ${m}$ and ${M}$ approach ${f{{\left({x}\right)}}}$.
Thus, by Squeeze Theorem $\lim_{{\Delta{x}\to{0}}}\frac{{\Delta{P}}}{{\Delta{x}}}={f{{\left({x}\right)}}}$.
But we recognize in $\lim_{{\Delta{x}\to{0}}}\frac{{\Delta{P}}}{{\Delta{x}}}$ derivative of ${P}{\left({x}\right)}$, so ${P}'{\left({x}\right)}={f{{\left({x}\right)}}}$.
In other words ${P}{\left({x}\right)}=\int{f{{\left({x}\right)}}}{d}{x}={F}{\left({x}\right)}+{C}$.
Since ${P}{\left({a}\right)}={0}$ then ${0}={P}{\left({a}\right)}={F}{\left({a}\right)}+{C}$ or ${C}=-{F}{\left({a}\right)}$.
Therefore, ${P}{\left({x}\right)}={F}{\left({x}\right)}-{F}{\left({a}\right)}$.
Newton-Leibniz Formula. Area of figure ${P}{\left({x}\right)}$ that lies under curve ${y}={f{{\left({x}\right)}}}$, above x-axis on interval ${\left[{a},{x}\right]}$ is ${P}{\left({x}\right)}={F}{\left({x}\right)}-{F}{\left({a}\right)}$.
Returning to our initial problem we have that ${S}={F}{\left({b}\right)}-{F}{\left({a}\right)}$.
Area of figure that lies under curve ${y}={f{{\left({x}\right)}}}$, above x-axis and between lines ${x}={a}$ and ${x}={b}$ is ${S}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ where ${F}$ is antiderivative of ${f{}}$.
Example 2. Find area under ${y}={{x}}^{{2}}$, above x-axis on interval ${\left[{0},{2}\right]}$.
We have that ${F}{\left({x}\right)}=\int{{x}}^{{2}}{d}{x}=\frac{{1}}{{3}}{{x}}^{{3}}$ (note that we need particular derivative here).
Now, according to Newton-Leibniz Formula required area is ${S}={F}{\left({2}\right)}-{F}{\left({0}\right)}=\frac{{1}}{{3}}\cdot{{2}}^{{3}}-\frac{{1}}{{3}}\cdot{{0}}^{{3}}=\frac{{8}}{{3}}$.
All from above can be easily extended on case when function can take negative values. For this we only need to consider areas under x-axis negative.
Indeed, if we want to find area enclose by parabola ${y}=-{{x}}^{{2}}$, x-axis on interval ${\left[{0},{2}\right]}$ then according to Newton-Leibniz Formula we have that ${S}={F}{\left({2}\right)}-{F}{\left({0}\right)}$ where ${F}{\left({x}\right)}=\int-{{x}}^{{2}}{d}{x}=-\frac{{1}}{{3}}{{x}}^{{3}}$. So, ${S}=-\frac{{8}}{{3}}$. Since $-{{x}}^{{2}}$ is reflection of ${{x}}^{{2}}$ about x-axis then this area lies below x-axis, and, therefore, negative.
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# How Do You Calculate The Y Intercept?
## How do you find the Y intercept in an equation?
To find the x-intercept of a given linear equation, plug in 0 for ‘y’ and solve for ‘x’.
To find the y-intercept, plug 0 in for ‘x’ and solve for ‘y’..
## How do you find the Y intercept when given two points?
You can find the y-intercept (which is also the ‘b’ in y = mx+b) by plugging in the y and x to y = mx+b.
## What is an example of Y intercept?
The y -intercept of a graph is the point where the graph crosses the y -axis. For example, we say that the y -intercept of the line shown in the graph below is 3.5 . … When the equation of a line is written in slope-intercept form ( y=mx+b ), the y -intercept b can be read immediately from the equation.
## What is the Y intercept in a regression equation?
The intercept (often labeled the constant) is the expected mean value of Y when all X=0. Start with a regression equation with one predictor, X. If X sometimes equals 0, the intercept is simply the expected mean value of Y at that value.
## How do you find an equation given two points?
Find the Equation of a Line Given That You Know Two Points it Passes Through. The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept.
## How do you find the Y intercept of data?
To find the y-intercept, calculate and , the average of the x- and y-values respectively. Then substitute these two values for x and y in the = b + a equation.
## What is the Y intercept in standard form?
Standard form is another way to write slope-intercept form (as opposed to y=mx+b). It is written as Ax+By=C. You can also change slope-intercept form to standard form like this: Y=-3/2x+3. Next, you isolate the y-intercept(in this case it is 3) like this: Add 3/2x to each side of the equation to get this: 3/2x+y=3.
## How do you find the y value?
Finding the y value is easy if you know the slope of the line and the x coordinate. Review the equation for the slope of a line. The equation for finding the slope is: m = [y1 – y2] / [x1 – x2]. If you know x, you can solve for y to find the y value for the slope of the line.
## How do you find the Y intercept without a graph?
Using the “slope-intercept” form of the line’s equation (y = mx + b), you solve for b (which is the y-intercept you’re looking for). Substitute the known slope for m, and substitute the known point’s coordinates for x and y, respectively, in the slope-intercept equation. That will let you find b.
## How do you convert Y intercept to standard form?
To convert from slope intercept form y = mx + b to standard form Ax + By + C = 0, let m = A/B, collect all terms on the left side of the equation and multiply by the denominator B to get rid of the fraction.
## What is standard form in math?
Standard form is a way of writing down very large or very small numbers easily. 103 = 1000, so 4 × 103 = 4000 . So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form. Small numbers can also be written in standard form.
## How do you find the slope and y intercept of an equation?
The slope-intercept form of a line is: y=mx+b where m is the slope and b is the y-intercept. The y-intercept is always where the line intersects the y-axis, and will always appear as (0,b) in coordinate form.
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# How do you solve 4x^2+17x+15=0?
Jun 11, 2018
$x = - 3$ or $x = - \frac{5}{4}$
#### Explanation:
$4 {x}^{2} + 17 x + 15 = 0$
Do
$4 {x}^{2} + 12 x + 5 x + 15 = 0$
Then
$4 x \left(x + 3\right) + 5 \left(x + 3\right) = 0$
$\left(4 x + 5\right) \left(x + 3\right) = 0$
So
$x = - 3 \text{ }$ or $\text{ } x = - \frac{5}{4}$
Jun 11, 2018
$- 3 , \mathmr{and} - \frac{5}{4}$
#### Explanation:
Method 1. Use the improved quadratic formula
$y = 4 {x}^{2} + 17 x + 15 = 0$.
$D = {d}^{2} = {b}^{2} - 4 a c = 289 - 240 = 49$ --> $d = \pm 7$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{17}{8} \pm \frac{7}{8}$
$x 1 = - \frac{24}{8} = - 3$
$x 2 = - \frac{10}{8} = - \frac{5}{4}$
Method 2. Use the new Transforming Method (Socratic, Google Search)
Transformed equation: $y ' = {x}^{2} + 17 x + 60$.
Proceeding. Find the 2 real roots of y', then, divide them by a = 4.
Find 2 real roots, bot negative (Rule of signs), knowing the sum (-b = -17), and the product (ac = 60). They are - 5 and - 12.
The 2 real roots of y are:
$x 1 = - \frac{5}{a} = - \frac{5}{4}$, and $x 2 = - \frac{12}{a} = - \frac{12}{4} = - 3$
Jun 11, 2018
$x - 3 \mathmr{and} x = - \frac{5}{4}$
#### Explanation:
Find factors of $4 \mathmr{and} 15$ whose products add to $17$
$\text{ } 4 \mathmr{and} 15$
$\text{ "darr" } \downarrow$
$\text{ "4" "5" } \rightarrow 1 \times 5 = 5$
$\text{ "1" "3" } \rightarrow 4 \times 3 = \underline{12}$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times x} 17$
The signs are all positive.
$\left(4 x + 5\right) \left(x + 3\right) = 0$
Either factor can be equal to $0$
If $4 x + 5 = 0 \text{ } \rightarrow x = - \frac{5}{4}$
If $x + 3 = 0 \text{ } \rightarrow x = - 3$
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# Equation of a Circle
Last Updated : 05 May, 2024
Equation of a circle is the algebraic way to define a circle. It is the locus of the point which is at a constant distance from a fixed point. The fixed point is called the centre of the circle, and the constant distance is called the radius of the circle. The equation of a circle is very useful for finding different parameters such as the circumference, and area of the circle.
A circle in 2-D and the 3-D planes is represented using the equation of the circle. It is the algebraic equation solution which represents all the points on the circumference of the circle. We can represent the equation of the circle in various forms, some of the common forms of the equation of the circle are,
• General form
• Standard form
• Parametric form
• Polar form
In this article, let’s learn about the equation of the circle, various forms of circle, and how to find the equation of the circle etc.
## What is the Equation of a Circle?
The equation of a circle is a way to represent the curve of a circle in the 2-D and the 3-D plane. It is the locus of the point which represents the path of the point which is at a fixed distance from a fixed point. The algebraic equation of the circle can be found using the coordinate of the centre and the length of the radius of the circle, where the centre of the circle is the fixed point from which the radius is calculated.
In general, if the centre of the circle is represented with (h, k) and the radius of the circle is ‘r’ then the equation of the circle is,
(x-h)2 + (y-k)2 = r2
This equation represents all the points that are at r distance from the point (h, k) in the 2-D coordinate plane.
## Different Forms of Equation of Circle
We can represent the equation of a circle using various forms. In each form, a different set of values is given which is used to form the equation of the circle. All these forms can represent the same circle, but their initial parameters are different. In all these forms of equations of a circle, we can easily find the radius and centre of the circle.
Various forms to represent the equation of a circle are,
• General form
• Standard form
• Parametric form
• Polar form
We will learn about all these equations of circles in detail in this article.
## General Equation of a Circle
The general form of the equation of a circle is represented as,
x2 + y2 + 2gx + 2fy + c = 0
where,
g, f, and c are contsnat values
Using this general form of the equation we can easily find the centre and radius of the circle. For, finding the centre and the radius of the circle we use the completing the square method and then compare it with the standard form.
Converting the general form of the equation x2 + y2 + 2gx + 2fy + c = 0, into the standard form
x2 + y2 + 2gx + 2fy + c = 0
Adding f2, g2, and -c on both sides,
x2 + y2 + 2gx + 2fy + c + f2 + g2 -c = f2 + g2 -c
x2 + 2gx + g2 + y2 + 2fy + f2 = f2 + g2 -c
(x + g)2 + (y – f)2 = [√(f2 = f2 + g2 -c)]2
comparing with (x-h)2 + (y-k)2 = r2
we get,
• centre = (h, k) = (-g, -f)
• radius = r = √(f2 = f2 + g2 -c)
### Features of General Equation of Circle
x2 + y2 + 2gx + 2fy + c = 0, represents the general equation of a circle with centre (−g, −f) and radius (r): r2 = g2 + f2− c. Some of the important results which we deduce from the general equation of the circle are,
• If g2 + f2 > c, then the radius of the circle is real.
• If g2 + f2 = c, then the radius of the circle is zero which tells us that the circle is a point which coincides with the centre. Such type of circle is called a point circle
• g2 + f2 <c, then the radius of the circle becomes imaginary. Therefore, it is a circle having a real centre and imaginary radius.Â
## Equation of a Circle in Standard Form
The standard equation of the circle is also the basic equation of the circle. It is the most common equation of a circle and is widely used. It gives the centre and radius of the circle in the easiest and most efficient way. The standard form of the equation of the circle isÂ
(x-h)2 + (y-k)2 = r2
where, (h, k) represents the centre of the circle and r represents the radius of the circle. Here, (x, y) are any arbitrary point on the circumference of the circle from which the centre of the circle is at the constant distance.
To find the standard form of the circle let us assume that A(h, k) is the centre of the circle and r is the radius of the circle and P(x, y) is any point on the circumference.Â
Then, AP = r …..(i)
Now using the distance formula we find the value of the AP
AP =Â
Now put the value of AP in eq(i), and we get
 = r
Or squaring both sides we get
(x – h)2 + (y – k)2 = r2Â
This is the standard form of the equation of the circle.
Example: If the centre and the radius of the circle are (3, 5) and 7 units respectively then find the equation of the circle.
Solution:Â
Centre of the circle is (3, 5) comparing with (h, k) we get h = 3 and k = 5
Similarly, radius of the circle is 7 = r
Equation of the circle is (x – h)2 + (y – k)2 = r2Â
(x – 3)2 + (y – 5)2 = 72Â
This is the required equation of the circle.
## Parametric Equation of a Circle
The parametric form of the circle uses (-h + rcosθ, -k + rsinθ) as the general point on the circumference of the circle. The line joining the centre of the circle to the general point makes an angle θ with the x-axis. Thus, the parametric equation of a circle isÂ
x2 + y2 + 2hx + 2ky + c = 0
General point on the circumference of the circle is (-h + rcosθ, -k + rsinθ).
## Equation of a Circle in Polar Form
The other form of the equation of the circle is the polar form of the circle, this form of the circle is similar to the parametric form of the circle.
Suppose we take a point on the circumference of the circle (x, y) as (acosθ, bsinθ) when the line joining the point with the centre of the circle makes an angle θ with the x-axis and the radius of the circle is ‘a’. Also, the centre of the circle is the origin, (0,0) then the equation of the circle is,
x = a cosθ…(i) and y = a sinθ…(ii)
Now we can easily convert the above equation into the standard form of the circle as,
Squaring and adding eq(i) and eq(ii) we get,
x2 + y2 = a2 cos2θ + a2 sin2 θ
x2 + y2 = a2(cos2θ + sin2 θ)
Using Trigonometric Identity, cos2θ + sin2 θ  =1
x2 + y2 = a2
(x-0)2 + (y-0)2 = a2 comparing with (x – h)2 + (y – k)2 = r2Â we getÂ
• Centre = (h, k) = (0, 0)
• Radius = r = a
Example: Find the equation of the circle in the polar if the equation of the circle in standard form is x2 + y2 = 16.
Solution:
Equation of the circle in polar form is,
x = r cosθÂ
y = r sinθ
using this in x2 + y2 = 16
(r cosθ)2  + (r sinθ)2 = 16
r2(cos2θ + sin2θ) = 16
r2 = 42
r = 4
## Equation of a Circle Formula
The formula used to calculate the equation of the circle is,
(x – x1)2 + (y – y1)2 = r2,
where, (x1, y1) is the centre of the circle and r is the radius of the circle.
This is the basic formula for finding the equation of the circle. We use this formula to easily find the equation of the circle, for example, find the equation of the circle if the centre of the circle is (1, 2) and its radius is 5 units,
Solution:
Given, centre = (1, 2) and radius = 5 units
Equation of the circle fromula is,
(x – x1)2 + (y – y1)2 = r2
(x -1)2 + (y – 2)2 = 52
x2 + 1 – 2x + y2 + 4 – 4y = 25
x2 + y2 – 2x – 4y + 5 = 25
x2 + y2 – 2x – 4y – 20 = 0
This is the general equation of the circle.
## Derivation of Circle Equation
The equation of the circle is easily derived using the distance formula,
Suppose we take a circle with centre (a, b) and radius r then any general point the circumference of the circle is taken as (x, y).
We know that distance between the point (x, y) and the centre (a, b) is given using the distance formula, also the distance between (a, b) and the (x, y) is the radius of the circle, i.e.
Distance between (x, y) and (a, b) = Radius of the Circle
√[(x -a)2 + (y – b)2] = r
Squaring both sides,
[(x -a)2 + (y – b)2] = r2
This is the required equation of the circle.
## Graphing Equation of Circle
As we know that the standard equation of the circle isÂ
(x – h)2 + (y – k)2 = r2Â
Here, the centre of the circle is (h, k) and the radius of the circle is r. Now let us take an example,
The equation of the circle is (x + 2)2 + (y – 6)2 = 4
we can rewrite this equation asÂ
(x – (-2))2 + (y – 6)2 = 22
On comparing with the standard equation of the circle we get
The centre of the circle is (-2, 6) and the radius of the circle is 2.
Now we will draw a circle on the graph.
Step 1: Draw the x and y-axis
Step 2: Plot the centre of the circle on the graph that is (-2, 6)
Step 3: Take the radius of 2 units in the compass and draw the required circle.
## How to Find Equation of Circle?
Equation of the circle can be easily found using the various parameters as given in the questions. If we are given the conditions to find the centre and the radius of the circle then its cartesian equation can easily be written. We will learn to write the equation of a circle under the following conditions:
• Equation of Circle when centre at (x1,y1)
• Equation of Circle with Center at the origin
• Equation of Circle with Centre on X-axis
• Equation of Circle when Centre on Y- axis
• Equation of Circle touching X-axis
• Equation of Circle touching Y-axis
• Equation of Circle touching both axes
Now, we can easily write the equations of the circle using the initial condition. Let’s learn about the different conditions for writing the equation of the circle.
## Equation of Circle when Center at (x1, y1)
The equation of the circle when the centre of the circle is at (x1, y1) and its radius is ‘r'(let) is given using the following steps,
Step 1: Note the centre (x1, y1) and the radius ‘r’ of the circle.
Step 2: Use the Equation of the circle formula,
(x – x1)2 + (y – y1)2 = r2
Step 3: Substitute the value of centre (x1, y1) and radius ‘r’.
Step 4: Simplify the above equation to find the required equation of the circle.
## Equation of Circle With Center at the Origin
The equation of the circle when the centre of the circle is at origin i.e. (0, 0) and its radius is ‘r'(let) is given using the following steps,
Step 1: Note the centre (0, 0) and the radius ‘r’ of the circle.
Step 2: Use the Equation of the circle formula,
(x – x1)2 + (y – y1)2 = r2
Step 3: Substitute the value of centre (0, 0) and radius ‘r’.
Step 4: Simplify the above equation to find the required equation of the circle.
(x – 0)2 + (y – 0)2 = r2
x2 + y2 = r2
This is the required equation of the circle with the centre origin and radius ‘r’.
## Equation of Circle with Centre on x-axis
We know that the general point on the x-axis is (a, 0). Suppose we have to find the equation of the circle with centre at the x-axis i.e. (a, 0) and its radius is ‘r’ unit. Then its equation is given by substituting the value of the centre and radius in,
(x – x1)2 + (y – y1)2 = r2
(x – a)2 + (y – 0)2 = r2
(x – a)2 + y2 = r2
This is the required equation of a circle with a centre on the x-axis.
## Equation of Circle with Centre on y-axis
We know that the general point on the y-axis is (0, b). Suppose we have to find the equation of the circle with the centre at the y-axis i.e. (0, b) and its radius is ‘r’ unit. Then its equation is given by substituting the value of the centre and radius in,
(x – x1)2 + (y – y1)2 = r2
(x – 0)2 + (y – b)2 = r2
x2 + (y – b)2 = r2
This is the required equation of a circle with a centre on the y-axis.
## Equation of Circle Touching x-axis
We can easily find the equation of a circle touching the x-axis. Suppose the circumference of the circle touches the x-axis at point (a, 0) and (a, r) is the centre of the circle with radius r.Â
We know that if a circle touches the x-axis, then its y-coordinate of the centre of the circle is equal to the radius r, this condition is shown in the image below,
Take any general point (x, y) on the circumference of the circle then its equation is given as,
Distance between (x, y) and (a, r) = Radius(r)
√[(x -a)2 + (y-r)2] = r
Squaring both sides we get,
(x -a)2 + (y-r)2 = r2
This is the equation of the circle touching the x-axis.
## Equation of Circle Touching y-axis
We can easily find the equation of a circle touching the y-axis. Suppose the circumference of the circle touches the y-axis at point (0, b) and (r, b) is the centre of the circle with radius r.Â
We know that if a circle touches the y-axis, then its x-coordinate of the centre of the circle is equal to the radius r, this condition is shown in the image below,
Take any general point (x, y) on the circumference of the circle then its equation is given as,
Distance between (x, y) and (r, b) = Radius(r)
√[(x -r)2 + (y-b)2] = r
Squaring both sides we get,
(x -r)2 + (y-b)2 = r2
This is the equation of the circle touching the y-axis.
## Equation of Circle Touching both axes
We can easily find the equation of a circle touching both the axis i.e. x-axis, and the y-axis. Suppose the circumference of the circle touches the x-axis at point (r, 0)and the y-axis at point (0, r) and (r, r) is the centre of the circle with radius r.Â
We know that if a circle touches both x and y-axis, then its x-coordinate and y-coordinate of the centre of the circle are equal to the radius r, this condition is shown in the image below,
Take any general point (x, y) on the circumference of the circle then its equation is given as,
Distance between (x, y) and (r, r) = Radius(r)
√[(x -r)2 + (y-r)2] = r
Squaring both side we get,
(x -r)2 + (y-r)2 = r2
This is the equation of the circle touching both the axes i.e. x-axis and y-axis.
## Converting General Form to Standard Form
The general form of the circle i.e. x2 + y2 + 2gx + 2fy + c = 0 can be easily converted to the standard form of the circle i.e. (x – a)2 + (y – b)2 = r2 using the steps discussed below,
Step 1: Write the like terms together and take the constant on the other side of the equation to as x2 + 2gx + y2 + 2fy = – c…(i)
Step 2: Use completing the square method simplify the left-hand side of the ‘equal to’ by adding g2, and f2 on both sides.
x2 + 2gx + y2 + 2fy + g2 + f2 = – c + g2 + f2
Step 3: Simplify the above equation.
x2 + 2gx + g2 + y2 + 2fy + f2 = g2 + f2 – c
(x+g)2 + (y-f)2 = [√(g2 + f2 – c)]2
Comparing the above equation with the standard equation of the circle as, (x – a)2 + (y – b)2 = r2 we get,
• Center = (-g, -f)
• Radius = √(g2 + f2 – c)
Thus, by using the above method general form of the is transformed into the standard form of the equation.
## Converting Standard Form to General Form
The standard form of the circle i.e. (x – a)2 + (y – b)2 = r2 Â can be easily converted to the general form of the circle i.e. x2 + y2 + 2gx + 2fy + c = 0 Â using the steps discussed below,
Step 1: Use the algebraic identity formula, (a-b)2 = a2 + b2 -2ab to simplify the given equation (x – a)2 + (y – b)2 = r2
x2 + a2 – 2ax + y2 + b2 – 2by = r2
Step 2: Arrange the equation with variable terms together and constant together as,
x2 – 2ax + y2 – 2by + a2 + b2 – r2 = 0
Comparing with x2 + y2 + 2gx + 2fy + c = 0 we get,
g = -a, f = -b, and c = a2 + b2 – r2Â
This is the required form of the equation of a circle in a general form.
## Features of Equation of Circle
The general form of the equation of the circle is x2 + y2 + 2gx + 2fy + c = 0. Some of the features of the equation of the circle are,
• It is quadratic in both x and y.
• Coefficient of x2 = y2. (It is advisable to keep the coefficient of x2 and y2 unity)
• There is no term containing xy i.e., the coefficient of xy is zero.
• It contains three arbitrary constants viz. g, f and c
## Examples on Circle Equations
Example 1: Find the centre and radius of the circle from the given equation: Â x2 + y2 + 6x – 4y + 4 = 0
Solution:
We have,
(x2 + 6x) + (y2 – 4y) = – 4
To make it a perfect square identity, add and subtract 9 and 4,
(x2 + 6x + 9) + (y2 – 4y + 4) – 9 – 4 = – 4
(x + 3)2 + (y – 2)2 = – 4 + 4 + 9
(x – (-3))2 + (y – 2)2 = 9
(x – (-3))2 + (y – 2)2 =32
On comparing with the standard equation of circle, we have
h = -3, k = 2 and r = 3
So the centre of the circle is (-3, 2)
and the radius of the circle = 3
Example 2: Find the equation of a circle when the endpoints of the diameter are (3, 2) and (5, 8).
Solution:Â
Given,
End points of the diameter are (3, 2) and (5, 8)
As we know radius is the mid point of the diameter then,
The coordinate of the radius are, [(3+5)/2, (2+8)/2] = (4, 5)
Length of radius = √[(4-3)2 + (5-2)2] = √(1+9)
              = √(10)
Equation of the circle is,
(x-a)2 + (y-b)2 = r2
where, (a, b) is centre and r is the radius,
Now,Â
(x-4)2 + (y-5)2 = [√(10)]2
The required equation of the circle is (x-4)2 + (y-5)2 = 10 whose end points of the diameter are (3, 2) and (5, 8).
Example 3: Find the centre and radius of the circle using the following equation: x2 + y2 – 4x + 6y = 12.
Solution:
Given: x2 + y2 – 4x + 6y = 12
We can write as
(x2 – 4x) + (y2 – 6y) = 12
By manipulating identity, we get
(x2 – 4x + 4) + (y2 6y + 9) – 9 – 4 = 12
(x2 – 4x + 4) + (y2 – 6y + 9) = 12 + 9 + 4
(x – 2)2 + (y – 3)2 = 25
(x – 2)2 + (y – 3)2 = 52
On comparing with the standard equation of circle, we have
h = 2, k = 3 and r = 5.
Example 4: Find the equation of a circle whose centre is (-3, -2) and whose radius is 6.
Solution:
Given,
h = -3, k = -2, and, r = 6
Using the standard equation of circle,Â
(x – h)2 + (y – k)2 = r2Â
(x – (-3))2 + (y – (-2))2 = 62
(x + 3)2 + (y + 2)2 = 62
x2 + 6x + 9 + y2 + 4y + 4 = 36
Hence, the required equation is
x2 + y2 + 6x + 4y – 23 = 0
Example 5: Find the equation of a circle whose radius is 7 and the centre is at the origin.
Solution:
Given : r = 7 and the centre = (0, 0)
Using the standard equation of circle,Â
(x – h)2 + (y – k)2 = r2Â
x2 + y2 = r2
x2 + y2 = 72
Hence, the required equation is
x2 + y2 – 49 = 0Â
Example 6: Find the equation of the circle that passed through the points (1, 0), (-1, 0), and (0, 1).
Solution:
Let the required circle beÂ
x2 + y2 + 2gx + 2fy + c = 0
As, it passes through (1, 0), (-1, 0), and (0, 1). Hence, they will satisfy the equation
By substituting them, we get
1 + 2g + c = 0 …(1)
1 – 2g + c = 0 …(2)
1 + 2f + c = 0 …(3)
From eq(1) and (2), we get
g = 0 and c = -1
Now, putting c = -1 in eq(3), we get
f = 0
Now, substituting the values of g, f and c in the main equation, we get
x2 + y2 = 1
Example 7: Find the equation of the image of the circle x2 + y2 + 16x – 8y + 64 = 0 in the line mirror x = 0.
Solution:
Given equation is,
x2 + y2 + 16x – 8y = – 64
(x2 + 16x) + (y2 – 8y) = – 64
On adding and subtracting 64 and 16, we get
(x2 + 16x + 64) + (y2 – 8y + 16) – 64 – 16 = – 64
(x2 + 16x + 64) + (y2 – 8y + 16) = – 64 + 64 + 16
(x + 8)2 + (y – 4)2 = 16
(x – (-8))2 + (y – 4)2 = 42
As, the radius of this circle is (-8, 4) and radius = 4.
The image of the circle in the line mirror will have its centre as (8, 4) and radius 4.
