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back to algebra2 video lesson
The question is how to move the graph of y = (1/2)x2 to get the graph of y = (1/2)(x - 3)2 - 2?
The function of y = (1/2)x2 can be written to y = (1/2)(x - 0)2 - 0, so its vertex is (0, 0), axis of symmetry is x = 0 which is y-axis, and the graph is open upward because of the coefficient of x square term, a > 0.
The vertex of y = (1/2)(x - 3)2 - 2 is (3, -2), the axis of symmetry is x = 3, and the graph is open upward because of the coefficient of the x square term, a > 0. Because both functions have the same coefficient of the x square term a, so we can move the graph of y = (1/2)x2 to get the graph of y = (1/2)(x - 3)2 - 2.
Note: If the quadratic function is written in the form y = a(x - h)2 + k, then the vertex is (h, k), the axis of symmetry is x = h.
The graph of the quadratic function y = ax2 + bx + x, where the coefficient a can not be zero, is the set of points (x, y) that satisfy the quadratic function y = ax2 + bx + c.
To draw the graph of y = (1/2)x2, because the coefficient a is larger than zero, so the graph is open upward. So x = 0 is the minimum value of the quadratic function. We let x = 0 be the middle value of the set of x.
x ... -2 -1 0 1 2 ...
y ... 2 1/2 0 1/2 2 ...
So we move the graph of y = (1/2)x2 right 3 units to get the graph of y = (x - 3)2, then move the graph down 2 units to get the graph of y = (1/2)(x - 3)2 - 2. |
# Thread: Question - tangents and their equation
1. ## Question - tangents and their equation
Hello everyone
I am studying tangents and come across this kind of problem where I am going to find the equation for one or more tangent with a specific directioncoefficient for an function.
Example:
Find the tangents for the funtion $\displaystyle y = 3x^2 - 2x + 1$ that has an directioncoefficient of 4.
How do I solve this kind of problem? I am halfway through, I think.
First we get the derivate for the function, which is:
$\displaystyle y = 6x - 2$
The we check and see for what X value Y equals 4, correct?
$\displaystyle 4 = 6x - 2$
$\displaystyle 1 = x$
And after that? Assuming that the above is correct.
2. Originally Posted by λιεҗąиđ€ŗ
Hello everyone
I am studying tangents and come across this kind of problem where I am going to find the equation for one or more tangent with a specific directioncoefficient for an function.
Example:
Find the tangents for the funtion $\displaystyle y = 3x^2 - 2x + 1$ that has an directioncoefficient of 4.
How do I solve this kind of problem? I am halfway through, I think.
First we get the derivate for the function, which is:
$\displaystyle y = 6x - 2$
The we check and see for what X value Y equals 4, correct?
$\displaystyle 4 = 6x - 2$
$\displaystyle 1 = x$
And after that? Assuming that the above is correct.
i have no idea what the direction coefficient is. but assuming what you did so far is correct, this is probably the way to go.
we know we want the tangent at x = 1. plugging x = 1 into the original equation we get y = 2. so we want the tangent at (1,2) with slope 4. use the point-slope form, and solve for y.
we have: $\displaystyle y - y_1 = m(x - x_1)$
where $\displaystyle (x_1,y_1) = (1,2)$ and $\displaystyle m = \mbox{ slope } = 4$
3. i have no idea what the direction coefficient is.
Heh, maybe it isn't called "direction coefficient" in english.
Ex. $\displaystyle y = 7x$ has an (something something) of seven. $\displaystyle \frac {\delta y} {\delta x} = 7$
4. Originally Posted by λιεҗąиđ€ŗ
Heh, maybe it isn't called "direction coefficient" in english.
Hej,
this is for your private math dictionary:
a) If you are refering to a straight line then the "direction coefficient" is called slope
b) if you are refering to the graph of a function then the "direction coefficient" is called gradient
5. Hej,
this is for your private math dictionary:
a) If you are refering to a straight line then the "direction coefficient" is called slope
b) if you are refering to the graph of a function then the "direction coefficient" is called gradient
Guten Tag
Thanks, ok so it's the slope we're after.
Direction coefficient sounds more scientific than slope though.
6. Originally Posted by λιεҗąиđ€ŗ
Direction coefficient sounds more scientific than slope though.
indeed it does! i'm going to start using that to confuse the people i tutor
Jhevon: Silly girl, just find the direction coefficient!
Silly Girl:
7. Hehe...
Right-o, back to the problem.
I have talked to an fellow and he expained the general formula for finding the tangents equation.
$\displaystyle t(x) = f(a) + f´(a)(a + x)$
T(x) = (3a^2 - 2a + 1) + (6a - 2)(a + x)
T(x) is the tangent of course.
$\displaystyle 4 = 6x - 2$
$\displaystyle 1 = x$
A is 1 in this specific problem.
Let's insert 1 and see what happens...
$\displaystyle T = (3(1)^2 - 2(1) + 1) + (6(1) - 2)((1) + x)$
$\displaystyle T = ( 3 - 2 + 1) + (6 + 6x - 2 - 2x)$
$\displaystyle T = 4x + 6$
Please correct me if you find any errors, this isn't what the book want for an answer. |
# If y varies inversely as x and y = 725 when x = 20, how do you find x when y is 50?
Jan 4, 2017
$x = 290$ when $y = 50$
#### Explanation:
When you are told that $y$ varies inversely as $x$, that means that
$x \cdot y = k$
where $k$ can be any single number. If you plug in the initially given numerical values for $x$ and $y$ you have
$20 \cdot 725 = 14500$
So your $k$-value becomes $14500$. Now you can say that, for this particular relation
$x y = 14500$
If you want to know your $x$-value when $y = 50$, you can plug $50$ in for $y$ and solve for $x$.
$x \cdot 50 = 14500$
so
$x = \frac{14500}{50}$
meaning that
$x = 290$. |
How do you find the x and y intercept of 4y=3x+12?
Dec 17, 2016
The x-intercept is: $x = - 4$ or (-4, 0).
The y-intercept is: $y = 3$ or (0, 3)
Explanation:
To find the x-intercept you set $y$ equal to $0$ and solve for $x$:
$4 \cdot 0 = 3 x + 12$
$0 = 3 x + 12$
$0 - 12 = 3 x + 12 - 12$
$- 12 = 3 x + 0$
$- 12 = 3 x$
$\frac{- 12}{3} = \frac{3 x}{3}$
$- 4 = x$ or $x = - 4$ or (-4, 0).
To find the x-intercept you set $x$ equal to $0$ and solve for $y$:
$4 y = 3 \cdot 0 + 12$
$4 y = 0 + 12$
$4 y = 12$
$\frac{4 y}{4} = \frac{12}{4}$
$y = 3$ or (0, 3) |
Chapter 2: Simple linear regression: The regression equation and the regression coefficient
Visual inspection of regression lines may be convenient, but their steepness and direction are usually indicated by numbers rather than figures. These numbers are called regression coefficients. This chapter will teach you how to compute them.
The regression equation
A good understanding of regression coefficients presupposes a basic knowledge of the mathematical representation of regression lines. The data and variables are the same as in the second example in chapter 1. The observed education length of person i (any person in the sample) is symbolised by yi, The value yi can be expressed as the sum of person i’s predicted education length (ŷi) and a residual (ei). We express this as follows:
yi = ŷi + ei
Furthermore, as we know from chapter 1, the regression line is a graphical expression of the association between people’s observed independent variable values and their predicted dependent variable values. Since the linear regression line is straight, this association can be expressed as a linear function. In such functions, the predicted education length is seen as changing in constant proportion to increases or decreases in people’s birth year variable values. We can express such functions as follows:
ŷi = a + b∙xi,
Here, xi is person i’s birth year, while a and b symbolise constants (fixed numbers). These constants are the regression coefficients, or, to be more exact, the a is often called the constant or the intercept, while the b is called variable x’s regression coefficient because it determines how the predicted y values (the ŷi) change as the value of xi changes. A more thorough exposition of these points can be found here.
If we let the right-hand side of the function ŷi = a + b∙xi replace ŷi in the function yi = ŷi + ei, we get:
yi = a + b∙xi. + ei
That is, the observed values of the dependent variable (in our example length of education) are conceived as being determined by four factors: The coefficients a and b, the independent variable xi (year of birth) and the residual ei. The latter is supposed to vary randomly and, hence, to be independent of xi. In Figure 6, this is illustrated for the persons with identification numbers 4 and 1892.
Figure 6. Graphical illustration of residuals and observed / predicted values
The line that slopes upwards from left to right is identical to the regression line we saw in Figure 2. The two circles that represent our selected persons are not positioned on the regression line, which means that their observed dependent variable values deviate from their predicted values. The observed values y4 and y1892 are marked on the right and left vertical axis, respectively. The predicted values ŷ4 and ŷ1892 are the y-variable values of the points on the regression line that lie immediately above person 4’s position and below person 1892’s position in the figure. These values are also marked on the vertical axes. The residuals, e4 and e1892, are the vertical distances between the persons’ circles and the regression line, i.e. they are the differences between the observed and the predicted dependent variable values (e4 = y4 - ŷ4.etc.). Persons whose circles are positioned below the regression line have negative residuals, while those whose circles lie above the regression line have positive residuals.
However, in order to know the sign and size of a person’s residual, we have to know where the regression line is positioned. Hence, we need a method for simultaneous computation of regression coefficients and residuals. In fact, as explained in chapter 1, the OLS method does just this by computing those a and b values that minimise the sum of squared residuals. We do not have to worry about the computational procedures. SPSS does all the necessary computations for us. Note, however, that this method always causes the sum and mean value of the residuals to be approximately 0.
Page 1
Performing ordinary linear regression analyses using SPSS
Follow the preparatory steps outlined in the first chapter, i.e. open the data set, turn on the design weight and select the Norwegian sample of persons born earlier than 1975. Then, run the regression analysis as follows:
• Click on ‘Regression’ and ‘Linear’ from the ‘Analyze’ menu.
• Find the dependent and the independent variables on the dialogue box’s list of variables.
• Select one of them and put it in its appropriate field. Then put the other variable in the other field. (See Figure 7.)
• Finally, click ‘OK’ and an output window will open.
You can also copy, paste and run this syntax
*Syntax for the example in chapter 2, the Norwegian sample. *The following command causes the cases to be weighted by the design weight variable 'dweight'.
WEIGHT BY dweight.
*The following commands cause SPSS to select for analysis those cases that belong to the Norwegian sample (value NO on country variable) and have lower values than 1975 on the birth year variable (& stands for AND, < stands for 'less than'). *In this process, the commands create a filter variable (filter_\$) with value 1 for the selected cases and value 0 for the non-selected cases. *Change the last part of line 2 (which starts after the first equals sign) if you wish to select other cases. If you do this, you should also change the variable label, which is in double quotation marks on line 3.
USE ALL.
COMPUTE filter_\$=cntry = 'NO' & yrbrn < 1975.
VARIABLE LABEL filter_\$ "cntry = 'NO' & yrbrn < 1975 (FILTER)".
VALUE LABELS filter_\$ 0 'Not Selected' 1 'Selected'.
FORMAT filter_\$ (f1.0).
FILTER BY filter_\$.
EXECUTE.
*The following commands cause a linear regression analysis to be performed on the selected data with dependent variable 'eduyrs' and independent variable 'yrbm'. *Change variable names in the last two lines if you wish to run the analysis with other dependent and independent variables.
REGRESSION
/MISSING LISTWISE
/STATISTICS COEFF OUTS R ANOVA
/CRITERIA=PIN(.05) POUT(.10)
/NOORIGIN
/DEPENDENT eduyrs
/METHOD=ENTER yrbrn.
Figure 7. Running a simple (bivariate) linear regression analysis
The output you get if you execute these commands correctly, contains the ‘Coefficients’ table shown here as Table 1.The computed values of a and b are shown in the B column. The item in the first row is the a-coefficient, which SPSS terms the ‘Constant’. The item in the second row is the birth year variable’s b-coefficient, which indicates the steepness of the regression line or, if you prefer, indicates how much the predicted value of the dependent variable (length of education) increases when the value of the independent birth year variable increases by one unit (one year). The coefficient’s value is 0.097 (or, more exactly, 0.096672408), which means that each new cohort’s predicted length of education is 0.097 years longer than that of the cohort that was born one year before it. The a-coefficient or ‘constant’ is identical to the predicted value of the dependent variable for those cases whose independent variable value is 0. But be careful when interpreting this coefficient. Here, its value is negative (-175.553), and surely no one has a negative length of education. The reason for this strange result is that the persons thus attributed a negative education length are supposed to have been born in year 0. But there are no survivors from year 0 in our sample, and the regression results only apply to persons whose x-variable values lie within the span used in the computations (values between 1910 and 1974). You should avoid making extrapolations beyond these limits, and extrapolations that extend far beyond these limits make little sense. Hence, the constant term has no substantial interpretation in this example, but we still need it for computations of predicted y values.
Table 1. SPSS output: Simple linear regression coefficients
The computed coefficient values may be seen as interesting in themselves. But they can also be used to compute predicted dependent variable values for particular persons or groups of persons. Such computations are done by inserting these persons’ independent variable values and the computed coefficient values into the right-hand side of the function ŷi = a + b∙xi. For example, person 4 in Figure 6 was born in 1954. If we insert this value into the function together with the coefficient values, we get:
ŷ4 = (-175.5535449315) + 0.096672408 ∙1954 = 13.34.
(Abnormal numbers of decimals are used for exact prediction.) In other words, we can predict this person’s length of education to be 13.34 years. But this person’s actual length of education is 11 years, so we get a residual value of -2.34 years, which accords with what Figure 6 shows us.
The coefficients presented in Table 1 pertain to those who are members of the ESS sample. The table also contains information about the accuracy of these coefficients regarded as indicators of the association between birth year and education length in the entire population of Norwegians born before 1975. Exactly what these columns (Std.Error, t and Sig.) tell us will, however, be explained in chapter 4.
The Beta column contains the regression coefficients one gets when the analysis is performed on standardised variables. You don’t have to know anything about them to perform ordinary regression analysis. In fact, in most cases you should avoid using them. Consult a regression analysis textbook if you want to know more about them.
Page 2
Interpretation of the Model summary table
The regression results comprise three tables in addition to the ‘Coefficients’ table, but we limit our interest to the ‘Model summary’ table, which provides information about the regression line’s ability to account for the total variation in the dependent variable. Figure 6 demonstrates that the observed y-values are highly dispersed around the regression line. Thus, as regression analysts often put it, the regression model only ‘explains’ a limited proportion of the dependent variable’s total variation. The dependent variable’s total variation can be measured by its variance. If the regression line is not completely horizontal (i.e. if the b coefficient is different from 0), then some of the total variance is accounted for by the regression line. This part of the variance is measured as the sum of the squared differences between the respondents’ predicted dependent variable values and the overall mean divided by the number of respondents. By dividing this explained variance by the total variance of the dependent variable, we arrive at the proportion of the total variance that is accounted for by the regression equation. This proportion varies between 0 and 1 and is symbolised by R2 (R Square). As can be seen from Table 2, the value of our R2 is 0.131, which means that 13.1 percent of the total variance in education length has been ‘explained’. Not very impressive, but not bad either compared with the R2 values one tends to get in analyses of social survey data. The R is the square root of R2. The Adjusted R2 will be discussed later.
Table 2. SPSS output: Simple linear regression goodness of fit
Exercises
1. Perform the same regression analysis as in the example presented above on data from the Polish (or another county’s) ESS sample.
2. Perform a regression analysis with ‘How happy are you’ as the dependent variable and ‘Subjective general health’ as the independent variable. (These variables are not metric, but they can, at least as an exercise, still be used in OLS regression.) Use data from a country of your own choice. What do the results tell you?
Go to next chapter >>
Page 3 |
# How do you find the asymptote(s) or holes of f(x) = (x^2-1) / ((x-1)(x^3-2x^2+2x-1)?
Feb 14, 2018
$f \left(x\right)$ has a horizontal asymptote $y = 0$ and a vertical asymptote $x = 1$. It has no slant asymptote or any holes.
#### Explanation:
Note that:
${x}^{3} - 2 {x}^{2} + 2 x - 1 = \left(x - 1\right) \left({x}^{2} - x + 1\right)$
The quadratic factor ${x}^{2} - x + 1$ is always positive, as you can see by checking its discriminant, or by completing the square:
${x}^{2} - x + 1 = {\left(x - \frac{1}{2}\right)}^{2} + \frac{3}{4}$
So we find:
$f \left(x\right) = \frac{{x}^{2} - 1}{\left(x - 1\right) \left({x}^{3} - 2 {x}^{2} + 2 x - 1\right)}$
$\textcolor{w h i t e}{f \left(x\right)} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(x + 1\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(x - 1\right) \left({x}^{2} - x + 1\right)}$
$\textcolor{w h i t e}{f \left(x\right)} = \frac{x + 1}{\left(x - 1\right) \left({x}^{2} - x + 1\right)}$
Note that the denominator is of higher degree than the numerator. Hence we can deduce that $f \left(x\right)$ has a horizontal asymptote $y = 0$.