So, the equation will be,
(x – 8)2 + (y – 4)2 = 42
x2 + y2 – 16x – 8y + 64 = 0
## Practice Problems on Equation of a Circle
1. Given the equation of a circle (x-2)2 + (y+3)2 = 49, determine the center and radius of the circle.
2. Write the equation of a circle with a center at (4, -1) and a radius of 5.
3. Find the equation of the line that is tangent to the circle x2 + y2 = 16 at the point (4, 0).
4. Convert the following general form of a circle equation to its standard form: x2 + y2 – 6x + 8y + 9 = 0. Then, identify the center and radius of the circle.
5. Determine if the line y = 2x + 3 intersects the circle (x – 3)2 + (y – 2)2 = 25. If it does, find the points of intersection.
## FAQs on Equation of a Circle
### What is the Equation for a Circle?
The standard equation for a circle is,
(x-h)2+(y-k)2 = r2
whereÂ
(h, k) is the center,
r‘ is the radius of the circle.
### What is the General Equation of Circle?
The general equation of the circle is, x2 + y2 + 2gx + 2fy + c = 0
### What is the Parametric Equation of Circle?
Parametric Equation of the circle is, x2 + y2 + 2hx + 2ky + c = 0 where x = -h + rcosθ and y = -k + rsinθ
### What is the Equation of Circle in Polar form?
The equation of circle in polar form is, x2 + y2 = a2
• x = a cosθ
• y = a sinθ
where,
a is the radius of the circle
θ is the angle made by the radius with the x-axis
### What is the Equation of Point Circle?
A point circle is the circle whose radius is zero. The equation of the point circle is,
(x-a)2 + (y-b)2 = 0
where, (a, b) is the radius of the circle.
### What are the Formulas for Circles?
Various formulas for the circle are,
• Circumference of a Circle = 2Ï€r
• Area of a Circle = Ï€r2
where,
Ï€ is the constant and its value is 3.1415,
r is the radius of the circle.
### What is the Equation of a Circle when the Centre is at the Origin?
When the centre of the circle is at origin (0,0) its equation is given as, (x -0)2 + (y-0)2 = a2
x2 + y2 = a2
where a is the radius of the circle.
### What is the equation of a circle when the endpoints of a diameter are given – A (1,3) and B (7,9)?
The equation of the circle when the endpoints of a diameter are given is (x-4)2 + (y-6)2 = 18.
### What is the equation of a circle given two points A (2,3) and B (6,7)?
The equation of the circle is (x – 4)2 + (y – 5)2 = 8.
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# When 3 dice are rolled what is the probability of getting a sum of 16?
Contents
Probability of a sum of 16: 6/216 = 2.8%
## When 3 dice are rolled what is the probability of getting a sum of 8?
The total number of ways to roll an 8 with 3 dice is therefore 21, and the probability of rolling an 8 is 21/216, which is less than 5/36.
## When 3 dice are rolled together the probability that the sum on the three dice is less than 17 is?
Probability for rolling three dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each (three) dies. When three dice are thrown simultaneously/randomly, thus number of event can be 63 = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its faces. (iii) getting a total of at least 5.
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## What is the probability of getting a sum as a 3 If a dice is thrown?
So, P(sum of 3) = 1/18.
## What is the chance of getting a total of 15 when 3 dice are thrown together?
If three dice are thrown simultaneously then probability of getting a sum of 15 is 5/108 .
## What is the probability of 3 dice?
Two (6-sided) dice roll probability table
Roll a… Probability
3 3/36 (8.333%)
4 6/36 (16.667%)
5 10/36 (27.778%)
6 15/36 (41.667%)
## How do you find the probability of 3 dice?
We divide the total number of ways to obtain each sum by the total number of outcomes in the sample space, or 216. The results are: Probability of a sum of 3: 1/216 = 0.5% Probability of a sum of 4: 3/216 = 1.4%
## How do you find the probability of a dice?
Probability = Number of desired outcomes ÷ Number of possible outcomes = 3 ÷ 36 = 0.0833. The percentage comes out to be 8.33 per cent. Also, 7 is the most likely result for two dice. Moreover, there are six ways to achieve it.
## What are the odds of rolling 3 of a kind with 3 dice?
The probability of getting the same number is 1/6. Throw the third die. The probability of getting the same number is again 1/6. So the probability of three numbers the same is 1/6×1/6.
## What is the probability of rolling a dice 3 times and getting a different number each time?
The probability that the number rolled on the third die is different from the first two dice is 4/6. Therefore, the probability that the three dice are all different on a single roll is 5/6 * 4/6 = 20/36 = 5/9.
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## What is the probability of getting sum 3 or 4 when 2 dice are rolled?
= 5/36. Hope this helps!
## What is the probability of getting an even number when a dice is rolled?
The probability of rolling an even number is three out of six, or three-sixths. This fraction can be simplified as the numerator and denominator are both divisible by three. Three divided by three is equal to one, and six divided by three is equal to two. This means that the answer, in its simplest form, is one-half.
## What is the probability of rolling doubles when rolling two dice?
Rolling doubles mean the outcomes of two dice are the same, such as 1&1 or 5&5. The probability of rolling doubles when rolling two dice is 6/36 = 1/6.
## How many ways can you roll 15 with three dice?
There are three ways to roll 15 with three dice: 663, 654 and 555.
## What is the probability of rolling a 6 with 3 dice?
That is 2 out of the total of 6*6*6 = 216 possibilities, so the probability is 2/216 = 1/108 = 0.9259% under the usual assumptions (unbiased dice, independent outcomes). Originally Answered: The experiment is rolling a fair die 3 times.
## What is the probability of rolling a sum of 8?
There are 36 outcomes in total. Five of them (2,6), (3,5), (4,4), (5,3) and (6,2) result in sum 8. So, assuming all outcomes are equiprobable, the answer is 5/36.
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# Completing the Square
Any quadratic equation can be solved using a technique called completing the square.
To use this method, we will rewrite the quadratic equation so that one side is a perfect square. Then we solve using the square root property.
Before we solve a quadratic equation in this way, let’s learn how to complete the square.
To “complete the square†means to transform a binomial of the form x2 + bx into a perfect square trinomial by adding a constant term.
For example, let’s complete the square for x2 + 6x.
We will use rectangles, called algebra tiles, to visualize the process.
• A square tile measuring x units on a side has area x2. We will use this tile to represent x2, the first term of x2 + 6x.
• A rectangular tile that is x units tall and 1 unit wide has an area of 1x. Since 6 · 1x is 6x, we will use six of these tiles to represent the second term of x2 + 6x.
Placed side by side, the tiles form a rectangle that represents x2 + 6x.
Now, we try to rearrange the tiles to form a square. To do this, we move one-half of the tall thin tiles. However, the result is not a complete square because the lower right portion is missing.
The missing piece is a 3-by-3 square.
Thus, to “complete the square,†we must add 3 · 3 = 9 unit tiles.
With the 9 new tiles, the area of the entire square is (x + 3)(x + 3) = (x + 3)2 = x2 + 6x + 9.
By adding 9 to x2 + 6x, we have created the perfect square (x + 3)2.
Let’s review the process we used to complete the square. We moved one-half of the x-tiles. x2 + 6x + ? Multiply the coefficient of x by Then we filled in the remaining space with a square of unit tiles. Square the result. 32 = 9 We added nine tiles to complete the square. x2 + 6x + 9 = (x + 3)(x + 3) = (x + 3)2
This process holds in general:
To find the number needed to complete the square, multiply the coefficient of the x-term by , and then square the result.
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# Equivalent Fractions – Explanation & Examples
In mathematics, equivalent fractions are simply fractions with different numerators and denominators but represent the same proportion of a whole. Equivalent fractions seem to be different at a look, but they have similar or equal value.
For example, the equivalent fractions for 1/4 are:
2/8, 3/12, 4/16, etc.
The equivalent fractions have an equal amount or value after simplification of both their numerator and denominators. The fractions will generate the same value if cancellation by a common factor is made on both the numerator and denominator.
## What are Equivalent Fractions?
Equivalent fractions are two or more fractions that result in the same value after simplification. Suppose a/b and c/d are two fractions, then the fractions are equivalent only if the simplification of each fraction results in e/f.
In other words,
a/b = c/d = e/f.
For example, a fraction 1/3 has an equivalent of 5/15 because of simplification of 5/15 results in the same value as 1/3.
Now the question arises for why these fractions are equal despite having different numbers. The answer to this query is that the fractions contain numerators and denominators that are not co-prime numbers. Therefore they have a common multiple which on division produces the same value.
Let’s take an example:
1/2 = 2/4 = 4/8
You can notice that yet the above two factions have different integers, but after dividing both the numerator and denominator by a common factor, the result is:
(4 ÷ 4)/(8 ÷ 4)
=1/2
In this case, if we simplify 2/4, the result 1/2 again.
(2 ÷ 2)/(4 ÷ 2)
= 1/2
It has been shown that either dividing the denominator or multiplying the numerator with the same factor does not alter the value of the fraction. And therefore, equivalent fractions have an equal value when simplified.
## How do you find equivalent fractions?
Consider a case with the fraction 1/5.
Multiplying both the numerator and denominator with 2, 3 and 4 gives:
1/5 x 2/2 = 2/10
1/5 x 3/3 = 3/15
1/5 x 4/4 = 4/20
Therefore, it can be concluded that:
1/5 = 2/10 = 3/15 = 4/20
The equivalent fraction can only be generated by multiplication or division by a common factor. Carrying out addition or subtraction on the fraction only changes the value of a fraction.
Example 1:
Given that the fractions 5/16 and x/12 are equivalent calculate the value of x.
Solution
Given that:
5/16 = x/12
x = (5 x 12)/16
x = 60/16
x =15/4
And thus, the value of x is 15/4.
Example 2:
Find the value of x if the fractions 3/5 and 4/x are equivalent.
Solution
Given that,
3/5 = 4/x
x = (4 x 5)/3
x = 20/3
## Practice Questions
1. Write up to 5 equivalent fractions for each of the following:
(i) 3/4
(ii) 4/5
(iii) 6/7
(iv) 4/5
2. Find the equivalent fractions having a denominator of 12 for each of the following fractions.
(i) 1/2
(ii) 1/3
(iii) 3/4
(iv) 5/6
3. Change the following fractions into equivalent fractions having a value of 24 as their denominator:
(i) 6/12
(ii) 3/8
(iii) 2/6
(iv) 4/6
4. Identify the pairs of fractions that are equivalent and which are not:
(i) 2/3 and 8/12
(ii) 3/7 and 12/28
(iii) 5/8 and 15/27
(iv) 36/44 and 9/11
(v) 4/5 and 5/4
(vi) 5/8 and 27/18
5. I think of an equivalent fraction to 10/15 with 2 as the numerator. What fraction with a numerator of 2 am I thinking of?
6. Erick notices that either 3/5 or 3/4 is equal to the fraction 12/20. Which fraction is equal to 12/20?
7. James is giving his brother 2/5 of her nut collection. Calculate how many of 1/5 s of his nut collection is he giving his brother?
8. Peter gave 1/4 and 3/12 of orange to Donald and Pedro, respectively. Determine if he gave out an equivalent fraction of an orange.
9. John conducted a survey in his class and discovered that 56/96 of the students sampled out participated in sports after school. Express the fraction in its simplest form?
10. 7 is a prime number in a fraction 7/x. What number can replace x in this fraction so that it is not in simplest form?
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### How to read fractions ?
$$\displaystyle \frac{1}{2}$$ "one half" or "a half"
$$\displaystyle \frac{1}{3}$$ "one third" or "a third"
$$\displaystyle \frac{1}{4}$$ "one quarter" or "a quarter"
$$\displaystyle \frac{2}{3}$$ "two thirds" not "two third" ( 2 is plural so we add "s" ) .
$$\displaystyle \frac{2}{5}$$ "two fifths"
$$\displaystyle \frac{3}{7}$$ " three sevenths"
$$\displaystyle \frac{5}{6}$$ " five sixths"
$$\displaystyle \frac{3}{4}$$ " three quarters"
$$\displaystyle \frac{11}{10}$$ "eleven tenths"
$$\displaystyle 2\frac{9}{10}$$ " two and nine tenths" is like $$\displaystyle \text 2 whole$$ and $$\displaystyle \text nine$$ $$\displaystyle \frac{1}{10}$$
Why do we read fractions in this way ?
## How to Read Fractions ?
We read fractions based on unit fractions. Unit fraction is shortly the smallest part of
fractions and all fractions are made of unit fractions.
Reading fraction is telling “the fraction consist of how many unit fraction”.
$$\displaystyle \frac{1}{2}$$ "a half" or "one half"
$$\displaystyle \frac{1}{3}$$ "one third"
$$\displaystyle \frac{2}{3}$$ "two thirds"
We add “s” to the end couse 2 is plural . Not one !
$$\displaystyle \frac{1}{4}$$ "a quarter" , or "one quarter
You may also read as “one fourth” but it is not preferred
$$\displaystyle \frac{3}{4}$$ "three quarters"
Or “three fourths” but not preferred.
$$\displaystyle \frac{1}{6}$$ " a sixth" or "one sixth"
$$\displaystyle \frac{5}{6}$$ "five sixths"
This fraction consist of 5 , $$\displaystyle \frac{1}{6}$$ th.
$$\displaystyle \frac{11}{10}$$ "eleven tenths"
$$\displaystyle 2\frac{4}{5}$$ " two and four fifths"
it is like “ two whole and four $$\displaystyle \frac{1}{5}$$ ths.
$$\displaystyle \frac{3}{5}$$
"3 divided by 5 " or "3 over 5"
This is one of the most common misconceptions in reading fractions .
a ) $$\displaystyle \frac{3}{5}$$ "3 divided by 5"
This reading style is valid for rational numbers, not fractions. There is a difference between
fractions and rational numbers.
b) $$\displaystyle \frac{3}{5}$$ "3 over 5"
It just describes its fractional form. it is like telling “there is a cat on the table”.
If you read fractions as above , students may understand what you are talking about, they may get the fraction form but they will fall in misconception as fractions are made up two numbers , numerator and dominator .They cannot get the concept of Fraction.
Best to avoid this kind of misconceptions by reading fractions based on unit fractions.
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# What is Probability?
Top
Sub Topics Probability is an important topic of mathematics that is widely used not only in mathematics but in several other fields. Probability of an event is the chances of that event to be happen. An event is supposed to be a process in the language of probability, such as - throwing a pair of dice, drawing a card from the deck of $52$, randomly selecting a marble from a jar of marbles of different colors etc.
## Definition
If there are n elementary events associated with an random experiment and m of them are favourable to an event E, then the probability of $E$ is denoted by $P(E)$ and is defined to the ratio $\frac{m}{n}$.
$\Rightarrow\ P(E)$ = $\frac{\text{Number of Favourable to Event A}}{\text{Total Number of Outcomes}}$
If an experiment has n equally likely outcomes in $S$ and $N$ of them are the event $A$, then the theoretical probability of event $A$ occurring is
$P(A)$ = $\frac{n}{N}$.
## Probability Multiplication Rule
If two events A and B from a sample S of a random experiment are mutually exclusive, then
$P(A\ \cup\ B)$ =$P(A) + P(B)$
In this section, we examine whether such a rule exists, if $\cup$ is replaced by $\cap$ and $'+'$ is replaced by $'x'$ in the above addition rule. If it does exist, what are the particular conditions restricted on the events $A$ and $B$. This leads us to understand the dependency and independence of the events.
The Probability Multiplication Rule:
Consider events $A$ and $B$.
then $P(A\ \cap\ B)$ = $P(A)\ \times\ P(B)$.
## Experimental Probability
The learning probability started with gambling, the probability mostly used in the field of physical science, commerce, biology science, medical science, weather forecasting, etc., an experimental having more than one possible outcome is called statistical experimental. Probability is a method which numerically measures the degree of uncertainty and, there, of certainty of the occurrences of an event. A statistical experimental is also known as a trial. In this article we shall learning an experimental probability.
Definition
Experimental probability of an event is the ratio of the number of times the event occurs to the total number of trials. The general meaning of the word probability is likelihood. If the event is absolutely certain the probability is said to be one or unity. If there is absolute impossibility of an event happening then probability is said to be zero. There are two ways of arriving at an actual measure of the probability of an event in the real life. They are mathematical probability and empirical probability, both of them falling under the category experimental probability, since they are based on the events and its performance.
Formula:
$Experimental\ Probability$ = $\frac{\text{Number of times the event happen}}{\text{Total number of trials}}$
## Probability Expected value
Expected values are calculated as the weighted average of all possible outcomes using the probability of the outcomes as weights. The mean of a random variable is more commonly referred to as its expected value. The value obtain from some experiment whose outcomes are represented by the random variable. The expected value of a random variable $X$ is denoted by $E(x)$.
The random variable $X$ is discrete and has a probability distribution $f(x)$, the expected value of the random variable is given by:
$E(x)$ = $\sum\ x\ f(x)$.
## Examples
Example 1:
A coin is tossed $50$ times. $25$ times head appear. Find the experimental probability of getting heads.
Solution:
Step 1: $Experimental\ probability$ = $\frac{\text{Number of Times the Event Happen}}{\text{Total Number of Trials}}$
Step 2: Number of times heads appeared = $25$.
Step 3: Total number of experiments = $50$.
Step 4: Therefore experimental probability of getting a head = $\frac{25}{50}$ = $\frac{1}{2}$
Example 2:
John plays game his college. He did not score in 29 out of 58 games he played. What is the experimental probability of john score in a game?
Solution:
Step 1: Number of game john make score = $58 – 29$ = $29$
Step 2: $P\ (\text{John make a score})$ = $\frac{\text{Number of games in which John make goals}}{\text{Total number of games John play}}$
Step 3: = $\frac{29}{58}$
= $\frac{1}{2}$
Step 4: Probability of john make a goal in a game is $\frac{1}{2}$.
Example 3:
A boy has four balls ($2$ black and $2$ red balls). What is the most likely mix of black and red ball.
What objects can be used to simulate the possible outcomes of the balls?
Find the theoretical probability that there are two red and two black balls.
Solution:
Assume that $P$(black ball) = $P$(red ball) = $\frac{1}{2}$
Objects can be used to simulate the possible outcomes of the balls?
Each ball can be black or red, so there are $2 \times 2 \times 2 \times 2$ or $16$ possible outcomes.
Use a simulation that also has $2$ outcomes for each of $4$ events. One possible simulation would be to toss four coins, one for each ball, with heads representing black balls and tails representing red balls.
The theoretical probability that there are two red and two black balls.
There are $16$ probable outcomes, and the number of combinations that have two red and two black ball is $4C2$ or $6$.
So, the theoretical probability is $\frac{6}{16}$ or $\frac{3}{8}$
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Saturday, June 29, 2024
TI 30Xa Algorithms: Solving Monic Quadratic Polynomials Quickly
TI 30Xa Algorithms: Solving Monic Quadratic Polynomials Quickly
We all know about the tried and true Quadratic Formula to solve quadratic equations. However, it is not the only way to tackle such problems.
Today’s blog is covers a way to quickly get the roots of the quadratic equation:
x^2 + p * x + q = 0
I am focusing on monic quadratic polynomials today. A polynomial is a monic polynomial if the leading coefficient is 1. In this instance, the coefficient of the x^2 term is 1. I’m also going to work with quadratic equations that have real roots.
Deviation
Please see the source article as the derivation and method is explained by the author Po-Shen Loh (https://www.poshenloh.com/quadraticdetail/). This method has also been developed by Viete and other classic mathematicians. Here I attempt to explain a derivation of this method.
Let r1 and r2 be the two roots of the polynomial and x^2 + p * x + q factors to:
x^2 + p * x + q = (x – r1) * (x – r2)
Expanding (x – r1) * (x – r2):
(x – r1) * (x – r2) = x^2 + (-r1 -r2) * x + r1 * r2
Then:
p = - r1 – r2 ⇒ r1 + r2 = -p
q = r1 * r2
Let one of the roots be defined as: [see Source]
r1 = -p/2 + u
Then:
r1 + r2 = -p
-p/2 + u + r2 = -p
r2 = -p/2 + u
And:
r1 * r2 = q
(-p/2 + u) * (-p/2 - u) = q
p^2/4 – u^2 = q
- u^2 = q – p^2/4
u^2 = p^2/4 – q
u = ±√(p^2/4 – q)
And the roots are:
r1 = -p/2 + √(p^2/4 – q)
r2 = -p/2 - √(p^2/4 – q)
x = -p/2 ± u
Note:
r1 = -p/2 + u
r1 – 2 * u = -p/2 + u – 2 * u
r1 – 2 * u = -p/2 – u
r1 – 2* u = r2
We can verify the above result with the quadratic equation:
x = (-p ± √( p^2 – 4 * q)) / 2
x = -p/2 ± √( p^2 – 4 * q) / 2
x = -p/2 ± √(( p^2 – 4 * q) / 4)
x = -p/2 ± √(( p^2/4 – q)
x = -p/2 ± u
Assumption: The roots are real (not complex).
Keystrokes:
p [ STO ] 1
q [ STO ] 2
[ ( ] [ RCL ] 1 [ x^2 ] [ ÷ ] 4 [ - ] [ RCL ] 2 [ ) ] [ √ ] (u)
[ - ] [ RCL ] 1 [ ÷ ] 2 [ = ] (r1, root 1)
[ ± ] [ - ] [ RCL ] 1 [ = ] (r2, root 2)
Memory registers used: M1 = p, M2 = q
Examples
Example 1: x^2 – 3*x – 4 = 0
p = -3, r = -4
3 [ ± ] [ STO ] 1
4 [ ± ] [ STO ] 2
[ ( ] [ RCL ] 1 [ x^2 ] [ ÷ ] 4 [ - ] [ RCL ] 2 [ ) ] [ √ ] (u = 2.5)
[ - ] [ RCL ] 1 [ ÷ ] 2 [ = ] (root 1, x = 4)
[ ± ] [ - ] [ RCL ] 1 [ = ] (root 2, x = -1)
x = 4, -1
Example 2: x^2 – 24*x + 135 = 0
p = -24, r = 135
24 [ ± ] [ STO ] 1
135 [ STO ] 2
[ ( ] [ RCL ] 1 [ x^2 ] [ ÷ ] 4 [ - ] [ RCL ] 2 [ ) ] [ √ ] (u = 3)
[ - ] [ RCL ] 1 [ ÷ ] 2 [ = ] (root 1, x = 15)
[ ± ] [ - ] [ RCL ] 1 [ = ] (root 2, x = 9)
x = 15, 9
Example 3: x^2 + 10*x + 24 = 0
p = 10, r = 24
10 [ STO ] 1
24 [ STO ] 2
[ ( ] [ RCL ] 1 [ x^2 ] [ ÷ ] 4 [ - ] [ RCL ] 2 [ ) ] [ √ ] (u = 1)
[ - ] [ RCL ] 1 [ ÷ ] 2 [ = ] (root 1, x = -4)
[ ± ] [ - ] [ RCL ] 1 [ = ] (root 2, x = -6)
x = -4, -6
Source
Loh, Po-Shen. “Quadratic Method: Detailed Explanation” Updated August 6, 2021. Retrieved May 19, 2024. https://www.poshenloh.com/quadraticdetail/
Until next time, have a great day. For the Americans, have a safe and sane Fourth of July. See you July 6!
Eddie
All original content copyright, © 2011-2024. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.
TI-84 Plus CE Python: Drawing Shapes with the ti_plotlib module
TI-84 Plus CE Python: Drawing Shapes with the ti_plotlib module Introduction Here are four scripts to draw shapes: RECT8: rec...
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# Precalculus (MATH-1300)
This course is basically a combination of College Algebra and Trigonometry. The course will have a great emphasis on functions. While you should have seen that concept before, you may not have concentrated on it very much. Now you will.
An expression like 3x + 2 literally stands for a real number; you simply don't know which number until you know which number the variable x stands for. But when we work with that expression, we often think about all of the values that it can take, as x varies, at once. For example, to graph that expression, we don't just draw a point (for a single value) or even really a bunch of points; we draw a line. Geometrically, that line is itself a single complete object in its own right, something more than a point. Algebraically, we can also think of an expression as describing a single complete object in its own right, something more than a number. That something is a function.
Much as we can write the solution set of an inequality as, for example, {x | x < 3}, so we can also write a function as, for example, (x ↦ 3x + 2). However, for historical reasons, that notation is not used in most Algebra books. Instead, it is common to give a function a name, just as we might give the value of an expression a name in a word problem, and there is special notation for that. If f stands for our function, then instead of writing f = (x ↦ 3x + 2), we write f(x) = 3x + 2 (stating it for all values of x), which means the same thing. This is convenient notation anyway, because with it we can calculate, for example, that f(5) = 3(5) + 2 = 17. (Remember that, while x here stands for a number, f stands for a function, which is not the same type of thing. So, 3(5) means 3 times 5, and even x(5) would mean x times 5, but f(5) does not work like that.)
Of course, we will also do more of solving equations, graphing expressions, applying Algebra in word problems, and everything else that's a part of Algebra. In particular, we will look at these with exponential, logarithmic, and trigonometric operations; that is, we will look at expressions like 2x (where x is allowed to be any real number, not just a rational number), log2x, and sin x. (If you don't know what those mean yet, that's OK; we'll cover that!)
Go back to the course homepage.
This web page was written by Toby Bartels, last edited on 2021 August 23. Toby reserves no legal rights to it.
The permanent URI of this web page is `http://tobybartels.name/MATH-1300/2021FA/introduction/`.
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# Algebra II : Polynomials
## Example Questions
1 2 11 12 13 14 15 16 17 19 Next →
### Example Question #10 : Write A Polynomial Function From Its Zeros
Which of the following equations belongs to a quadratic with zeros at x = 2 and x = 7?
Explanation:
A quadratic with zeros at x = 2 and x = 7 would factor to , where it is important to recognize that a is the coefficient. So while you might be looking to simply expand to , note that none of the options with a simple term (and not ) directly equal that simple quadratic when set to 0.
However, if you multiply by 2, you get . That’s a simple addition of 18x to both sides of the equation to get to the correct answer: .
### Example Question #11 : Write A Polynomial Function From Its Zeros
Which of the following equations represents a quadratic equation with zeros at x = 3 and x = -6, and that passes through point (2, -24)?
Explanation:
When finding the equation of a quadratic from its zeros, the natural first step is to recreate the factored form of a simple quadratic with those zeros. Putting aside the coefficient, a quadratic with zeros at 3 and -6 would factor to:
So you know that a possible quadratic for these zeros would be:
Now you need to determine the coefficient of the quadratic, and that’s where the point (2, -24) comes in. That means that if you plug in x = 2, the result of the quadratic will be -24. So you can set up the equation:
, where a is the coefficient you’re solving for. And you know that this is true when x = 2, so if you plug in x = 2 you can solve for a:
So a = 3. When you then distribute that coefficient of 3 across the original simple quadratic, you have:
### Example Question #11 : Write A Polynomial Function From Its Zeros
Which of the following equations belongs to a quadratic with zeros at and ?
Explanation:
A quadratic with zeros at 2 and 7 would factor to , where it is important to recognize that is the coefficient. So while you might be looking to simply expand to , note that none of the options with a simple term (and not ) directly equal that simple quadratic when set to 0.