The only factor in the denominator that can be zero is $\left(x - 1\right)$, occurring when $x = 1$, when the numerator (in the simplified expression) is non-zero. We can deduce that $f \left(x\right)$ has a vertical asymptote $x = 1$.
The factors that we cancelled were also at $x = 1$, so would have resulted in a hole instead of an asymptote if the extra factor in the denominator were not present. As it is, we have a vertical asymptote there.
There are no slant (a.k.a. oblique) asymptotes. Those only happen when the degree of the numerator exceeds the denominator by exactly $1$.
graph{(x^2-1)/((x-1)(x^3-2x^2+2x-1)) [-12.66, 12.65, -6.33, 6.33]} |
Graphing Hyperbolas
# Graphing Hyperbolas
• PRACTICE (online exercises and printable worksheets)
Hyperbolas were introduced in three prior lessons:
The purpose of this current section is to go from the equation of a hyperbola to its graph.
In particular, we'll see that every hyperbola has a so-called central box that is helpful in determining its shape.
Short on time? Jump right to the summary!
This lesson includes lots of details.
Once you understand things, it can be written more compactly.
For your convenience, the standard equations of hyperbolas
(that were derived in the preceding section) are repeated here:
## Equations of Hyperbolas: Center at the Origin
In both equations, $\,a > 0\,$ and $\,b > 0\,$.
Also, $\,c\,$ is the positive number for which $\,c^2 = a^2 + b^2\,$.
With the equations in standard form, $\,a^2\,$ is the denominator of the positive term on the left side.
## Foci on the $x$-axis
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ Vertices: $\,(-a,0)\,$ and $\,(a,0)\,$
Foci: $\,(-c,0)\,$ and $\,(c,0)\,$
## Foci on the $y$-axis
$$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$ Vertices: $\,(0,-a)\,$ and $\,(0,a)\,$
Foci: $\,(0,-c)\,$ and $\,(0,c)\,$
## Review:
The following facts are used in this lesson:
• The square root function is increasing:
In particular, if $\,0 \le x \le y\,$, then $\,0 \le \sqrt{x} \le \sqrt{y}\,$.
• Renaming the square root of a square:
For all real numbers $\,x\,$, $\sqrt{x^2} = |x|\,$.
• The absolute value of a nonnegative number is itself:
If $\,x \ge 0\,$, then $\,|x| = x\,$.
• Squared quantities are nonnegative:
For all real numbers $\,x\,$, $\,x^2 \ge 0\,$.
• Solving an absolute value inequality:
For all real numbers $\,x\,$ and for $\,k > 0\,$: $$|x| \ge k\ \ \ \ \text{is equivalent to}\ \ \ \ (x\ge k\ \ \text{ or }\ \ x\le -k)$$
• The square root of a product is the product of the square roots:
Providing both $\,x\ge 0\,$ and $\,y\ge 0\,$, then $\,\sqrt{xy} = \sqrt{x}\sqrt{y}\,$.
• Solving an absolute value equation:
For all real numbers $\,x\,$ and for $\,k \ge 0\,$: $$|x| = k\ \ \ \ \text{is equivalent to}\ \ \ \ x = \pm k$$
## Important Information Given by the Equations of Hyperbolas
Many equations easily give you lots of information about their graphs, without any memorizing!
This is certainly true with hyperbolas, as follows.
## What part of the plane do hyperbolas live in?
Let's rewrite and analyze the equation $\,\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\,$:
\displaystyle \begin{alignat}{2} \frac{x^2}{a^2} &= 1 + \frac{y^2}{b^2} &\qquad&\text{(solve for the term containing\,x\,$)}\cr &\ge 1 &&\text{(since$\,\frac{y^2}{b^2} \ge 0\,)} \end{alignat}
Thus:
\displaystyle \begin{alignat}{2} &\frac{x^2}{a^2}\ge 1&&\cr\cr &x^2\ge a^2 &\qquad&\text{(multiply both sides by\,a^2 > 0\,)} \end{alignat}
Take note of this intermediate result:
it follows that for any point on the hyperbola, $\,x^2 - a^2 \ge 0\,$.
We'll use this fact below.
Continuing:
\displaystyle \begin{alignat}{2} &\sqrt{\strut x^2} \ge \sqrt{\strut a^2}&&\text{(the square root function is increasing)}\cr\cr &|x| \ge |a|&&\text{(\,\sqrt{x^2} = |x|\,$)}\cr\cr &|x| \ge a&&\text{(since$\,a > 0\,$,$\,|a| = a\,)}\cr\cr &x \ge a\ \ \ \text{ or }\ \ \ x\le -a&\qquad&\text{(solving the absolute value inequality)} \end{alignat}
This confirms that any point $\,(x,y)\,$ satisfying $\,\displaystyle\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\,$
must lie in one of the two half-planes, $\,x \ge a\,$ or $\,x\le -a\,$, shaded green below:
## What do hyperbolas look like when $\,x\,$ is big?
What happens to the graph of $\,\displaystyle\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\,$ for large values of $\,x\,$?
To answer this question, first ‘re-name’ the equation by solving for $\,y\,$:
\begin{alignat}{2} &\frac{y^2}{b^2} = \frac{x^2}{a^2} - 1&\qquad&\text{(re-arrange)}\cr\cr &y^2 = b^2\left(\frac{x^2}{a^2} - 1\right)&&\text{(multiply both sides by \,b^2\,)}\cr\cr &y^2 = b^2\left(\frac{x^2}{a^2} - \frac{a^2}{a^2}\right)&&\text{(rename \,1\, as \,\frac{a^2}{a^2})}\cr\cr &y^2 = \frac{b^2}{a^2}(x^2 - a^2)&&\text{(factor out \,\frac{1}{a^2})}\cr\cr &\sqrt{y^2} = \sqrt{\frac{b^2}{a^2}(x^2 - a^2)}&\qquad&\text{(take square roots of both sides)}\cr\cr &|y| = \sqrt{\frac{b^2}{a^2}(x^2 - a^2)}&\qquad&\text{(\sqrt{y^2} = |y|\,)}\cr\cr &|y| = \sqrt{\frac{b^2}{a^2}} \sqrt{\strut x^2 - a^2}&\qquad&\text{(as shown above, \,x^2 - a^2\ge 0\, for all points on the hyperbola)}\cr\cr &|y| = \frac{b}{a}\sqrt{\strut x^2 - a^2}&&\text{(both \,a > 0\, and \,b > 0\,)}\cr\cr &|y| = \frac{b}{a}\sqrt{\strut x^2\bigl(1 - \frac{a^2}{x^2}\bigr)}&&\text{(factor out \,x^2\,; we're exploring this equation for BIG \,x\, right now, so \,x\ne 0\,)}\cr\cr &y = \pm\frac{b}{a}\sqrt{\strut x^2\bigl(1 - \frac{a^2}{x^2}\bigr)}&&\text{(solve the absolute value equation; notice that the prior right-hand side is nonnegative)}\cr\cr \end{alignat}
This version of the equation might look harder than what we started with ($\displaystyle\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$),
but it's better for understanding what happens when $\,x\,$ gets big.
Here's the key idea (which will be firmed up in Calculus):
As $\,x\,$ gets bigger and bigger, $\,\displaystyle\frac{a^2}{x^2}\,$ looks more and more like $\,0\,$.
More precisely:
We can get $\,\displaystyle\frac{a^2}{x^2}\,$ as close to $\,0\,$ as we want,
by making $\,x\,$ sufficiently large.
As an example, let $\,x = 1000a\,$.
(Here, $\,x\,$ isn't particularly big—it's just one thousand times bigger than $\,a\,$.)
Already, though: $$\frac{a^2}{x^2} = \frac{a^2}{(1000a)^2} = \frac{1}{1{,}000{,}000}$$ That's pretty close to zero!
So, here's what happens to the equation $\displaystyle\,y = \pm\frac{b}{a}\sqrt{\strut x^2\bigl(1 - \frac{a^2}{x^2}\bigr)}\,$ as $\,x\,$ gets big (big positive, or big negative):
• As $\,x\,$ gets big, $\,\displaystyle\frac{a^2}{x^2}\,$ approaches $\,0\,$.
• As $\,\displaystyle\frac{a^2}{x^2}\,$ approaches $\,0\,$, $\,\displaystyle(1-\frac{a^2}{x^2})\,$ approaches $\,1\,$.
• As $\,\displaystyle(1-\frac{a^2}{x^2})\,$ approaches $\,1\,$, the expression $\displaystyle\,x^2(1-\frac{a^2}{x^2})\,$ looks more and more like $\,x^2\,$.
• Thus, $\,y = \pm\frac{b}{a}\sqrt{\strut x^2\bigl(1 - \frac{a^2}{x^2}\bigr)}\,$ looks more and more like $\,y = \pm\frac{b}{a}\sqrt{x^2}\,$, which simplifies to $\,y = \pm\frac{b}{a}|x|\,$.
• This cleans up even more!
Because of the ‘plus or minus’ sign, we can lose the absolute value.
Here are the details:
• If $\,x\ge 0\,$, then $\,|x| = x\,$.
In this case, $\,y = \pm \frac{b}{a}|x|\,$ becomes $\,y = \pm \frac{b}{a}x\,$.
• If $\,x < 0\,$, then $\,|x| = -x\,$.
In this case, $\,y = \pm \frac{b}{a}|x|\,$ becomes $\,y = \pm \frac{b}{a}(-x) = \mp \frac{b}{a}x\,$.
• ‘$\displaystyle\,y = \pm \frac{b}{a}x\,$’ is a shorthand for ‘$\,y = \frac bax \ \text{ or } \ y = -\frac bax\,$’.
‘$\displaystyle\,y = \mp \frac{b}{a}x\,$’ is a shorthand for ‘$\,y = -\frac bax \ \text{ or }\ y = \frac bax\,$’.
• The sentence ‘$\,A \text{ or } B\,$’ is equivalent to ‘$\,B\text{ or } A\,$’.
• Thus, $\,y = \pm \frac{b}{a}|x|\,$ is equivalent to $\,y = \pm\frac bax\,$.
## Asymptotes for Hyperbolas
The lines $\displaystyle\,y = \pm\frac bax\,$ are asymptotes for the hyperbola $\displaystyle\,\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\,$. As $\,x\,$ gets big (positive or negative), the graph of $\displaystyle\,\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\,$ gets closer and closer to these lines, as shown at right. The asymptotes form an ‘envelope’ inside which the hyperbola lives. The envelope gets tighter and tighter as $\,x\,$ gets bigger and bigger! The green box is called the central box of the hyperbola, and is a useful graphing aid: its sides are parallel to the axes it passes through the vertices of the hyperbola it crosses the other axis at $\,\pm b\,$ The diagonals of the central box are the asymptotes for the hyperbola!
Similarly, the lines $\displaystyle\,y = \pm\frac abx\,$ are asymptotes for the hyperbola $\displaystyle\,\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\,$: Compare $\displaystyle\,\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\,$ with the previously-discussed equation $\displaystyle\,\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\,$: the variables $\,x\,$ and $\,y\,$ have just been switched. Thus, the exact same derivation as before (with variables switched) yields the asymptotes $\displaystyle\,x = \pm \frac ba y\,$. Solve for $\,y\,$ to get the asymptotes $\displaystyle\,y = \pm\frac ab x\,$. Alternately, solve for $\,y\,$ in $\displaystyle\,\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\,$ to get:$\displaystyle\,y = \pm \frac ab\sqrt{\strut x^2\bigl(1 + \frac{b^2}{x^2}\bigr)}$ For large $\,x\,$, the graph looks like: $\displaystyle\,y = \pm \frac ab x\,$
## Getting the Central Box Graphically
If you have a hyperbola graphed with the foci already located, then it's easy to get the central box graphically: Make sure that ‘$\,1\,$’ on $x$-axis is the same as ‘$\,1\,$’ on the $y$-axis (so that a circle actually looks like a circle). Sweep out (part of) the circle with center at the origin and radius $\,c\,$,as shown at right. Mark where the circle intersects the vertical line through the vertex. The green triangle has hypotenuse $\,c\,$ and bottom leg $\,a\,$. Since $\,c^2 = a^2 + b^2\,$, the remaining leg must have length $\,b\,$. Of course, you can use this information in a different way: if you already have $\,a\,$ and $\,b\,$ marked, then just rotate the hypotenuse to locate the focus!
## Summary: Graphing Hyperbolas
$$\color{green}{\frac{x^2}{a^2}} - \color{red}{\frac{y^2}{b^2}} = 1$$ Vertices: Set $\,y = 0\,$, solve for $\,x\,$, to get $\,x = \pm a\,$. The positive term ($\displaystyle\color{green}{\frac{x^2}{a^2}}\,$) determines the vertices! $$\color{green}{\frac{y^2}{a^2}} - \color{red}{\frac{x^2}{b^2}} = 1$$ Vertices: Set $\,x = 0\,$, solve for $\,y\,$, to get $\,y = \pm a\,$. The positive term ($\displaystyle\color{green}{\frac{y^2}{a^2}}\,$) determines the vertices! Use the other term to find $\,b\,$; draw in the central box. The diagonals of the central box are the asymptotes for the hyperbola.
## Steps to Graph a Hyperbola:
• [Vertices] Find the vertices.
• [Central Box] Use the other term to find $\,b\,$; draw in the central box. (See the examples below.)
• [Diagonals] The diagonals of the central box are the asymptotes of the hyperbola.
• [Draw Hyperbola] Draw in the hyperbola, using the asymptotes as an ‘envelope’.
• [Foci] If needed, use $\,c^2 = a^2 + b^2\,$ to locate the foci.
## Examples: Graphing Hyperbolas
In both examples below, there are only $\,x^2\,$, $\,y^2\,$, and constant terms.
When the variable terms are on the same side, they have different signs.
So, we know we're dealing with hyperbolas!
## Example #1: Put the Equation in Standard Form
Graph: $\,9x^2 - 4y^2 = 36\,$
Find the vertices, central box, asymptotes, and foci.
When all the numbers ‘work out nicely’, it may be easiest to put the equation in standard form.
Here, both $\,9\,$ and $\,4\,$ go into $\,36\,$ evenly (resulting in perfect square denominators).
Put the equation in standard form: \displaystyle \begin{alignat}{2} &\frac{9x^2}{36} - \frac{4y^2}{36} = \frac{36}{36}&\qquad&\text{(get a\,1\,on the right-hand side)}\cr\cr &\frac{x^2}{4} - \color{red}{\frac{y^2}{9}} = 1&&\text{(simplify)}\cr \end{alignat} (This is an equation of the form $\displaystyle\,\frac{x^2}{a^2} - \color{red}{\frac{y^2}{b^2}} = 1\,$.) Find the vertices: Set $\,y = 0\,$: $\displaystyle \frac{x^2}{4} = 1\,$, $\displaystyle x^2 = 4\,$, $\displaystyle x = \pm 2$ Use the other term to find $\,b\,$. Draw the central box, its diagonals, and the hyperbola: Compare $\displaystyle\,\frac{y^2}{b^2}\,$ with $\displaystyle\,\frac{y^2}{9}\,$ to see that $\,b = 3\,$. Mark $\,\pm 3\,$ on the $y$-axis. Sketch in the central box and its diagonals (which are the asymptotes). Sketch the hyperbola inside the ‘envelope’ formed by the diagonals. Foci: We have $\,a = 2\,$ and $\,b = 3\,$: $c^2 = a^2 + b^2 = 4 + 9 = 13\,$, $\,c = \sqrt{13} \approx 3.6\,$ Report results: vertices: $\,(\pm 2,0)\,$ central box: has corners $\,(\pm 2,\pm 3)\,$ asymptotes: $\,y = \pm \frac 32x\,$ foci: $\,(\pm\sqrt{13},0)\,$
You can also head up to WolframAlpha and type in: graph 9x^2 - 4y^2 = 36
You'll get everything: graph, vertices, foci, asymptotes, and more!
## Example #2: Don't Put the Equation (Completely) in Standard Form
If the numbers don't work out nicely, then you don't have to write it completely in standard form.
As long as you get the variable terms on the left, and the ‘$\,1\,$’ on the right, you're good to go!
Graph: $\,9y^2 = 7 + 14x^2\,$
Find the vertices, central box, asymptotes, and foci.
Get the variable terms on the left, ‘$\,1\,$’ on the right: \displaystyle \begin{alignat}{2} &9y^2 - 14x^2 = 7&\qquad&\text{variable terms on left, constant on right}\cr\cr &\frac{9y^2}{7} - \frac{14x^2}{7} = \frac{7}{7}&\qquad&\text{(get a\,1\,on the right-hand side)}\cr\cr &\frac{9y^2}{7} - \color{red}{2x^2} = 1 \end{alignat} (This is an equation of the form $\displaystyle\,\frac{y^2}{a^2} - \color{red}{\frac{x^2}{b^2}} = 1\,$.) To go from here to ‘complete’ standard form would be a bit ugly, resulting in: $\displaystyle\frac{y^2}{7/9} - \frac{x^2}{1/2} = 1$ As you'll see, this isn't needed! Find the vertices: Set $\,x = 0\,$: $\displaystyle \frac{9y^2}{7} = 1\,$, $\displaystyle y^2 = \frac{7}{9}\,$, $\displaystyle y = \pm \sqrt{\frac79} = \pm\frac{\sqrt 7}3 \approx \pm 0.88$ So: $\displaystyle \,a = \frac{\sqrt 7}{3}$ Use the other term to find $\,b\,$. Draw the central box, its diagonals, and the hyperbola: Compare $\displaystyle\,\frac{x^2}{b^2}\,$ with $\,2x^2\,$ to find $\,b\,$: \displaystyle \begin{alignat}{2} &\frac{x^2}{b^2} = 2x^2&\qquad&\text{(compare the term to the desired form)}\cr\cr &\frac{1}{b^2} = 2&&\text{(cancel\,x^2\,$)}\cr\cr &b^2 = \frac 12&&\text{(solve for$\,b^2\,)}\cr\cr &b = \pm \frac{1}{\sqrt 2} \approx \pm 0.71 \end{alignat} Mark $\displaystyle\,\pm \frac{1}{\sqrt 2}\,$ on the $x$-axis. Sketch in the central box and its diagonals (which are the asymptotes). Sketch the hyperbola inside the ‘envelope’ formed by the diagonals. Foci: We have $\displaystyle\,a = \frac{\sqrt 7}{3}\,$ and $\displaystyle\,b = \frac{1}{\sqrt 2}\,$: $\displaystyle c^2 = a^2 + b^2 = \bigl(\frac{\sqrt 7}{3}\bigr)^2 + \bigl(\frac{1}{\sqrt 2}\bigr)^2 = \frac 79 + \frac 12 = \frac{14}{18} + \frac{9}{18} = \frac{23}{18}\,$ $\displaystyle c = \frac{\sqrt{23}}{3\sqrt{2}} \approx 1.13$ Report results: Note that: $\displaystyle \frac ab \ \ =\ \ \frac{\frac{\sqrt 7}{3}}{\frac{1}{\sqrt 2}} \ \ =\ \ \frac{\sqrt 7}{3}\cdot \frac{\sqrt 2}{1} \ \ =\ \ \frac{\sqrt{14}}{3}$ vertices: $\displaystyle\,\left(0 \ ,\ \pm \frac{\sqrt{7}}3\right)\,$ central box: has corners $\displaystyle\,\left(\pm \frac{1}{\sqrt{2}} \ ,\ \pm \frac{\sqrt{7}}{3}\right)\,$ asymptotes: $\displaystyle\,y = \pm \frac{\sqrt{14}}{3}x\,$ foci: $\displaystyle\,\left(0 \ ,\ \pm\frac{\sqrt{23}}{3\sqrt 2}\right)\,$
Master the ideas from this section |
## Precalculus (6th Edition) Blitzer
The arc length is $\frac{20\pi }{5}\text{ inches or }20.94\,\text{inches}$
We know that the formula which connects the arc length $s$, the angle intercepted by the arc to the center of the circle $\theta$, and the radius of the circle or arc $r$ is: $s=r\theta$ Where $\theta$ is expressed in radians. To convert the angle mentioned in degrees to radians, multiply the angle in degree with $\frac{\pi }{180}.$ In the provided problem, r is 8 inches and the angle that is intercepted by the arc is 150 degrees. When converting the angle in degree to radians: \begin{align} & \theta =150{}^\circ \times \frac{\pi }{180}\text{ radians} \\ & =\frac{5\pi }{6}\text{ radians} \end{align} So, the length of the arc is: \begin{align} & s=r\theta \\ & =8\times \frac{5\pi }{6} \\ & =\frac{20\pi }{3}\text{ inches}\text{.} \end{align} Round off the length value to two decimal places; $\frac{20\pi }{3}\text{ inches}\approx \text{20}\text{.94 inches}\text{.}$ |
# MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3
In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 14 Statistics Ex 14.3 Pdf, These solutions are solved subject experts from the latest edition books.
## MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3
Percentage Decrease Calculator is a free online tool that displays the percentage decrease for the given amount.
Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Solution:
Median:
Let us prepare a cumulative frequency table:
Now, we have N = 68 ⇒ $$\frac{N}{2}=\frac{68}{2}$$ = 34
The cumulative frequency just greater than 34 is 42 and it corresponds to the class 125 – 145.
∴ 125 – 145 is the median class.
∴ l = 125, cf = 22, f= 20 and h = 20
Using the formula,
Median = l + $$\left[\frac{\frac{N}{2}-c f}{f}\right]$$ × h
= 125 + $$\left[\frac{34-22}{20}\right]$$ × 20
= 125 + $$\frac{12}{20}$$ × 20 = 125 + 12 = 137 units.
Mean: Let assumed mean, a = 135
∵ Class size, h = 20
∴ ui = $$\frac{x_{i}-a}{h}=\frac{x_{i}-135}{20}$$
Now, we have the following table:
∴ $$\overline{x}$$ = a + h × [$$\frac{1}{N}$$ Σfiui] = 135 + 20 × $$\frac{7}{68}$$
= 135 + 2.05 = 137.05 units.
Mode:
∵ Class 125 – 145 has the highest frequency i.e., 20.
∴ 125 – 145 is the modal class.
We have: h = 20, l = 125 , f1 = 20, f0 = 13, f2 = 14
We observe that the three measures are approximately equal.
Use this median calculator to quickly find the median of a set of numbers. Just enter your numbers in the box and click on the button that says calculate.
Learn how to find Recursive Sequences Formula Calculator for arithmetic sequences.
Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.
Solution:
Here, we have N = 60
Now, cumulative frequency table is:
Since, median = 28.5 (Given)
∴ Median class is 20 – 30 and l = 20, f = 20, cf = 5 + x, N = 60
∴ l + $$\left[\frac{\frac{N}{2}-c f}{f}\right]$$ × h
⇒ 28.5 = 20 + $$\left[\frac{30-(5+x)}{20}\right]$$ × 10
⇒ 28.5 = 20 + $$\frac{25-x}{2}$$
⇒ 57 = 40 + 25 – x
⇒ x = 40 + 25 – 57 = 8
Also, 45 + x + y = 60
⇒ 45 + 8 + y = 60
⇒ y = 60 – 45 – 8 = 7.
Thus x = 8, y = 7
The percentage difference calculator is here to help you compare two numbers.
Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
Solution:
The given table is cumulative frequency distribution. We write the frequency distribution as given below :
∵ The cumulative frequency just greater than 50 is 78.
∴ The median class is 35 – 40.
Now, $$\frac{N}{2}$$ = 50, l = 35, cf = 45, f = 33 and h = 5
Thus, the median age = 35.76 years.
Also you can check the age difference between your loved ones, friends using the Age difference Calculator.
When the positive number a is rounded to the nearest tenth, the result is the number b.
Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
Find the median length of the leaves.
[Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5-126.5,
126.5 – 135.5 ………… 171.5 – 180.5.]
Solution:
After changing the given table as continuous classes we prepare the cumulative frequency table as follows:
The cumulative frequency just above 20 is 29 and it corresponds to the class 144.5 – 153.5.
So, 144.5 – 153.5 is the median class.
We have: $$\frac{N}{2}$$ = 20, l = 144.5, f= 12, cf = 17 and h = 9
∴ Median = l + $$\left[\frac{\frac{N}{2}-c f}{f}\right]$$ × h
= 144.5 + $$\left[\frac{20-17}{12}\right]$$ × 9
= 144.5 + $$\frac{3}{12}$$ × 9 = 144.5 + $$\frac{9}{4}$$
= 144.5 + 2.25 = 146.75
Median length of leaves = 146.75 mm.
The percentage off calculator generates lightning-fast results without asking the users to follow intricate procedures.
Question 5.
The following table gives the distribution of the life time of 400 neon lamps:
Find the median life time of a lamp.
Solution:
To compute the median, let us write the cumulative frequency distribution as given below:
Since, the cumulative frequency just greater than 200 is 216.
∴ The median class is 3000-3500 and so l = 3000, cf= 130, f = 86, h = 500
∴ Median = l + $$\left[\frac{\frac{N}{2}-c f}{f}\right]$$ × h
= 3000 + $$\left[\frac{200-130}{86}\right]$$ × 500
= 3000 + $$\frac{70}{86}$$ × 500 = 3000 + $$\frac{35000}{86}$$
= 3000 + 406.98 = 3406.98
Thus, median life time of a lamp = 3406.98 hours.
Online frequency distribution calculator tool makes the calculation faster and it displays the frequency distribution in a fraction of seconds.
Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Solution:
Median: The cumulative frequency distribution table is as follows:
Since, the cumulative frequency just greater than 50 is 76.
∴ The class 7-10 is the median class.
We have, $$\frac{N}{2}$$ = 50 , f = 7, cf = 36, f = 40 and h = 3
Mode:
Since the class 7 – 10 has the maximum frequency i.e., 40.
∴ The modal class is 7 – 10.
So, we have l = 7,h = 3, f1 = 40, f0 = 30, f2 = 16
Thus, the required median = 8.05, mean = 8.32 and mode = 7.88.
Guidelines for conversion of CGPA into percentage.
Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Solution:
We have cumulative frequency table as follows:
The cumulative frequency just greater than 15 is 19, which corresponds to the class 55 – 60.
So, median class is 55-60 and we have $$\frac{N}{2}$$ = 15,
l = 55, f = 6, cf = 13 and h = 5
Thus, the required median weight of the students = 56.67 kg. |
# Video: Understanding the Commutative Property of Multiplication in Word Problems
Kathryn Kingham
Halle noticed that for the shown rectangle, she could either multiply 6 by 7 or 7 by 6 to get its area, which is 42 square inches. Identify the property which allows her to do this.
01:38
### Video Transcript
Halle noticed that for the shown rectangle, she could either multiply six by seven or seven by six to find the area, which is forty-two inches squared. Identify the property which allows her to do that.
We are looking for a property. First, we know that it’s a property that deals with multiplication. Property of multiplication is the identity property of multiplication; this is the property that says if you multiply a number by one, the result is the original number.
The identity property will not help us do what Halle did here. It would be the associative property of multiplication. The associative property looks like this and it deals with multiplying three values and how we group them. The associative property does not apply here because we’re only working with two values.
The commutative property of multiplication is the property that allows us to multiply six by seven or seven by six to find that the area of this rectangle is forty-two inches squared. The commutative property states that two numbers can be multiplied in either order; in our case, six times seven equals seven times six. |
# Factors of 954: Prime Factorization, Methods, and Examples
The factors of 954 are numbers that are divisible by 954 without leaving a remainder behind. The number 954 is an even composite as it has twelve factors in total.
Let us explore more about the factors of 954.
### Factors of 954
Here are the factors of number 954.
Factors of 954: 1, 2, 3, 6, 9, 18, 53, 106, 159, 318, 477, and 954
### Negative Factors of 954
The negative factors of 954 are similar to their positive aspects, just with a negative sign.
Negative Factors of 954: –1, -2, -3, -6, -9, -18, -53, -106, -159, -318, -477, and -954
### Prime Factorization of 954
The prime factorization of 954 is the way of expressing its prime factors in the product form.
Prime Factorization: 2 x 3 x 3 x 53
In this article, we will learn about the factors of 954 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 954?
The factors of 954 are 1, 2, 3, 6, 9, 18, 53, 106, 159, 318, 477, and 954. These numbers are the factors as they do not leave any remainder when divided by 954.
The factors of 954 are classified as prime numbers and composite numbers. The prime factors of the number 954 can be determined using the prime factorization technique.
## How To Find the Factors of 954?
You can find the factors of 954 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 954, create a list containing the numbers that are exactly divisible by 954 with zero remainders. One important thing to note is that 1 and 954 are 954’s factors, as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 954 are determined as follows:
$\dfrac{954}{1} = 954$
$\dfrac{954}{2} = 477$
$\dfrac{954}{3} = 318$
$\dfrac{954}{6} = 159$
$\dfrac{954}{9} = 106$
$\dfrac{954}{18} = 53$
Therefore, 1, 2, 3, 6, 9, 18, 53, 106, 159, 318, 477, and 954 are the factors of 954.
### Total Number of Factors of 954
For 954, there are twelve positive factors and twelve negative ones. So in total, there are twenty-four factors of 954.
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of 954 is given as follows:
The factorization of 954 is 1 x 2 x 3$^2$ x 53.
The exponent of 1, 2, and 53 is 1. The exponent of 3 is 2.
Adding 1 to each and multiplying them together results in twenty-four.
Therefore, the total number of factors of 954 is 24. Twelve are positive, and twelve factors are negative.
### Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor, the smallest prime factor.
## Factors of 954 by Prime Factorization
The number 954 is a composite. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 954 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 954, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 954 can be expressed as:
954 = 1 x 2 x 3$^2$ x 53.
## Factors of 954 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.
For 954, the factor pairs can be found as follows:
1 x 954 = 954
2 x 477 = 954
3 x 318 = 954
6 x 159 = 954
9 x 106 = 954
18 x 53 = 954
The possible factor pairs of 954 are given as (1, 954), (2, 477), (3, 318), (6, 159), (9, 106), and (18, 53).
All these numbers in pairs, when multiplied, give 954 as the product.
The negative factor pairs of 954 are given as:
-1 x -954 = 954
-2 x -477 = 954
-3 x -318 = 954
-6 x -159 = 954
-9 x -106 = 954
-18 x -53 = 954
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, 1, -2, -3, -6, -9, -18, -53, -106, -159, -318, -477, and -954 are called negative factors of 954.
The list of all the factors of 954, including positive as well as negative numbers, is given below.
Factor list of 954: 1, -1, 2, -2, 3, -3, 6, -6, 9, -9, 18, -18, 53, -53, 106, -106, 159, -159, 318, -318, 477, -477, 954, and -954
## Factors of 954 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 954 are there?
### Solution
The total number of Factors of 954 is twelve.
Factors of 954 are 1, 2, 3, 6, 9, 18, 53, 106, 159, 318, 477, and 954.
### Example 2
Find the factors of 954 using prime factorization.
### Solution
The prime factorization of 954 is given as:
954 $\div$ 2 = 477
477 $\div$ 3 = 159
159 $\div$ 3 = 53
53 $\div$ 53 = 1
So the prime factorization of 954 can be written as:
2 x 3 x 3 x 53 = 954 |
# Fraction calculator
This calculator adds two fractions. When fractions have the same denominators calculator simply adds the numerators and place the result over the common denominator. Then simplify the result to the lowest terms or a mixed number.
## The result:
### 5/8 + 2/8 = 7/8 = 0.875
Spelled result in words is seven eighths.
### How do we solve fractions step by step?
1. Add: 5/8 + 2/8 = 5 + 2/8 = 7/8
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(8, 8) = 8. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 8 × 8 = 64. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - five eighths plus two eighths is seven eighths.
#### Rules for expressions with fractions:
Fractions - use a forward slash to divide 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 numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right. |
# how to write decimals in standard form
Step 2: Replace the decimal point with Welcome to Decimal Expanded Form, Word Form, and Standard Form with Mr. J! Step #3.
The way you wrote it is the standard notation. Word Form With Decimals Worksheets. Source: www.youtube.com. Solution: Place the decimal point 6.78120009; Count the digits after the decimal place. The standard form of a decimal number in Britain is known as Scientific notation, where the number is written in the following way: 4527.7 = 4.5277 x 10. 24.
- whole-number part -.
If the number was given in the expanded form, perform the addition to If it is a whole number, add the decimal point after the last digit. Therefore, every number written in standard form will start by moving the decimal place so that the base remains between 1 and 10.
Then take the number of spaces you counted between the 2 decimal points and write it as an exponent. The definition for the standard form of decimal numbers is the same as that of whole numbers, Any number that we can write as a decimal number, between 1.0 and 10.0, multiplied by a power of 10, is said to be in standard form.
Worksheets are Decimals work, Decimal work, Decimals practice booklet table of contents, Writing and reading decimals, Standard form 1, Writing decimals in word form and expanded form work, Decimals work converting expanded form decimals, Standard and expanded form t1l1s1. First write 10.
Below, the steps are explained through an example.
Example 1: Write each mixed number as a decimal. Hence, the standard form of 42300000000 is 4.23 10 10. In this example, 4527.7 can be written as 4.5277 10 in scientific notation. Read the digits to the right of the decimal point as a whole number. Expert-verified answer question. Word Form Worksheets 4th Grade Expanded Form Math Worksheets Decimals.
X 10 4.
If the number was given in words, convert these to numbers. After converting to standard form the answer will be written as 4.22.
Note where the decimal point in the number is.
First, read the digits to the left of the decimal point as a whole number. Write 10 raised to the power of the spaces between the decimal points.
1 10.
Need help with putting decimals in expanded form, word form, or standard form? Given : 8.234 E14. Writing Decimals in Standard and Decomposed Form. = (12346 / 10000) 10000.
Here are the steps to follow when writing decimals in words.
It will become 9.