However, if you multiply by 2, you get . That’s a simple addition of 18x to both sides of the equation to get to the correct answer: .
### Example Question #721 : Intermediate Single Variable Algebra
Which of the following represents a quadratic with zeros at 5 and 2 and that passes through point (1, 16)?
Explanation:
When you're finding the equation of a quadratic given its zeros, a good start is to ignore the coefficient (you'll return to it later) and to construct the factored form of the quadratic using the zeros. Here with zeros of 5 and 2, that factored form would be:
That then means you can expand it to:
From here, you can see that the answer choices all maintain the same general relationship as this simple quadratic, but have different coefficients. This is where you can use the point (1, 16) to your advantage. This means that if you plug in to the quadratic, it will produce an answer of . To solve this algebraically, return to using a coefficient in front of your entire quadratic. You'll then have:
Where is the coefficient you can solve for knowing that the equation will produce a value of when you input . So plug that in and you can solve for the coefficient:
Meaning that:
So you know that .
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# How to solve for radius
The radius of a circle is the distance from the center to the edge of the circle. It is often used to calculate the diameter of a circle, or used as a reference for measuring the size of an object. It is measured in either millimeters or inches.
## How can we solve for radius
The formula for radius is: The quick and simple way to solve for radius using our online calculator is: R> = (A2 − B2) / (C2 + D2) Where R> is the radius, A, B, C and D are any of the four sides of the rectangle, and A2> - B2> - C2> - D2> are the lengths of those sides. So if we have a square with side length 4cm and want to find its radius value, we would enter formula as 4 cm − 4 cm − 4 cm − 4 cm = 0 cm For example R> = (0cm) / (4 cm + 2cm) = 0.5cm In this case we would know that our square has an area of 1.5cm² and a radius of 0.5cm From here it is easy to calculate the area of a circle as well: (radius)(diameter) = πR>A>² ... where A> is
When calculating a circle’s radius, you need to take into account both the radius of the circle’s circumference and the radius of its diameter. You can use this formula to solve for either or both: With these formulas, all you have to do is find the radius of each side in relation to the other one. You should also remember that the radius increases as your circle gets larger. If a circle has a radius of 1 unit, then its radius will double (or triple) as it grows from 1 unit in size. Once you know how much bigger a circle is than another one, you can calculate its diameter. Divide the first circle’s circumference by the second one’s diameter and multiply by pi to get the answer.
R is a useful tool for solving for radius. Think of it like a ruler. If someone is standing in front of you, you can use your hand to measure their height and then use the same measurement to determine the radius of their arm. For example, if someone is 5 feet tall and has an arm that is 6 inches long, their radius would be 5 inches. The formula for calculating radius looks like this: [ ext{radius} = ext{length} imes ext{9} ] It's really just making the length times 9. So, if they're 6 inches tall and their arm is 6 inches long, their radius would be 36 inches. Using R makes sense when you are trying to solve for any other dimension besides length - such as width or depth. If a chair is 4 feet wide and 3 feet deep, then its width would be equal to half its depth (2 x 3 = 6), so you could easily calculate its width by dividing 2 by 1.5 (6 ÷ 2). But if you were trying to figure out the chair's height instead of its width, you would need an actual ruler to measure the distance between the ground and the seat. The solution to this problem would be easier with R than without it.
BEST TEACHER - It's true that a lot of teachers in my high school hate this app/ban it (if they know about it) because the majority of students just copy down answers. But this app is also near perfect at teaching you the steps, their order, and how to do each step in both written and visual elements. Just like what a great teacher would do except you get to do it at your pace.
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# How To Solve Pythagorean Theorem Problems
Because of these angles it´s easier to find out the sides of a right triangle.Let´s try to understand this by using two examples.Pythagoras or the disciples to him constructed the first known algebraic proof of the theorem and famous writers such as Plutarch and Ciceron acclaimed him for discovering this proof.
Among these, the Primitive Pythagorean Triples, those in which the three numbers have no common divisor, are most interesting.
A few of them are: Also Pythagorean Triples can be created with the a Pythagorean triple by multiplying the lengths by any integer. We see looks like the legs of a right triangle with a multiplication factor of 111.
A Pythagorean Triple is a set of 3 positive integers such that , i.e.
the 3 numbers can be the lengths of the sides of a right triangle.
For example, Right triangle has legs of length and .
One of the most used and beautiful theorems in math is the Pythagorean theorem.Lets call the hypotenuse $c$ which gives us that $c^2=1^2 1^2$ $c^2=2$ Take the square root $c=\sqrt ≈ 1,414$ There exists a couple of special types (or cases) of right triangles.Two of them being $30°-60°$ right triangles and °-45°$right triangles.This theorem has been know since antiquity and is a classic to prove; hundreds of proofs have been published and many can be demonstrated entirely visually(the book The Pythagorean Proposition alone consists of more than 370).The Pythagorean Theorem is one of the most frequently used theorems in geometry, and is one of the many tools in a good geometer's arsenal.In this type of triangle the opposite side of the °$ angle is half of the hypotenuse: $b=\frac12 ·c = 0,5·c$ In a 45-45 degree right triangle we can get the length of the hypotenuse by multiplying the length of one leg by $\sqrt$ to get the length of the hypotenuse: $c = \sqrt·a = \sqrt·b$ In this section you will find examples and solutions where we use the pythagorean theorem to solve these problems, even if they aren´t directly connected to a right triangle. : The blue point has the coordinates (1, 1) and the red point has the coordinates (5, 3).Now let´s draw lines to form a right triangle where we use the two points as corners.1 Understanding the Pythagorean theorem 1.1 The theorem 1.2 A brief history 2 Basic examples where Pythagoras theorem is used 2.1 Find the length of the hypotenuse in a right triangle 2.2 Find the length of a leg in a right triangle 2.3 Is it a right triangle? 2.5 Special types of right triangles, 30°-60° and 45°-45° right triangles 3 More advanced examples 3.1 Distance between two points 3.2 The distance formula 3.3 The diagonal distance in a cube 4 Continue to learn more about the Pythagorean theorem The Pythagorean theorem or Pythagoras’ theorem is a relationship between the sides in a right triangle.A right triangle is a triangle where one of the three angles is an 90-degree angle.Let´s call the hypotenuse $c$ and calculate the distance.$c^2=4^2 2^2$ $c^2=16 4$ $c^2=20$ $c=\sqrt ≈ 4,47$ This format will always hold true and because of that we can form a formula called the distance formula.
## Comments How To Solve Pythagorean Theorem Problems
• ###### Learn Pythagorean Theorem Problems Solving Right Triangles
Pythagorean Theorem Examples Solving Right Triangle Problems. Pythagorean theorem problems start by giving you the length of two of the sides of a right triangle. Using the Pythagorean formula, it is possible to calculate the length of the third side. Because you are using squares and square roots, you may need the help of a calculator.…
• ###### Ways to Solve Pythagoras Theorem Questions - wikiHow
When doing word problems about traveling, if you are meant to find the shortest distance you will likely use the Pythagorean Theorem. The shortest distance will be the length of the hypotenuse of a triangle superimposed over the area. Learn the most common Pythagorean triples by heart.…
• ###### Pythagorean theorem word problems - Basic mathematics
Pythagorean theorem word problems. Let the length of the ladder represents the length of the hypotenuse or c = 13 and a = 5 the distance from the ladder to the wall. The ladder will never reach the top since it will only reach 12 feet high from the ground yet the top is 14 feet high.…
• ###### Using the Pythagorean Theorem to Solve Problems Prealgebra
To solve problems that use the Pythagorean Theorem, we will need to find square roots. In Simplify and Use Square Roots we introduced the notation latex\sqrt{m}/latex and defined it in this way…
• ###### Solving Problems Involving the Pythagorean Theorem
Draw a right triangle and then read through the problems again to determine the length of the legs and the hypotenuse. Step 2 Use the Pythagorean Theorem a 2 + b 2 = c 2 to write an equation to be solved. Remember that a and b are the legs and c is the hypotenuse the longest side or the side opposite the 90º angle.…
• ###### Pythagorean Theorem calculator - Basic mathematics
Pythagorean Theorem calculator. The Pythagorean Theorem calculator will help you to solve Pythagorean problems with ease. Note that the triangle below is only a representation of a triangle. Your triangle may have a different shape, but it has to be a right triangle. If you looking for either a or b, make sure that the value you enter for a or b is not bigger or equal to c.…
• ###### How to Use the Pythagorean Theorem. Step By Step Examples and Practice
Use the Pythagorean theorem to calculate the value of X. Round your answer to the nearest hundredth. Remember our steps for how to use this theorem. This problems is like example 2 because we are solving for one of the legs.…
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Question
Introduction
In simple cubic lattice, given two cubical voids, find the shortest distance between them. We’re going to use a simple heuristic approach for this problem.
How to find the shortest distance between two cubical voids in a simple cubic lattice
Find the shortest distance between two cubical voids in a simple cubic lattice.
This is quite an easy problem, but it’s good practice to get used to writing down what you’re doing and thinking about your work.
Problem Statement
In this problem, you will be given the coordinates of two cubical voids in a simple cubic lattice. Your task is to find the shortest distance between these two voids.
A cubical void is an empty space surrounded by six faces (which are themselves cubes). Each face shares an edge with exactly three other faces; this makes it easy to count the number of steps required to get from one face to another along any given path through all six faces, because each time we pass through one edge, we must go through exactly two more edges before exiting back into our original position at one end or another (as shown below).
Assumptions
This solution assumes that:
• The lattice is simple cubic (i.e., all points can be described by the same set of three Cartesian coordinates).
• The lattice is infinite, i.e., there are no boundaries or other obstacles that would prevent a void from moving in any direction.
• There are only two voids to consider; this may seem like an arbitrary restriction but it’s actually necessary for the solution to work out in general (see below).
Methodology
The first step is to simplify the problem. We can do this by assuming that every void has a radius equal to 1 and that all of them are connected, which means we only have to consider one sphere with center at (0,0) and radius 1.
Next we use an algorithm:
• Calculate the distance from (0,0) to all points on the surface of our sphere.
• For each point whose distance from (0,0) was calculated in step 2 above: find its nearest neighbor on this list; call this point P1
Heuristic for solution
This is a heuristic for finding the shortest distance between two cubical voids in a simple cubic lattice.
The algorithm starts by finding the center of mass of one of the voids. This can be done by taking all lines connecting adjacent vertices and treating them as springs, so that they’re pushed apart by their masses (i=0..n-1). Then consider this spring system in equilibrium, where each vertex moves along its shortest path from its neighbors’ centers of mass; these are called “relaxed” positions for each vertex. Now take any other point p_2 within some radius r_2 from p_1’s relaxed position; if it lies on one side or another of this line connecting them then we know that p_1 must have been closer than r_2 to p_2 when they were relaxed!
A new record is set for the shortest distance between two cubical voids in a simple cubic lattice
The shortest distance between two cubical voids in a simple cubic lattice is 1.431 nm, achieved by Xiang Li at the University of Science and Technology Beijing (China) and colleagues. They used an atomic force microscope to measure the distance between two holes in an ultrathin silicon film. The previous record was held by Jun Zhang, who measured a distance of 2.50 nm using scanning tunnelling microscopy at his lab at Nankai University (China).
We hope that you enjoyed reading this article as much as we did writing it. We would also like to thank all those who helped us with their valuable suggestions and feedbacks.
1. Shortest Distance Between Two Cubical Voids In Simple Cubic Lattice
Introduction
In this article, we are going to explore a simple but interesting problem that has a shortest distance between two cubes in a cubic lattice. We will use a graphical notation to represent the problem and solve it numerically. This problem can be used as an introduction to various topics in algebra such as linear equations, matrix operations and Vandermonde determinants.
The Problem
The shortest distance between two cubic voids in a simple cubic lattice is 8.8 centimeters. This is the result of solving the equation d(x, y) = 0 for x and y in the coordinate system that passes through the centers of the voids.
The Solution
There is no perfect solution to the problem of finding the shortest distance between two cubical voids in a simple cubic lattice. However, some methods can be used to approximate the distance between the two voids.
One method that can be used is called the Voronoi diagram. The Voronoi diagram shows the distance between points in a given shape by coloring them according to their relative distance from the point of interest. For our purposes, we will only need to know about two points – the first void and the second void. We will color each point based on how close it is to one of these two points.
The first void will be colored black if it is closer to the second void than any other point, and white if it is not closer to any other point. Similarly, the secondvoid will be colored black if it is closer to the firstvoid than any other point, and white if it is not closer to any other point.
Now all we have to do is figure out which points are closest to both voids and color them accordingly. This can be done using simple algebraic techniques or even just trial and error until we get a result that looks good. In general, getting better results requires more iterations but eventually you’ll find a value for nearest neighbor that gives you good accuracy for your given shapes.
Discussion
The shortest distance between two voids in a simple cubic lattice is determined by the sum of the distances between each of the voids. This distance is smaller than the distance between any other two points on the lattice, because it takes into account the fact that one void overlaps another.
2. The shortest distance between two cubical voids in a simple cubic lattice is 1.00, and the shortest distance between two cubical voids in a body-centered cubic lattice is 0.9737.
The shortest distance between two cubical voids in a simple cubic lattice is
The shortest distance between two cubical voids in a simple cubic lattice is a = 4a0/3. The shortest distance between two cubical voids in a body-centered cubic lattice is a = 2a0/3.
The shortest distance between two cubical voids in a body-centered cubic lattice is
The shortest distance between two cubical voids in a body-centered cubic lattice is 0.601531 nm, which is also the length of one side of an atom.
Takeaway:
• The shortest distance between two cubical voids in a simple cubic lattice is 1/2a.
• The shortest distance between two cubical voids in a body-centered cubic lattice is 3/4a, where “a” stands for the length of one side of one cube (and not its diagonal).
The shortest distance between two cubical voids in a simple cubic lattice is 1/2a. The shortest distance between two cubical voids in a body-centered cubic lattice is 3/4a, where “a” stands for the length of one side of one cube (and not its diagonal).
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# Functions Worksheet Domain Range and Function Notation Answers
The first step to learning about functions is to define what a domain is. A domain is the set of all possible input values for a given function, while the range is the range of all possible output values. The difference between a domain and a span can be illustrated by graphs, which can serve as a visual representation of a function’s behavior. The interval and domain notation of a function are usually the same, but there are some important differences.
Domain and range of graphs worksheet doc Myscres from functions worksheet domain range and function notation answers , source:myscres.com
The domain and range of a function are two completely different types of expressions. The domain is always the smaller of the two. The range is the opposite. The two are always written from left to right. When the range is greater than zero, the domain is a square, and vice versa. The domain of a function is the area where its solution lies. The domain and range of a function are the values of a tangent, or the area under the tangent.
The domain of a function is the set of all its solutions. For even roots, use the interval form. A negative number in the radicand does not belong in the even root. The radicand must either be greater or equal to zero. Hence, the domain of a piecewise function includes the values of all the inputs that are real. To determine the domain of a piecewise function, list all the possible input values of an ordered pair.
Day 05 HW Functions Domain Range Increasing Decreasing from functions worksheet domain range and function notation answers , source:youtube.com
The domain of a function is the set of all its values. This can be found in any standard notation. A range, on the other hand, is an arbitrary space. If the radicand has a negative value, it is called a tangent space. The domain of a trig function is the range of all the values of the radicand.
In addition to the domain of a function, the domain of a radicand is the set of all positive and negative real numbers. The radicand is the set of all negative real numbers. The function’s inputs and outputs can be in different categories. This makes it easy for students to learn the different types of functions and understand how they apply to their lives. These exercises are a great way to review the different aspects of functions.
â Finding the Domain of a Function Made Easy â from functions worksheet domain range and function notation answers , source:youtube.com
In order to solve equations using domain and range, the students must identify the radicand. They need to know that the radicand can contain negative numbers. However, the radicand cannot be negative. If the radicand is negative, the function should be written in interval form. The corresponding equation is the domain of the function. If the radicand is positive, the resulting equation is the radicand.
The domain of a function is the set of all possible inputs and outputs. In addition, a domain can be defined as an even or a negative real number. The radicand must be a positive number. Then, the radicand should be a minus number. In this case, the radicand is the area of the problem. If the radicand is positive, then it is the domain of the function. If it is negative, it should be represented as a decimal.
Identifying Functions From Graphs Worksheet Determining the Domain from functions worksheet domain range and function notation answers , source:ning-guo.com
The domain of a function is a set of real numbers. The domain of a given function is the set of all the positive and negative numbers. Depending on the function, the domain can be a single variable or multiple variables. The range of a given function is a set of objects. Then, the range and domain of a given variable is the range of all the values in the specified interval.
The domain of a function is the domain of the variable. The domain is a range of the variables that are within the domain of a function. A range is the range of the variables. The function can also be a constant. If a range is not specified, then it is the region in which the variables are. The range is the domain of a function. Regardless of its name, the radicand is the variable’s radicand.
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Day 10 HW 1 to 31 Domain Range End Behavior Increasing from functions worksheet domain range and function notation answers , source:youtube.com
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Domain and Range Worksheet 2 Answers Beautiful Domain and Range from functions worksheet domain range and function notation answers , source:therlsh.net
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# Divide Paper into nths Divisions
For the most part, knowing how to divide a sheet of paper into equal thirds, fourths, or fifths is enough. On a rare occasion you may need to divide a sheet into equal sevenths, or ninths. Divisions smaller than that are quite specialized and few "everyday" paper folders will need that type of skill.
However, if you are one of those people who enjoy origami and math, or if you design your own origami, you may wish to learn how to divide a square sheet of paper into any number of divisions. It is surprisingly easy - you only need to know 3 things:
• the General Idea,
• the Powers of Two, and
• the Methodology
#### Divide Paper into nths Divisions: The General Idea
The general idea is this: you will make two diagonal folds and the intersection of the two folds is your target division.
The first diagonal fold is easy: fold in half corner to corner and then unfold. The second fold requires a bit of calculation and an understanding of the "power of two".
The Powers of Two
The "power of two" simply means:
two to the power of one equals two,
two to the power of two equals four,
two to the power of three equals eight,...
This can be summarize in a table ------------>
21 = 2 22 = 4 23 = 8 24 = 16 25 = 32 . . .
The Methodology
To determine the location of the second diagonal fold you need to find the greatest "power of two" that is less than the number of divisions you want (n).
21 = 2 22 = 4 23 = 8 24 = 16 25 = 32
So for example, if you want to divide a square sheet of paper into 5 equal segments then n = 5. The greatest "power of two" that is less than 5 is 4, as shown on the table. The difference between 5 and 4 is 1. The distance from the bottom edge of the paper should be 1/4.
Fold your second diagonal from the bottom-left corner to the 1/4 mark; the intersection (dot) is the location of the 1/5 division.
The sequence of folds to achieve 1/5 division is shown here. Notice that finding the 1/4 mark along the right edge of the paper is easy (fold in half, then fold in half again). Indeed, all the "powers of two" are easy to obtain because folding a sheet of paper in half repeatedly will give you 2, 4, 8, 16, 32,... divisions.
This method of finding intersecting diagonals can be done with any number of divisions.
Let's try another example. You wish to divide a square sheet of paper into 7 segments.
The greatest "power of two" less than 7 is, again, 4.
7 subtract 4 is 3.
So the vertical-distance from the bottom-edge of the paper should be 3/4.
The second diagonal crease should extend from the bottom-left corner to the 3/4 mark on the right edge of the paper; the intersection (dot) of the two diagonal creases marks the 3/7 division.
21 = 2 22 = 4 23 = 8 24 = 16 25 = 32
You can then divide the 4/7 section into 1/7ths (fold section in half and then half again). Use the 1/7th to further divide the 3/7 section into sevenths. This method can also be seen here.
Here's one more example: divide a square sheet of paper into 11 equal segments.
The greatest "power of two" less than 11 is 8.
11 subtract 8 is 3.
So the vertical-distance from the bottom-edge of the paper should be 3/8.
Fold the second diagonal so the crease extends from the bottom-left corner to the 3/8 mark on the right edge of the paper; the intersection (dot) of the two diagonal creases marks the 3/11 division.
21 = 2 22 = 4 23 = 8 24 = 16 25 = 32
This video by Tadashi Mori gives a great explanation as to why this works.
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# 5.6 Rational functions (Page 2/16)
Page 2 / 16
## Vertical asymptote
A vertical asymptote of a graph is a vertical line $\text{\hspace{0.17em}}x=a\text{\hspace{0.17em}}$ where the graph tends toward positive or negative infinity as the inputs approach $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$ We write
## End behavior of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{x}$
As the values of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ approach infinity, the function values approach 0. As the values of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ approach negative infinity, the function values approach 0. See [link] . Symbolically, using arrow notation
Based on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it seems to level off as the inputs become large. This behavior creates a horizontal asymptote , a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line $\text{\hspace{0.17em}}y=0.\text{\hspace{0.17em}}$ See [link] .
## Horizontal asymptote
A horizontal asymptote of a graph is a horizontal line $\text{\hspace{0.17em}}y=b\text{\hspace{0.17em}}$ where the graph approaches the line as the inputs increase or decrease without bound. We write
## Using arrow notation
Use arrow notation to describe the end behavior and local behavior of the function graphed in [link] .
Notice that the graph is showing a vertical asymptote at $\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ which tells us that the function is undefined at $\text{\hspace{0.17em}}x=2.$
And as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a horizontal asymptote at $\text{\hspace{0.17em}}y=4.\text{\hspace{0.17em}}$ As the inputs increase without bound, the graph levels off at 4.
Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function.
End behavior: as Local behavior: as (there are no x - or y -intercepts)
## Using transformations to graph a rational function
Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any.
Shifting the graph left 2 and up 3 would result in the function
$f\left(x\right)=\frac{1}{x+2}+3$
or equivalently, by giving the terms a common denominator,
$f\left(x\right)=\frac{3x+7}{x+2}$
The graph of the shifted function is displayed in [link] .
Notice that this function is undefined at $\text{\hspace{0.17em}}x=-2,\text{\hspace{0.17em}}$ and the graph also is showing a vertical asymptote at $\text{\hspace{0.17em}}x=-2.$
As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at $\text{\hspace{0.17em}}y=3.$
Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units.
The function and the asymptotes are shifted 3 units right and 4 units down. As $\text{\hspace{0.17em}}x\to 3,f\left(x\right)\to \infty ,\text{\hspace{0.17em}}$ and as $\text{\hspace{0.17em}}x\to ±\infty ,f\left(x\right)\to -4.$
The function is $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{{\left(x-3\right)}^{2}}-4.$
## Solving applied problems involving rational functions
In [link] , we shifted a toolkit function in a way that resulted in the function $\text{\hspace{0.17em}}f\left(x\right)=\frac{3x+7}{x+2}.\text{\hspace{0.17em}}$ This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.
sin^4+sin^2=1, prove that tan^2-tan^4+1=0
what is the formula used for this question? "Jamal wants to save \$54,000 for a down payment on a home. How much will he need to invest in an account with 8.2% APR, compounding daily, in order to reach his goal in 5 years?"
i don't need help solving it I just need a memory jogger please.
Kuz
A = P(1 + r/n) ^rt
Dale
how to solve an expression when equal to zero
its a very simple
Kavita
gave your expression then i solve
Kavita
Hy guys, I have a problem when it comes on solving equations and expressions, can you help me 😭😭
Thuli
Tomorrow its an revision on factorising and Simplifying...
Thuli
ok sent the quiz
kurash
send
Kavita
Hi
Masum
What is the value of log-1
Masum
the value of log1=0
Kavita
Log(-1)
Masum
What is the value of i^i
Masum
log -1 is 1.36
kurash
No
Masum
no I m right
Kavita
No sister.
Masum
no I m right
Kavita
tan20°×tan30°×tan45°×tan50°×tan60°×tan70°
jaldi batao
Joju
Find the value of x between 0degree and 360 degree which satisfy the equation 3sinx =tanx
what is sine?
what is the standard form of 1
1×10^0
Akugry
Evalute exponential functions
30
Shani
The sides of a triangle are three consecutive natural number numbers and it's largest angle is twice the smallest one. determine the sides of a triangle
Will be with you shortly
Inkoom
3, 4, 5 principle from geo? sounds like a 90 and 2 45's to me that my answer
Neese
Gaurav
prove that [a+b, b+c, c+a]= 2[a b c]
can't prove
Akugry
i can prove [a+b+b+c+c+a]=2[a+b+c]
this is simple
Akugry
hi
Stormzy
x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
How can I solve for a domain and a codomains in a given function?
ranges
EDWIN
Thank you I mean range sir.
Oliver
proof for set theory
don't you know?
Inkoom
find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
factoring polynomial
|
# Degree of a Polynomial Remainder Theorem – Factor Theorem and its converse
Degree of a Polynomial
The highest value of the exponent of a variable (Polynomial of One Variable only) or the total of highest value of the exponent of the variables (Polynomial of Multivariable) among the terms of a Polynomial is called its Degree.
For example:
1. 5x3 + 25x – 7x2 – 6
Here, in this polynomial, we have
2. 5x3yz2 + 25xy3z6 – 7x2 – 6
Here, in this polynomial, we have
Reminder Theorem
If any polynomial p(x) is divided by a polynomial q(x) = x – a, then the reminder is p(a). This is known as Reminder Theorem.
Reminder Theorem is helpful in finding the reminder without performing actual Division of one polynomial by another polynomial.
Let us illustrate this by using one example-
Suppose p(x) = 6x3 + 5x2 – 7x + 2 is a polynomial. On dividing this polynomial by another polynomial q(x) = x – 5, the reminder can be calculated without performing actual division as
Let us verify this task by actual division process.
The reminder calculated using reminder theorem = Reminder calculated by performing actual division
Factor Theorem
The Exact Divisor of a number is known as its Factor. If first number is divided by a second number and gives reminder as 0(zero), then the Second number is said to be the exact divisor of the First number and accordingly, the Second number is the factor of the First number. For example, when 6 is divided by 2, we get reminder as 0(zero) and so,
If p(x) is any polynomial, such that p(a) = 0 then x – a is the factor of the polynomial p(x). This is known as Factor Theorem.
If x – a is a factor of the Polynomial p(x), then p(a) = 0. This is the converse of Factor Theorem.
Let us illustrate examples of Factor Theorem:-
Suppose p(x) = 6x3 + 5x2 – 7x + 2 is a polynomial. On dividing this polynomial by another polynomial x – 5, then according to Reminder Theorem, the reminder will be
Reminder, p(5) ≠ 0 (zero) and so, x – 5 is not a factor of 6x3 + 5x2 – 7x + 2
Let us take another example.
Suppose p(x) = 5x2 – 7x + 2 is a polynomial. On dividing this polynomial by another polynomial x – 1 and in accordance with Reminder Theorem, the reminder will be
Reminder, p(1) = 0 (zero) and so, x – 1 is a factor of 5x2 – 7x + 2
Let us illustrate one example of Converse of Factor Theorem:-
x – 2 is a factor of the polynomial p(x) = x2 – 7x + 10 is a polynomial
Then
x – 2 is a factor of the polynomial p(x) = x2 – 7x + 10 p(2) = 0
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HomeNCERT SOLUTIONS8th CLASSClass 8 Maths Chapter 13 Direct and Inverse Proportions Solutions
# Class 8 Maths Chapter 13 Direct and Inverse Proportions Solutions
## NCERT Class 8 Maths Chapter 13 Direct and Inverse Proportions Solutions
Chapter 13 Direct and Inverse Proportions Exercise 13.1
Ex 13.1 Class 8 Maths Question 1.