= (5 x 10) + ( 7 x 1) + (4 x ) + (9 x ) = 57.49. Standard Form >. Thus the standard form of a number is obtained as follows: Step 1: The first number is 4. To read and write decimals, use the following steps: Step #1. Learn to write the expanded form of decimal numbers in an easy way with 2 methods explained thoroughly.
Final Words: According to the mathematicians, standard form is indicated as the system of writing numbers which can be useful when working with large and small numbers. Slope intercept form x intercept.
22. How to write in standard form?
= 1.2346 10 4 {Here only one number will be kept before decimal point that is 1} Therefore the standard form of In this lesson, you will learn how to read and write decimals. Say "and" for the decimal point. With our free printable worksheets on writing decimals in standard form, converting expanded form of decimal numbers into standard form is no longer a tall order. So, a number such as 150 is simply written as 150 in standard notation. While you typically use numbers to write decimals in mathematics, there might be situations when it's more appropriate to use words. If you moved it to the right, append "x 10 -n ", using the same logic. To change 500 to scientific notation, move the decimal (understood to be at the end of the number) two spaces to the left so that only the 5 is to the left of the decimal. How to write Word 2007 in Japanese; Solving for x In Standard Form of a Linear Equation; Vertex Form to Standard Form; Promo Writing Word End - Effect for Crazytalk Animator; How to write a bog standard essay; WDSF Word Open Standard | Final | Tilburg Moves 2013; 30-0 standoff ffa decimation; 301A1 Point Slope Form Writing Decimals In Standard Form Without any approximations they contain fractions within any number!
By definition, any number that can be written as the decimal number, between the numbers 1.01.0 and 10.010.0, and then multiplied by the power of the number 1010, is known to be in the standard form. 4942 views |.
In tutorial 3.5 we discussed decimals as an alternative form to fractions for expressing the non-whole numbers. A way of writing small and large numbers with one number before the decimal point and multiplied by a power of 10.
To Find : Write in Standard form. Worksheets are Writing and reading decimals, Decimals hundredths, Expanded form with decimals a, Expanded form with decimals a, Reading and writing decimals, Standard and expanded form t1l2s1, Writing scientific notation, Decimals work. Step 2: Identify the decimal point in the number. If you moved the decimal point to the left, append "x 10 n " to the number, where n is the number of positions you moved the point.
For .034002, write 3.4002.
Standard form format: The general format in which we write numbers to represent them in standard form is: a b 10 a times b^{10} a b 1 0. Here, a is some number that is greater than or equal to 1 and is smaller than 10. While b is the number that is the power of 10. we will get the original number if we multiply a with the 0.005463 = 5.463 x 10^-3 Standard form, or standard index form, is a system of writing numbers which can be particularly useful for working with very large or very small numbers.It is based on using powers.
If by "standard form" you mean what people in the US call "scientific notation," it's exactly like whole numbers except the exponents are negative. How to convert a decimal number into standard form. Decimal Form.
The standard form of a number is a way of writing the number in a form that follows certain rules.
Convert to standard form: It can be written as 11.49 X 10 12.
You may use our Place Value and Decimals Chart (PDF) as a visual reference for the examples presented in this lesson.
Click on the image to view or download the image. In this video, I talk about how to use the word form and expanded form of decimal numbers to write them in standard form. It might be helpful to understand how to pronounce and discuss these numbers when writing or speaking about a decimal.
Free quadratic equation calculator - Solve quadratic equations using factoring, complete the square and the quadratic formula step-by-step
Being savvier about the place value of decimals is the key here. To determine the expanded form with decimals, simply write the digits in the spaces at the bottom of the chart, then read off the resulting combination.
This online calculator can find the distance between a given line and a given point. Standard notation means to write the number in its standard form. Decimals can be represented in standard form or scientific form as: I m a g e w i l l b e U p l o a d e d S o o n. Decimals can be written in expanded form and further in exponential form: Let us consider the number 256.5.
Step #2. Write this number in standard form | The number MUST be between 1-10. Displaying all worksheets related to - Writing Decimals In Standard Form. 42300000000 is the number.
7i(5 +2i) 7 i ( 5 + 2 i)(15i)(9+2i) ( 1 5 i) ( 9 + 2 i)(4+i)(2+3i) ( 4 + i) ( 2 + 3 i)(18i)(1 +8i) ( 1 8 i) ( 1 + 8 i) How do you write 8.234 E14 in standard form?
Step 1: Write down the number. These Free Word Form With Decimals Worksheets exercises will have your kids engaged and entertained while they improve their skills.
Place Value >. Watch more videos on decimals. Step #4. Refer to a place value chart so you understand the value of decimals - from tenths to hundredths to all the Step 3: After identifying the decimal point, move the decimal to the first non-zero digit in the number. How do you put an equation in point slope form into standard or slope intercept form virtual nerd.
(Week 4, Day 4 of e-learning)
Divide 5.23 X 108by 1.24 X 10 4. Standard Form of a decimal. So put the decimal point after the first number | The power of 10 reflects the number of places the decimal point has moved | . 1 million.
For Example 290000 is a number which is written in standard form as 2.9 X 10 5.Use an online standard form calculator that helps to convert the number in standard form and also measure the Scientific E-Notation, Engineering notation, and Real numbers of the given 3. Example: Convert 678120009 into standard form.
[Writing Decimals In Standard Form] - 17 images - grade 6 math worksheets pdf sixth grade math worksheets with answers, converting decimals from written to standard numerical form youtube, writing decimals in standard form jack frost, writing decimals in standard form jack frost, Step 1: Write the number out as a series of digits. Decimals. The same applies to decimals. To use this standard notation calculator, follow the below steps:Enter the number in the given input box.Press the Calculate button to see the result.You can reset the values by using the Reset For Example.
A Number In Scientific Notation. 0.0009. The number in the expanded form will be: 256.5 = (2 x 100) + (5 x 10) + (6 x 1) + (5 x (. Step 1: Read the number to the left of the decimal point and write it in word form. For example, the number 5410000 in Step 3: The number of digits after 4 is 10. Step 2: Adding the decimal point after 4, it becomes 4.. You need a common denominator. 100 is divisible by both 10 and 100, so multiply both the numerator and denominator of this character by 10. You get 10 on the top and 100 on the bottom. 1/10 is the same thing as 10 over 100. 10/100 plus 5/100 is equal to 15 over 100, so this piece right here is equal to 15/100. It is written as ax+by=c. To find, the standard form of the number 12346 = ?
To convert a number to the standard form, it is important to understand the process properly in a stepwise manner. You can see the decimal point is lying after 4 digits from the left side.
The scoop about the standard form of
This step is the same as when you convert a large number to standard form. 7.1365 x 10 ^2 When to write a decimal in word form. How To Write Decimals In Expanded Form - 9 images - grade 6 place value worksheet numbers in expanded form, expanded form,
Because, 4527.7 = 4.5277 1000 = 4.5277 10.
Displaying all worksheets related to - Write Decimal In Standard Form. Steps to Write Decimals in Word Form. Example 2: Write 2550 in standard form. - fractional part -. By Multiplying and dividing 12346 by 10000, we get. For example, let us write the number 12,340,000 in normalized standard form.
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Proof: By Induction
Because the addition of natural numbers "$$+$$" is commutative, it is without loss of generality sufficient to show the right cancellation property, i.e. $x+z=y+z\Rightarrow x=y,~~~~~~(x,y,z\in\mathbb N).$ The proposition can be proven by induction.
Base case Base case $$z=0$$.
For arbitrary $$x,y\in\mathbb N$$, it follows from the definition of addition that $x+0=y+0\Rightarrow x=y.$
Note that $$x+0=x$$ and $$y+0=y$$ are both equalities being equivalence relations. Thus, we can replace the implication sign "$$\Rightarrow$$" by the equivalence sign "$$\Leftrightarrow$$", and the reasoning is still logically correct:
$x+0=y+0\Leftrightarrow x=y.$
Induction step $$z\to z^+:= z+1$$.
Now, let assume that the inclusion $$x+a_0=y+a_0\Leftrightarrow x=y$$ has been proven for all $$z_0\le z$$, where we use "$$\le$$" as the order relation of natural numbers. Then it follows again from the definition of addition that $x+z^+=y+z^+\Leftrightarrow (x+z)^+=(y+z)^+.$ Because both successors are unique and because $$x+z=y+z$$ is equivalent to $$x=y$$ by assumption, it follows that $(x+z)^+=(y+z)^+\Leftrightarrow x=y.$ Altogether, it follows from $$x + z=y + z$$ that $$x=y$$ for all natural numbers $$x,y,z$$. Thus, the addition of natural numbers is cancellative. $x+z=y+z \Rightarrow x=y,$
and we have proven the conversion of the cancellative law:
$x=y\Rightarrow x+z=y+z$
for all natural numbers $$x,y,z$$.
Thank you to the contributors under CC BY-SA 4.0!
Github:
References
Bibliography
1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013 |
# Find the equation of a line drawn perpendicular to the line $\large\frac{x}{4}$$+\large\frac{y}{6}$$=1$ through the point, where it meets the $y$ - axis
$\begin {array} {1 1} (A)\;2x-3y+18=0 & \quad (B)\;3y-2x+18=0 \\ (C)\;2x-3y-18=0 & \quad (D)\;3y-2x-18=0 \end {array}$
Toolbox:
• If two lines are perpendicular, then the product of their slopes is -1. i.e., $m_1m_2=-1$
• Equation of the line with slope $m$ and passing through $(x_1, y_1)$ is $y-y_1=m(x-x_1)$
Given equation of the line is $\large\frac{x}{4}+\large\frac{y}{6}$$=1 This can be written as 3x+2y-12=0 This can be written in the form of y=mx+c (i.e.,) y= \bigg( \large\frac{-3}{2} \bigg)$$x+6$
Hence the slope of the given line $= \large\frac{-3}{2}$
Slope of line perpendicular of the given line
$= \large\frac{-1}{-\bigg(\Large\frac{3}{2}\bigg)}$$= \large\frac{2}{3} It is given that the line intersect the y - axis. Let this point be (0, y) On substituting x with 0 in the equation of the given line, we obtain \large\frac{y}{6}$$=1 \Rightarrow y=6$
Now the equation of the line whose slope is $\large\frac{2}{3}$ and point (0,6) is
$y-6=\large\frac{2}{3} (x-0)$
$\Rightarrow 3y-18=2x$
$\Rightarrow 2x-3y+18=0$
Hence the required equation is $2x-3y+18=0$ |
# Go Math Grade 3 Answer Key Chapter 5 Use Multiplication Facts Extra Practice
Go Math Grade 3 Answer Key Chapter 5 Use Multiplication Facts Extra Practice gives your preparation a head start. Thus, students who wish to prepare different questions of Chapter 5 Extra Practice can refer to HMH Go Math Grade 3 Answer Key Chapter 5 Use Multiplication Facts Extra Practice. Resolve all your doubts on the concepts by checking the step by step solutions provided for the 3rd Grade Go Math Answer Key Ch 5 Use Multiplication Facts Extra Practice.
## Go Math Grade 3 Answer Key Chapter 5 Use Multiplication Facts Extra Practice
Learn all basics regarding Multiplication taking the help of the Go Math Answer Key Chapter 5 Extra Practice. You will have the basics of multiplication using the line plot, number line, and graphs. Get acquainted with tips and tricks to solve various problems on Multiplication easily by referring to Examples over here. You will achieve better grades after practicing 3rd Grade Go Math Answer Key Ch 5 Use Multiplication Facts Extra Practice only a daily basis.
### Common Core – Page No. 101000
Lesson 5.1
Describe a pattern for the table. Then complete the table.
Question 1.
Teams 2 3 4 5 6 Players 12 18 24 _________ _________
Teams 2 3 4 5 6 Players 12 18 24 30 36
Explanation:
Multiply 6 with a number of tables.
Multiply 6 with 5 teams = 6 × 5 = 30
Multiply 6 with 6 teams = 6 × 6 = 36
Question 2.
Tables 4 5 6 7 8 Chairs 16 20 _________ 28 _________
Tables 4 5 6 7 8 Chairs 16 20 24 28 32
Explanation:
Multiply 4 with number of tables.
Number of chairs for 6 tables = x
Number of chairs for 8 tables = y
Now multiply number of tables with 4 = 6 × 4 = 24
And then multiply 8 tables with 4 = 8 × 4 = 32
Therefore the missing numbers in the table are 24 and 32
Lesson 5.2
Find the unknown factor.
Question 3.
72 = 9 × t
t = _______
Explanation:
t × 9 = 72
t = 72/9 = 8
The unknown factor t is 8.
Question 4.
4 × ★ = 28
★ = _______
Explanation:
4 × ★ = 28
★ = 28/4 = 7
★ = 7
Question 5.
b × 5 = 30
b = _______
Explanation:
b × 5 = 30
b = 30/5 = 6
Thus the unknown factor b is 6.
Question 6.
d × 3 = 24
d = _______
Explanation:
d × 3 = 24
d = 24/3 = 8
Therefore the unknown factor d is 8.
Question 7.
48 = 8 × p
p = _______
Explanation:
8 × p = 48
p = 48/8
p = 6
Thus the unknown factor p is 6.
Question 8.
6 × ▲ = 24
▲= _______
6 × ▲= 24
▲= 24/6
▲= 4
So the unknown factor▲is 4.
Question 9.
56 = 7 × ■
■ = _______
Explanation:
7 × ■ = 56
■ = 56/7
7 divides 56 eight times.
So the unknown factor ■ is 8.
Question 10.
2 × g = 20
g = _______
Explanation:
2 × g = 20
g = 20/2 = 10
Therefore the unknown factor g is 10.
Question 11.
h × 7 = 35
h = _______
Explanation:
h × 7 = 35
h = 35/7
h = 5
Thus the unknown factor h is 5.
Question 12.
9 = 9 × a
a = _______
Explanation:
9 × a = 9
a = 9/9
a = 1
So the unknown factor a is 1.
Question 13.
c × 4 = 36
c = _______
Explanation:
c × 4 = 36
c = 36/4
4 divides 36 nine times.
c = 9
Therefore the unknown factor is 9.
Question 14.
5 × y = 40
y = _______
Explanation:
5 × y = 40
y = 40/5
y = 8
Thus the unknown factor is 8.
### Common Core – Page No. 102000
Lesson 5.3
Solve.
Question 1.
Hailey plants 6 rows of marigolds. Each row has 20 marigolds. How many marigolds does Hailey plant in all?
_______ marigolds
Explanation:
Given that, Hailey plants 6 rows of marigolds.
Each row contains 20 marigolds.
Total number of marigolds that Hailey planted in all = x
x = 20 × 6 = 120
Therefore Hailey planted 120 marigolds.
Question 2.
A meeting room has 8 rows of chairs. Each row has 10 chairs. The first people to arrive fill 2 rows. How many chairs are not filled?
_______ chairs
Explanation:
Given, A meeting room has 8 rows of chairs.
Each row has 10 chairs.
Total number of chairs = 8 × 10 = 80 chairs
The first people to arrive fill 2 rows.
That means 2 × 10 = 20 chairs
Number of chairs that are not filled = total number of chairs – number of filled chairs
= 80 – 20 = 60 chairs.
Lesson 5.4
Question 3.
1. Use a number line to find the product.
4 × 30 = _______
Step 1:
Starts at 0.
Step 2:
Make a jump of 30s until you reach 120.
Step 3:
Count the number of jumps till you reach 120.
Number of jumps = 4
4 × 30 = 120
Use place value to find the product.
Question 4.
40 × 8 = _______ tens × 8
= _______ tens = _______
i. 4
ii. 32
iii. 320
Explanation:
4 × tens = 4 tens = 40
4 tens × 8 = 32 tens
32 tens = 32 × 10 = 320
Question 5.
5 × 60 = 5 × _______ tens
= _______ tens = _______
i. 6
ii. 30 tens
iii. 300
Explanation:
60 = 6 × tens = 6 tens
5 × 6 tens = 30 tens
30 tens = 30 × 10 = 300
Lesson 5.5
Find the product.
Question 6.
9 0
× 3
——
_______
Explanation:
First multiply 3 with ones = 3 × 0 = 0
Next multiply 3 with tens = 3 × 90 = 270
So the product of 90 and 3 is 270.
Question 7.
5 0
× 8
——
_______
Explanation:
First multiply 8 with ones = 8 × 0 = 0
Now multiply 8 with tens = 8 × 50 = 400
The product of 50 and 8 is 400.
Question 8.
7 0
× 9
——
_______
Explanation:
Multiply 9 with ones = 9 × 0 = 0
Multiply 9 with tens = 9 × 70 = 630
The product of 9 and 70 is 630.
Question 9.
8 0
× 7
——
_______
Explanation:
Multiply 7 with ones = 7 × 0 = 0
And then multiply 7 with tens = 7 × 80 = 560
Thus the product of 80 and 7 is 560.
Solve.
Question 10.
During the summer, Jayden volunteers at the library for 20 hours each week for 7 weeks. How many hours does Jayden volunteer in all?
_______ hours
Explanation:
During the summer, Jayden volunteers at the library for 20 hours each week for 7 weeks.
For each week he worked 20 hours
Number of hours he worked for 7 weeks = y
y = 7 × 20 = 140 hours
Therefore Jayden volunteers at the library for 20 hours.
Question 11.
Trisha teaches 8 different cooking classes. There are 20 students in each class. How many students in all are in Trisha’s cooking classes?
_______ students |
What Is 3/7 as a Decimal + Solution With Free Steps
The fraction 3/7 as a decimal is equal to 0.428.
An expression in mathematics that demonstrates how many parts a number can be divided into is known as a Fraction. Its constituents include a numerator and a denominator separated by a line. The Numerator is the number present above the line, whereas the Denominator is a number below the line.
Here, we will explain the Long Division method to solve a fraction.
Solution
To solve a fraction, we have to start by transforming it into division. Since components of the division include Dividend and Divisor, so the numerator of the fraction becomes dividend and the denominator becomes divisor. In the example to solve, we get 3 as a dividend and 7 as a divisor. This can be mathematically represented as:
Dividend = 3
Divisor = 7
Fraction of 3/7 means the division of 3 into 7 equal parts. When solving this fraction we get the magnitude of 1 part as the Quotient, which is known as the final result of division. However, if a fraction is not fully divided, we get some quantity left behind. This is known as Remainder.
Quotient = Dividend $\div$ Divisor = 3 $\div$7
The given fraction of 3/7 is solved using Long Division and the solution is presented below:
Figure 1
3/7 Long Division Method
Below is a step-by-step explanation to solve the given fraction. We have:
3 $\div$ 7
While solving a division sum or fraction, the first step is to find, whether it is a Proper or an Improper Fraction. In the given fraction, we have 3 as a dividend, which is smaller than 7, the divisor. So this is a proper fraction. Hence, we have a requirement of a Decimal Point to complete our calculations. We can do this by adding a zero to the right of our dividend. By doing this, we get 30, which will now be divided by 7.
30 $\div$ 7 $\approx$ 4
Where:
7 x 4 = 28
The remainder is 30 – 28 = 2, which is greater than zero. So, we again add a zero to its right but without any decimal point and make it 20. Further calculations are presented as:
20 $\div$ 7 $\approx$ 2
Where:
7 x 2 = 14
This time the remainder is 20 – 14 = 6. Again 6 is smaller than 7, so we make it 60 by inserting a zero to its right. Now, 60 is divided by 7.
60 $\div$ 7 $\approx$ 8
Where:
7 x 8 = 56
Now, the remainder is:
60 – 56 = 4
Again, there is a non-zero remainder produced. This shows that the fraction is partially divided and we get a Quotient of 0.428 with a Remainder equal to 4. We solve it up to more decimal places to get a more accurate answer.
Images/mathematical drawings are created with GeoGebra. |
Paul's Online Notes
Home / Calculus I / Applications of Integrals / Area Between Curves
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### Section 6.2 : Area Between Curves
6. Determine the area of the region bounded by $$x = {y^2} - y - 6$$ and $$x = 2y + 4$$.
Show All Steps Hide All Steps
Hint : It’s generally best to sketch the bounded region that we want to find the area of before starting the actual problem. Having the sketch of the graph will usually help with determining the right/left functions and the limits for the integral.
Start Solution
Let’s start off with getting a sketch of the region we want to find the area of.
We are assuming that, at this point, you are capable of graphing most of the basic functions that we’re dealing with in these problems and so we won’t be showing any of the graphing work here.
Here is a sketch of the bounded region we want to find the area of.
Note that we won’t include any portion of the region above the top intersection point or below the bottom intersection point. The region needs to be bounded by one of the given curves on each boundary. If we went past the top intersection point we would not have an upper bound on the region. Likewise, if we went past the bottom intersection point we would not have a lower bound on the region.
Show Step 2
It should be clear from the graph that the right function is $$x = 2y + 4$$ and the left function is $$x = {y^2} - y - 6$$.
Since we weren’t given any limits on $$y$$ in the problem statement we’ll need to get those. However, we should never just assume that our graph is accurate or that we will be able to read it accurately enough to guess the coordinates from the graph. This is especially true when the intersection points of the two curves (i.e. the limits on $$y$$ that we need) do not occur on an axis (as they don’t in this case).
So, to determine the intersection points correctly we’ll need to find them directly. The intersection points are where the two curves intersect and so all we need to do is set the two equations equal and solve. Doing this gives,
${y^2} - y - 6 = 2y + 4\hspace{0.25in} \to \hspace{0.25in}{y^2} - 3y - 10 = \left( {y - 5} \right)\left( {y + 2} \right) = 0\hspace{0.25in} \to \hspace{0.25in}y = - 2,\,\,\,\,y = 5$
Therefore the limits on $$y$$ are : $- 2 \le y \le 5$.
Note that you may well have found the intersection points in the first step to help with the graph if you were graphing by hand which is not a bad idea with faced with graphing this kind of region.
Show Step 3
At this point there isn’t much to do other than step up the integral and evaluate it.
We are assuming that you are comfortable with basic integration techniques so we’ll not be including any discussion of the actual integration process here and we will be skipping some of the intermediate steps.
The area is,
$A = \int_{{ - 2}}^{5}{{2y + 4 - \left( {{y^2} - y - 6} \right)\,dy}} = \int_{{ - 2}}^{5}{{10 + 3y - {y^2}\,dy}} = \left. {\left( {10y + \frac{3}{2}{y^2} - \frac{1}{3}{y^3}} \right)} \right|_{ - 2}^5 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{343}}{6}}}$ |
# How Do I Calculate A Tip?
Updated on August 8, 2009
## How much is the tip?
Ahhh. The end of a great dining experience draws to a close. The food was fine, the ambiance great, your partner never looked better, and the wine choice...spectacular.
"You're check sir." breaks the spell of the moment, and panic sets in. You are left with the task of figuring out the tip. Does this sound familiar?
Well panic no more. Here is a simple and quick technique for figuring out the tip amount for any bill, and you can do it without a calculator.
## Calculating a 10% Tip
I will first explain the process using simple numbers and then I will show you how to deal with more complex. For this example, you will be handed a bill for \$10.00, and since the service was marginal, you think 10% will be the suitable tip amount.
Follow along using Fig. 1 below. Find the decimal point in the original amount, in this case \$10.00. Move the decimal point one spot to the left. You are left with \$1.00, since the third place zero can be dropped. \$1.00 is the 10% tip you would leave.
To help remember which way to move the decimal, remember the saying "Less is left."
## Calculating a 15% and 20% Tip
Using the numbers from the previous example, you could also find a 15% and 20% tip amount .
Since 20% is twice as much as 10%, the 20% amount can be found by doubling \$1.00. So 20% of \$10.00 will be \$2.00.
15% is found exactly half-way between 10% and 20%. So what number falls exactly half-way between \$1.00 and \$2.00? A 15% tip on \$10.00 would be \$1.50.
## Calculating Another Tip
For this example, the check amount will be a more realistic amount. The process remains the same, and I will offer you tips to make it a little easier.
The check you are handed is for \$53.76. This time you are feeling generous and decide to leave a 20% tip.
Follow along using Fig. 2 below. Find the decimal point in the original amount, in this case \$53.76. Move the decimal point one spot to the left. You now have \$5.37, since the third place digit (6) can be dropped. \$5.37 is 10% of the total bill.
## Calculating a 20% Tip
You have now found that \$5.37 is the 10% tip amount, but you wish to leave 20%. Doubling the amount will give you \$10.74 as the 20% tip amount. You now know what to tip your server.
If \$5.37 is makes the calculator in your head overheat, here are a couple of ideas to make the math easier.
Round the numbers to something more manageable is the easiest fix. \$5.37 can be rounded to \$5.40, or \$5.50 if that will make the math easier to handle. Remember, we are not dealing with exact science, when it comes to leaving a tip. \$5.37 could even be rounded off to \$5.00 if it will help. If you round the number to \$5.00, you would be leaving \$10.00 instead of \$10.74.
Estimating is another easy method of reducing the math. If you know that 10% of \$50.00 is \$5.00, and 10% of \$60.00 is \$6.00, then 10% of \$53.76 must fall in between, and a little closer to \$5.00. Double the amount and you have 20%.
So enjoy your next meal and the next time the bill comes, don't stress...Impress...
## Other Information
These examples used a dinner check for the calculations, but this can be applied to discounts, sales tax and other situations when percentages are used to describe amounts.
12
0
69
## Question or Comment?
0 of 8192 characters used |
# Translation by a Vector
Lesson
Translation by a vector refers to the movement of one or more points of a space in a particular direction by a specified amount.
A vector is often represented geometrically as an arrow with a particular direction and length. We can slide the arrow to any location in the space, without changing its direction or length, and it is considered to be the same vector.
So, to translate a point in the space by a given vector, we place the tail of the arrow at the point. The translated point is then at the head of the arrow. This process is called vector addition. It is illustrated in the diagram below.
To accomplish the same result algebraically, we represent the points by their coordinates. The translation vector is also represented by its coordinates. These are the distances by which the points are to be moved in the directions of each of the coordinate axes.
For example, the coordinate pair $\left(a,b\right)$(a,b) represents a translation vector where points are to be moved $a$a units horizontally and $b$b units vertically. When applied to any point $\left(x,y\right)$(x,y), the result is the translated point $\left(x+a,y+b\right)$(x+a,y+b).
Vector addition to translate points is also valid in spaces of three (or more) dimensions. For example, an arbitrary point $\left(x,y,z\right)$(x,y,z) in ordinary three-dimensional space is translated by the vector $\left(a,b,c\right)$(a,b,c) to the new point $\left(x+a,y+b,z+c\right)$(x+a,y+b,z+c)
#### Example 1
The points $(1,-2)$(1,2), $(2.5,3)$(2.5,3) and $(-1,0)$(1,0) in the Cartesian plane represent the corners of a triangular shape. A copy of this shape, such that all three corners have only positive coordinates, can be obtained by many different translations of the original triangle. One such translation is by the vector $\left(2,3\right)$(2,3), Find the coordinates of the corners of the copy under this translation.
We apply the translation to each of the corner points by vector addition. Thus, the corners of the copy are at $\left(1+2,-2+3\right)=(3,1)$(1+2,2+3)=(3,1), $\left(2.5+2,3+3\right)=(5.5,6)$(2.5+2,3+3)=(5.5,6) and $\left(-1+2,0+3\right)=(1,3)$(1+2,0+3)=(1,3).
#### Example 2
What translation would be needed to bring the point $(-1,0)$(1,0) in Example $1$1 to the origin of the coordinate axes? Find the new locations of the corners under this translation.
We need to add the vector $(1,0)$(1,0) since $(-1,0)+(1,0)=(0,0)$(1,0)+(1,0)=(0,0). The other two corners are at $\left(2.5+1,3+0\right)=(3.5,3)$(2.5+1,3+0)=(3.5,3) and $\left(1+1,-2+0\right)=(2,-2)$(1+1,2+0)=(2,2). |
3 Tutor System
Starting just at 265/hour
A train travels 360 km at a uniform speed. If, the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Let the speed of the train = x km/hr
If, speed had been 5 km/hr more, train would have taken 1 hour less.
So, according to this condition
$$\Rightarrow {{360} \over {x}} = {{360} \over {x + 5}} + 1$$
$$\Rightarrow 360 ( {{1}\over{x}} - {{1}\over {x + 5}}) = 1$$
$$\Rightarrow 360 ( {{x + 5 - x}\over{x(x + 5)}}) = 1$$
$$\Rightarrow (360)(5) = x^2 + 5x$$
$$\Rightarrow x^2 + 5x - 1800 = 0$$
Comparing equation $$x^2 + 5x - 1800 = 0$$ with general equation $$ax^2 + bx + c = 0$$,
We get a = 1, b = 5 and c = -1800
Applying quadratic formula = $$x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}$$
$$\Rightarrow x = {{-5 ± \sqrt{(5)^2 - 4(1)(-1800)}} \over {(2)(1)}}$$
$$\Rightarrow x = {{-5 ± \sqrt{25 + 7200}} \over {2}} = {{-5 ± \sqrt{7225}} \over {2}}$$
$$\Rightarrow x = {{-5 + 85} \over {2}} , {{-5 - 85} \over {2}}$$
$$\Rightarrow x = 40,-45$$
Since speed of train cannot be in negative. Therefore, we discard x = -45
Therefore, speed of train = 40 km/hr |
# Fraction (mathematics)
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In mathematics, a fraction (from the Latin fractus, meaning broken) is a concept used to convey a proportional relation between a part and the whole. It consists of a numerator (an integer — the part) and a denominator (a natural number — the whole). For instance, the fraction can represent three equal parts of a whole object, if the object is divided into five equal parts. Any rational number can be written as a fraction.
Fractions are a special case of ratios. For instance, is a valid ratio, but it is not a fraction since we cannot compute an equivalent fraction with an integer numerator and a natural number denominator. A fraction with equal numerator and denominator is equal to one (e.g., ). Because the division by zero is undefined, zero should never be the denominator of a fraction.
Due to tradition and conventions, there are at least two ways to write a fraction. The numerator and the denominator may be separated by a slash (e.g., 3/4), or by a vinculum (e.g., ). Since we can compute the quotient from a fraction, we can represent any fraction with a decimal numeral (e.g., ).
In higher mathematics, a fraction is viewed as an element of a field of fractions, the smallest field of the integral domain.
## Forms
A vulgar fraction (or common fraction) simply refers to a numerator divided by a denominator (e.g., and ). It is said to be a proper fraction if the absolute value of the numerator is less than the absolute value of the denominator (e.g. ). An improper fraction (in the United Kingdom, top-heavy fraction) is said if the absolute value of the numerator is greater than or equal to the absolute value of the denominator (e.g. ). All integers different from -1, 0, and 1 can be represented by an improper fraction, since for example . The 1 at the denominator is sometimes called an "invisible denominator".
A mixed number is the sum of an integer and a proper fraction (e.g., ). An improper fraction can be transformed into a mixed number and vice-versa.
## Special cases
• A vulgar fraction with a numerator of 1, e.g. , is a unit fraction.
• An Egyptian fraction is the sum of distinct unit fractions, e.g. .
• A decimal fraction is a vulgar fraction in which the denominator is a power of ten, e.g. .
• A dyadic fraction is a vulgar fraction in which the denominator is a power of two, e.g. .
• An expression that has the form of a fraction but actually represents division by or into an irrational number is sometimes called an "irrational fraction". A common example found in trigonometry is , the measure of a right angle in radians.
• A complex fraction (or compound fraction) is a fraction in which the numerator and denominator contain a fraction (e.g., ). To simplify, divide the numerator by the denominator.
• A continued fraction is an expression such as , where the are integers.
• Rational functions are represented in the form of a fraction, where the numerator and denominator are polynomials. They are the quotient field of the polynomials.
• In algebra, some rational expressions (a fraction with an algebraic expression in the denominator) are written as the sum of other rational expressions with denominators of lesser degree. For instance, the rational expression can be rewritten as the sum of two fractions : and . The decomposition is made of partial fractions.
## Notation
A cake with one quarter removed. There is a proportional relation between the removed quarter and the whole cake.
There are many equivalent notations for a fraction.
The numerator and the denominator may be separated by a slanted line, the slash or the solidus (e.g., 3/4), or by an horizontal line, the vinculum (e.g., ). In some contexts (like road signs in some countries), it is clear that two numerals are the numerator and the denominator. They are written without any separator : a b (e.g., 3 4). The colon separator is usually found in written ratios (e.g., 1.5 : 2 = 3 : 4). The "÷" symbol is sometimes used to express a fraction (e.g., 3 ÷ 4), but should not.
Percentages ("%") allow to write a numeral as a fraction, the denominator being implicitly 100 (e.g., 14.5% = ). Per mills ("‰") allow to write a numeral as a fraction, the denominator being implicitly 1,000 (e.g., 22.3‰ = ). Per cent mille (pcm) allow to write a numeral as a fraction, the denominator being implicitly 100,000 (e.g., 78.7 pcm = ). Parts per million (ppm), parts per billion (ppb) and parts per trillion (ppt) are others way to write a numeral as a fraction with the implicit denominators 1 million, 1 billion, and 1 trillion.
In pie charts, any portion convey the proportional relation between a part and the whole.
A fraction is sometimes represented by a rectangular grid. We could represent by this grid :
## Arithmetic operations
The most common arithmetic operations on fractions are addition, subtraction, multiplication, and division. When adding and subtracting, we must often compute the equivalent fractions. When dividing, we usually compute the multiplicative inverse. After any computation, the end result should be an irreducible fraction.
In this section, each subsection builds upon knowledge acquired in previous subsections. For instance, the addition subsection does not come before the multiplication subsection, since fraction addition usually involves fraction multiplication.
Any integer can be represented by a fraction (e.g., and ). Thus, if an operation is applied to an integer and a fraction, convert the integer into a fraction and apply the appropriate algorithm.
For the remainder of this section, it is understood that and .
### Equivalent fractions
Multiplying (or integer dividing) the numerator and the denominator of a fraction by the same non-zero integer results in a new fraction that is said to be equivalent to the original fraction. For instance, and are equivalent, since .