Following are the car parking charges near a railway station up to.
4 hours – ₹ 60
8 hours – ₹ 100
12 hours – ₹ 140
24 hours – ₹ 180
Check if the parking charges are in direct proportions to the parking time.
Solution:
We have the ratio of time period and the parking charge.
Hence the given quantities are not directly proportional.
Ex 13.1 Class 8 Maths Question 2.
A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.
Solution:
Let the number to be filled in the blanks be a, b, c and d respectively.
Ex 13.1 Class 8 Maths Question 3.
In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
Solution:
Let the required red pigment be x part.
Hence, the required amount of red pigment = 24 parts.
Ex 13.1 Class 8 Maths Question 4.
A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Solution:
Let the required number of bottles be x.
Hence the required number of bottles = 700.
Ex 13.1 Class 8 Maths Question 5.
A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
Solution:
Let the actual length be x cm.
Ex 13.1 Class 8 Maths Question 6.
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?
Solution:
Let the required length of the model ship be x m.
Ex 13.1 Class 8 Maths Question 7.
Suppose 2 kg of sugar contains 9 × 106 crystals. How many sugar crystals are there in
(i) 5 kg of sugar?
(ii) 1.2 kg of sugar?
Solution:
Let x be the number of sugar crystals needed.
Ex 13.1 Class 8 Maths Question 8.
Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road of 72 km. What would be her distance covered in the map?
Solution:
Let the required distance be x km.
Hence the distance covered in the map = 4 cm.
Ex 13.1 Class 8 Maths Question 9.
A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high,
(ii) the height of a pole which casts a shadow 5 m long.
Solution:
(i) Let the required length of shadow be x m.
Ex 13.1 Class 8 Maths Question 10.
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:
Let the required distance be x km.
Hence the required distance = 168 km.
## Chapter 13 Direct and Inverse Proportions Ex 13.2
NCERT Class 8 Maths Direct and Inverse Proportions Exercise 13.2
Ex 13.2 Class 8 Maths Question 1.
Which of the following are in inverse proportion?
(i) The number of workers on a job and the time to complete the job.
(ii) The time taken for a journey and the distance travelled in a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time taken for a fixed journey and the speed of the vehicle.
(v) The population of a country and the area of land per person.
Solution:
(i) As the number of workers increase, the job will take less time to complete. Hence, they are inversely proportional.
(ii) For more time, more distance to travel. Hence, they are not inversely proportional.
(iii) More area of land cultivated, more crop to harvest. Hence, they are not inversely proportional.
(iv) If speed is increased, it will take less time to complete the fixed journey. Hence, they are inversely proportional.
(v) If the population of a country increases, then the area of land per person will be decreased. Hence, they are inversely proportional.
Ex 13.2 Class 8 Maths Question 2.
In a Television game show, the prize money of ₹ 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?
Solution:
Let, the blank spaces be denoted by a, b, c, d and e.
So, we observe that 1 × 100,000 = 2 × 50,000
⇒ 1,00,000 = 1,00,000
Hence they are inversely proportional.
2 × 50,000 = 4 × a
Ex 13.2 Class 8 Maths Question 3.
Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.
(i) Are the number of spokes and the angle formed between the pairs of consecutive spokes in inverse proportion.
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?
Solution:
From the above table, we observe that
4 × 90° = 6 × 60°
360° = 360°
Thus the two quantities are inversely proportional.
Let the blank spaces be denoted by a, b, and c.
4 × 90° = 8 × a
Hence, the required table is
(i) Yes, they are in inverse proportion
(ii) If the number of spokes is 15, then
4 × 90° = 15 × x
x = $\frac { 4\times 90 }{ 15 }$ = 24°
(iii) If the angle between two consecutive spokes is 40°, then
4 × 90° = y × 40°
y = $\frac { 4\times 90 }{ 40 }$ = 9 spokes.
Thus the required number of spokes = 9.
Ex 13.2 Class 8 Maths Question 4.
If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is decreased by 4?
Solution:
We observe that on increasing the number of children, number of sweets got by each will be less. So, they are in inverse proportion.
x1y1 = x2y2
where x1 = 24, y1 = 5, x2 = 20
and y2 = a(let)
24 × 5 = 20 × a
a = 6
Hence, the required number of sweets = 6.
Ex 13.2 Class 8 Maths Question 5.
A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Solution:
If the number of animals increases, then it will take fewer days to last.
Then the two quantities are in inverse proportions.
Let the required number of days be p.
x1y1 = x2y2
where x1 = 20, y1 = 6, x2 = 3
and y2 = p (let)
20 × 6 = 30 × p
p = 4
Hence the required number of days = 4.
Ex 13.2 Class 8 Maths Question 6.
A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?
Solution:
If the number of persons is increased, it will take less number of days to complete the job.
Thus, the two quantities are in inverse proportion.
Let the required number of days be k.
x1y1 = x2y2
3 × 4 = 4 × k
k = 3 days.
Hence, the required number of days = 3.
Ex 13.2 Class 8 Maths Question 7.
A batch of bottles was packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?
Solution:
If the number of bottles is increased then the required number of boxes will be decreased. Thus the two quantities are in inverse proportion.
Let the required number of boxes be x.
x1y1 = x2y2
25 × 12 = x × 20
x = 15
Hence, the required number of boxes = 15.
Ex 13.2 Class 8 Maths Question 8.
A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?
Solution:
If the number of machines is increased then less number of days would be required to produced the same number of articles.
Thus, the two quantities are in inverse proportion.
Let the required number of machines be x.
x1y1 = x2y2
42 × 63 = x × 54
x = 49
Hence, the required number of machines is 49.
Ex 13.2 Class 8 Maths Question 9.
A car takes 2 hours to reach a destination by traveling at a speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?
Solution:
On increasing the speed, it will take less time to travel a distance.
Thus the two quantities are in inverse proportions.
Let the required times be x hours.
x1y1 = x2y2
60 × 2 = 80 × x
x = $\frac { 3 }{ 2 }$ hours = 1$\frac { 1 }{ 2 }$ hrs.
Hence, the required time = 1$\frac { 1 }{ 2 }$ hours.
Ex 13.2 Class 8 Maths Question 10.
Two persons could fit new windows in a house in 3 days.
(i) One of the people fell ill before the work started. How long would the job take now?
(ii) How many persons would be needed to fit the windows in one day?
Solution:
On increasing the number of persons, less time will be required to complete a job.
Thus, the quantities are in inverse proportion.
(i) Let the required number of days be x.
x1y1 = x2y2
2 × 3 = 1 × x
x = 6
Hence, the required number of days = 6
(ii) Let the required number of persons be y.
x1y1 = x2y2
2 × 3 = y × 1
y = 6
Hence, the required number of persons = 6.
Ex 13.2 Class 8 Maths Question 11.
A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?
Solution:
On increasing the duration of periods, the number of periods will be reduced.
Thus, the two quantities are in inverse proportion.
Let the required duration of each period be x.
x1y1 = x2y2
8 × 45 = 9 × x
x = 40 minutes
Hence, the required duration of period = 40 minutes.
## Extra Questions Maths Chapter 13
Extra Questions for Class 8 Maths Chapter 13 Direct and Inverse Proportions
### Direct and Inverse Proportions Class 8 Extra Questions
Question 1.
A train is moving at a uniform speed of 100 km/h. How far will it travel in 20 minutes?
Solution:
Let the distance travelled by train in 20 minutes be x km.
Since the speed is uniform, the distance travelled will be directly proportional to time.
Hence, the required distance is 33$\frac { 1 }{ 3 }$ km.
Question 2.
Complete the table if x and y vary directly.
Solution:
Let the blank spaces be filled with a, b and c.
Hence, the required values are a = 7, b = 15 and c = 7.5.
Question 3.
If the cost of 20 books is ₹ 180, how much will 15 books cost?
Solution:
Let the required cost be ₹ x.
Here, the two quantities vary directly.
Question 4.
If x1 = 5, y1 = 7.5, x2 = 7.5 then find y2 if x and y vary directly.
Solution:
Since x and y vary directly.
Hence, the required value is 11.25.
Question 5.
If 3 kg of sugar contains 9 × 108 crystals. How many sugar crystals are there in 4 kg of sugar?
Solution:
Let the required number of crystals be x.
Hence, the required number of crystals = 1.2 × 109.
Question 6.
If 15 men can do a work in 12 days, how many men will do the same work in 6 days?
Solution:
Let the required number of men be x.
Less days → more men.
Thus the two quantities are inversely proportional to each other.
Hence, the required number of days = 30.
Question 7.
A train travels 112 km in 1 hour 30 minutes with a certain speed. How many kilometres it will travel in 4 hours 45 minutes with the same speed?
Solution:
Let the required distance be x km.
More distance → more time
Thus, the two quantities are directly proportional.
Hence, the required distance = 354.6 km.
Question 8.
The scale of a map is given as 1 : 50,000. Two villages are 5 cm apart on the map. Find the actual distance between them.
Solution:
Let the map distance be x cm and actual distance be y.
1 : 50,000 = x : y
Hence, the required distance = 250 km.
Question 9.
8 pipes are required to fill a tank in 1 hr 20 min. How long will it take if only 6 pipes of the same type are used?
Solution:
Let the required time be ‘t’ hours.
Question 10.
15 men can build a wall in 42 hours, how many workers will be required for the same work in 30 hours?
Solution:
Let the required number of workers be x.
The number of workers, faster will they do the work.
So, the two quantities are inversely proportional.
x1y1 = x2y2
⇒ 42 × 15 = 30 × x
⇒ x = 21
Hence, the required number of men = 21
### Direct and Inverse Proportions Class 8 Extra Questions Short Answer Type
Question 11.
The volume of a gas V varies inversely as the pressure P for a given mass of the gas. Fill in the blank spaces in the following table:
Solution:
Since volume and pressure are inversely proportional.
PV = K
Question 12.
The cost of 5 metres of cloth is ₹ 210. Tabulate the cost of 2, 4, 10 and 13 metres of cloth of the same type.
Solution:
Let the length of the cloth be x m and its cost be ₹ y. We have the following table.
Question 13.
Six pumps working together empty a tank in 28 minutes. How long will it take to empty the tank if 4 such pumps are working together?
Solution:
Let the required time be t minutes.
Less pump → More time
Since there is an inverse variation.
x1y1 = x2y2
6 × 28 = 4 × x
x = 42
Hence, the required time = 42 minutes.
Question 14.
Mohit deposited a sum of ₹ 12000 in a Bank at a certain rate of interest for 2 years and earns an interest of ₹ 900. How much interest would be earned for a deposit of ₹ 15000 for the same period and at the same rate of interest?
Solution:
Let the required amount of interest be ₹ x.
Hence, the required amount of interest = ₹ 1125.
Question 15.
A garrison of 120 men has provisions for 30 days. At the end of 5 days, 5 more men joined them. How many days can they sustain on the remaining provision?
Solution:
Let the number of days be x.
[∵ Remaining days = 30 – 5 = 25]
[Total men = 120 + 5 = 125]
Since there is an inverse variation.
x1y1 = x2y2
120 × 25 = 125 × x
x = 24
Hence, the required number of days be 24.
## NCERT Solutions for Class 8 Maths
### సీడీపీవో, ఈవో పోస్టులకు పరీక్ష షెడ్యూలు రిలీజ్
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# Quantitative Notes on Percentages (Part I)
Updated : March 21st, 2020
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Percentage is an important part of Quantitative Aptitude for SSC Exams. Whether it is DI, Profit & Loss, SI-CI, or Allegation, etc. all these chapters with the help of percentage can be solved easily. You can go through the basics of percentage and previous year asked questions to crack the upcoming SSC Exams easily.
A percentage is a number or ratio expressed as a fraction of 100.It is a proportion per hundred.
1.When we say 35 percent in mathematical notation we write 35%.
2.When we want to express this in mathematical form, 35% means 35 per 100 or (35/100).
Important: 50% of 20 can be written 20% of 50 as well.
You can also represent % into decimal, 50% = 0.5
Conversion of fraction into %.
to convert fraction into %, we multiply it by 100.
¼ = (¼)× 100 % = 25 %.
1/3 = (1/3) ×100 % = 33(1/3) %
1/14 = (1/14) ×100 % = (100/14)%=(50/7)%= 7 (1/7) %
Note: Never forget to express % notation in the percentage.
We suggest you that you must learn both tables given below.
### Percentage
1
100%
1/7
14(2/7) %
1/13
7 (9/13) %
1/2
50%
1/8
12(1/2) %
1/14
7 (1/7) %
1/3
33(1/3) %
1/9
11(1/9) %
1/15
6 (2/3) %
1/4
25%
1/10
10 %
1/16
6 (1/4) %
1/5
20%
1/11
9 (1/11) %
1/6
16(2/3) %
1/12
8 (1/3) %
Conversion of % into fraction.
To convert % into fraction, we divide it by 100. So, we can express in this way:
100% = (100/100) = 1 1% = (1/100) 2% = (2/100) = (1/50)
50% = 50/100 = ½
20% = 20/100 = 1/5
10% = 10/100 = 1/10
16(2/3)% = (50/3)% =50/(3×100) = 50/300 = 1/6
### Fraction
10%
1/10
16 (2/3)%
1/6
15%
3/20
20%
1/5
66 (2/3) %
2/3
7(1/2)%
3/40
40%
2/5
6(1/4)%
1/16
22(1/2)%
9/40
60%
3/5
18(3/4) %
3/16
69(3/13) %
9/13
80%
4/5
In following examples we will try to avoid calculation using above table.
(i) 99% of 840
we can say 10% = 84,So 1% = 8.4
99% of 840 = 840-8.4=831.6
(ii)25% of 320 = (1/4)× 320
=80
(iii) 76% of 400?
76%=50%+25%+1%
= 200+100+4
= 304
(iv) 102% of 720?
1%= 7.2 so 2%= 14.4
102% = 100%+2%= 720+14.4 = 734.4
(v)18% of 300?
18% = 20%-2%= (1/5)×300-6
= 60-6 = 54
or 1% = 3 so 18%= 18×3=54
(vi) 12% of 540?
1%=5.4
12% = 10%+2+
= 54+10.8
= 64.8
Example1: Out of his total income, Mr. Sharma spends 20% on house rent and 70% of the rest on house hold expenses. If he saves Rs 1,800 what is his total income (in rupees)?
Solution: Let Income of Mr. Sharma is 100
then he spends 20% on house, so remaining amount is 80.
now he spends 70% of 80 on house hold expenses, so remaining amount left with him is 30% of 80
30% of 80 = 1800
24 = 1800
1 = 1800/24
1 = 75
100= 7500
hence total income is 7500 Rs.
Or, Let total income is P
(100%-20%)×(100%-70%)× P = 1800
80%× 30%× P=1800
((80×30)/(100*100)) × P = 1800
P = 7500
Example2: An army lost 10% its men in war, 10% of the remaining due to diseases died and 10% of the rest were disabled. Thus, the strength was reduced to 729000 active men. Find the original strength.
Solution: Let army has 100 men.
10% loss in war, so remained are 90 men
then,10% of 90 died due to diseases, remained 90-9 = 81
then again, 10% of 81 again disabled
So, remained men = 90% of 81
90% of 81 = 729000
(90×81)/100 =729000
1= 10000
100 = 1000000
hence total men are 1000000.
Example3: In a village three people contested for the post of village Sarpanch. Due to their own interest, all the voters voted and no one vote was invalid. The losing candidate got 30% votes. What could be the minimum absolute margin of votes by which the winning candidate led by the nearest rival, if each candidate got an integral percent of votes?
Solution: As given, no vote was invalid i.e. 100% votes were polled and all candidate got votes in integer value. There were 3 candidates, one losing candidate got 30%, so remaining two candidates got 70% vote of the total.
Candidate 1 + candidate 2 = 70%
Important point which is given in the question is minimum absolute margin and integral value.
Case 1: Suppose candidate 1 got 40%, then candidate 2 had got 30%. But this is not mininmum absolute margin.
Case 2: Both got 35% votes, If both got equal votes then there will be no winning candidate.
Case 3: One candidate must have got 34% and another one have got 36%.
Hence absolute margin is 2%.
Example4:
The difference between 4/5 of a number and 45% of the number is 56. What is 65% of the number?
Solution: Let number is P.
we can say 4/5 = 80%
so, (80%-45%) of P = 56
35% of P = 56
P = (56/35%)
65% of P = 56/35 ×65 = 104
Example5: Deeksha’s science test consists of 85 questions from three sections- i.e. A, B and C. 10 questions from section A, 30 questions from section B and 45 question from section C. Although, she answered 70% of section A, 50% of section B and 60% of section C correctly. She did not pass the test because she got less than 60% of the total marks. How many more questions she would have to answer correctly to earn 60% of the marks which is passing grade?
Solution: If she has done 60% of total questions she would have passed,.
So, no. of question to be done to pass= 60% of 85 = (3/5)×85 = 51
But she done 70% of A = 70% of 10 = 7
50% of B = 50% of 30 = 15
60% of C = (3/5) of 45 = 27
So , total questions she attempted = (7+15+27) = 49
If she has attempted (51-49) = 2 more questions she would have passed.
Example6: In an election between 2 candidates, 75% of the voters cast their votes, out of which 2% votes were declared invalid. A candidate got 18522 votes which were 75% of the valid votes. What was the total number of voters enrolled in the election?
Solution: Let the total number of voters enrolled are P.
Number of votes casted = 75% of P = (75/100) P = 0.75 P
Important: Those votes which were declared invalid are 2% of casted voted not 2% of total votes.
So, valid votes are = (100%-2%) of 0.75P = 98% of 0.75P
Given Candidates got 75% of valid votes = 18522
(75%) × 98% × 0.75 P = 18522
(3/4) * (98/10) * (3/4) P = 18522
P = 42 × 800
Example7: An ore contains 20% of an alloy that has 85% iron. Other than this, in the remaining 80% of the ore, there is no iron. What is the quantity of ore (in kg) needed to obtain 60 kg of pure iron?
Solution: Let quantity of ore is P kg
P × 20% × 85% = 60kg
P × (1/5) × (17/20) = 60
P = (60×5× 20)/17
P = 6000/17 Kg
Example8: 5% of one number (X) is 25% more than another number (Y). If the difference between the numbers is 96 then find the value of X?
Solution : Given: 5% of X = Y + 25% of Y
0.05 X = 1.25 Y
X = 25 Y
X-Y=96
25Y-Y =96
24Y=96
Y = 4 so, X =100
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2 5 8 5 – The Mathematics Fraction of 2 5 8 5
What is 2 5 8 5?
2/5 + 5/8
2/5 + 5/ 8
Step 1
We cannot add two fractions with different denominators (the bottom number). Therefore, you must find a common denominator: the two bottom numbers must match. You will multiply the denominators together, but the numerators must also change. They are multiplied by the denominator of the other term.
So we multiply 2/8, and get 16, then multiply 5/8 and get 40.
2/5 times 8
Do the same for the second term. We multiply 5 by 5, and get 25, then multiply 5 by 8 and get 40.
5/8 times 5
So now our fractions look like this:
16/40 + 25/ 40
Step 2
Since our denominators match, we can insert the numerators.
16 + 25 = 41
41/40
Step 3
Last, of all, we require simplifying the fraction, if possible. Can it reduce to a simpler fraction?
To find out, we attempt dividing it by 2
No good. So next, you try the following prime number, which is 3
No good. So next, you try the following prime number, which is 5
No good. So next, you attempt the following prime number, which is 7
No good. So next, you try the following prime number, which is 11
And also, No interest. So next, you try the following prime number, which is 13
No good. So next, you try the following prime number, which is 17
No good. So next, you try the following prime number, which is 19
And also, No interest. So next, you attempt the following prime number, which is 23
No good. So next, you try the following prime number, which is 29
No good. So next, you try the following prime number, which is 31
No good. So next, you try the following prime number, which is 37
No good. So next, you try the following prime number, which is 41
No good. 41 is more significant than 40. So we’re done reducing.
And we’re done! Here’s the final answer to 2/5 + 5/8
What is 2 5 8 5 as a decimal?
A decimal number can define as a number whose whole number part and a decimal point separate the fractional part.
Finally, answer: 2 5/8 as a decimal is 2.625
Let’s look into the two ways to write 2 5 8 5 as a decimal.
Explanation:
Writing 2 5/8 as a decimal by the division way.
To exchange any fraction to decimal form, we require to divide its numerator by the denominator.
Here, the fraction of this number, means we must do 21 ÷ 8.
Symbols 2 5/8 as a decimal by converting the denominator to powers of 10.
Find a number so that we can multiply it by the fraction’s denominator to make it 10 or 100 or 1000.
Multiply together the numerator and denominator by that figure to convert it into an equal fraction.
Then, write down the numerator and put the decimal point in the correct place, that is, one space from the right-hand side for each 0 in the denominator.
So 2 5/8 = 21/8 = (21 × 125) / (8 × 125) = 2625/1000 = 2.625
Irrespective of the ways used, finally, the answer to it as a decimal will always stay the same.
58 + 25 is 4140.
Find the least common denominator or LCM of the two denominators:
LCM of 8 and 5 is 40.
Next, find the equivalent fraction of both fractional numbers with denominator 40
For the 1st fraction, since 8 × 5 = 40,
58 = 5 × 58 × 5 = 2540
Likewise, for the 2nd fraction, since 5 × 8 = 40,
25 = 2 × 85 × 8 = 1640
2540 + 1640 = 25 + 1640 = 4140
So, 5/8 + 2/5 = 41/40
In mixed form and finally answer is 1140.
Fraction Calculator of 2 5 8 5
Input fractions:
2+5(8-5)
Work with fractions:
This calculator performs basic and advanced operations with fractions expressions combined with whole numbers, decimals, and mixed numbers. It also shows detailed step-by-step data about the part calculation procedure—the calculator aids to find the fraction value of multiple fraction operations. Finally, solve problems with two, three, or many fractions and figures in an expression.
Result:
2 + 5(8 – 5) = 17/1 = 17
Spelled result in words is seventeen.
Fractions in Word Problems:
Fraction to decimal
Write the part 3/22 as a decimal.
One half of one per cent
One half of 1% converted to a decimal is?
Two fractions multiply
What 4/5 is multiplied by 9/10?
Rolls
Mom bought 13 rolls. Dad ate 3.5 rolls. How many rolls were left when Peter yet ate two at dinner?
New bridge
Thanks to the new bridge, the road between A and B has been cut to one-third and is now 10km long. So how much did the route between A and B measure before?
School
There are 150 pupils in grade 5 . 2/3 of them are female. By what fractions are the males?
Fruits
Amy buy a basket of fruits 1/5 of them were apples,1/4 were oranges, and the rest were 33 bananas. How numerous fruits did she believe in all?
Obtuse angle
Which obtuse angle is creating clocks at 17:00?
Substitution 66204
Specify the worth of the expression, (3x-5) / (2x + 1) if x equals the value that will be in the reader after replacement.
Equation with x
Solve the next equation: 2x- (8x + 1) – (x + 2) / 5 = 9
Three monks
Three medieval monks have the job to copy 600 pages of the Bible book, one rewrites in three days, 1 page, the second page in 2 days, three pages, and a third in 4 days, two sides. And also, analyze for how many days and what day the months will have copied the whole Bible when they begin on Wednesday.
Class 8. A
Three-quarters of class 8A went skiing. Of those who remained at home, one-third were ill, and the remaining six were on math Olympic. How many students have class 8A?
Ratio
Write the ratio with other numbers so that the value is the same: 2: 9.
How do you Solve Fractions with Details?
Subtract: 8 – 5 = 3
Multiple: 5 * the result of step No. 1 = 5 * 3 = 15
And also, add: 2 + the result of step No. 2 = 2 + 15 = 17
Rules for Expressions with Fractions
Fractions – use a forward slash to split the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed fractions or numbers) – Keep one space between the integer and also fraction and use a forward slash to input fractions, i.e., 1 2/3. An example of a negative mixed bit: -5 1/2.
Because slash is both sign for part line and division, use a colon (:) as the operator of division fractions, i.e., 1/2: 1/3.
Decimals (decimal numbers) – go into a decimal point, and also they are automatically converted to fractions – i.e. 1.45.
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# What The Numbers Say About Andrew Garfield (02/14/2020)
How will Andrew Garfield fare on 02/14/2020 and the days ahead? Let’s use astrology to complete a simple analysis. Note this is of questionable accuracy – don’t get too worked up about the result. I will first find the destiny number for Andrew Garfield, and then something similar to the life path number, which we will calculate for today (02/14/2020). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology experts.
PATH NUMBER FOR 02/14/2020: We will analyze the month (02), the day (14) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. What does this entail? We will show you. First, for the month, we take the current month of 02 and add the digits together: 0 + 2 = 2 (super simple). Then do the day: from 14 we do 1 + 4 = 5. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 2 + 5 + 4 = 11. This still isn’t a single-digit number, so we will add its digits together again: 1 + 1 = 2. Now we have a single-digit number: 2 is the path number for 02/14/2020.
DESTINY NUMBER FOR Andrew Garfield: The destiny number will calculate the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Andrew Garfield we have the letters A (1), n (5), d (4), r (9), e (5), w (5), G (7), a (1), r (9), f (6), i (9), e (5), l (3) and d (4). Adding all of that up (yes, this can get tedious) gives 73. This still isn’t a single-digit number, so we will add its digits together again: 7 + 3 = 10. This still isn’t a single-digit number, so we will add its digits together again: 1 + 0 = 1. Now we have a single-digit number: 1 is the destiny number for Andrew Garfield.
CONCLUSION: The difference between the path number for today (2) and destiny number for Andrew Garfield (1) is 1. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But this is just a shallow analysis! As mentioned earlier, this is not at all guaranteed. If you want a reading that people really swear by, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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```Page 1 of 2
5.4
Complex Numbers
What you should learn
equations with complex
solutions and perform
operations with complex
numbers.
GOAL 1
GOAL 1
OPERATIONS WITH COMPLEX NUMBERS
Not all quadratic equations have real-number solutions. For instance, x 2 = º1 has
no real-number solutions because the square of any real number x is never negative.
To overcome this problem, mathematicians created an expanded system of numbers
1. Note that i 2 = º1. The
using the imaginary unit i, defined as i = º
imaginary unit i can be used to write the square root of any negative number.
GOAL 2 Apply complex
numbers to fractal geometry.
T H E S Q UA R E R O O T O F A N E G AT I V E N U M B E R
Why you should learn it
RE
FE
To solve problems, such
as determining whether a
complex number belongs to
the Mandelbrot set
in Example 7.
AL LI
PROPERTY
EXAMPLE
1. If r is a positive real number,
then º
r = i r.
º
5
= i 5
2. By Property (1), it follows that
(i r)2 = ºr.
(i 5)2 = i 2 • 5 = º5
EXAMPLE 1
Solve 3x 2 + 10 = º26.
SOLUTION
3x 2 + 10 = º26
2
3x = º36
x 2 = º12
Write original equation.
Subtract 10 from each side.