A fraction where the numerator and the denominator do not have any common factor, 1 excepted, is said irreducible (or in its lowest terms). If it is not the case, then divide its numerator and its denominator by their gcd. For instance, is not in lowest terms because both 4 and 20 can be exactly divided by 4, giving . In contrast, is in lowest terms.
### Comparison
To compare fractions with different denominators, find their equivalent fraction with the same denominator. The fraction with highest numerator in absolute value is the greatest.
For instance, is greater than ?
The "cross-multiply" method say to multiply the top and bottom numbers crosswise. The product of the denominators is used as a common (but not necessary the least common) denominator. The highest numerator identifies the largest fraction. Since both denominators are the same, they can be dropped.
For instance, is greater than ?
### Multiplication
Formally, apply this algorithm to multiply the fractions and :
1.
2.
By hands, the multiplication is done like this.
1. For the resulting fraction,
1. Set its numerator to the product of both numerators.
2. Set its numerator to the product of both denominators.
2. Reduce the resulting fraction if you need to.
For instance, what is the result of ?
Since the result is not an irreducible fraction, we may reduce it. We divide the numerator and the denominator by their gcd, 2 :
.
### Multiplicative inverse
The multiplicative inverse of a fraction is :
.
### Division
Dividing by a fraction is the same as multiplying by its inverse.
Formally, apply this algorithm to divide the fractions and :
1.
2.
By hands, the division is done like this.
1. Exchange the numerator and the denominator in the second fraction (equivalent to computing the multiplicative inverse).
2. For the resulting fraction,
1. Set its numerator to the product of both numerators.
2. Set its numerator to the product of both denominators.
3. Reduce the resulting fraction if you need to.
For instance, what is the result of ?
The result is an irreducible fraction.
The additive inverse of a fraction is :
Formally, apply this algorithm to add the fractions and :
1.
2.
By hands, the addition is done like this.
1. Compute an equivalent fraction of and , making sure both have the same denominator.
2. For the resulting fraction,
1. Set its numerator to the addition of the numerators.
2. Set its denominator to the computed denominator (the three fractions have the same denominator).
3. Reduce the resulting fraction if you need to.
For instance, what is the result of ?
Let's find a number that both denominators will divide : It is 12. We are ready to compute the equivalent fractions :
This is the final answer since it is an irreducible fraction.
### Subtraction
Formally, apply this algorithm to subtract the fractions and :
1.
2.
By hands, the subtraction is done like this.
1. Compute an equivalent fraction of and , making sure both have the same denominator.
2. For the resulting fraction,
1. Set its numerator to the subtraction of the numerators.
2. Set its denominator to the computed denominator (the three fractions have the same denominator).
3. Reduce the resulting fraction if you need to.
Since this algorithm is very similar to the addition algorithm, we do not give any example.
## Mixed numerals
A mixed numeral has the form , where A is the integer part and is the fractional part. Implicitly, there is a plus symbol between the parts (e.g., ).
Just like any fraction, we can add, subtract, multiply, and divide mixed numerals. However, before applying the operation, convert the mixed numeral to improper fraction, or there can be wrong results.
For instance, what is the sum of and ? The minus sign applies to the fractional part : . The answer is or .
### Mixed numeral to improper fraction
A mixed numeral can be converted to an improper fraction with this algorithm :
1. Insert a plus symbol between the integer and the fraction.
2. Replace the integer with its equivalent fraction on 1.
For instance, transform to an improper fraction.
### Improper fraction to mixed numeral
An improper fraction can be converted to a mixed numeral with this algorithm :
1. Integer divide the numerator by the denominator.
2. The quotient becomes the whole part and the remainder becomes the numerator of the fractional part.
3. The fraction has the same denominator.
For instance, transform to a mixed number.
## Decimal numerals
A decimal numeral has the form , where the are base ten digits (0, 1, 2,... 9).
### Fraction to decimal numeral
To get the equivalent decimal numeral of a fraction, divide the numerator by the denominator (e.g., ).
### Decimal numeral to fraction
In many cases, it is easier to work with decimal numerals, but they lack precision compared to fractions. Sometimes an infinite number of decimals is required to convey the same precision. Thus, it is often useful to convert decimal numerals into fractions.
Before going any further in this section, we need to observe a property of the decimal numerals. For instance, The infinite expansion is composed of 3s. For square root of 2, we have . The infinite expansion is composed of different digits without any repeated pattern. For 4.35, we have The infinite expansion is composed of zeroes, but they are not written down by convention. Thus, all numbers written in decimal notation have an infinite decimal expansion.
Because of this observation, we only need to use two algorithms to convert decimal numerals to fractions. In order to choose the appropriate algorithm, we need to find if there is a repeating pattern, or not, in the decimal expansion.
#### Repeating pattern algorithm
The idea behind the following algorithm is to compute two different multiples of the same number in such a way as to align the repeated pattern after the decimal separator.
1. Excluding any repeated pattern, count the number of digits after the decimal separator (p) in the decimal numeral n.
2. Compute
3. Including the first repeated pattern, count the number of digits after the decimal separator (q) in the decimal numeral n.
4. Compute
5. Write the equation , where a is unknown.
6. Isolate a, the fraction to find.
For instance, convert 7.85891891891... to a fraction.
m is equal to 2
102 = 100
7.85891891891... × 100 = 785.891891891...
n is equal to 5
105 = 100000
7.85891891891... × 100000 = 785891.891891...
100000 × a - 100 × a = 785891.891891... - 785.891891891...
99900 × a = 785106
...
For instance, convert 4.37 to a fraction.
m is equal to 2.
102 = 100
4.37000... × 100 = 437.000...
n is equal to 3.
103 = 1000
4.37000... × 1000 = 4370.00...
1000 × a - 100 × a = 4370.00... - 437.000...
900 × a = 3933
...
#### Non-repeating pattern algorithm
The conversion is done using observation and needs.
If the decimal numeral n is a multiple of (a truncated value of ), then solve , where k is the multiple to find. If the constant is unknown, then truncate the decimal expansion to needed precision. Convert the new numeral using repeating pattern algorithm.
## History
Ancient Egyptians used what is called today Egyptian fractions. In China, they were in use around the first century of our era.
Some problems from Antiquity explicitly request fraction use :
1. I found a stone, but I did not weigh it. After adding a seventh of its weight and added the eleventh of the new result, it weighs 1 ma-na [mass unit]. What is the stone weight ? (Babylonian problem, tablet YBC 4652, problem 7)
2. A number added to its seventh gives 19. What is the number ? (Rhind Mathematical Papyrus, problem 24)
3. A number added to its quarter gives 15. What is the number ? (Rhind Mathematical Papyrus, problem 26)
4. Suppose we have 9 golden rods and 11 white silver rods which, when they are weighed, have exactly the same weight. If we replace one golden rod with one white silver rod, gold is lighter by 13 liang [mass unit]. What is the weight of one golden rod and of one white silver rod ? (The Nine Chapters on the Mathematical Art, problem 7.17)
Historically, any numeral that did not represent a whole number was called a "fraction". The numerals that we now call "decimals" were originally called "decimal fractions"; the numeral we now call "fractions" were called "vulgar fractions", meaning a "commonplace fraction". |
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# What is the ratio between times taken by a train 240 m long to cross an electric pole and a bridge of 80 m length ?A.2 : 3B.3 : 4C.4 : 5D.5 : 6
Last updated date: 13th Jun 2024
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Hint: At first the train needs to cross an electric pole and the distance travelled to cross a pole is the length of the train . Next the train needs to cross a bridge of 80 m , so the distance travelled by the train to cross the bridge is the sum of the length of the bridge and length of the train. The ratio of these two distances give the ratio of the time as distance is proportional to time.
Step 1:
In the first case the train has to cross an electric pole .
Given that the length of the train is 240 m long .
We know that the length of the train is equal to the distance travelled by the train to cross the electric pole.
Therefore the distance travelled by the train to cross the electric pole = 240 m
Step 2 :
In the second case the train has to cross a bridge of 80 m long
Given that the length of the train is 240 m and length of the train is 80 m
Therefore the distance travelled by the train to cross the bridge is the sum of the length of the train and length of the bridge.
The distance travelled by the train to cross the bridge = 240 + 80 = 320 m
Step 3 :
We know that distance is proportional to time
Hence the ratio of the distances is proportional to the ratio of the time asked.
The ratio is 240 : 320
Simplifying the ratio to bring it to its simplest form we get 3 : 4.
The correct option is B.
Note: In solving ratio problems we need to make sure that both the quantities are of the same units.
If one quantity is given in metres and another in centimetres then we need to convert both to either metres or centimetres |
Online Calculator Resource
Difference of Two Squares Calculator
Difference of 2 Squares
$a^2 - b^2 = \; ?$
$4(x + 3y^{2})(x - 3y^{2})$
Solution:
Factor the equation$4x^{2} - 36y^{4}$using the identity$a^2 - b^2 = (a + b)(a - b)$First factor out the GCF:$4(x^{2} - 9y^{4})$Both terms are perfect squares so
from a2 - b2 we can find a and b.$a = \sqrt[]{x^{2}} = x$$b = \sqrt[]{9y^{4}} = 3y^{2}$Therefore$a^2 - b^2 = (x)^2 - (3y^{2})^2$Complete the factoring of a2 - b2
to (a + b)(a - b)$4(x + 3y^{2})(x - 3y^{2})$Final Answer:$4(x + 3y^{2})(x - 3y^{2})$
Calculator Use
This is a factoring calculator if specifically for the factorization of the difference of two squares. If the input equation can be put in the form of a2 - b2 it will be factored. The work for the solution will be shown for factoring out any greatest common factors then calculating a difference of 2 squares using the idenity:
$$a^2 - b^2 = (a + b)(a - b)$$
Factored terms that contain additional differences of two squares will also be factored.
Difference of Two Squares when a is Negative
If both terms a and b are negative such that we have -a2 - b2 the equation is not in the form of a2 - b2and cannot be rearranged into this form.
If a is negative and we have addition such that we have -a2 + b2 the equation can be rearranged to the form of b2 - a2which is the correct equation only the letters a and b are switched; we can just rename our terms.
For example, factor the equation
$$-4y^{2} + 36$$
We can rearrange this equation to
$$36 - 4y^{2}$$
and now solve the difference of two squares with a = 36 and b = 4y2
Solution:
Factor the equation (rearranged)
$$36 - 4y^{2}$$
using the identity
$$a^2 - b^2 = (a + b)(a - b)$$
First factor out the GCF:
$$4(9 - y^{2})$$
Both terms are perfect squares so from a2 - b2 we can find a and b.
$$a = \sqrt[]{9} = 3$$
$$b = \sqrt[]{y^{2}} = y$$
Therefore
$$a^2 - b^2 = (3)^2 - (y)^2$$
Complete the factoring of a2 - b2 to (a + b)(a - b)
$$4(3 + y)(3 - y)$$
$$4(3 + y)(3 - y)$$ |
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# WORKBOOK
IN MATHEMATICS
VI
( First Quarter )
## May Ester M. Rubio
Master Teacher I
CHAPTER ONE
Whole Numbers
Lesson No. 1 ..
Expression or Equation
Lesson No. 2
## Base, Exponent and Power
Lesson No. 3.
Power
Lesson No. 4
Evaluating Expression
## Two to Three Step Word
Problem Involving Whole Numbers
Lesson No. 1
EXPRESSION OR EQUATION
Expression it may consist of one number of more numbers with operation and
2
## grouping symbols without the equals (=) symbol
Equation it may consist of numbers with operation and grouping symbols with equals
( -) symbol.
## Identify which set is an expression and which set is an equation.
Set A
5147 + 75
1663
547 + (1 843 985 )
Set B
5 147 + 75 = 5 222
1 843 985 = 858
( 2 147 X 50 ) 50 = 2 147
PRACTICE
Directions: Write EX if it is an expression or EQ if it is an equation.
_____ 1. 82 X 4 + ( 28 7)
_____ 2. ( 9510 10 ) + ( 2 848 4 ) = N
_____ 3 . ( 2 147 X 50 ) 50
_____ 4. 5 202 X 4
_____ 5. 47 + 32 = 79
_____ 6. ( 1 X 3 ) + 8 = 12
_____ 7. 95 ( 2 + 5 )
_____ 8. ( 6 + 7 ) + ( 100 4 ) = 38
_____ 9. 9510 10
_____ 10. 305
Lesson No. 2
BASE, EXPONENT AND POWER
3
2 =8
BASE is the number that is multiplied repeatedly or the
factor
Exponent is a number written at the upper right hand of a
number. It tells the number of times the number
is multiplied by itself .
Power is the product obtained after multiplying a number
by itself as many times as indicated by its exponent
PRACTICE
Directions: Identify the base, exponent and the power
1)
2)
3)
4)
5)
Equation
5 = 25
72 = 49
24 = 16
( 4+5)2 = 81
(15 9)4 = 1296
Base
Exponent
Power
Exponent
2
3
2
5
3
Power
49
64
36
32
1000
MORE PRACTICE
Directions: Write the equation.
Equation
1.
2.
3.
4.
5.
Base
7
4
6
2
( 20 2 )
Lesson No. 3
POWER
4
To find the power, multiply the base by itself as many times as indicated by its
exponent.
Example
23 = 2 X 2 X 2
=
42 = 4 X 4
X2
16
6
PRACTICE
Directions: Find the power.
1. 36 = _____
6. ( 10 7 ) 2 = _____
2. 45 = _____
7. ( 9 5 ) 2 = _____
3. 210 = _____
8. 252 = _____
4. 112 = _____
9. 93 = _____
5. ( 3 + 4 ) 3 = _____
10. 54 = _____
MORE PRACTICE
Directions: Match column B to column A.
Column A
Column B
_____ 1. 1 024
a) 12 2
_____ 2. 125
b) ( 45 5 ) 3
_____ 3. 144
c) ( 2 + 2 ) 5
_____ 4. 1 000
d) ( 8 3 ) 3
_____ 5. 729
e) ( 2 X 5 ) 3
Lesson No. 4
EVALUATING EXPRESSION
5
## To evaluate an expression means to find its value in the simplest form by
following the PEMDAS rule.
## simplify the expression enclosed by the parenthesis ( ) or other grouping symbols
like the brace [ ] and the bracket { }
## perform the operation of subtraction
Example:
( 20 16 ) 2 +
10
(82)=
42
10
4X4
10
16
10
26
22
More Examples
Evaluate
1.
Solution
92 = N
9X9=
2. 2 + 33
81
2+(3X3X3)
2+
27
29
3. ( 5 1 ) 22
4(2X2)
4
4. 15 X 2 ( 32 3 )
15 X 2 [ ( 3 X 3 ) 3 ]
15 X 2 ( 9 3 )
6
15 X 2
30
5
PRACTICE
Directions: Evaluate the following expressions.
1. 24 = _____
2. 32 + 22 = _____
3. ( 45 8 ) + 42 = _____
4. 38 + 12 ( 42 + 12 ) = _____
5. 10 X 12 6 + 52 = _____
MORE PRACTICE
Directions: Evaluate the following expressions.
1. 28 2 + 7 32 = _____
2. 82 X 4 + ( 28 7 ) = _____
3. ( 69 12 ) + ( 83 47 ) 2 = _____
4. ( 150 75 ) 2 + ( 290 5 ) = _____
5. 5 X ( 13 7 )2 + [ ( 4 2 ) + 5 ] = _____
6. 43 + 34 52
= _____
7. ( 5 3 ) 3 + ( 2 1 ) 4 ( 4 2 ) 3 = _____
8. 6 + ( 12 5 ) 2 ( 9 3 ) 3 = _____
9. 18 X ( 15 9 ) 2 17 ( 9 5 ) 2 = _____
10. ( 516 412 ) ( 15 ) ( 45 36 ) 2 + ( 8 5 ) 3 = _____
HOMEWORK
7
## Evaluate the following expressions. Follow the PEMDAS rule.
1. ( 47 321 + 27 385 ) x 21 6 = N
2. [ ( 96 000 + 282 000 ) 2 ] X 36 = N
3. 90 5 + ( 63 800 20 000 ) = N
4. ( 37 254 X 20 14 252 ) 2 = N
5. ( 85 200 + 9 250 ) X ( 47 243 8 198 ) = N
MIND ENHANCER
Change the following descriptions into numerical expressions then evaluate.
1. The second power of 10 plus 15
2. Fourteen cubed minus 28
3. The second power of the sum of 250 and 25
4. The difference between 1 000 and the third power of 10
5. The product of 12 and the third power of the quotient of 9 and 3
6. The square of the sum of 20 and 3
7. The product of 6 and 5 raised to the third power
8. The square of the quotient of 15 divided by 3
9. The sun of the square of 100 and the square of 10
10. The difference between the fourth power of 3 and 3 squared
11. The sun of 6 and 9 added to the difference of 19 and 16
12. Twice the difference of 15 and 11 subtracted from the product of 14 and 5
13. The quotient of 48 and 8 multiplied by the sum of 4 and 8
14. The difference of 114 and 86 divided by the difference of 56 and 49
15. Twice the sum of 8 and 9 subtracted from the product of 9 and 8
Lesson No. 5
Two Three Step Word Problems
8
## Involving Whole Numbers
When solving word problems on whole numbers, use the RSTUVW Approach.