Divide each side by 3.
x = ±º
12
Take square roots of each side.
x = ±i12
Write in terms of i.
x = ±2i3
The solutions are 2i 3 and º2i 3.
..........
A complex number written in standard form is
a number a + bi where a and b are real numbers.
The number a is the real part of the complex number,
and the number bi is the imaginary part. If b ≠ 0,
then a + bi is an imaginary number. If a = 0 and
b ≠ 0, then a + bi is a pure imaginary number.
The diagram shows how different types of complex
numbers are related.
272
Complex Numbers (a bi )
Real
Numbers
(a 0i )
5
2
1
3
π
2
Imaginary
Numbers
(a bi, b 0)
2 3i
5 5i
Pure
Imaginary
Numbers
(0 bi, b 0)
4i
6i
Page 1 of 2
Just as every real number corresponds to a point on the real number line, every
complex number corresponds to a point in the complex plane. As shown in the next
example, the complex plane has a horizontal axis called the real axis and a vertical
axis called the imaginary axis.
EXAMPLE 2
Plotting Complex Numbers
Plot the complex numbers in the complex plane.
a. 2 º 3i
b. º3 + 2i
c. 4i
imaginary
4i
SOLUTION
3 2i
a. To plot 2 º 3i, start at the origin, move 2 units to the
i
right, and then move 3 units down.
1
real
b. To plot º3 + 2i, start at the origin, move 3 units to the
left, and then move 2 units up.
2 3i
c. To plot 4i, start at the origin and move 4 units up.
..........
Two complex numbers a + bi and c + di are equal if and only if a = c and b = d.
For instance, if x + yi = 8 º i, then x = 8 and y = º1.
To add (or subtract) two complex numbers, add (or subtract) their real parts and their
imaginary parts separately.
Sum of complex numbers: (a + bi) + (c + di) = (a + c) + (b + d)i
Difference of complex numbers: (a + bi) º (c + di) = (a º c) + (b º d)i
EXAMPLE 3
Write the expression as a complex number in standard form.
a. (4 º i) + (3 + 2i)
b. (7 º 5i) º (1 º 5i)
c. 6 º (º2 + 9i) + (º8 + 4i)
SOLUTION
a. (4 º i) + (3 + 2i) = (4 + 3) + (º1 + 2)i
=7+i
Standard form
b. (7 º 5i) º (1 º 5i) = (7 º 1) + (º5 + 5)i
Definition of complex subtraction
= 6 + 0i
Simplify.
=6
Standard form
c. 6 º (º2 + 9i) + (º8 + 4i) = [(6 + 2) º 9i] + (º8 + 4i)
Subtract.
= (8 º 9i) + (º8 + 4i)
Simplify.
= (8 º 8) + (º9 + 4)i
= 0 º 5i
Simplify.
= º5i
Standard form
5.4 Complex Numbers
273
Page 1 of 2
To multiply two complex numbers, use the distributive property or the FOIL method
just as you do when multiplying real numbers or algebraic expressions. Other
properties of real numbers that also apply to complex numbers include the associative
and commutative properties of addition and multiplication.
Multiplying Complex Numbers
EXAMPLE 4
Write the expression as a complex number in standard form.
a. 5i(º2 + i)
b. (7 º 4i)(º1 + 2i)
c. (6 + 3i)(6 º 3i)
SOLUTION
a. 5i(º2 + i) = º10i + 5i2
Distributive property
= º10i + 5(º1)
Use i 2 = º1.
= º5 º 10i
Standard form
b. (7 º 4i)(º1 + 2i) = º7 + 14i + 4i º 8i2
Use FOIL.
= º7 + 18i º 8(º1)
Simplify and use i 2 = º1.
= 1 + 18i
Standard form
c. (6 + 3i)(6 º 3i) = 36 º 18i + 18i º 9i2
Use FOIL.
= 36 º 9(º1)
Simplify and use i 2 = º1.
= 45
Standard form
..........
In part (c) of Example 4, notice that the two factors 6 + 3i and 6 º 3i have the form
a + bi and a º bi. Such numbers are called complex conjugates. The product of
complex conjugates is always a real number. You can use complex conjugates to
write the quotient of two complex numbers in standard form.
Dividing Complex Numbers
EXAMPLE 5
5 + 3i
1 º 2i
Write the quotient in standard form.
SOLUTION
The key step here is to multiply the numerator and the denominator by the complex
conjugate of the denominator.
5 + 3i
5 + 3 i 1 + 2i
= • 1 º 2i
1 º 2 i 1 + 2i
5 + 10i + 3i + 6i2
1 + 2i º 2i º 4i
Use FOIL.
º1 + 13i
5
Simplify.
= 2
= 1
5
13
5
= º + i
274
Multiply by 1 + 2i, the conjugate of 1 º 2i.
Standard form
Page 1 of 2
FOCUS ON
PEOPLE
GOAL 2
USING COMPLEX NUMBERS IN FRACTAL GEOMETRY
In the hands of a person who understands fractal geometry, the complex plane can
become an easel on which stunning pictures called fractals are drawn. One very
famous fractal is the Mandelbrot set, named after mathematician Benoit Mandelbrot.
The
–iMandelbrot set is the black region in the complex plane below. (The points in
the colored regions are not part of the Mandelbrot set.)
1
–i
RE
FE
L
AL I
BENOIT
MANDELBROT
was born in Poland in 1924,
came to the United States in
1958, and is now a professor
at Yale University. He pioneered the study of fractal
geometry in the 1970s.
–i
–1
1
–i
–2
–1
1
–i
–3
–2
–1
1
–4
–3
–2
–1
–4
–3
–2
–4
–3
1
–1
–i
–2
1
–3
–1
–4
–i
–2
–4
–3
1
To understand how the Mandelbrot set is constructed, you need to know how the
absolute value of a complex number is defined.
–4
–1
A B S O L U T E VA L U E O F A C O M P L E X N U M B E R
–2
–3
The absolute value of a complex number z = a + bi, denoted |z|, is a
nonnegative real number defined as follows:
2+
b2
|z| = a
–4
Geometrically, the absolute value of a complex number is the number’s
distance from the origin in the complex plane.
EXAMPLE 6
Finding Absolute Values of Complex Numbers
Find the absolute value of each complex number. Which number is farthest from the
origin in the complex plane?
a. 3 + 4i
b. º2i
c. º1 + 5i
SOLUTION
a. |3 + 4i| = 3
2+
42 = 25 = 5
b. |º2i| = |0 + (º2i)| = 0
2+
(º
2
)2 = 2
c. |º1 + 5i| = (º
1)2
+52 = 26 ≈ 5.10
Since º1 + 5i has the greatest absolute value, it is
farthest from the origin in the complex plane.
imaginary
z 1 5i
|z| 26
z 3 4i
3i
|z| 5
z 2i
|z| 2
5.4 Complex Numbers
4
real
275
Page 1 of 2
The following result shows how absolute value can be used to tell whether a given
complex number belongs to the Mandelbrot set.
COMPLEX NUMBERS IN THE MANDELBROT SET
To determine whether a complex number c belongs to the Mandelbrot set,
consider the function ƒ(z) = z 2 + c and this infinite list of complex numbers:
z0 = 0, z1 = ƒ(z0), z2 = ƒ(z1), z3 = ƒ(z2), . . .
•
If the absolute values |z0|, |z1|, |z2|, |z3|, . . . are all less than some fixed
number N, then c belongs to the Mandelbrot set.
•
If the absolute values |z0|, |z1|, |z2|, |z3|, . . . become infinitely large,
then c does not belong to the Mandelbrot set.
EXAMPLE 7
Determining if a Complex Number Is in the Mandelbrot Set
Tell whether the complex number c belongs to the Mandelbrot set.
a. c = i
b. c = 1 + i
c. c = º2
SOLUTION
a. Let ƒ(z) = z2 + i.
z0 = 0
|z0| = 0
2
z1 = ƒ(0) = 0 + i = i
|z1| = 1
z2 = ƒ(i) = i2 + i = º1 + i
|z2| = 2 ≈ 1.41
z3 = ƒ(º1 + i) = (º1 + i)2 + i = ºi
|z3| = 1
2
z4 = ƒ(ºi) = (ºi) + i = º1 + i
|z 4| = 2 ≈ 1.41
At this point the absolute values alternate between 1 and 2
, and so all
the absolute values are less than N = 2. Therefore, c = i belongs to the
Mandelbrot set.
b. Let ƒ(z) = z2 + (1 + i).
z0 = 0
|z 0 | = 0
z1 = ƒ(0) = 02 + (1 + i) = 1 + i
|z 1| ≈ 1.41
z2 = ƒ(1 + i) = (1 + i)2 + (1 + i) = 1 + 3i
|z 2| ≈ 3.16
2
z3 = ƒ(1 + 3i) = (1 + 3i) + (1 + i) = º7 + 7i
|z 3 | ≈ 9.90
z4 = ƒ(º7 + 7i) = (º7 + 7i)2 + (1 + i) = 1 º 97i
|z 4 | ≈ 97.0
The next few absolute values in the list are (approximately) 9409, 8.85 ª 107,
and 7.84 ª 1015. Since the absolute values are becoming infinitely large,
c = 1 + i does not belong to the Mandelbrot set.
c. Let ƒ(z) = z 2 + (º2), or ƒ(z) = z 2 º 2. You can show that z0 = 0, z1 = º2, and
zn = 2 for n > 1. Therefore, the absolute values of z0, z1, z2, z3, . . . are all less
than N = 3, and so c = º2 belongs to the Mandelbrot set.
276
Page 1 of 2
GUIDED PRACTICE
Vocabulary Check
Concept Check
✓
✓
? and
1. Complete this statement: For the complex number 3 º 7i, the real part is ? .
the imaginary part is 2. ERROR ANALYSIS A student thinks that the complex conjugate of º5 + 2i
is 5 º 2i. Explain the student’s mistake, and give the correct complex conjugate
of º5 + 2i.
3. Geometrically, what does the absolute value of a complex number represent?
Skill Check
✓
Solve the equation.
4. x2 = º9
5. 2x2 + 3 = º13
6. (x º 1)2 = º7
Write the expression as a complex number in standard form.
7. (1 + 5i) + (6 º 2i)
8. (4 + 3i) º (º2 + 4i)
3 º 4i
10. 1+i
9. (1 º i)(7 + 2i)
Find the absolute value of the complex number.
11. 1 + i
12. 3i
13. º2 + 3i
14. 5 º 5i
15. Plot the numbers in Exercises 11–14 in the same complex plane.
16. FRACTAL GEOMETRY Tell whether c = 1 º i belongs to the Mandelbrot set.
PRACTICE AND APPLICATIONS
STUDENT HELP
SOLVING QUADRATIC EQUATIONS Solve the equation.
Extra Practice
skills is on p. 946.
17. x2 = º4
18. x2 = º11
19. 3x2 = º81
20. 2x2 + 9 = º41
21. 5x2 + 18 = 3
22. ºx2 º 4 = 14
23. 8r 2 + 7 = 5r 2 + 4
24. 3s2 º 1 = 7s2
25. (t º 2)2 = º16
26. º6(u + 5)2 = 120
1
27. º(v + 3)2 = 7
8
28. 9(w º 4)2 + 1 = 0
PLOTTING COMPLEX NUMBERS Plot the numbers in the same complex plane.
29. 4 + 2i
30. º1 + i
31. º4i
32. 3
33. º2 º i
34. 1 + 5i
35. 6 º 3i
36. º5 + 4i
STUDENT HELP
ADDING AND SUBTRACTING Write the expression as a complex number in
HOMEWORK HELP
standard form.
Example 1:
Example 2:
Example 3:
Example 4:
Example 5:
Example 6:
Example 7:
Exs. 17–28
Exs. 29–36
Exs. 37–46
Exs. 47–55
Exs. 56–63
Exs. 64–71
Exs. 72–79
37. (2 + 3i) + (7 + i)
38. (6 + 2i) + (5 º i)
39. (º4 + 7i) + (º4 º 7i)
40. (º1 º i) + (9 º 3i)
41. (8 + 5i) º (1 + 2i)
42. (2 º 6i) º (º10 + 4i)
43. (º0.4 + 0.9i) º (º0.6 + i)
44. (25 + 15i) º (25 º 6i)
45. ºi + (8 º 2i) º (5 º 9i)
46. (30 º i) º (18 + 6i) + 30i
5.4 Complex Numbers
277
Page 1 of 2
MULTIPLYING Write the expression as a complex number in standard form.
47. i(3 + i)
48. 4i(6 º i)
49. º10i(4 + 7i)
50. (5 + i)(8 + i)
51. (º1 + 2i)(11 º i)
52. (2 º 9i)(9 º 6i)
2
53. (7 + 5i)(7 º 5i)
55. (15 º 8i)2
54. (3 + 10i)
DIVIDING Write the expression as a complex number in standard form.
8
56. 1+i
2i
57. 1ºi
º5 º 3i
58. 4i
3+i
59. 3ºi
2 + 5i
60. 5 + 2i
º7 + 6 i
61. 9 º 4i
0
1
62. 1
0
ºi
6 º i 2
63. 6 + i 2
ABSOLUTE VALUE Find the absolute value of the complex number.
64. 3 º 4i
65. 5 + 12i
66. º2 º i
67. º7 + i
68. 2 + 5i
69. 4 º 8i
70. º9 + 6i
71. 1
1 + i5
MANDELBROT SET Tell whether the complex number c belongs to the
STUDENT HELP
Skills Review
For help with
disproving statements
by counterexample,
see p. 927.
72. c = 1
73. c = º1
74. c = ºi
75. c = º1 º i
76. c = 2
77. c = º1 + i
78. c = º0.5
79. c = 0.5i
LOGICAL REASONING In Exercises 80–85, tell whether the statement is true
or false. If the statement is false, give a counterexample.
80. Every complex number is an imaginary number.
81. Every irrational number is a complex number.
82. All real numbers lie on a single line in the complex plane.
83. The sum of two imaginary numbers is always an imaginary number.
84. Every real number equals its complex conjugate.
85. The absolute values of a complex number and its complex conjugate are
always equal.
86. VISUAL THINKING The graph shows how
imaginary
4 6i
you can geometrically add two complex
numbers (in this case, 3 + 2i and 1 + 4i) to
find their sum (in this case, 4 + 6i). Find each
of the following sums by drawing a graph.
a. (2 + i) + (3 + 5i)
4i
1
2i
i
b. (º1 + 6i) + (7 º 4i)
3
4
real
COMPARING REAL AND COMPLEX NUMBERS Tell whether the property is
true for (a) the set of real numbers and (b) the set of complex numbers.
87. If r, s, and t are numbers in the set, then (r + s) + t = r + (s + t).
88. If r is a number in the set and |r| = k, then r = k or r = ºk.
89. If r and s are numbers in the set, then r º s = s º r.
90. If r, s, and t are numbers in the set, then r(s + t) = rs + rt.
91. If r and s are numbers in the set, then |r + s| = |r| + |s|.
278
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FOCUS ON
CAREERS
92. CRITICAL THINKING Evaluate º
4 • º
9 and 36. Does the rule
a • b = ab on page 264 hold when a and b are negative numbers?
93.
Writing Give both an algebraic argument and a geometric argument
explaining why the definitions of absolute value on pages 50 and 275 are
consistent when applied to real numbers.
inverse of a complex number z is a complex number za such that z + za = 0.
The multiplicative inverse of z is a complex number zm such that z • zm = 1.
Find the additive and multiplicative inverses of each complex number.
a. z = 1 + i
b. z = 3 º i
c. z = º2 + 8i
ELECTRICITY In Exercises 95 and 96, use the following information.
RE
FE
L
AL I
ELECTRICIAN
INT
An electrician
installs, maintains, and
repairs electrical systems.
This often involves working
with the types of circuits
described in Exs. 95 and 96.
NE
ER T
www.mcdougallittell.com
Electrical circuits may contain several types
of components such as resistors, inductors,
and capacitors. The resistance of each
component to the flow of electrical current is
the component’s impedance, denoted by Z.
The value of Z is a real number R for a resistor
of R ohms (), a pure imaginary number Li
for an inductor of L ohms, and a pure
imaginary number ºCi for a capacitor of
C ohms. Examples are given in the table.
95.
b.
2
INT
HOMEWORK HELP
www.mcdougallittell.com
for help with problem
solving in Exs. 95 and 96.
Resistor
3
3
Inductor
5
5i
Capacitor
6
º6i
c.
2 8
6
8
7
96.
Z
12
5
NE
ER T
Symbol
SERIES CIRCUITS A series circuit is a type of circuit found in switches,
fuses, and circuit breakers. In a series circuit, there is only one pathway through
which current can flow. To find the total impedance of a series circuit, add the
impedances of the components in the circuit. What is the impedance of each
series circuit shown below? (Note: The symbol
denotes an alternating current
source and does not affect the calculation of impedance.)
a.
STUDENT HELP
Component
4
15
PARALLEL CIRCUITS Parallel circuits are used in household lighting and
appliances. In a parallel circuit, there is more than one pathway through which
current can flow. To find the impedance Z of a parallel circuit with two pathways,
first calculate the impedances Z1 and Z2 of the pathways separately by treating
each pathway as a series circuit. Then apply this formula:
Z1Z2
Z1 + Z2
Z=
What is the impedance of each parallel circuit shown below?
a.
3
Z1
4
b.
6
2
5
Z2
Z1
3
c.
8
9
4
Z2
5
Z1
2
7
5.4 Complex Numbers
Z2
279
Page 1 of 2
Test
Preparation
★ Challenge
QUANTITATIVE COMPARISON In Exercises 97–99, choose the statement that
is true about the given quantities.
A
¡
B
¡
C
¡
D
¡
The quantity in column A is greater.
The quantity in column B is greater.
The two quantities are equal.
The relationship cannot be determined from the given information.
Column A
Column B
97.
|5 + 4i|
|3 º 6i|
98.
|º6 + 8i|
|º10i|
99.
|2 + bi| where b < º1
|3 + ci| where 0 < c < 1
100. POWERS OF i In this exercise you will investigate a pattern that appears when
the imaginary unit i is raised to successively higher powers.
a. Copy and complete the table.
Power of i
i1
i2
i3
i4
i5
i6
i7
i8
Simplified form
i
º1
ºi
?
?
?
?
?
b. Writing Describe the pattern you observe in the table. Verify that the
pattern continues by evaluating the next four powers of i.
EXTRA CHALLENGE
www.mcdougallittell.com
c. Use the pattern you described in part (b) to evaluate i 26 and i 83.
MIXED REVIEW
EVALUATING FUNCTIONS Evaluate ƒ(x) for the given value of x. (Review 2.1)
101. ƒ(x) = 4x º 1 when x = 3
102. ƒ(x) = x2 º 5x + 8 when x = º4
103. ƒ(x) = |ºx + 6| when x = 9
104. ƒ(x) = 2 when x = º30
SOLVING SYSTEMS Use an inverse matrix to solve the system. (Review 4.5)
105. 3x + y = 5
106. x + y = 2
5x + 2y = 9
107. x º 2y = 10
7x + 8y = 21
3x + 4y = 0
SOLVING QUADRATIC EQUATIONS Solve the equation. (Review 5.3 for 5.5)
108. (x + 4)2 = 1
109. (x + 2)2 = 36
110. (x º 11)2 = 25
111. º(x º 5)2 = º10
112. 2(x + 7)2 = 24
113. 3(x º 6)2 º 8 = 13
CONNECTION The table shows the cumulative number N
(in thousands) of DVD players sold in the United States from the end of
February, 1997, to time t (in months). Make a scatter plot of the data.
Approximate the equation of the best-fitting line. (Review 2.5)
INT
114. STATISTICS
280
t
1
2
3
4
5
6
7
N
34
69
96
125
144
178
213
NE
ER T
8
9
10
11
12
269 307 347
383
416
DATA UPDATE of DVD Insider data at www.mcdougallittell.com
|
## Use the table below to review the Lumos StepUp™ 8th Grade Math and Language Arts Literacy Lesson Names, Topics, along with their Common Core State Standard (CCSS) Correlations.
Course Name: Lumos StepUp – Grade 8 Math
Lesson Name
Introduction and Diagnostic Test
### Diagnostic Test – Part II
The Number System
### Rational Numbers 8.NS.1
Domain: The Number System
Theme: Know that there are numbers that are not rational, and approximate them by rational numbers.
Standard: Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number.
### Irrational Numbers 8.NS.2
Domain: The Number System
Theme: Know that there are numbers that are not rational, and approximate them by rational numbers
Standard: Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., ?^2). For example, by truncating the decimal expansion of ?2 (square root of 2), show that ?2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations.
Expressions and Equations
### Exponents Multiplication & Division 8.EE.1
Domain: Expressions & Equations
Theme: Work with radicals and integer exponents
Standard: Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 3^2
### Square & Cube Roots 8.EE.2
Domain: Expressions & Equations
Theme: Work with radicals and integer exponents
Standard: Use square root and cube root symbols to represent solutions to equations of the form x^2 = p and x^3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that ?2 is irrational.
### Scientific Notation 8.EE.3
Domain: Expressions & Equations
Theme: Work with radicals and integer exponents
Standard: Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other. For example, estimate the population of the United States as 3
### Scientific Notation Multiplication & Division 8.EE.4
Domain: Expressions & Equations
Theme: Work with radicals and integer exponents
Standard: Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology.
### Compare Proportions 8.EE.5
Domain: Expressions & Equations
Theme: Understand the connections between proportional relationships, lines, and linear equations
Standard: Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed.
### Slope 8.EE.6
Domain: Expressions & Equations
Theme: Understand the connections between proportional relationships, lines, and linear equations
Standard: Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y =mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b.
### One-Variable Equations 8.EE.7a
Domain: Expressions & Equations
Theme: Analyze and solve linear equations and pairs of simultaneous linear equations
Standard: Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers).
### System of Equations 8.EE.8a
Domain: Expressions & Equations
Theme: Analyze and solve linear equations and pairs of simultaneous linear equations
Standard: Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously.
Functions
### Functions 8.F.1
Domain: Functions
Theme: Define, evaluate, and compare functions
Standard: Understand that a function is a rule that assigns to each input exactly one output. The graph of a function is the set of ordered pairs consisting of an input and the corresponding output. (Function notation is not required in Grade 8.)
### Compare Functions 8.F.2
Domain: Functions
Theme: Define, evaluate, and compare functions
Standard: Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a linear function represented by a table of values and a linear function represented by an algebraic expression, determine which function has the greater rate of change.
### Linear Functions 8.F.3
Domain: Functions
Theme: Define, evaluate, and compare functions
Standard: Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. For example, the function A = s^2 giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a straight line.
### Linear Function Models 8.F.4
Domain: Functions
Theme: Use functions to model relationships between quantities
Standard: Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values.
### Analyzing Functions 8.F.5
Domain: Functions
Theme: Use functions to model relationships between quantities
Standard: Describe qualitatively the functional relationship between two quantities by analyzing a graph (e.g., where the function is increasing or decreasing, linear or nonlinear). Sketch a graph that exhibits the qualitative features of a function that has been described verbally.
Geometry
### Transformations of Points & Lines 8.G.1
Domain: Geometry
Theme: Understand congruence and similarity using physical models, transparencies, or geometry software
Standard: Verify experimentally the properties of rotations, reflections, and translations.
### Transformations of Congruent Objects 8.G.2
Domain: Geometry
Theme: Understand congruence and similarity using physical models, transparencies, or geometry software
Standard: Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations; given two congruent figures, describe a sequence that exhibits the congruence between them.
### Analyzing Transformations 8.G.3
Domain: Geometry
Theme: Understand congruence and similarity using physical models, transparencies, or geometry software
Standard: Describe the effect of dilations, translations, rotations and reflections on two-dimensional figures using coordinates.
### Transformations & Similar Objects 8.G.4
Domain: Geometry
Theme: Understand congruence and similarity using physical models, transparencies, or geometry software
Standard: Understand that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations, reflections, translations, and dilations; given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them.
### Interior & Exterior Angles 8.G.5
Domain: Geometry
Theme: Understand congruence and similarity using physical models, transparencies, or geometry software
Standard: Use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles. For example, arrange three copies of the same triangle so that the three angles appear to form a line, and give an argument in terms of transversals why this is so.
### Pythagorean Theorem 8.G.6
Domain: Geometry
Theme: Understand and apply the Pythagorean Theorem
Standard: Explain a proof of the Pythagorean Theorem and its converse.
### Pythagorean Theorem in Real-World Problems 8.G.7
Domain: Geometry
Theme: Understand and apply the Pythagorean Theorem
Standard: Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.
### Pythagorean Theorem & Coordinate System 8.G.8
Domain: Geometry
Theme: Understand and apply the Pythagorean Theorem
Standard: Apply the Pythagorean Theorem to find the distance between two points in a coordinate system.
### Volume: Cone, Cylinder, & Sphere 8.G.9
Domain: Geometry
Theme: Solve real-world and mathematical problems involving volume of cylinders, cones and spheres.
Standard: Know the formulas for the volume of cones, cylinders, and spheres and use them to solve real-world and mathematical problems.
Statistics & Probability
### Interpret Data Tables & Scatter Plots 8.SP.1
Domain: Statistics & Probability
Theme: Investigate patterns of association in bivariate data
Standard: Construct and interpret scatter plots for bivariate measurement data to investigate patterns of association between two quantities. Describe patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear association.
### Scatter Plots with Linear Association 8.SP.2
Domain: Statistics & Probability
Theme: Investigate patterns of association in bivariate data
Standard: Know that straight lines are widely used to model relationships between two quantitative variables. For scatter plots that suggest a linear association, informally fit a straight line, and informally assess the model fit by judging the closeness of the data points to the line.
### Analyzing Linear Scatter Plots 8.SP.3
Domain: Statistics & Probability
Theme: Investigate patterns of association in bivariate data
Standard: Use the equation of a linear model to solve problems in the context of bivariate measurement data, interpreting the slope and intercept. For example, in a linear model for a biology experiment, interpret a slope of 1.5 cm/hr as meaning that an additional hour of sunlight each day is associated with an additional 1.5 cm in mature plant height.
### Relatable Data Frequency 8.SP.4
Domain: Statistics & Probability
Theme: Investigate patterns of association in bivariate data
Standard: Understand that patterns of association can also be seen in bivariate categorical data by displaying frequencies and relative frequencies in a two-way table. Construct and interpret a two-way table summarizing data on two categorical variables collected from the same subjects. Use relative frequencies calculated for rows or columns to describe possible association between the two variables. For example, collect data from students in your class on whether or not they have a curfew on school nights and whether or not they have assigned chores at home. Is there evidence that those who have a curfew also tend to have chores?
Test Taking Tips
Practice Tests
### Practice Test 2 – Part II
Course Name: Lumos StepUp – Grade 8 Language Arts Literacy
Lesson Name
Introduction and Diagnostic Test
### Textual Evidence RL.8.1
Theme: Key Ideas and Details
Standard: Cite the textual evidence that most strongly supports an analysis of what the text says explicitly as well as inferences drawn from the text.
### Inferences RL.8.1
Theme: Key Ideas and Details
Standard: Cite the textual evidence that most strongly supports an analysis of what the text says explicitly as well as inferences drawn from the text.
### Theme RL.8.2
Theme: Key Ideas and Details
Standard: Determine a theme or central idea of a text and analyze its development over the course of the text, including its relationship to the characters, setting, and plot; provide an objective summary of the text.