The letters stand for :
R Read the problem carefully
S Study the given facts
T Think of the operation/s to be used
U Use the operation/s to solve the problem
W Write your answer to the problem in a complete sentence
Example:
A five star hotel charges Php 38 000 a month from a foreigner who stays
in one unit of the hotel. If the foreigner stays in the hotel for two years, how much
should be charged from him ?
## Now analyze the problem. Follow the RSTUVW Approach
Step 1 : Read the problem carefully.
Step 2 : Study the given facts.
Given :
_______________________________________
_______________________________________
________
Operation/s:
To be used.
## Step 4 : Use the operation/s
to solve the problem
_______________________________________
_______________________________________
________
Solve :
_______________________________________
_______________________________________
________
## Step 6 : Write the answer in
PRACTICE
Analyze and then solve the problems following the steps in the RSTUVW Approach.
1. Mother earns Php 38 000 less than fathers monthly salary. If father earns
Php 22 000 a month, how much is the combines salary of mother and father ?
2. James gives Php 12 000 to her mother every month. He has been giving this
amount for 18 months. If mother spends only Php 10 000 of James money and
saves the rest, how much would mother save after 18 months ?
3. The serial number of a bundle of Php 1 000 bills starts from MN 370003 and
ends at MN 370255. How much is the total amount of the bundle ?
4. Mr. Reyes bought 30 boxes of ponkan. Each box costs Php 200. If there are 50
pieces of ponkans per box, how much did Mr. Reyes earn from selling all the
ponkans at 3 for Php 25 ?
5. A house is rented at Php 2 500 a month. If a family rented it for 2 and a half
years, how much did they pay ?
MORE PRACTICE
Analyze and then solve the problems following the steps in the RSTUVW Approach.
1. A fisherman caught 134 kls of bangus, if a kilo cost Php 65.00 how much did he
sold if there are 29 kls left ?
2. Cyril took note of the scores of his five friends in a 100-item test in spelling. Raffy
scored 84; Louie, 88; Mac, 81; Bogie, 85; and Tony, 87. If Cyril scored 91; what
was the average score of the six boys ?
3. Mr. Hernandez paid Php 1, 720, 000 for a house and lot last year. This year he
sold it for Php 2, 488, 500. How much profit did he make after paying Php 18,
742 to the sales agent ?
4. The area of the Pacific Ocean is about 165 760 000 square kilometers ( km 2 )
and that of the Atlantic Ocean is about 82 400 000 km2. How much bigger is the
Pacific Ocean than the Atlantic Ocean ?
5. The reading on an electric meter for the month of September was 5 124 kWh.
The next month, the reading was 5 332 kWh. How much electricity was used ?
10
HOMEWORK
Analyze and then solve the problems following the steps in the RSTUVW Approach.
1. Nida took 36 pictures with her new camera. Four of the pictures were
overexposed and were not developed. If it cost Php 4 to print each picture, what
was the total cost of the pictures ?
2. A parking lot has 28 rows with 25 car spaces in each row. If 3 rows are removed
and used as a driveway, how many cars can be parked in the lot ?
3. Find Jennas average grade if her grades were 86, 89, 92, 94, 88, 93, 87, 90 and
91.
4. A university charges Php 136 tuition fee per unit in college. If Kate takes 21 units
and is given a discount of Php 278, how much should she pay ?
5. Eight parishes collected a total of Php 95 488 for the fire victims. What was the
average collection from each parish ?
11
CHAPTER TWO
DECIMALS
Lesson No. 1 ..
Lesson No. 2
Lesson No. 3.
Lesson No. 4
## Writing Decimals in Standard and
Expanded Notation
Lesson No. 6 .
Thousandths
Lesson No. 1
12
## Study the numbers inside the box. What are they ?
1.0
0.1
0.01
0.001
0.0001
0.10001
Lets visualize these decimals. Which figure represents 0.01 , 1.0 and 0.1.
a)
b)
c)
Decimal
writing tenths,
of ten parts of a unit.
## fraction is just another way of
hundredths and other powers
1
10
= 0.1
1
100
= 0.01
1
1 000
= 0.001
1
10 000 =0.0001
## - read as one- ten thousandth
PRACTICE
Complete the table below.
Fractions
Decimal
13
1.
1
10
2.
Five - tenths
3.
0.8
4.
3
100
5.
0.08
6.
0.19
7.
Fifteen thousandths
8.
Twenty five thousandths
9.
125
1 000
10.
0.0120
MORE PRACTICE
Do you know the name of the spaceship used by Neil A. Armstrong, Edwin Aldrin
and Michael Collins during their flight to the moon called Apollo 11 ?
Find out the name of the spaceship by changing the fractions below into
decimals. After getting the answer, change it to its corresponding letter. Then form
those letters to know the name of the spaceship.
1. 15
100
4. 347
= _____
2. 24
100
1 000
= _____
= _____
182
10 000
= _____
= _____
6. 4 734
10 000
= _____
5.
3. 124
1000
7. 5 432
10 000
A = 0.24
R = 0.0182
U = 0.347
N = 0.4734
B = 0.04734
S = 0.15
V = 0.5432
T = 0.124
W = 0.05432
Lesson No. 2
14
= ___
## Look at the figure and study the place value of decimals.
12.5682
Tenths
Hundredths
Thousandths
Ten thousandths
In a decimal, the value of the digits to the right of the decimal point is always less
than one. The value of a digit starts from tenths then hundredths, thousandths , tenthousandths , and so on.
PRACTICE
Identify the place value of the underlined digit.
1) 98.2035
= _________________
6) 32.8116
= ____________________
2) 5.3910
= _________________
7) 5.2
= ____________________
3) 6348.9045 = _________________
8) 62.3887
= ____________________
4) 1000.3857= _________________
9) 3.5276
= ____________________
5) 638.5294 = _________________
10) 6.3421
= ____________________
MORE PRACTICE
Identify the place value of the number three digit in the following expressions.
1) 8.314 ____________________
6) 1.5283 ________________________
2) 295.6378 _________________
7) 58.4385 _______________________
3) 125.9563 ________________
8) 98.7635 _______________________
4) 85.6372 _________________
9) 18.347 ________________________
5) 1.5234 __________________
HOMEWORK
15
1) 267.249
6) 725 703.825 2
2) 1 388.561 3
7) 2 140.725 3
3) 39 347.06
8) 65.182 9
4) 811 329.502 6
9) 52 385.042 1
5) 15 347.403 9
10) 629.315 6
Lesson No. 3
## Reading and Writing Decimals
16
The decimal fraction has a value that is always less than one. Let us examine the
value of each digit in this number.
Ten
Thousands
Thousands
hundreds
tens
ones
tenths
hundredths
thousandths
Ten
thousandths
9
1 000
6
10 000
Whole Number
Decimal point
4
10
3
100
Decimal Fraction
Note that as we go to the right, the value of the digit becomes smaller.
Lets read more decimals.
Decimal
5.2058
3.0035
4.0005
0.3002
Five and two thousand fifty-eight ten thousandths
Three and thirty-five ten thousandths
Four and five ten thousandths
Three thousand two ten thousandths
PRACTICE
Write in words the decimal numbers below.
1) 12.5438_____________________________________________________________
2) 40.0538_____________________________________________________________
3) 19.0054_____________________________________________________________
4) 254.0309____________________________________________________________
5) 143.2800____________________________________________________________
6) 15.29 _______________________________________________________________
7) .006 ________________________________________________________________
8) 85.3000 _____________________________________________________________
9) 52.33781 ____________________________________________________________
10) 5.00005 ____________________________________________________________
MORE PRACTICE
Write these words in symbols.
17
## 1.Five hundred two and two ten-thousandths _____________________________
2. three and forty-five ten-thousandths __________________________________
3. twenty two and one hundred fifteen ten-thousandths ____________________
4. four and five hundred forty-two ten-thousandths_________________________
5. eight ten thousandths _____________________________________________
6. eleven and thirteen ten-thousandths _________________________________
7. one hundred forty and seventeen thousandths __________________________
8. five hundred six and seven hundred eight thousandths ____________________
9. one thousand four and five hundred eight thousandths _____________________
10. eight hundred one and eighteen ten-thousandths ________________________
11. one thousand thirty-seven ten thousandths _____________________________
12. sixty-four and one hundred forty-six thousandths ________________________
13. four thousand five and six thousand one ten thousandths __________________
14. twenty-seven and seven hundred two thousandths _______________________
15. sixteen and seven thousand forty-eight ten thousandths ___________________
HOMEWORK
Write the decimals for these fractions then write in words how you read them.
1) 5
10 000
Write: _____________________________________________________
2)
52
10 000
Write: _____________________________________________________
3)
143
10 000
Write: ______________________________________________________
4) 2048
10 000
Write: ______________________________________________________
5) 1 002
10 000
Write: ______________________________________________________
Lesson No. 4
Writing Decimals in Standard and Expanded Notations
18
Study how the decimals written in standard notation can be written in expanded
notation.
Standard Notation
1) 87.97
Expanded Notation
8 X 101 + 7 X 100 + 9 X 10-1 + 7 X 10-2
where:
8 X 101 = 80
7 X 100 = 7
9 X 10-1 = .9
7 X 10-2 = .07
2) 265.256
## 2 X 102 + 6 X 101 + 5 X 100 + 2 X 10-1 + 5 X 10-2 +6
X 10-3
Where:
2 X 102 = 200
6 X 101 = 60
5 X 100 = 5
2 X 10-1 = .2
5 X 10-2 = .05
6 X 10-3 = .006
PRACTICE
Write the following decimals in expanded notation.
1) 47.8234 = __________________________________________________________
2) 108.9672 = __________________________________________________________
3) 347.8371 = __________________________________________________________
4) 638.2439 = __________________________________________________________
5) 901.4318 = __________________________________________________________
6) 1 892.7325 = ________________________________________________________
7) 6 934.8253 = ________________________________________________________
8) 9 849.4772 = ________________________________________________________
9) 89 415.3473 = _______________________________________________________
10) 183 219.1827 = _____________________________________________________
MORE PRACTICE
Write these decimals written in words in their standard and expanded notations.
Decimals in Words
Standard Notation
## 1) eight ten thousandths
19
Expanded Notation
2) twenty-three ten
thousandths
thousandths
## 4) one and seven hundred
eighteen ten thousandths
thousandths
ten thousandths
## 8) ninety-four and one
thousand eight ten
thousandths
## 9) one hundred four and
fifty-six thousandths
## 10) four hundred seventeen
and ninety-five ten
thousandths
HOMEWORK
Write the following in expanded notation.
1) 125.698
4) 6.3840
2) 6 284.08
5) 8 230.239 5
3) 64 238.1658
Lesson No. 5
## Comparing and Ordering Decimals
20
In the decimal 11.86 which has a larger place value, the digit 8 or 6 ? Study the
figure below to find the answer.
## The shaded part is 8
10
The dotted part is 6
100
Now, can you tell
or 0.8
or 0.06.
which is bigger, 0.8 or 0.06 ?
______________________________
## Study more examples on how to compare decimals.
1) 4.9 > 4.09 > 4.009 > 4.0009
2) 1.1273 < 1.1274 < 1.1275 < 1.1276
3) 1.1 = 1.10 = 1.100 = 1.1000
PRACTICE
A. Compare the decimals by writing > , < or =.
1) 14.07 _____ 14.17
## B. Arrange the given decimals from greatest to least.
1.238
2.476
8.983
21 1.2
2.4760
1)
2)
9.002
8.102
9.200
8.0102
8.1
3)
12.0001
12.0010
12.0100
12.1000
12.1
25.1235
25.2235
25.0125
25.1
25.123
40.05
40.04
40.041
40.055
40.045
4)
5)
MORE PRACTICE
A. Compare these decimals by writing < , > or = in the blank.
1) 0.162 _____ 0.106
2) 0.036 _____ 0.031
3) 0.4
_____ 0.40
## 4) 3.53 _____ 3.59
5) 7.01 _____ 7.103
6) 0.61 _____ 0.601
7) 9.2
_____ 9.200
## 8) 10.021 _____ 0.045
9) 0.7562 _____ 0.7559
10) 8.627 _____ 8.649
## B. Order the decimals from the least to the greatest.
1) 0.5
0.49
0.53
0.51
2) 0.03
0.029
0.3
0.305 __________________________________
3) 1.237
1.273
1.027
1.230 __________________________________
4) 7.0
7.326
7.3
7.320 __________________________________
0.011
0.0019 __________________________________
5) 0.0101 0.0099
__________________________________
HOMEWORK
A. Compare these decimals by writing < , > or = in the blank.
22
## 1) 0.008 _____ 0.0009
2) 0.19321 _____ 0.19231
3) 0.0019 _____ 0.0002
4) 12.7 _____ 12.7000
5) 0.999 _____ 0.99
6) 0.1 _____ 0.01
7) 0.2 _____ 0.02
8) 0.03 _____ 0.04
9) 0.21 _____ 0.021
10) 0.003 _____ 0.0003
## B. Order the decimals from the least to the greatest.
1) 0.07
0.64
0.73
0.7
____________________________________
2) 0.45
0.405
0.5
0.54
____________________________________
3) 0.031
0.301
0.310 0.03
____________________________________
4) 1.628
1.476
5) 4.285
4.258
Lesson No. 6
23
## Here are the rules in rounding off decimals.
1) If the nearest digit at the right of the digit to be rounded off is 5 or above, add 1 to
the digit to be rounded off and drop all the other digits to its right.
2) If the nearest digit at the right of the digit to be rounded off is 4 and below retain
the digit to be rounded off and drop all the digits at the right of the digit to be
rounded off.
Here are examples
Rounded to the Nearest
Number
2.01234
5.12344
3.43211
7.56789
8.98765
tenths
hundredths
thousandths
Ten thousandths
2.0
5.1
3.4
7.6
9.0
2.01
5.12
3.43
7.57
8.99
2.012
5.123
3.432
7.568
8.988
2.0123
5.1234
3.4321
7.5679
8.9877
PRACTICE
Complete the table below.
Rounded to the Nearest
Number
1) 4.381576
2) 2.914325
3) 3.435438
4) 9.023542
5) 10.020387
6) 15.354032
7) 18.439002
8) 20.354389
9) 25.030105
10) 30.438298
tenths
hundredths
MORE PRACTICE
A. Round off the following decimals to the nearest tenth.
1) 13.726 4
_____________________
2) 280.094 7
_____________________
3) 311.150 0
_____________________
4) 526.949 8
_____________________
5) 419.899 1
_____________________
6) 2 114.981
_______________________
7) 3 760.075 1
_______________________
8) 825.346 7
_______________________
9) 4 008.019 9
_______________________
24
thousandths
Ten thousandths
10) 15 903.158 2
_______________________
## B. Round off the above decimals to the nearest hundredth.
1) __________________________
6) _________________________
2) __________________________
7) _________________________
3) __________________________
8) _________________________
4) __________________________
9) _________________________
5) __________________________
10) ________________________
HOMEWORK
Complete the chart.
Number
Whole
Number
Tenth
Hundredth
Thousandth
1) 4.237 69
2) 28.912 47
3) 88.004 56
4) 92.308 47
5) 108.206 39
6) 207.159 53
7) 398.204 18
8) 562.555 55
9) 712.897 64
10) 899.125 30
25
Ten
Thousandth
CHAPTER THREE
Decimals
Lesson No. 1 ..
## Estimate Sums and Differences of
Whole Numbers and Decimals
Lesson No. 2
Decimals
Lesson No. 3.
Decimals
## Problem Solving Involving Addition
and Subtraction of Decimals
Lesson No. 1
## ESTIMATE SUMS AND DIFFERENCES OF WHOLE
NUMBERS AND DECIMALS
26
Study how to estimate the sum of my money in the bank to the nearest hundredths.
Initial deposit
P 15 278 . 4386
253 . 2500
= P 15 278 . 44
=
________________________________
253 . 25
__________________________
P 15 531 . 69
Study how to estimate the sum to the nearest tenths.
Initial deposit
P 15 278 . 4386
253 . 25
= P 15 278 . 40
=
________________________________
253 . 30
__________________________
P 15 531 . 70
Study how to estimate the sum to the nearest ones.
Initial deposit
P 15 278 . 4386
253 . 25
= P 15 278 . 00
=
________________________________
253 . 00
__________________________
P 15 531 . 00
When estimating sums and differences, round off the given numbers
first to the desired place value.
PRACTICE
Estimate the sum and difference to the indicated place value.
Given
1) 24.84397
+ 123.29824
Ones
2) 476.19438
+ 1 937.99825
3) 143.25376
+ 22.14398
4) 6 725.58354
+ 1 276.52145
5) 10 478.08238
+ 11 725.97438
6)
473.8279
- 121.2781
7) 921.8198
- 141.3124
8) 1 763.9254
651.9835
9) 4 381.1494
- 1 211.9999
10) 14 738.9154
- 1 488.1815
27
Tenths
MORE PRACTICE
Estimate the sum and difference to the indicated place value.