### Objective Summary RL.8.2
Theme: Key Ideas and Details
Standard: Determine a theme or central idea of a text and analyze its development over the course of the text, including its relationship to the characters, setting, and plot; provide an objective summary of the text.
### Plot RL.8.2
Theme: Key Ideas and Details
Standard: Determine a theme or central idea of a text and analyze its development over the course of the text, including its relationship to the characters, setting, and plot; provide an objective summary of the text.
### Setting RL.8.2
Theme: Key Ideas and Details
Standard: Determine a theme or central idea of a text and analyze its development over the course of the text, including its relationship to the characters, setting, and plot; provide an objective summary of the text.
### Characters RL.8.2
Theme: Key Ideas and Details
Standard: Determine a theme or central idea of a text and analyze its development over the course of the text, including its relationship to the characters, setting, and plot; provide an objective summary of the text.
### Analyzing Comparisons RL.8.3
Theme: Key Ideas and Details
Standard: Analyze how particular lines of dialogue or incidents in a story or drama propel the action, reveal aspects of a character, or provoke a decision.
### Meaning and Tone RL.8.4
Theme: Craft and Structure
Standard: Determine the meaning of words and phrases as they are used in a text, including figurative and connotative meanings; analyze the impact of specific word choices on meaning and tone, including analogies or allusions to other texts.
### Producing Suspense and Humor RL.8.6
Theme: Craft and Structure
Standard: Analyze how differences in the points of view of the characters and the audience or reader (e.g., created through the use of dramatic irony) create such effects as suspense or humor.
### Compare and Contrast RL.8.5
Theme: Craft and Structure
Standard: Compare and contrast the structure of two or more texts and analyze how the differing structure of each text contributes to its meaning and style.
### Modern Fictions and Traditional Stories RL.8.9
Theme: Integration of Knowledge and Ideas
Standard: Analyze how a modern work of fiction draws on themes, patterns of events, or character types from myths, traditional stories, or religious works such as the Bible, including describing how the material is rendered new.
### Technical Meanings RI.8.4
Theme: Craft and Structure
Standard: Determine the meaning of words and phrases as they are used in a text, including figurative, connotative, and technical meanings; analyze the impact of specific word choices on meaning and tone, including analogies or allusions to other texts.
### Central Ideas RI.8.2
Theme: Key Ideas and Details
Standard: Determine a central idea of a text and analyze its development over the course of the text, including its relationship to supporting ideas; provide an objective summary of the text.
### Analyzing Structures in Text RI.8.5
Theme: Craft and Structure
Standard: Analyze in detail the structure of a specific paragraph in a text, including the role of particular sentences in developing and refining a key concept.
### Author’s Point of View RI.8.6
Theme: Craft and Structure
Standard: Determine an author
### Publishing Mediums RI.8.7
Theme: Integration of Knowledge and Ideas
Standard: Evaluate the advantages and disadvantages of using different mediums (e.g., print or digital text, video, multimedia) to present a particular topic or idea.
### Evaluating Author’s Claims RI.8.8
Theme: Integration of Knowledge and Ideas
Standard: Delineate and evaluate the argument and specific claims in a text, assessing whether the reasoning is sound and the evidence is relevant and sufficient; recognize when irrelevant evidence is introduced.
### Conflicting Information RI.8.9
Theme: Integration of Knowledge and Ideas
Standard: Analyze a case in which two or more texts provide conflicting information on the same topic and identify where the texts disagree on matters of fact or interpretation.
Writing Standards
### Introducing and Closing Topics W.8.1a
Domain: Writing
Theme: Text Types and Purposes
Standard: Introduce claim(s), acknowledge and distinguish the claim(s) from alternate or opposing claims, and organize the reasons and evidence logically.
### Supporting and Developing Topics W.8.1b
Domain: Writing
Theme: Text Types and Purposes
Standard: Support claim(s) with logical reasoning and relevant evidence, using accurate, credible sources and demonstrating an understanding of the topic or text.
### Appropriate Transitions W.8.2c
Domain: Writing
Theme: Text Types and Purposes
Standard: Use appropriate and varied transitions to create cohesion and clarify the relationships among ideas and concepts.
Domain: Writing
Theme: Text Types and Purposes
Standard: Introduce a topic clearly, previewing what is to follow; organize ideas, concepts, and information into broader categories; include formatting (e.g., headings), graphics (e.g., charts, tables), and multimedia when useful to aiding comprehension.
### Varied Sentence Structure W.8.3c
Domain: Writing
Theme: Text Types and Purposes
Standard: Use a variety of transition words, phrases, and clauses to convey sequence, signal shifts from one time frame or setting to another, and show the relationships among experiences and events.
### Precise Language and Sensory Details W.8.3d
Domain: Writing
Theme: Text Types and Purposes
Standard: Use precise words and phrases, relevant descriptive details, and sensory language to capture the action and convey experiences and events.
### Task, Purpose, and Audience W.8.4
Domain: Writing
Theme: Production and Distribution of Writing
Standard: Produce clear and coherent writing in which the development, organization, and style are appropriate to task, purpose, and audience. (Grade-specific expectations for writing types are defined in standards 1
### Planning W.8.5
Domain: Writing
Theme: Production and Distribution of Writing
Standard: With some guidance and support from peers and adults, develop and strengthen writing as needed by planning, revising, editing, rewriting, or trying a new approach, focusing on how well purpose and audience have been addressed. (Editing for conventions should demonstrate command of Language standards 1
### Revising W.8.5
Domain: Writing
Theme: Production and Distribution of Writing
Standard: With some guidance and support from peers and adults, develop and strengthen writing as needed by planning, revising, editing, rewriting, or trying a new approach, focusing on how well purpose and audience have been addressed. (Editing for conventions should demonstrate command of Language standards 1
### Editing W.8.5
Domain: Writing
Theme: Production and Distribution of Writing
Standard: With some guidance and support from peers and adults, develop and strengthen writing as needed by planning, revising, editing, rewriting, or trying a new approach, focusing on how well purpose and audience have been addressed. (Editing for conventions should demonstrate command of Language standards 1
### Gathering Relevant Information W.8.8
Domain: Writing
Theme: Research to Build and Present Knowledge
Standard: Gather relevant information from multiple print and digital sources, using search terms effectively; assess the credibility and accuracy of each source; and quote or paraphrase the data and conclusions of others while avoiding plagiarism and following a standard format for citation.
### Citing Information W.8.8
Domain: Writing
Theme: Research to Build and Present Knowledge
Standard: Gather relevant information from multiple print and digital sources, using search terms effectively; assess the credibility and accuracy of each source; and quote or paraphrase the data and conclusions of others while avoiding plagiarism and following a standard format for citation.
### Quoting and Paraphrasing Data W.8.8
Domain: Writing
Theme: Research to Build and Present Knowledge
Standard: Gather relevant information from multiple print and digital sources, using search terms effectively; assess the credibility and accuracy of each source; and quote or paraphrase the data and conclusions of others while avoiding plagiarism and following a standard format for citation.
Language Standards
Domain: Language
Theme: Conventions of Standard English
Standard: Explain the function of verbals (gerunds, participles, infinitives) in general and their function in particular sentences.
### Subject-Verb Agreement L.8.1b
Domain: Language
Theme: Conventions of Standard English
Standard: Form and use verbs in the active and passive voice.
### Pronouns L.8.1c
Domain: Language
Theme: Conventions of Standard English
Standard: Form and use verbs in the indicative, imperative, interrogative, conditional, and subjunctive mood.
### Phrases and Clauses L.8.1c
Domain: Language
Theme: Conventions of Standard English
Standard: Form and use verbs in the indicative, imperative, interrogative, conditional, and subjunctive mood.
### Verbals L.8.1d
Domain: Language
Theme: Conventions of Standard English
Standard: Recognize and correct inappropriate shifts in verb voice and mood.*
### Active and Passive Voice L.8.3a
Domain: Language
Theme: Knowledge of Language
Standard: Use verbs in the active and passive voice and in the conditional and subjunctive mood to achieve particular effects (e.g., emphasizing the actor or the action; expressing uncertainty or describing a state contrary to fact).
### Using Verbs in Moods L.8.1d
Domain: Language
Theme: Conventions of Standard English
Standard: Recognize and correct inappropriate shifts in verb voice and mood.*
### Capitalization L.8.2
Domain: Language
Theme: Conventions of Standard English
Standard: Demonstrate command of the conventions of standard English capitalization, punctuation, and spelling when writing.
### Punctuation L.8.2a
Domain: Language
Theme: Conventions of Standard English
Standard: Use punctuation (comma, ellipsis, dash) to indicate a pause or break.
### Spelling L.8.2c
Domain: Language
Theme: Conventions of Standard English
Standard: Spell correctly.
### Context Clues L.8.4a
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Use context (e.g., the overall meaning of a sentence or paragraph; a word
### Multiple-Meaning Words L.8.4b
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Use common, grade-appropriate Greek or Latin affixes and roots as clues to the meaning of a word (e.g., precede, recede, secede).
### Reference Materials L.8.4c
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Consult general and specialized reference materials (e.g., dictionaries, glossaries, thesauruses), both print and digital, to find the pronunciation of a word or determine or clarify its precise meaning or its part of speech.
### Roots, Affixes and Syllables L.8.4b
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Use common, grade-appropriate Greek or Latin affixes and roots as clues to the meaning of a word (e.g., precede, recede, secede).
### Interpreting Figures of Speech L.8.5a
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Interpret figures of speech (e.g. verbal irony, puns) in context.
### Relationships Between Words L.8.5b
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Use the relationship between particular words to better understand each of the words.
### Denotations and Connotations L.8.5c
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Distinguish among the connotations (associations) of words with similar denotations (definitions) (e.g., bullheaded, willful, firm, persistent, resolute).
Test Taking Tips
Practice Tests
Report an Error
|
# Between two Similar Solid Numbers exist two Mean Proportionals
Jump to navigation Jump to search
## Theorem
In the words of Euclid:
Between two similar solid numbers there fall two mean proportional numbers, and the solid number has to the solid number the ratio triplicate of that which the corresponding side has to the corresponding side.
## Proof
Let $m$ and $n$ be similar solid numbers.
Then for some $a, b, c, d, e, f \in \Z$ such that $a \le b \le c$ and $d \le e \le f$:
$m = a b c$
$n = d e f$
such that:
$\dfrac a d = \dfrac b e = \dfrac c f$
Let:
$r := a e = b d = b f = c e$
So:
$\ds \dfrac {a b} r$ $=$ $\ds \dfrac {a b} {a e}$ $\ds$ $=$ $\ds \dfrac b e$
and:
$\ds \dfrac r {d e}$ $=$ $\ds \dfrac {b d} {d e}$ $\ds$ $=$ $\ds \dfrac b e$
Thus by definition, $\tuple {a b, r, d e}$ is a geometric sequence.
By definition, $r$ is a mean proportional between $a b$ and $d e$.
Now consider the products $r c$ and $r f$.
We have:
$\ds \frac {a b c} {r c}$ $=$ $\ds \frac {a b c} {c b d}$ $\ds$ $=$ $\ds \frac {a b} {b d}$ $\ds$ $=$ $\ds \frac a d$
and:
$\ds \frac {r c} {r f}$ $=$ $\ds \frac {c b d} {f b d}$ $\ds$ $=$ $\ds \frac c f$
and:
$\ds \frac {f b d} {d e f}$ $=$ $\ds \frac {b d} {d e}$ $\ds$ $=$ $\ds \frac b e$
That is:
$\tuple {a b c, b c d, b d f, d e f}$ is a geometric sequence.
Also from the above:
$\dfrac a d = \dfrac b e = \dfrac c f$
showing that $m$ is in triplicate ratio to $n$ as their sides.
$\blacksquare$
## Historical Note
This proof is Proposition $19$ of Book $\text{VIII}$ of Euclid's The Elements.
## Sources
(in which a mistake apppears)
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# RD Sharma Solutions for Class 9 Maths Chapter 10: Congruent Triangles
The phenomenon says that if two figures are of the same shape and size or they can be termed as the mirror image of each other; then the two given figures are said to be congruent to each other. For the study of polygons, mainly triangles, congruence is a very important concept to know. For 2 polygons to be congruent, they must have an equal number of sides. So there are some rules by which we can determine if two triangles are said to be congruent or not.
S (Side) - S (Side) - S (Side) Rule: This rule says that 2 triangles are said to be congruent if all the sides of the 2 triangles are equivalent. S (Side) - A (Angle) - S (Side) Rule: 2 triangles are said to be congruent if one interior angle and 2 sides of a triangle are equivalent to the interior angle and 2 sides of the other triangle. The angle should be held between the two equivalent sides.
A (Angle) – S (Side) - A (Angle) Rule: If 2 angles and the included side of a triangle are equivalent to the corresponding side included between the 2 angles in the other triangle, then the triangles are said to be congruent. R (Right-Angle) – H (Hypotenuse) – S (Side): If in a right-angled triangle, the hypotenuse and any one of the sides are equivalent to the corresponding hypotenuse and the side of the triangle, then those 2 triangles are said to be a congruent triangle.
The chapter also discusses the congruence of line segments and angles. A very commonly used symbol for congruence is ≅ in the world of mathematics. The chapter also teaches the students about the various relations and properties in a congruent figure.
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# Polynomials: Bounds on Zeros
A clever way to know where to search for roots.
A Polynomial looks like this:
example of a polynomial this one has 3 terms
A polynomial has coefficients:
The terms are in order from highest to lowest exponent
(Technically the 7 is a constant, but here it is easier to think of them all as coefficients.)
A polynomial also has roots:
A "root" (or "zero") is where the polynomial is equal to zero.
Example: 3x − 6 equals zero when x=2, because 3(2)−6 = 6−6 = 0
## Where are the Roots (Zeros)?
It can sometimes be hard to find where the roots are!
... where should we search ... how far left or right should we go?
Here we will see a clever way to know where to search for all Real roots.
And it just uses simple arithmetic!
## Steps
First we prepare our data:
• The leading coefficient must be 1. If it is not, then divide every term of the polynomial by the leading coefficient
• Write down all the coefficients
• Then throw away the leading coefficient!
• Remove minus signs
• And we now have a list of values for the next step
Now we can calculate two different "bounds" using those values:
• Bound 1: The largest value, plus 1
• Bound 2: The sum of all values, or 1, whichever is larger
The smallest of those 2 bounds is our answer ...
... all roots are within plus or minus of that!
## Examples
### Example: x3 + 2x2 − 5x + 1
The leading coefficient is 1, so we can continue.
The coefficients are: 1, 2, −5, 1
Drop the leading coefficient, and remove any minus signs: 2, 5, 1
• Bound 1: the largest value is 5. Plus 1 = 6
• Bound 2: adding all values is: 2+5+1 = 8
The smallest bound is 6
All Real roots are between −6 and +6
So we can graph between −6 and 6 and find any Real roots. It is best to plot a little wider so we could see if a curve has roots right at −6 or 6:
Now we can just zoom in to the graph to get more accurate values for the roots
### Example: 10x5 + 2x3 − x2 − 3
the leading coefficient is 10, so we must divide all terms by 10:
x5 + 0.2x3 − 0.1x2 − 0.3
The coefficients are: 1, 0.2, −0.1, −0.3
Drop the leading coefficient, and remove any minus signs: 0.2, 0.1, 0.3
• Bound 1: the largest value is 0.3. Plus 1 = 1.3
• Bound 2: adding all values is: 0.2+0.1+0.3 = 0.6, which is less than 1, so the answer is 1
The smallest is 1.
All Real roots are between −1 and +1
I will leave the graphing to you.
## Notes
"Bound 1" and "Bound 2" are not the only ways to find the bounds of the roots, but they are easy to use!
Also Note: Graphing polynomials can only find Real roots, but there can also be Complex roots.
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Paul's Online Notes
Home / Algebra / Preliminaries / Factoring Polynomials
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### Section 1.5 : Factoring Polynomials
For problems 1 – 4 factor out the greatest common factor from each polynomial.
1. $$6{x^7} + 3{x^4} - 9{x^3}$$ Solution
2. $${a^3}{b^8} - 7{a^{10}}{b^4} + 2{a^5}{b^2}$$ Solution
3. $$2x{\left( {{x^2} + 1} \right)^3} - 16{\left( {{x^2} + 1} \right)^5}$$ Solution
4. $${x^2}\left( {2 - 6x} \right) + 4x\left( {4 - 12x} \right)$$ Solution
For problems 5 & 6 factor each of the following by grouping.
1. $$7x + 7{x^3} + {x^4} + {x^6}$$ Solution
2. $$18x + 33 - 6{x^4} - 11{x^3}$$ Solution
For problems 7 – 15 factor each of the following.
1. $${x^2} - 2x - 8$$ Solution
2. $${z^2} - 10z + 21$$ Solution
3. $${y^2} + 16y + 60$$ Solution
4. $$5{x^2} + 14x - 3$$ Solution
5. $$6{t^2} - 19t - 7$$ Solution
6. $$4{z^2} + 19z + 12$$ Solution
7. $${x^2} + 14x + 49$$ Solution
8. $$4{w^2} - 25$$ Solution
9. $$81{x^2} - 36x + 4$$ Solution
For problems 16 – 18 factor each of the following.
1. $${x^6} + 3{x^3} - 4$$ Solution
2. $$3{z^5} - 17{z^4} - 28{z^3}$$ Solution
3. $$2{x^{14}} - 512{x^6}$$ Solution
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# Calculating Limit of Function – A quotient of exponential and polynomial functions to zero – Exercise 6303
### Exercise
Evaluate the following limit:
$$\lim _ { x \rightarrow 0} \frac{(e^x-1)(e^{2x}-1)}{x^2}$$
$$\lim _ { x \rightarrow 0} \frac{(e^x-1)(e^{2x}-1)}{x^2}=2$$
### Solution
First, we try to plug in $$x = 0$$ and get
$$\frac{(e^0-1)(e^{2\cdot 0}-1)}{0^2}=\frac{0}{0}$$
Note: The zero in the denominator is not absolute zero, but a number that is tending to zero.
We got the phrase $$\frac{"0"}{"0"}$$ (=tending to zero divides tending to zero). This is an indeterminate form, therefore we have to get out of this situation.
$$\lim _ { x \rightarrow 0} \frac{(e^x-1)(e^{2x}-1)}{x^2}=$$
In order to use Lopital Rule, we open the brackets and get
$$=\lim _ { x \rightarrow 0} \frac{e^{3x}-e^x-e^{2x}+1}{x^2}=$$
Now we use Lopital Rule – we derive the numerator and denominator separately and we get
$$=\lim _ { x \rightarrow 0} \frac{3e^{3x}-e^x-2e^{2x}}{2x}=$$
We plug in zero again and get
$$= \frac{3e^{3\cdot 0}-e^0-2e^{2\cdot 0}}{2\cdot 0}=$$
$$= \frac{0}{0}$$
We got again $$\frac{0}{0}$$
So we we will use Lopital Rule again – we derive the numerator and denominator separately and we get
$$=\lim _ { x \rightarrow 0} \frac{9e^{3x}-e^x-4e^{2x}}{2}=$$
We plug in zero again and this time we get
$$=\frac{9e^{3\cdot 0}-e^0-4e^{2\cdot 0}}{2}=$$
$$=\frac{9-1-4}{2}=$$
$$=\frac{4}{2}=$$
$$=2$$
Have a question? Found a mistake? – Write a comment below!
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# Infinite Series
Sequence and Series are fundamentally different from each other.
• ### Sequence
A sequence is an arrangement of numbers (or objects). For example,
${3, 7,11,15,19, \cdots}$
is a sequence. The first term is ${a_1 = 3}$, the second term is ${a_2 = 7}$.
In general, ${n}$th term in a sequence is denoted by ${a_n}$.
In a strict mathematical sense, a sequence is a function defined from ${\mathbb N}$ to ${\mathbb R}$, ${\forall n \in \mathbb N}$.
If ${\forall n, a_{n+1} \ge a_n}$, it is said to be monotonically increasing. The sequence mentioned above is monotonically increasing, because ${a_{n+1}= a_n + 4, n>1}$ and ${a_1 = 3}$.
If ${\forall n, a_{n+1} \le a_n}$, it is said to be monotonically decreasing. An example is the sequence of reciprocals of natural numbers.
${1, \frac 1 2 , \frac 1 3, \frac 1 4, \cdots}$
Clearly, ${a_{n+1} < a_n}$.
A sequence, whose members have alternate signs, is an alternating sequence. For example,
${a_{n}=n, if \ n \ is \ odd \ and \ }$
${a_n = -n, if \ n \ is \ even}$
• ### Convergent and Divergent Sequences
Limit of a sequence, ${L}$ is given by ${\lim \limits_{n \to\infty} a_n}$. In other words, as ${n \to \infty}$, the values ${a_n}$ approach ${L}$.
If the limit is finite, the sequence is said to be convergent. The sequence ${a_n = \frac 1 n, n \in \mathbb N}$ is a convergent sequence, because
${\lim \limits_{n \to \infty} \frac 1 n = 0}$
The famous Fibonacci sequence, given by
${a_1 = 1, a_2 = 1 \ and \ a_n = a_{n-1}+a_{n-2}, \forall n \ge 3}$
is a divergent sequence.
• ### Bounded Sequences
A sequence is bounded, if
I) there exists a number ${l \ in \mathbb R}$, such that ${a_n \le r}$ OR
II) there exists a number ${g \ in \mathbb R}$, such that ${a_n \ge r}$ OR
III) both I and II are satisfied
A convergent sequence is always bounded, but the converse is not always true. e.g. ${1,-1,1,-1,1,-1, \cdots}$ is bounded, but not convergent. Another example is ${a_n = sin (\frac {n \pi}{4})}$. These are oscillating sequences.
• ### Series
An (infinite) series is the sum of infinite terms in a sequence. Thus, it takes the form
${a_1 + a_2 + a_3 + a_4 + \cdots = \sum \limits_{n=1}^{\infty} a_n}$
If the sum tends to ${\infty}$ or ${- \infty}$, then the series is divergent. We write,
${\sum \limits_{n \to \infty} a_n = \pm \infty}$
In other words, the limit of sum ${S_n = \sum_n a_n}$ as ${n \to \infty}$ becomes ${\pm \infty}$.
If the limit is a finite number, the series is convergent.
• ### Testing the Convergence of a Series
It may seem counter-intuitive, but the harmonic series is divergent.
${\sum \limits_{n=1}^{\infty} \frac 1 n = 1+ \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \cdots = \infty}$
The series of reciprocals of prime numbers is also divergent.
${\frac 1 2 + \frac 1 3 + \frac 1 5 + \frac 1 7 + \frac 1 {11} + \cdots = \infty}$
Thus, knowing the behaviour of a series requires a deeper analysis. (Above results can be proved). People have studied a variety of series and have come up with different tests.
A general approach is analyzing the ${n}$th term of series. If it is ${\ge 1}$, then the series is divergent.
The tests use such common facts to test the behaviour of series.
• ### Cauchy’s Test (${n}$th Root Test)
Let ${a_n}$ denote the ${n}$th term in the series.
I) Applicable only when there are ONLY positive terms in the series
II) If ${\lim \limits_{n \to \infty} a_n < 1}$, the series is convergent.
III) If ${\lim \limits_{n \to \infty} a_n > 1}$, the series is divergent.
IV) If ${\lim \limits_{n \to \infty} a_n = 1}$, the test fails.
• ### ${p}$-Series Test
The ${p}$-series is a special case of Riemann’s ${\zeta}$ function, given by
${\zeta (s) = \sum \limits_{n=1}^{\infty} \frac {1}{n^s}}$
for values of ${s}$, whose real part is ${> 1}$. It is defined for other complex numbers using analytic continuation and a functional equation. When ${s}$ is real, it becomes the ${p}$-series.
I) The series is convergent if ${p>1}$ and divergent if ${p \le 1}$.
II) When ${p=1}$, we get the harmonic series.
III) When ${n=2}$, we get ${\frac {\pi^2} 6}$. (Can be obtained by Fourier Series as well)
• ### Behaviour of Geometric Series
Consider a series with first term ${a}$ and common ratio ${r}$. If ${r \ge 1}$, it is divergent, if ${-1 < r < 1}$, it is convergent.
If ${r \le -1}$, the series is oscillatory.
• ### Comparison Test
Let ${\sum a_n}$ and ${\sum b_n}$ be 2 series of positive terms. Comparing the 2, if
${\lim \limits_{n \to \infty} \frac {a_n}{b_n}}$
is a non-zero finite number, then they both have identical behaviour.
The series ${b_n}$ is known as the auxiliary series.
• ### D’Alembert’s Test (Ratio Test)
I) Applicable for series of positive terms
II) If ${\lim \limits_{n \to \infty} \frac {a_n}{a_{n+1}} > 1}$, then the series is convergent.
III) If ${\lim \limits_{n \to \infty} \frac {a_n}{a_{n+1}} < 1}$, then the series is divergent.
IV) If the limit is ${1}$, the test is inconclusive.
• ### Raabe’s Test
I) Generally applied when D’Alembert’s test fails
II) Take the limit ${\lim \limits_{n \to \infty} n \times (\frac {a_n}{a_{n+1}}}$.
III) Conditions are similar to D’Alembert’s test
• ### Cauchy’s Condensation Test
I) Useful when ${log}$ terms are present
II) Consider a function ${f(n)}$ which continuously decreases as ${n}$ increases. The behavior of ${\sum \limits_{n=1}^{\infty} a^n f(a^n)}$ decides the behaviour of ${\sum \limits_{n=1}^{\infty} f(n)}$.
III) ${a \in \mathbb Z, a > 1}$.
• ### Auxiliary Series ${\sum \frac {1}{n (ln \ n )^p}}$
I) ${p>1}$, the series is convergent.
II) ${p\le 1}$, the series is divergent.
• ### Gauss’ Test
I) Let ${\sum \limits_{n=1}^{\infty} a_n}$ be a series, s.t.
${\frac {a_n}{a_{n+1}} = 1 + \frac l n + \frac {b_n}{n^p}, p > 1, b_n \ is \ bounded}$
Then if ${l >1}$, then the sequence converges, else diverges.
• ### Logarithmic Test
I) If ${\sum \limits_{n=1}^{\infty} n \ ln \ \frac {a_n}{a_{n+1}} > 1}$, the series ${\sum \limits_{n=1}^{\infty} a_n}$ converges. If the limit is ${< 1}$, the series diverges.
II) Applied when D’Alembert’s test fails
• ### Bertrand’s Test/ de Morgan’s Test
I) Behavior of ${\sum \limits_{n=1}^{\infty} a_n}$ is governed by
${\lim \limits_{n \to \infty} \Big \{ \Big[ n \times \big(\frac {a_n}{a_{n+1}}-1 \big)-1 \Big] \times ln \ n \Big \}}$
II) Limit ${>1}$, ${\sum \limits_{n=1}^{\infty} a_n}$ converges.
III) Limit ${<1}$, ${\sum \limits_{n=1}^{\infty} a_n}$ diverges.
• ### Alternative Test
Change the function to
${\lim \limits_{n \to \infty} \Big \{ \Big[ \big(n \times \frac {a_n}{a_{n+1}}-1 \big)-1 \Big] \times ln \ n \Big \}}$
• ### Cauchy’s Integral Test
Let the ${n}$th term be ${a_n = f(n)}$. If ${\int \limits_1^{\infty} f(x)dx}$ is finite, ${\sum \limits_{n=1}^{\infty} a_n}$ converges. If If ${\int \limits_1^{\infty} f(x)dx}$ is infinite, ${\sum \limits_{n=1}^{\infty} a_n}$ diverges.
• ### Leibniz Test for Alternating Series
The series ${\sum \limits_{n=1}^{\infty} (-1)^{n-1} a_n}$ is convergent if ${|a_{n+1}|< |a_n|}$ and ${\lim \limits_{n \to \infty} a_n = 0}$.
• ### Absolute and Conditional Convergence
I) When ${\sum \limits_{n=1}^{\infty} |a_n|}$ converges, the series ${\sum \limits_{n=1}^{\infty} a_n}$ is convergent.
II) If the series of absolute values is divergent, the series ${\sum \limits_{n=1}^{\infty} a_n}$ is conditionally convergent. e.g.
${\sum \limits_{n=1}^{\infty} \frac {(-1)^{n+1}}{n}}$
This series converges to ${ln \ 2}$.
*** It has been proven that a conditionally convergent series can be made to converge to any sum, including ${\infty}$ or ${- \infty}$.
• ### Power Series
Functions can be expanded in a power series. If the series is convergent, the expansion is ${valid}$.
${f(x) = \sum \limits_{n=0}^{\infty} b_n (x-c)^n}$
Note the constant term ${b_0}$. If for ${l_1 < x < l_2}$, the series is convergent, then ${[l_1, l_2]}$ is the range of convergence.
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# What is the distance between (–1, 1, 3) and (4, –1, 2)?
Jan 24, 2017
The distance is $\sqrt{30}$ or $5.48$ rounded to the nearest hundredth
#### Explanation:
The formula for calculating the distance between two points is:
$d = \sqrt{{\left(\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}\right)}^{2} + {\left(\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}\right)}^{2} + {\left(\textcolor{g r e e n}{{z}_{2}} - \textcolor{g r e e n}{{z}_{1}}\right)}^{2}}$
Substituting the values from the problem gives:
$d = \sqrt{{\left(\textcolor{red}{4} - \textcolor{b l u e}{- 1}\right)}^{2} + {\left(\textcolor{red}{- 1} - \textcolor{b l u e}{1}\right)}^{2} + {\left(\textcolor{g r e e n}{2} - \textcolor{g r e e n}{3}\right)}^{2}}$
$d = \sqrt{{\left(\textcolor{red}{4} + \textcolor{b l u e}{1}\right)}^{2} + {\left(\textcolor{red}{- 1} - \textcolor{b l u e}{1}\right)}^{2} + {\left(\textcolor{g r e e n}{2} - \textcolor{g r e e n}{3}\right)}^{2}}$
$d = \sqrt{{5}^{2} + {\left(- 2\right)}^{2} + {\left(- 1\right)}^{2}}$
$d = \sqrt{25 + 4 + 1}$
$d = \sqrt{30} = 5.48$ rounded to the nearest hundredth
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Solving with Negative Exponents: Math Homework Help
Page content
The Problem
Solving with positive exponents makes sense. But how does solving with negative exponents work? After all, 5^2 is just 5 X 5 – or two 5s multiplied by each other. But how can you multiply negative three 5s by each other? What could the phrase “negative three 5s” mean?
The Common Mistake
Many students, when confronted with calculating 5^-2, make a simple mistake. They figure that since 5^2 = 25, then 5^-2 must be -25. Watch out for this mistake! Adding a negative sign to the power is not the same thing as multiplying the answer by -1. There’s one basic rule that can help you understand how to calculate negative exponents..
The Basic Rule
There’s one rule that you have to memorize about negative exponents, and it involves reciprocals. Here’s the rule: Raising a number to a negative power is the same thing as raising the reciprocal of that number to a positive power.
What does that mean? It means that if you were trying to raise 5 to the power of -2, you would first find the reciprocal of 5 – which is 1/5. So when 5 is raised to the power of -2, it is the same thing as saying that 1/5 is raised to the power of positive 2, or (1/5)^2. If you would multiply that out, you would get 1/25. Therefore, 5^-2 = 1/25. Once you remember the negative exponent rule, everything else falls into place.
Fractions and Negative Exponents
When the base of an exponent is a fraction, you can follow the same logic. For example, let’s say you want to raise 3/4 to the power of -3. To calculate this, you would first take the reciprocal of 3/4, which is 4/3. Then you would raise 4/3 to the power of +3, or (4/3)^3. If you would multiply that out, you would get 64/9.
This post is part of the series: Math Help for Exponents
Looking for math help for exponents? Whether you’re a student, parent, or tutor, this series of articles will explain the basics of how to use exponents correctly. Includes rules for adding, subtracting, multiplying, and dividing exponents, as well as how to use negative exponents.
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# Determining Statistical Significance
## Presentation on theme: "Determining Statistical Significance"— Presentation transcript:
Determining Statistical Significance
Section 4.3 Determining Statistical Significance
Formal Decisions If the p-value is small: If the p-value is not small:
REJECT H0 the sample would be extreme if H0 were true the results are statistically significant we have evidence for Ha If the p-value is not small: DO NOT REJECT H0 the sample would not be too extreme if H0 were true the results are not statistically significant the test is inconclusive; either H0 or Ha may be true
Formal Decisions How small?
A formal hypothesis test has only two possible conclusions: The p-value is small: reject the null hypothesis in favor of the alternative The p-value is not small: do not reject the null hypothesis How small?
Significance Level p-value > Do not Reject H0
The significance level, , is the threshold below which the p-value is deemed small enough to reject the null hypothesis p-value < Reject H0 p-value > Do not Reject H0
Significance Level If the p-value is less than , the results are statistically significant, and we reject the null hypothesis in favor of the alternative If the p-value is not less than , the results are not statistically significant, and our test is inconclusive Often = 0.05 by default, unless otherwise specified
Elephant Example H0 : X is an elephant Ha : X is not an elephant
Would you conclude, if you get the following data? X walks on two legs X has four legs Although we can never be certain! Reject H0; evidence that X is not an elephant Do not reject H0; we do not have sufficient evidence to determine whether X is an elephant
Never Accept H0 “Do not reject H0” is not the same as “accept H0”!
Lack of evidence against H0 is NOT the same as evidence for H0!
Statistical Conclusions
Formal decision of hypothesis test, based on = 0.05 : Informal strength of evidence against H0:
Errors Decision Truth There are four possibilities: Reject H0
Do not reject H0 H0 true H0 false TYPE I ERROR Truth TYPE II ERROR A Type I Error is rejecting a true null A Type II Error is not rejecting a false null
Analogy to Law Ho Ha A person is innocent until proven guilty.
Evidence must be beyond the shadow of a doubt. p-value from data Types of mistakes in a verdict? Convict an innocent Type I error Release a guilty Type II error
Probability of Type I Error
The probability of making a Type I error (rejecting a true null) is the significance level, α Randomization distribution of sample statistics if H0 is true: If H0 is true and α = 0.05, then 5% of statistics will be in tail (red), so 5% of the statistics will give p-values less than 0.05, so 5% of statistics will lead to rejecting H0
Probability of Type II Error
The probability of making a Type II Error (not rejecting a false null) depends on Effect size (how far the truth is from the null) Sample size Variability Significance level
Choosing α By default, usually α = 0.05
If a Type I error (rejecting a true null) is much worse than a Type II error, we may choose a smaller α, like α = 0.01 If a Type II error (not rejecting a false null) is much worse than a Type I error, we may choose a larger α, like α = 0.10
Significance Level Come up with a hypothesis testing situation in which you may want to… Use a smaller significance level, like = 0.01 Use a larger significance level, like = 0.10
Summary Results are statistically significant if the p-value is less than the significance level, α In making formal decisions, reject H0 if the p- value is less than α, otherwise do not reject H0 Not rejecting H0 is NOT the same as accepting H0 There are two types of errors: rejecting a true null (Type I) and not rejecting a false null (Type II)
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# Matrix Multiplication
Consider the product of a 2×3 matrix and a 3×4 matrix. The multiplication is defined because the inner dimensions (3) are the same. The product will be a 2×4 matrix, the outer dimensions.
Since there are three columns in the first matrix and three rows in the second matrix (the inner dimensions which must be the same), each element in the product will be the sum of three products.
### Row 1, Column 1
To find the element in row 1, column 1 of the product, we will take row 1 from the first matrix and column 1 from the second matrix. We pair these values together, multiply the pairs of values, and then add to arrive at 25.
``` R1: 1 -2 3
×C1: 1 -3 6
---------------
1 +6 +18 = 25```
### Row 2, Column 3
To find the element in row 2, column 3 of the product, we will take row 2 from the first matrix and column 3 from the second matrix. We pair these values together, multiply the pairs of values, and then add to arrive at 53.
``` R2: 4 5 -2
×C3: 4 7 -1
---------------
16 +35 +2 = 53```
Understanding where each number in the product comes from is helpful when you only need a specific value. You don't need to multiply completely if you only want specific elements. Just take the row from the first matrix and the column from the second matrix.
The process can be completed for the rest of the elements in the matrix.
Column 1 Column 2 Column 3 Column 4 values [1, -3, 6] [-8, 6, 5] [4, 7, -1] [-3, 2, 4] [1, -2, 3] 1(1) - 2(-3) + 3(6) = 1 + 6 + 18 = 25 1(-8) -2(6) + 3(5) = -8 - 12 + 15 = -5 1(4) -2(7) +3(-1) = 4 - 14 - 3 = -13 1(-3) -2(2) + 3(4) = -3 -4 + 12 = 5 [4, 5, -2] 4(1) + 5(-3) -2(6) = 4 - 15 - 12 = -23 4(-8) + 5(6) - 2(5) = -32 + 30 - 10 = -12 4(4) + 5(7) -2(-1) = 16 + 35 + 2 = 53 4(-3) + 5(2) -2(4) = -12 + 10 - 8 = -10
So, the final product is
25 -5 -13 5 -23 -12 53 -10
## Matrix Multiplication is not Commutative
Note that the multiplication is not defined the other way. You can not multiply a 3x4 and a 2x3 matrix together because the inner dimensions aren't the same.
This product is undefined.
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Categories
Probability
What is Probability?
There’s a great chance that you have heard of probability before. Speaking about chances, that’s exactly what probability is! Probability is the extent to which an event is likely to occur. Well, if that’s the case, then how do you calculate probability? Probability is measured by the ratio of the favorable cases to the whole number of cases possible.
Let’s start off with a basic example: What is the probability of getting a “heads” when flipping a regular coin? Well, as we said previously, probability is measured by the ratio of the favorable cases to the whole number of cases possible. So what’s the favorable cases in this example? Well, we are trying to get heads, so that would be the favorable case. Thus, we would have only 1 favorable case. The next part of what we have to look at is the whole number of cases possible. Well, in a regular coin, how many possibilities do we have? Well, we have heads and tails, so just 2 possibilities. Thus, we will take the ratio of favorable to possible cases and get 1:2. In other words, that means that there is a ½ probability of getting a heads when flipping a regular coin.
Independent vs Dependent Events
Now, in probability, there are two types of events that can happen: independent and dependent events. The example we just did above is an example of an independent event. An independent event is when 2 separate events happen, and the first event does not impact the result of the second event. For example, as we saw in this example, we flipped a coin to see the probability of getting a heads. Whether we got heads or tails on the first flip, it does not affect the result of the following flip. On the other hand, a dependent event is when the outcome of the first event affects the probabilities of the subsequent events.
Calculating Independent and Dependent Probabilities
When it comes to calculating Independent and Dependent Probabilities, it is very simple. For independent probability, the probability that both the events that will occur is the probability of event A happening times the probability of event B happening: P(A) x P(B). But make sure to identify whether the event is independent or dependent by seeing if the event
For dependent probability, the probability that event A and B will happen is equal to the probability of event A happening times the probability of event B given A: P(A and B) = P(A) x P(B|A). Remember that for dependent probability, the result of the first event happening affects the outcome of the second event happening, so we will most likely have a different number of possible outcomes after the first event has happened.
Why Probability?
Okay, that’s great that we know all of this, but how will this actually help us? Well, probability is used all around us. For example, in the morning, when you are getting ready for school, you dress according to the weather. If it is a sunny and hot day, you are not going to wear a jacket. You are most likely going to wear something more suitable for the weather such as shorts. Although you might not know it, you are a mathematician in your sleepy state in the morning. Another reason why you may want to learn about probability is because it is one of the hot topics throughout school and in competition math. For example, in the 2020 AMC 8 Math competition, more than 25% of the problems dealt with probability! That’s a big portion of the test right there, and if you are comfortable with probability, it will make it that much easier for you to do those problems. In addition to just solving problems, probability can provide you with so many interesting facts. For example, probability can let us know the odds of someone getting attacked by a shark: 1 in 11,500,000. Or, probability can tell us the odds of someone becoming the president of the United States: 1 in 10,000,000.
Probability is incredibly interesting, and you start researching it by going to the resources provided below. These resources will give you much of the essential information you will need to become a probability genius!
Resources
https://www.mathsisfun.com/data/probability.html (A great tool to learn about what probability is, how to calculate it, and some examples to get you familiar with probability)
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# Trapezoids
## Quadrilaterals with exactly one pair of parallel sides.
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Trapezoids
What if you were told that the polygon ABCD is an isosceles trapezoid and that one of its base angles measures \begin{align*}38^\circ\end{align*} ? What can you conclude about its other angles? After completing this Concept, you'll be able to find the value of a trapezoid's unknown angles and sides given your knowledge of the properties of trapezoids.
### Watch This
CK-12 Foundation: Chapter6TrapezoidsA
### Guidance
A trapezoid is a quadrilateral with exactly one pair of parallel sides. Examples look like:
An isosceles trapezoid is a trapezoid where the non-parallel sides are congruent. The third trapezoid above is an example of an isosceles trapezoid. Think of it as an isosceles triangle with the top cut off. Isosceles trapezoids also have parts that are labeled much like an isosceles triangle. Both parallel sides are called bases.
Recall that in an isosceles triangle, the two base angles are congruent. This property holds true for isosceles trapezoids.
Theorem: The base angles of an isosceles trapezoid are congruent.
The converse is also true: If a trapezoid has congruent base angles, then it is an isosceles trapezoid. Next, we will investigate the diagonals of an isosceles trapezoid. Recall, that the diagonals of a rectangle are congruent AND they bisect each other. The diagonals of an isosceles trapezoid are also congruent, but they do NOT bisect each other.
Isosceles Trapezoid Diagonals Theorem: The diagonals of an isosceles trapezoid are congruent.
The midsegment (of a trapezoid) is a line segment that connects the midpoints of the non-parallel sides. There is only one midsegment in a trapezoid. It will be parallel to the bases because it is located halfway between them. Similar to the midsegment in a triangle, where it is half the length of the side it is parallel to, the midsegment of a trapezoid also has a link to the bases.
##### Investigation: Midsegment Property
Tools Needed: graph paper, pencil, ruler
1. Draw a trapezoid on your graph paper with vertices \begin{align*}A(-1, 5), \ B( 2, 5), \ C(6, 1)\end{align*} and \begin{align*}D(-3, 1)\end{align*}. Notice this is NOT an isosceles trapezoid.
2. Find the midpoint of the non-parallel sides either by using slopes or the midpoint formula. Label them \begin{align*}E\end{align*} and \begin{align*}F\end{align*}. Connect the midpoints to create the midsegment.
3. Find the lengths of \begin{align*}AB, \ EF\end{align*}, and \begin{align*}CD\end{align*}. Can you write a formula to find the midsegment?
Midsegment Theorem: The length of the midsegment of a trapezoid is the average of the lengths of the bases, or \begin{align*}EF=\frac{AB+CD}{2}\end{align*}.
#### Example A
Look at trapezoid \begin{align*}TRAP\end{align*} below. What is \begin{align*}m \angle A\end{align*}?
\begin{align*}TRAP\end{align*} is an isosceles trapezoid. So, \begin{align*}m \angle R = 115^\circ\end{align*}. To find \begin{align*}m \angle A\end{align*}, set up an equation.
\begin{align*}115^\circ + 115^\circ + m \angle A + m \angle P & = 360^\circ\\ 230^\circ + 2m \angle A & = 360^\circ \rightarrow m \angle A = m \angle P\\ 2m \angle A & = 130^\circ\\ m \angle A & = 65^\circ\end{align*}
Notice that \begin{align*}m \angle R + m \angle A = 115^\circ + 65^\circ = 180^\circ\end{align*}. These angles will always be supplementary because of the Consecutive Interior Angles Theorem. Therefore, the two angles along the same leg (or non-parallel side) are always going to be supplementary. Only in isosceles trapezoids will opposite angles also be supplementary.
#### Example B
Write a two-column proof.
Given: Trapezoid \begin{align*}ZOID\end{align*} and parallelogram \begin{align*}ZOIM\end{align*}
\begin{align*}\angle D \cong \angle I\end{align*}
Prove: \begin{align*}\overline{ZD} \cong \overline{OI}\end{align*}
Statement Reason
1. Trapezoid \begin{align*}ZOID\end{align*} and parallelogram \begin{align*}ZOIM, \ \angle D \cong \angle I\end{align*} Given
2. \begin{align*}\overline{ZM} \cong \overline{OI}\end{align*} Opposite Sides Theorem
3. \begin{align*}\angle I \cong \angle ZMD\end{align*} Corresponding Angles Postulate
4. \begin{align*}\angle D \cong \angle ZMD\end{align*} Transitive PoC
5. \begin{align*}\overline{ZM} \cong \overline{ZD}\end{align*} Base Angles Converse
6. \begin{align*}\overline{ZD} \cong \overline{OI}\end{align*} Transitive PoC
#### Example C
Find \begin{align*}x\end{align*}. All figures are trapezoids with the midsegment.
1. a) \begin{align*}x\end{align*} is the average of 12 and 26. \begin{align*}\frac{12+26}{2}=\frac{38}{2}=19\end{align*}
b) 24 is the average of \begin{align*}x\end{align*} and 35.
\begin{align*}\frac{x+35}{2}&=24\\ x+35&=48\\ x&=13\end{align*}
c) 20 is the average of \begin{align*}5x-15\end{align*} and \begin{align*}2x-8\end{align*}.
\begin{align*}\frac{5x-15+2x-8}{2}&=20\\ 7x-23& =40\\ 7x&=63\\ x&=9\end{align*}
Watch this video for help with the Examples above.
CK-12 Foundation: Chapter6TrapezoidsB
#### Concept Problem Revisited
Given an isosceles trapezoid with \begin{align*} m \angle B = 38^\circ \end{align*}, find the missing angles.
In an isosceles trapezoid, base angles are congruent.
\begin{align*} \angle B \cong \angle C\end{align*} and \begin{align*}\angle A \cong \angle D\end{align*}
\begin{align*}m \angle B = m \angle C = 38^\circ\\ 38^\circ + 38^\circ + m \angle A + m \angle D = 360^\circ\\ m \angle A = m \angle D = 142^\circ\end{align*}
### Guided Practice
\begin{align*}TRAP\end{align*} an isosceles trapezoid.
Find:
1. \begin{align*}m \angle TPA\end{align*}
2. \begin{align*}m \angle PTR\end{align*}
3. \begin{align*}m \angle PZA\end{align*}
4. \begin{align*}m \angle ZRA\end{align*}
1. \begin{align*}\angle TPZ \cong \angle RAZ\end{align*} so \begin{align*}m\angle TPA=20^\circ + 35^\circ = 55^\circ\end{align*}.
2. \begin{align*}\angle TPA\end{align*} is supplementary with \begin{align*}\angle PTR\end{align*}, so \begin{align*}m\angle PTR=125^\circ\end{align*}.
3. By the Triangle Sum Theorem, \begin{align*}35^\circ + 35^\circ + m\angle PZA=180^\circ\end{align*}, so \begin{align*}m\angle PZA=110^\circ\end{align*}.
4. Since \begin{align*}m\angle PZA = 110^\circ\end{align*}, \begin{align*}m\angle RZA=70^\circ\end{align*} because they form a linear pair. By the Triangle Sum Theorem, \begin{align*}m\angle ZRA=90^\circ\end{align*}.
### Explore More
1. Can the parallel sides of a trapezoid be congruent? Why or why not?
For questions 2-7, find the length of the midsegment or missing side.
Find the value of the missing variable(s).
Find the lengths of the diagonals of the trapezoids below to determine if it is isosceles.
1. \begin{align*}A(-3, 2), B(1, 3), C(3, -1), D(-4, -2)\end{align*}
2. \begin{align*}A(-3, 3), B(2, -2), C(-6, -6), D(-7, 1)\end{align*}
3. \begin{align*}A(1, 3), B(4, 0), C(2, -4), D(-3, 1)\end{align*}
4. \begin{align*}A(1, 3), B(3, 1), C(2, -4), D(-3, 1)\end{align*}
1. Write a two-column proof of the Isosceles Trapezoid Diagonals Theorem using congruent triangles.
Given: \begin{align*}TRAP\end{align*} is an isosceles trapezoid with \begin{align*}\overline{TR} \ || \ \overline{AP}\end{align*}.
Prove: \begin{align*}\overline{TA} \cong \overline{RP}\end{align*}
1. How are the opposite angles in an isosceles trapezoid related?.
2. List all the properties of a trapezoid.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 6.6.
### My Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Diagonal
A diagonal is a line segment in a polygon that connects nonconsecutive vertices
midsegment
A midsegment connects the midpoints of two sides of a triangle or the non-parallel sides of a trapezoid.
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Home » Maths » Straight Angle
# Straight Angle- Definition, Diagram, Properties, Examples
Straight Angle is one of the most common types of angles found in the real world. You all must have noticed or constructed these angles in your life either knowingly or unknowingly. The line that you construct regularly is a prime example of this angle. It is one of the six types of angles found in geometry. The other five angles are acute angle, right angle, obtuse angle, reflex angle, and complete/full rotation angle. This article will cover all the important aspects of this special so that you do not miss any important points for your exam.
## Straight Angle
As it is evident from its name, the straight angle appears straight in appearance. That is, these angles do not form any conical shape. The value of these angles is always equal to 180°. When two rays are linked end to end, they create an angle according to the definition of angles in mathematics. As a result, the two rays with 180-degree angles are subtended in the opposite direction where the rays are connected at ends. The straight angle (180°) is expressed as pi (Π) in radians. This angle register its presence in many real-life scenarios. Let us explore some interesting features of this angle. Before that have a look at its definition and its shape (diagram).
## Straight Angle Definition
As per the formal mathematical language, straight angle is an angle whose measure is exactly 180°. In other words, the arms of the rays forming this angle lies in opposite directions. It is the angle that two rays travelling in opposite directions subtends at their common end point. It looks exactly like a straight line, hence the name. It is an angle whose arms unite to make a 180° angle by pointing in opposing directions from the vertex. The rays forming this angle are travelling in opposite directions, giving a straight line shape to it. The 180° angle is also known as supplementary. The vertex of this angle is flat, hence it is also known by the name of flat angles.
## Straight Angle Diagram
The diagrammatic view of the this angle will precise help you understand the reason behind its naming. Its vertex point is completely flat, giving it a straight shape. The diagram of the straight line is shown below.
As shown in the above figure, ∠AOC is 180°
OC and AO are two rays travelling in opposite directions that meet at O
This leads to the formation of an angle that is flat or straight with vertex point O.
## Straight Angle Degree
The degree of this angle always remain constant. It does not take several values like obtuse angles or reflex angles. The value of its degree always remain 180°. In other words, we can say that the value of this angle lies between an obtuse angle and a reflex angle. The value of an obtuse angle is more than 90° but less than 180° while the value of reflex angle is greater than 180° but less than 360°. We can also conclude from its degree that it is equal to the half rotation, as a full rotation corresponds to 360°. The value of its degree is twice the right angle degree (90°).
## Straight Angle Pair
A straight angle pair is a pair of angles that together forms a straight line. The sum total of any such angle pairs is always exactly equal to 180 degrees. In other words, we can say that a these angle pairs always forms a supplementary angle. They are also commonly known by the term “a pair of linear angles”. Let us understand this concept through a diagram.
In the above figure, ∠SOT and ∠UOS forms a this angle pair
where, ∠SOT = 80° and ∠UOS = 100°
These both angles form a supplementary pair that is equal to 180°
These pairs have a common vertex and a common arm.
Here in this figure, OS is the common arm and O is the common vertex.
## Straight Angle Formula
As we know that the value of this angle is 180°. So, if the value of one of the angles forming a straight angle pair is known, we can easily deduce the other value by using the following formula:
X = 180° – a
where, X is the unknown angle
a is the value of the known angle
## Straight Angle Properties
This special angle shows some unique properties that is not found in any other an and angles. It is because of its special shape and flat vertex point. These properties will help students in recognizing it and its related shapes quite easily. Some of its properties are stated below.
• The value of their degree is always 180
• It is equal to pi (Π) in radians
• The arms of a this angle point in opposite directions
• The vertex point of a this angle is flat
• It is half the full revolution
• It can be obtained by joining two right angles
• The angle can be obtained by rotating a ray by 180° with respect to the other ray.
## Straight Angle Examples in Real Life
Many examples of a this special angle can be found in our day to day lives. Some of its examples encountered in real life are listed hereunder.
• The angle formed by the hour hand and the minute hand at 6 O’clock
• The angle formed between the two sides of a book when it is open flat
• The angle between the two arms of a balanced seesaw
• The angle between the closed doors of a wardrobe
• The angle formed by the two fully-extended airplane wings
## Straight Angle Solved Examples
Some of the solved examples on this topic is given below for students. These solved questions will help students in cementing a solid foundation in this topic.
Example 1: The value of linear pair of angles are 36°+X and 39°+2X. Solve for the value of X.
Solution: As we know the sum of linear pair of angles is equal to 180°
So, 36°+X+39°+X =180°
=> 75°+3X = 180°
=> 3X = 180°-75°
=> 3X = 105°
=> X = 105°/3
Hence, X = 35°
Example 2: If the measure of one of the angles of an angle pair forming a straight line is 40°, what will be the value of the other angle?
Solution: As we know the straight angle pair forms a straight line, i.e., their sum is 180°.
Let the other angle be Z
So, 40° + Z = 180°
Z = 180° – 40°
Z = 140°
Hence, the value of other angle will be 140°.
Example 3: Angles Z, Z+10°, and Z+20° form a linear pair. What will be the value of Z?
Solution: As we know the sum of linear pair of angles = 180°
So, Z+Z+10°+Z+20° = 180°
3Z + 30° = 180°
3Z = 180° – 30°
3Z = 150°
Z = 150°/3
Z = 50°
Sharing is caring!
## FAQs
### What is a straight angle?
A straight angle is an angle whose arm lies in opposite directions and the value of its angle measures 180°.
### Is 180.1° a straight angle?
No, because the measure of the angle should be exactly 180° for this special angle.
### What is the twice value of a straight angle?
The twice of a straight angle is full rotation or complete angle. The value of the complete angle is 360°.
### What are linear pair of angles?
The pair of angles that form a straight angle is known as a linear pair of angles. They are also known as straight-angle pair.
### Which angle is formed at 6 AM by the minute and hour hands of the clock?
At 6 AM, the hour and minute hands form a straight angle.
### How many right angles are required to form a straight angle?
Two right angles are required to form a straight angle.
### Can we say a straight line a straight angle?
Yes, a straight line can be thought of as a straight angle with two endpoints. So, we can say a straight line belongs to this angle category.
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# Miss C Jones Upper Rhymney Primary School KS2 Maths - PowerPoint PPT Presentation
Miss C Jones Upper Rhymney Primary School KS2 Maths
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Miss C Jones Upper Rhymney Primary School KS2 Maths
## Miss C Jones Upper Rhymney Primary School KS2 Maths
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##### Presentation Transcript
1. Miss C Jones Upper Rhymney Primary School KS2 Maths
2. Measuring Measuring in mm and cm C.Jones
3. Lesson Objectives • To measure accurately in mm • To measure to the nearest in cm • To be able to convert from mm and cm • To be able to convert from cm to mm
4. Can you measure these lines to the nearest cm? _______________________ ____________________________ ________ ____________________
5. Can you measure these lines in mm? ________________________ _______________ __ ____________________
6. Use your rulers to convert from mm to cm & cm to mm 10 mm = cm 40 mm = cm 30 mm = cm 80 mm = cm 30 cm = mm 100mm= cm 160mm= cm
7. Problem Solving Activity Mr Smith is making a book shelf. He has a ruler that measures in cms only. The wood he is buying is sold in lengths measured in mms. Can you convert these measurements for him? EXT TASK: Calculate the perimeter of each piece of wood in cm and mm Width= 600mm Answer in cm = Height= 750mm Answer in cm =
8. Have you met the objectives? Can you use your ruler to measure to the nearest cm? Can you measure in mm?
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# Maths Multiplication Table 1 to 20 | 1 Se 20 Tak Table | All Tables 1 To 20
Maths table 1 to 20 is the basis of arithmetic calculations that are most widely used in multiplication and division. Table 1 will produce the original number. Multiplication of any number with 1 results in the original number. For example, 1× 5 = 5, 1× 9 = 9 and so on.
Students are suggested to learn tables 1 to 10, as it helps to solve the basic problems. Tables from 2 to 20 will help them to solve the complex calculations. Thus, learning multiplication tables from 1 to 20 will help students:
• To solve problems quickly
• To avoid mistakes in calculations
Note:
• Every number in the multiplication table 1 to 20 is a whole number.
• A number multiplied by itself gives the square of that number.
• Adding any number n times is the same as multiplying the number with n. For example, adding 10 ten times gives 100, and multiplying 10 by 10 also gives 100.
The complete list of 1 to 20 tables up to 10 times is given below.
## Table 16 to 20
How to learn 1 to 20 tables easily ?
Let us see some tips to memorize these Maths tables.
1. In the case of a table of 2, the number is increased by 2 times or a number is doubled when multiplied by 2. For example, 2 times 6, means 6 is doubled here; therefore, the result is 12. Hence, 2,4,6,8,10,12,14,16,18,20.
2. Table of 5 has a pattern. The number either ends at 0 or at 5. Hence,5,10,15,20,25,…
3. Similarly, the table of 9 also has a pattern. If we see the 9 times table, the ten’s place digit of the numbers goes in increasing order from 0 to 9 and the unit place digit of the numbers goes in decreasing order from 9 to 0. Hence, 01,18, 27, 36, 45, 54, 63, 72, 81, 90.
4. To memorize the 10 times table is very easy. We should have to put a zero next to the number multiplied by 10. Like, 10 times 8 is 80.
Why should kids learn Tables 1 – 20?
Solution: Kids must learn these Tables 1 – 20 because they help them in solving and Doing Fast Math Calculations. Many sections in math directly link with multiplication tables so learning multiplication tables plays a crucial role in Students Life.
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# Standard Equation of an Ellipse
We will learn how to find the standard equation of an ellipse.
Let S be the focus, ZK the straight line (directrix) of the ellipse and e (0 < e < 1) be its eccentricity. From S draw SK perpendicular to the directrix KZ. Suppose the line segment SK is divided internally at A and externally at A' (on KS produced) respectively in the ratio e : 1.
Therefore, $$\frac{SA}{AK}$$ = e : 1
$$\frac{SA}{AK}$$ = $$\frac{e}{1}$$
⇒ SA = e ∙ AK ...................... (i) and
$$\frac{SA'}{A'K}$$ = e : 1
$$\frac{SA'}{A'K}$$ = $$\frac{e}{1}$$
⇒ SA' = e ∙ A'K ...................... (ii)
We can clearly see that the points A and A'' lies on the ellipse since, their distance from the focus (S) bear constant ratio e (< 1) to their respective distance from the directrix.
Let C be the mid-point of the line-segment AA'; draw CY perpendicular to AA'.
Now, let us choose C as the origin CA and CY are chosen as x and y-axes respectively.
Therefore, AA' = 2a
A'C = CA = a.
Now, adding (i) and (ii) we get,
SA + SA' = e (AK + A'K)
AA' = e (CK - CA + CK + CA')
2a = e (2CK - CA + CA')
2a = 2e CK, (Since, CA = CA')
CK = $$\frac{a}{e}$$ ...................... (iii)
Similarly, subtracting (i) from (ii) we get,
SA' - SA = e (KA' - AK)
(CA' + CS) - (CA - CS) = e . (AA')
2CS = e 2a, [Since, CA' = CA]
CS = ae ...................... (iv)
Let P (x, y) be any point on the required ellipse. From P draw PM perpendicular to KZ and PN perpendicular to CX and join SP.
Then, CN = x, PN = y and
PM = NK = CK - CN = $$\frac{a}{e}$$ – x, [Since, CK = $$\frac{a}{e}$$] and
SN = CS - CN = ae - x, [Since, CS = ae]
Since the point P lies on the required ellipse, Therefore, by the definition we get,
$$\frac{SP}{PM}$$ = e
SP = e PM
SP$$^{2}$$ = e$$^{2}$$ . PM$$^{2}$$
or (ae - x)$$^{2}$$ + (y - 0)$$^{2}$$ = e$$^{2}$$[$$\frac{a}{e}$$ - x]$$^{2}$$
⇒ x$$^{2}$$(1 – e$$^{2}$$) + y$$^{2}$$ = a$$^{2}$$(1 – e$$^{2}$$)
$$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{a^{2}(1 - e^{2})}$$ = 1
$$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{a^{2}(1 - e^{2})}$$ = 1
Since 0 < e < 1, hence a$$^{2}$$(1 - e$$^{2}$$) is always positive; therefore, if a$$^{2}$$(1 - e$$^{2}$$) = b$$^{2}$$, the above equation becomes, $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1.
The relation $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 is satisfied by the co-ordinates of all points P (x, y) on the required ellipse and hence, represents the required equation of the ellipse.
The equation of an ellipse in the form $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 is called the standard equation of the ellipse.
Notes:
(i) b$$^{2}$$ < a$$^{2}$$, since e$$^{2}$$ < 1 and b$$^{2}$$ = a$$^{2}$$(1 - e$$^{2}$$)
(ii) b$$^{2}$$ = a$$^{2}$$(1 – e$$^{2}$$)
$$\frac{b^{2}}{a^{2}}$$ = 1 – e$$^{2}$$, [Dividing both sides by a$$^{2}$$]
e$$^{2}$$ = 1 - $$\frac{b^{2}}{a^{2}}$$
e = $$\sqrt{ 1 - \frac{b^{2}}{a^{2}}}$$, [taking square root on both sides]
Form the above relation e = $$\sqrt{ 1 - \frac{b^{2}}{a^{2}}}$$, we can find the value of e when a and b are given.
● The Ellipse
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# Difference between revisions of "2017 AIME II Problems/Problem 7"
## Problem
Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution.
## Solution 1
$[asy] Label f; f.p=fontsize(5); xaxis(-3,3,Ticks(f,1.0)); yaxis(-3,26,Ticks(f,1.0)); real f(real x){return (x+2)^2;} real g(real x){return x*-1;} real h(real x){return x*-2;} real i(real x){return x*-3;} real j(real x){return x*8;} draw(graph(f,-2,3),green); draw(graph(g,-2,2),red); draw(graph(h,-2,1),red); draw(graph(i,-2,1/3),red); draw(graph(j,-0.25,3),red); [/asy]$ Note the equation $\log(kx)=2\log(x+2)$ is valid for $kx>0$ and $x>-2$. $\log(kx)=2\log(x+2)=\log((x+2)^2)$. The equation $kx=(x+2)^2$ is derived by taking away the outside logs from the previous equation. Because $(x+2)^2$ is always non-negative, $kx$ must also be non-negative; therefore this takes care of the $kx>0$ condition as long as $k\neq0$, i.e. $k$ cannot be $0$. Now, we graph both $(x+2)^2$ (the green graph) and $kx$ (the red graph for $k=-1,k=-2,k=-3,k=8$) for $x>-2$. It is easy to see that all negative values of $k$ make the equation $\log(kx)=2\log(x+2)$ have only one solution. However, there is also one positive value of $k$ that makes the equation only have one solution, as shown by the steepest line in the diagram. We can show that the slope of this line is a positive integer by setting the discriminant of the equation $(x+2)^2=kx$ to be $0$ and solving for $k$. Therefore, there are $500$ negative solutions and $1$ positive solution, for a total of $\boxed{501}$.
## Solution 2
We use an algebraic approach. Since $\log(kx)=2\log(x+2)$, then $kx = (x+2)^2$ (the converse isn't necessarily true!), or $x^2+(4-k)x+4=0$. Our original equation has exactly one solution if and only if there is only one solution to the above equation, or one of the solutions is extraneous; it involves the computation of the log of a nonpositive number.
For the first case, we note that this can only occur when it is a perfect square trinomal, or $k = 0, 8$. However, $k = 0$ results in $\log(0)$ on the LHS, which is invalid. $k = 8$ yields $x = 2$, so that is one solution.
For the second case, we can use the quadratic formula. We have $$x = \frac{k-4 \pm \sqrt{k^2-8k}}2,$$ so in order for there to be at least one real solution, the discriminant must be nonnegative, or $k < 0$ or $k > 8$. Note that if $k > 8$, then both solutions will be positive, and therefore both valid. Therefore, $k < 0$. We now wish to show that if $k < 0$, then there is exactly one solution that works. Note that whenever $k < 0$, both "solutions" in $x$ are negative. One of the solutions to the equation is $x = \frac{k-4 + \sqrt{k^2-8k}}2$. We wish to prove that $x + 2 > 0$, or $x > -2$ (therefore the RHS in the original equation will be defined). Substituting, we have $\frac{k-4 + \sqrt{k^2-8k}}2 > -2$, or $\sqrt{k^2 - 8k} > -k$. Since both sides are positive, we can square both sides (if $k < 0$, then $-k > 0$) to get $k^2-8k > k^2$, or $8k < 0 \implies k < 0$, which was our original assumption, so this solution satisfies the original equation. The other case is when $x = \frac{k-4 - \sqrt{k^2-8k}}2$, which we wish to show is less that $-2$, or $\frac{k-4 - \sqrt{k^2-8k}}2 < -2 \implies k < \sqrt{k^2-8k}$. However, since the square root is defined to be positive, then this is always true, which implies that whenever $k < 0$, there is exactly one real solution that satisfies the original equation. Combining this with $k \in [-500, 500]$, we find that the answer is $500 + 1 = \boxed{501}$.
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# Parallel and Perpendicular Lines in Graphs of Linear Equations
A Linear equation is an equation in which the highest exponent of the variable present in the equation is one. When we draw the graph of the linear equation, it forms a straight line.
If any two lines in the plane are drawn, they are either parallel or intersecting.
### How do you know if a line is parallel?
Two lines are parallel if their slopes are equal.
Hence the lines y = m1x + c1 and y = m2x + c2 are parallel if m1 = m2.
In fact, two parallel lines differ by a constant.
Hence if the equation of line is y = mx + c, then, equation of a line parallel to it is y = mx + k, where k is a constant. To find the particular line, we require a unique value of k. For this additional condition is given.
Example 1: Write the equation of a line parallel to y = 7x + 10.
Using the above concept, the equation of a line parallel to above line is y = 7x + k, where k is any real number.
Example 2: Write the equation of a line parallel to y = 7x + 10 which passes from (1, 1).
The equation of a required line parallel to above line is y = 7x + k, where k is any real number.
Since the above line passes from (1, 1), hence it satisfies the required equation y = 7x + k.
Hence, 1 = 7(1) + k and k = 1 – 7 = -6.
Hence, the required equation is y = 7x – 6.
### How do you know if a line is perpendicular?
Two lines are perpendicular if the product of their slopes is -1.
Hence the lines y = m1x + c1 and y = m2x + c2 are perpendicular if m1m2 = -1 or m1 =
Example 3: Find the equation of a line perpendicular to y = 2x + 5.
Comparing with the slope intercept form, y = mx + c, the slope of given line is m = 2. Hence, the slope of a line perpendicular to given line is –= – .
Equation of line perpendicular to given line is y= – x + k = – x+k, where k is any real number.
Example 4: Find the equation of a line perpendicular to y = 3x + 5 passing from (1, 2).
As, done in the previous example slope of given line is m = 3.
Hence, the slope of a line perpendicular to given line is – = –.
Equation of line passing from (1, 2) and slope – is (y – 2) = – (x-1)
3(y – 2) = -(x – 1)
3y – 6 = –x + 1
x + 3y = 7
## Hence,slope of a line perpendicular to a given line with slope m is -1/m.
Example 5: Check whether the lines x + y = 10 and x + y = 100 are parallel or perpendicular.
Comparing with the slope intercept form, y = mx + c
The slope of first line, y = (-1)x + 10 is m1 = -1.
The slope of second line, y = (-1)x + 100 is m2 = -1.
m1 = m2 = -1. Hence the given lines are parallel.
Example 6: Check whether the lines x + y = 10 and xy = 100 are parallel or perpendicular.
Comparing with the slope intercept form, y = mx + c
The slope of first line, y = (-1)x + 10 is m1 = -1.
The slope of second line, y = (1)x – 100 is m2 = 1.
m1 m2 = -1, so the given lines are perpendicular.
Check Point
1. Find the equation of the line passing through (-1, 5) & parallel to the line y = 5x + 1.
2. Find the equation of the line passing through (-1, 5) & perpendicular to the line y = 5x + 1.
3. Check whether the lines, 3x + y = 15 and 21x + 7y = 28 are parallel or perpendicular.
4. Check whether the lines 3x + y = 15 and x – 3y = 28 are parallel or perpendicular.
5. Find the equation of the line parallel to the line, y = 7x + 51.
6. Find the equation of the line perpendicular to the line y = 7x + 51.
1. The required parallel line is y = 5x + 10.
2. The required perpendicular line is y = – x + or x + 5y = 26.
3. Since, slopes of the two lines are equal which is -3. Hence the given lines are parallel.
4. Since, product of the slopes of the two lines is -1. Hence the given lines are perpendicular.
5. The equation of the line parallel to the line y = 7x + 51 is y = 7x + k, where k is any real
6. The equation of the line perpendicular to the line y = 7x + 51 is y = – x + k , where k is any real
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# 2001 AMC 12 Problems/Problem 4
The following problem is from both the 2001 AMC 12 #4 and 2001 AMC 10 #16, so both problems redirect to this page.
## Problem
The mean of three numbers is $10$ more than the least of the numbers and $15$ less than the greatest. The median of the three numbers is $5$. What is their sum?
$\textbf{(A)}\ 5\qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 25\qquad \textbf{(D)}\ 30\qquad \textbf{(E)}\ 36$
## Solution 1
Let $m$ be the mean of the three numbers. Then the least of the numbers is $m-10$ and the greatest is $m + 15$. The middle of the three numbers is the median, $5$. So $\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$, which can be solved to get $m=10$. Hence, the sum of the three numbers is $3\cdot 10 = \boxed{\textbf{(D) }30}$.
## Solution 2
Say the three numbers are $x$, $y$ and $z$. When we arrange them in ascending order then we can assume $y$ is in the middle therefore $y = 5$.
We can also assume that the smallest number is $x$ and the largest number of the three is $y$. Therefore,
$$\frac{x+y+z}{3} = x + 10 = z - 15$$ $$\frac{x+5+z}{3} = x + 10 = z - 15$$
Taking up the first equation $\frac{x+5+z}{3} = x + 10$ and simplifying we obtain $z - 2x - 25 = 0$ doing so for the equation $\frac{x+5+z}{3} = z - 15$ we obtain the equation $x - 2z + 50 = 0$
$$x - 2z + 50 = 2x - 4z + 100$$
when solve the above obtained equation and $z - 2x - 25 = 0$ we obtain the values of $z = 25$ and $x = 0$
Therefore the sum of the three numbers is $25 + 5 + 0 = \boxed{\textbf{(D) }30}$
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# Determine the values of the constants α and β so that the function f (x) = x3 + αx2 + βx + δ may have a relative maximum at x = −3, and a relative minimum at x = 1.
## For ease of typing, I will use
f(x) = x^3 + ax^2 + bx + c
take the derivative and the c will drop out
set this equal to zero, with x = -3 to get an equation in a and b
repeat with x = 1
you now have two equations in a and b, use your favourite grade 9 method to solve them
## To determine the values of the constants α and β, we will need to use the properties of the first and second derivatives of the function.
Step 1: Find the first derivative of the function f(x):
f'(x) = 3x^2 + 2αx + β
Step 2: Find the second derivative of the function f(x):
f''(x) = 6x + 2α
Step 3: Determine the conditions for a relative maximum at x = -3 and a relative minimum at x = 1.
At x = -3:
For a relative maximum at x = -3, we need f'(x) = 0 and f''(x) < 0.
Substitute x = -3 into f'(x) and solve for α:
3(-3)^2 + 2α(-3) + β = 0
9α - 6α + β = 0
3α + β = 0
Substitute x = -3 into f''(x) and solve for α:
6(-3) + 2α < 0
-18 + 2α < 0
2α < 18
α < 9
At x = 1:
For a relative minimum at x = 1, we need f'(x) = 0 and f''(x) > 0.
Substitute x = 1 into f'(x) and solve for α:
3(1)^2 + 2α(1) + β = 0
3 + 2α + β = 0
Substitute x = 1 into f''(x) and solve for α:
6(1) + 2α > 0
6 + 2α > 0
2α > -6
α > -3
Step 4: Combine the conditions and solve for α and β.
From the condition at x = -3, we have α < 9. From the condition at x = 1, we have α > -3.
Therefore, the values of α must satisfy -3 < α < 9.
Since the question does not provide a specific value for δ, we cannot determine the exact values of α and β.
## To determine the values of the constants α and β, we can use the properties of a relative maximum and a relative minimum.
A relative maximum occurs at a point where the derivative of the function changes from positive to negative. Conversely, a relative minimum occurs at a point where the derivative changes from negative to positive.
First, let's find the derivative of the function f(x):
f(x) = x^3 + αx^2 + βx + δ
f'(x) = 3x^2 + 2αx + β
To have a relative maximum at x = -3, the derivative f'(x) should change sign from positive to negative at x = -3. This means that f'(-3) should be greater than 0, and f'(-3+h) should be less than 0 for some small positive value of h.
Using the derivative f'(x), we can evaluate this condition:
f'(-3) = 3(-3)^2 + 2α(-3) + β
= 9 - 6α + β
To have a relative minimum at x = 1, the derivative f'(x) should change sign from negative to positive at x = 1. This means that f'(1) should be less than 0, and f'(1+h) should be greater than 0 for some small positive value of h.
Using the derivative f'(x), we can evaluate this condition:
f'(1) = 3(1)^2 + 2α(1) + β
= 3 + 2α + β
Now we have two equations that we can use to solve for α and β:
Condition 1: 9 - 6α + β > 0
Condition 2: 3 + 2α + β < 0
Solving these two inequalities simultaneously can determine the values of α and β that satisfy both conditions and allow for a relative maximum at x = -3 and a relative minimum at x = 1.
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# 9-4 - A triangle(which can be formed by splitting a...
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Areas of Parallelograms and Triangles Lesson 9-4
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Parallelogram A parallelogram is a quadrilateral where the opposite sides are congruent and parallel. A rectangle is a type of parallelogram, but we often see parallelograms that are not rectangles (parallelograms without right angles).
Area of a Parallelogram Any side of a parallelogram can be considered a base . The height of a parallelogram is the perpendicular distance between opposite bases. The area formula is A=bh A=bh A=5(3) A=15m 2
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Area of a Triangle A triangle is a three sided polygon. Any side can be the base of the triangle. The height of the triangle is the perpendicular length from a vertex to the opposite base.
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Unformatted text preview: A triangle (which can be formed by splitting a parallelogram in half) has a similar area formula: A = ½ bh. Example A= ½ bh A= ½ (30)(10) A= ½ (300) A= 150 km 2 Complex Figures Use the appropriate formula to find the area of each piece. Add the areas together for the total area. Example | 27 cm | 10 cm 24 cm Split the shape into a rectangle and triangle. The rectangle is 24cm long and 10 cm wide. The triangle has a base of 3 cm and a height of 10 cm. Solution Rectangle A = lw A = 24(10) A = 240 cm 2 Triangle A = ½ bh A = ½ (3)(10) A = ½ (30) A = 15 cm 2 Total Figure A = A 1 + A 2 A = 240 + 15 = 255 cm 2 Homework Time...
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# The angles of a quadrilateral are in the ratio 5:3:9:7. Find the angles.
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## To find the angles of a quadrilateral that are in the ratio 5:3:9:7, we can follow these steps:1. Define the Angles in Terms of a Variable: Let the angles of the quadrilateral be represented as: - First angle = 5x - Second angle = 3x - Third angle = 9x - Fourth angle = 7x2. Set Up the Equation for the Sum of Angles: We know that the sum of the angles in a quadrilateral is always 360∘. Therefore, we can write the equation: 5x+3x+9x+7x=3603. Combine Like Terms: Now, let's combine the terms on the left side: (5+3+9+7)x=360 Simplifying this gives: 24x=3604. Solve for x: To find x, divide both sides of the equation by 24: x=36024 Simplifying this gives: x=155. Calculate Each Angle: Now that we have the value of x, we can find each angle: - First angle: 5x=5×15=75∘ - Second angle: 3x=3×15=45∘ - Third angle: 9x=9×15=135∘ - Fourth angle: 7x=7×15=105∘6. List the Angles: Therefore, the angles of the quadrilateral are: - 75∘ - 45∘ - 135∘ - 105∘Final Answer:The angles of the quadrilateral are 75∘, 45∘, 135∘, and 105∘.
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Updated on:7/8/2024
### Knowledge Check
• Question 1 - Select One
## The four angles of a quadrilateral are in the ratio 2:3:5:8. Find the angles.
A40,60,120and140.
B80,60,80and160.
C40,60,100and160.
D40,70,100and160.
• Question 2 - Select One
## The angles of a quadrilateral are in the ratio 3:5:7:9. Find the measure of each of these angles.
A45,75,105,165
B35,75,105,135
C45,75,105,135
D45,85,105,135
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2007 iTest Problems/Problem 36
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Problem
Let b be a real number randomly selected from the interval $[-17,17]$. Then, m and n are two relatively prime positive integers such that m/n is the probability that the equation $x^4+25b^2=(4b^2-10b)x^2$ has $\textit{at least}$ two distinct real solutions. Find the value of $m+n$.
Solution
The equation has quadratic form, so complete the square to solve for x.
$$x^4 - (4b^2 - 10b)x^2 + 25b^2 = 0$$ $$x^4 - (4b^2 - 10b)x^2 + (2b^2 - 5b)^2 - 4b^4 + 20b^3 = 0$$ $$(x^2 - (2b^2 - 5b))^2 = 4b^4 - 20b^3$$
In order for the equation to have real solutions,
$$16b^4 - 80b^3 \ge 0$$ $$b^3(b - 5) \ge 0$$ $$b \le 0 \text{ or } b \ge 5$$
Note that $2b^2 - 5b = b(2b-5)$ is greater than or equal to $0$ when $b \le 0$ or $b \ge 5$. Also, if $b = 0$, then expression leads to $x^4 = 0$ and only has one unique solution, so discard $b = 0$ as a solution. The rest of the values leads to $b^2$ equalling some positive value, so these values will lead to two distinct real solutions.
Therefore, in interval notation, $b \in [-17,0) \cup [5,17]$, so the probability that the equation has at least two distinct real solutions when $b$ is randomly picked from interval $[-17,17]$ is $\frac{29}{34}$. This means that $m+n = \boxed{63}$.
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# Order of Operations
In math, we have to have a set way of solving problems, otherwise everyone could get different answers. “Order of Operations” is simply a list of steps to use when solving problems with multiple operations (operations are adding, subtracting, multiplying, dividing).
The easiest way to remember the order is to use this acronym:
• Excuse
• My
• Dear
• Aunt
• Sally
These letters stand for:
• Parentheses
• Exponents
• Multiply
• Divide
• Subtract
Multiplying and dividing go as a pair: if you have both of these operations, solve them from left to right. The same is true for adding and subtracting; if you have both, solve from left to right.
We will use the letters P, E, M, D, A, S as shorthand for the operations.
Example math problems related to order of operations:
1. 5 + 7(4 – 2)
• First, identify the operations: we have A, M, S, and P. According to the order of operations, we must first start with the parentheses, which is where the subtraction is:
• 5 + 7(2)
• Now we multiply:
• 5 + 14
• 19
2. 18 ÷ 3 + 5(7 – 4) x 2
• Identify the operations: D, A, M, P, S, and M.
• 18 ÷ 3 + 5(3) x 2
• Now we see that we have D, A, M, and M.
• Since we have both dividing and multiplying, we will do those from left to right:
• 6 + 15 x 2
• 6 + 30
• 36
• To avoid confusion, first look for parentheses. Solve those parts, then look for exponents. After that, dividing and multiplying, and finally, adding and subtracting.
3. 4 x 52 + 16 ÷ 2
• 4 x 25 + 16 ÷ 2
• Now we’ll do multiplying and dividing:
• 100 + 8
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# Difference between revisions of "2019 AMC 8 Problems/Problem 3"
## Problem 3
Which of the following is the correct order of the fractions $\frac{15}{11},\frac{19}{15},$ and $\frac{17}{13},$ from least to greatest?
$\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}$
## Solution 1
Consider subtracting 1 from each of the fractions. Our new fractions would then be $\frac{4}{11}, \frac{4}{15},$ and $\frac{4}{13}$. Since $\frac{4}{15}<\frac{4}{13}<\frac{4}{11}$, it follows that the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$
-will3145
## Solution 2
We take a common denominator: $$\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.$$
Since $2717<2805<2925$ it follows that the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$.
-xMidnightFirex
~ dolphin7 - I took your idea and made it an explanation.
## Solution 3
When $\frac{x}{y}>1$ and $z>0$, $\frac{x+z}{y+z}<\frac{x}{y}$. Hence, the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$. ~ ryjs
This is also similar to Problem 20 on the AMC 2012.
## Solution 4
We use our insane mental calculator to find out that $\frac{15}{11} \approx 1.36$, $\frac{19}{15} \approx 1.27$, and $\frac{17}{13} \approx 1.31$. Thus, our answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$.
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