1)
Given
10 478.08238
+ 11 725.97438
Ones
Tenths
2) 63.1345
+ 94.76315
3) 51.048
+ 36.78904
4) 6 391.4132
+ 3 955.84761
5) 86 325.147
+ 75 431.85623
6) 24.4812
- 18.5769
7) 286.5718
- 269.7923
8) 42 800.1443
- 39 765.2576
9) 391 865.4579
- 287 648.3651
10) 82 101.63578
- 79 143.54651
Lesson No. 2
28
## In adding or subtracting decimals, remember to align the decimal points.
To add or subtract, follow the process of adding or subtracting whole numbers.
Example:
0.5 + 0.687 + 0.16 + 0.8932 = n
## 0 . 8 0 0 0 ------- insert 3 zeros
0 . 6 8 7 0 ------- insert 1 zero
+
0.8932
2.5402
Example:
0.5 -0.2873=n
## 0 . 5 0 0 0 ----- insert 3 zeros
-
0.2873
0.2127
PRACTICE
A. Add or subtract.
1) 0.38 + 0.47 = n
6) 0.23 0.16 = n
2) 0.412 + 0.638 = n
7) 0.97 0.178 = n
3) 0.529 + 0.646 = n
8) 0.232 0.046 = n
4) 0.4129 + 0.4738 = n
9) 0.76 0.057 = n
5) 0.284 + 0.325 = n
## B. Arrange the numbers vertically then add or subtract.
1) 0.048 + 0.027 + 0.049
2) 0.78 + 0.3621 + 0.691 + 0.3
3) 0.349 + 0.7005 + 0.2421 + 0.8 + 0.32
4) 0.45 0.2185
5) 0.83 0.52983
1) 0.203 0.16
6) 5.7 3.276
2) 0.133 0.0857
7) 6
29
- 4.767
3) 0.9 0.41078
8) 9.0 4.73
4) 6.4 4.32
9) 3.4 1.26
5) 8.27 4.163
## 10) 47.6 9.06
MORE PRACTICE
A. Add or subtract the following.
1) 0.8756 + 0.6872
## 11) 0.9265 0.7157
2) 0.7393+ 0.1877
## 12) 0.2457 0.2349
3) 0.3576 + 0.8546
## 13) 0.8765 0.5678
4) 0.2917 + +0.6358
## 14) 0.1542 0.1435
5) 0.6483 + 0.2529
## 15) 0.5937 0.3788
6) 0.5365 + 0.7325
## 16) 0.6911 0.3695
7) 0.5176 + 0.4958
## 17) 0.3276 0.1178
8) 0.3985 + 0.6358
## 18) 0.3361 0.1163
9) 0.3961 + 0.5559
## B. Add or subtract the following.
1) 0.3749
2) 0.6085
3) 0.7583
+ 0.4572
+ 0.3937
4) 0.5056
5) 0.3835
0.3935
6) 0.9651
7) 0.8570
0.5237
0.5356
30
0.2658
0.2786
8) 0.6593
- 0.3247
9) 0.5635
-
10) 0.7651
0.1418
0.4239
HOMEWORK
Find the sum or difference.
1) 0.684
+ 0.295
5) 0.84
+ 0.8056
9) 0.6
- 0.1858
2) 0.83
+ 0.567
6) 0.84
- 0.6358
3) 0.73
+ 0.3073
7) 0.540
- 0.2365
4)
0.9002
+ 0.8634
8) 0.3
- 0.076
10) 0.9702
-
0.1694
Lesson No. 3
## Addition and Subtraction of Mixed Decimals
A. Addition or subtraction of whole numbers and decimals.
31
+
0. 3 2 5
8 5 1. 3 2 5
1 2.0 0
-
.61
1 1.3 9
## B. Addition or subtraction of mixed decimals
In adding and subtracting mixed decimals, remember to align the decimal points
and regroup when necessary.
PRACTICE
A Find the sum.
1) 38.07 + 2 = _____
2) 9.721 + 0.015 = _____
3) 9.4 + 11.2 + 8.75 = _____
4) 3.21 + 1.7 +6 + 3.749 = _____
5) 21.7 + 10.05 + 6.3 + 0.43 = _____
6) 453.029 + 319.84 + 95.23 = _____
7) 24.640 3 + 9.408 + 36.3 = _____
8) 16.52 + 4.311 + 8.591 + 8.846 = _____
9) 317.48 + 46.238 9 + 445 + 39.482 2 = _____
10) 143.217 + 9.45 + 307.020 + 258.726 = _____
11) 29.18 0.12 = _____
12) 8 7.256 1 = _____
## 13) 254.604 38 = _____
14) 342.16 79.7 = _____
15) 715.08 546.9 = _____
16) 237.2 84.317 = _____
17) 363 4.281 = _____
32
## 18) 157.546 148.347 = _____
19) 354.36 93.208 6 = _____
20) 1.000 43 0.065 18 = _____
## B. Add or subtract these mixed decimals.
1) 82.4385
2) 5.435
3) 637.4
4) 82.19
+ 65.9685
+ 6.762
+ 389.85
+ 45.02
48.3914
5) 43.16
+
9.9
3.84
6) 891.7
+ 188.43
16.432
9) 2.893
- 1.8432
13) 9.36
- 7.1316
781.432
7) 38.6709
+
870.108
10) 523.4
-
79.329
14) 15.84
- 12.635
.4388
8.9
8) 15.84
+ 1.467
5.67
87.432
11) 392
12) 3.806
82.685
- 1.481
15) 324.5
- 168.95
MORE PRACTICE
Add and subtract these decimals.
1) 14.39847
2) 38.217601
3) 7.38254
4) 9.76582
+35.10121
+31.421354
+2.10300
+ 3.20000
6) 9.36782
7) 18.25835
8) 67.43284
+ 4.13299
+ 21.23149
+ 39.00355
5) 8.2
+ 5.73824
33
9) 25.49003
10) 189.2542
11) 62.35
+89.25405
+ 352.53589
+ 84.23617
13) 25.29685
14) 187.38415
15) 128.39084
+ 60.0362
17) 35.17849
100.06398
+ 657.67604
18) 245.65085
19) 63.385268
+ 34.067581
21) 4.38295
-
22.14398
30) 63.9
-
1.8345
34) 639.1432
3.6789
37) 78.5682
-
26) 143.25376
12.41254
33) 8.5104
-
- 24.317101
61.32248
29) 243.82
-
22) 43.438251
2.14174
25) 76.43479
-
+ 334.409574
- 395.5548
38) 40.208
69.47931
- 24.9564
+169.871834
23) 87.43825
-
174.12432
31) 32.56878
-
+ 69.56376
16) 256.7834
+
17.45986
35) 68.235
- 59.3596
39) 568.391
- 487.45881
50.76
20) 761.16658
+
88.08288
24) 83.70385
21.21432
27) 176.34321
-
12) 87.218
- 44.28
28) 1 276.35439
-
192.19999
32) 435.631
-
315.9476
36) 824.31
- 487.6529
40) 735.38
- 187.93345
HOMEWORK
A. Add the following.
## B. Subtract the following
1) 7.6458 + 2.4936 =
1) 47.0143 - 19.2336 =
34
2) 38.256 + 27.967 =
2) 26.9206 - 9.8769 =
3) 48.7435 + 21.6950 =
3) 8.0008 - 5.4569 =
4) 15.614 + 7.9965 =
4) 10.9200 - 4.7513 =
5) 26.4579 + 48.9536 =
5) 35.4070 - 16.2135 =
6) 7.9192 + 4.1918 =
6) 58.0675 - 52.3679 =
7) 3.7586 + 2.5965=
7) 7.6000 - 3.5947 =
8) 54.6059 + 17.5064 =
8) 19.2143 - 13.3465 =
9) 76.5432 + 12.3654 =
9) 22.190 - 18.621 =
12) 36.0070
13) 6.9786 +
2.6143 =
- 23.2159 =
Lesson No. 4
## Word Problems Involving Decimals
Problem No. 1
35
If mothers money of P 102 130.53 earned an interest of P 617.04 on November
30, how much is her balance on that date after it was deducted P 123.41 for the tax ?
## Let us analyze the problem together using the RSTUVW Approach.
Problem: Mothers balance on
November 30.
Given:
P 102 130.53
P 617.04
to be used.
P 123.41
Solution:
P 102 130.53
+
617.04
P 102 747.57
## Use the operation or the process
thought of to answer the problem.
P 102 747.57
123.41
102 624.16
## Mother has a remaining balance of
P 102 624.16 in the bank on Nov. 30
## Write the final answer to the problem.
PRACTICE
Answer the word problems.
1) During a vacation, Bens records showed gasoline purchases of 19.75 gallons , 15.4
gallons, 13.85 gallons and 21.06 gallons. How many gallons of gasoline did he buy ?
Given : ____________________________________________________________
Operation: __________________________________________________________
Number Sentence: ___________________________________________________
36
2) The perimeter of a triangle is equal to the sum of the length of its sides. Find the
perimeter of a triangle whose sides are 8.75 cm, 9.6 cm and 10.375 cm.
Given: ___________________________________________________________
Operation : ________________________________________________________
Number Sentence: _________________________________________________
3) Rachel has P 3 316.40 in her savings account. If she made withdrawals of P 285.00,
P 472.46 and P 1 042.25, how much money is left in her account ?
Given: _____________________________________________________________
Operation: __________________________________________________________
Number Sentence: ___________________________________________________
4) One week a jogger ran the following distances : 15.3 km , 18.75 km , 19 km , 21.5 km
,25.375 km and 30.25 km. If his weekly average is 150 km, did he run less or more ?
5) Romeo has P 20 000 in available credit on his Visa charge card. If she purchases a
portable CD player for P 8 139.55, a boom box for P 399.95 and two way speakers
for P 3 425.05, how much available credit does she have left ?
Given: ______________________________________________________________
Operation: ___________________________________________________________
Number Sentence: ____________________________________________________
37
MORE PRACTICE
Analyze and solve the problems below. Choose the letter of the correct answer.
1) Two tourists traveled a total of 6 102.67 km during a tour. If they traveled 3 713.79
km by airplane and completed the rest of the tour by car, how many kilometers did
they travel by car ?
A) 9816.46 km
B) 98164.6 km
C) 2388.88 km
D) 238.888 km
2) What is the perimeter of a triangle whose sides measure 27.15 cm, 36.8 cm and 39
cm ?
A) 102.95 cm
B) 1029.5 cm
C) 38965.68 cm
D) 3896.568 cm
3) Mrs. Sta. Agueda spent the following amounts while shopping for her children: P
246.75 for an umbrella, P384.95 for a pair of shoes, P306.25 for a bag and P 311.10
for some small items.If she initially had P 1 500.00, how much money does she have
now ?
A) P 1 249.05
B) P 250.95
C) P 2 749.05
D) P 562.05
4) A barangay has a total road length of 184.75 km. If 99.54 km of this road had been
paved by the previous administration and 36.8 km by the present administration, how
much more remains to be paved ?
A) 321.09 km
B) 173.14 km
C) 136.34 km
D) 48.41 km
5) Rosa wanted to give her mother a Bible which costs P 195.75 but she has saved
P 150.00 only. If her older sister gives her P 30.00 more, would Rosa be able to buy
the Bible ?
A) YES
B) NO
C) MAYBE
D) CANNOT SAY
6) Junjun bought a backpack for P 165.50, notebooks for P 55.75 and other school
things for P 68.60. How much change should he get from P 500 ?
A) P 210.15
B) P 289.85
C) P 789.85
38
D) P 289.05
7) A triangle has sides that measure 26.718 m and 28.6 m. If its perimeter is 85.608 m,
what is the measure of the third side ?
A) 55.318 m
B) 678.5268 m
C) 30.29 m
D) 64.62 m
8) Mac had two baggage that weighed 20.75 kg and 13.8 kg. The airplane he would
take allowed only 30 kg of baggage per person. By how much were his baggage
overweight ?
A) 6.95 kg
B) 4.55 kg
C) 64.55 kg
D) 34.55 kg
9) Jill bought a kilo of rice at P 21.50 and 2.5 kg of sugar at P 55.90. How much change
should she get from P 100.00 ?
A) P 77.40
B) P 20.10
C) P 22.37
D) P 22.60
10) Lucille paid P 98.95 for a book, P 65.90 for a ream of bond paper, P 256.90 for a
pen set, and P 179.75 for a drawing set. How much change should she get if she
gave the salesclerk P 650 ?
A) P 601.50
B) P 1 201.50
C) P 48.50
D) P 327.20
HOMEWORK
1) How much should the following pay for their orders ?
Mc Jays Native Kakanin
kutsinta
P 7.75
banana cue
P 5.30
coffee
P 11.25
puto
P 4.50
soft drink
P 9.50
milk
P 10.75
suman
P 5.25
tea
P 10.85
## A) Rafael 2 kutsinta, tea
B) Tedjoy 1 suman, soft drink
C) Macky 2 banana cue, milk
D) Bernie 3 puto, coffee
39
## E) Dawn 3 kutsinta, coffee
2) This year, a school parade was 4.06 km. Last year, the route was 3.97 km. How
much shorter was the parade last year ?
3) Edna saved P 0.95 on Friday and P 0.87 on Saturday. How much did Edna save in
two days ?
4) Father gave mother five hundred peso bill for marketing. Mother spent P 112.50 for
rice and P 161.75 for fish and meat. How much was left of the five hundred pesos ?
5) Aubrey had P 86.70 in savings. Her mom gave her P 50.00 more. After buying
materials for her project she had P 26.75 left. How much did the material cost ?
CHAPTER FOUR
40
Multiplying Decimals
Lesson No. 1 ..
Multiplying Decimals
Lesson No. 2
Whole Numbers
Lesson No. 3.
Decimals
Lesson No. 4
1 000
and 0.001
Lesson No. 6 .
## Problem Solving Involving
Multiplication of Decimals
Lesson No. 1
## Multiplying Decimals Through Ten Thousandths
41
An easy way of multiplying decimals is to multiply the factors as if they are whole
numbers. Then count the number of decimal places of the factors and the total is the
number of the decimal places in the product.
Here is an example of multiplying decimals :
.45
2 decimal places
X .38
2 decimal places
360
135
0. 1 7 1 0
4 decimal places
PRACTICE
Multiply these.
1) 0.83
2) 0.38
3) 0.23
4) 0.49
0.7
0.2
0.3
6) 0.374
7) 0.827
8) 0.395
0.9
11) 0.9855
X
0.8
0.8
12) 0.8246
X
0.3
0.8
9) 0.826
0.6
13) 0.2796
X
5) 0.54
0.4
0.4
0.4
10) 0.934
X
0.2
14) 0.9288
15) 0.2541
0.6
0.7
16) 0.64
17) 0.35
18) 0.89
19) 0.75
20) 0.27
X 0.84
X 0.78
X 0.26
X 0.98
X 0.61
21) 0.583
X
0.15
26) 0.1257
22) 0.614
23) 0.279
24) 0.428
25) 0.549
X 0.34
X 0.25
X 0.84
X 0.62
27) 0.8249
28) 0.3277
42
29) 0.3641
30) 0.9421
0.42
0.82
31) 0.95
32) 0.65
0.218
0.79
33) 0.41
0.32
34) 0.78
0.54
35) 0.64
0.279
X 0.824
X 0.527
X 0.348
36) 0.826
37) 0.729
38) 0.342
39) 0.518
40) 0.394
X 0.349
X 0.378
41) 0.8942
42) 0.6499
43) 0.4782
44) 0.5297
45) 0.5279
0.984
0.824
0.961
0.341
0.824
0.671
0.964
0.946
46) 0.2846
47) 0.9277
48) 0.6382
49) 0.8266
50) 0.8394
X 0.8247
X 0.9352
X 0.8645
X 0.3213
X 0.6152
MORE PRACTICE
Multiply the following decimals.
1) 0.23
2) 0.59
3) 0.89
4) 0.91
5) 0.57
X 0.08
X 0.06
X 0.05
X 0.07
X 0.09
6) 0.17
7) 0.97
8) 0.64
9) 0.63
10) 0.58
X 0.84
X 0.31
X 0.87
X 0.42
X 0.46
11) 0.589
X
0.37
11) 0.641
X
0.507
16) 0.8441
12) 0.284
X
0.45
13) 0.349
X
14) 0.389
0.37
0.21
15) 0.642
X 0.96
12) 0.272
13) 0.645
14) 0.542
15) 0.319
X 0.219
X 0.197
X 0.342
X 0.247
17) 0.5186
18) 0.1224
19) 0.2135
43
20) 0.2187
0.2184
X 0.2581
X 0.8156
0.2415
X 0.5318
HOMEWORK
Multiply the following decimals.
1) 0.32
2) 0.826
3) 0.9578
4) 0.3527
5) 0.1852
X 0.5
X 0.27
X 0.824
X 0.2759
0.52
Lesson No. 2
44
X
## 1 6.2 0 ( two decimal places )
An easy way of multiplying decimals is to multiply the factors as if they are whole
numbers. Then count the number of decimal places of the factors and the total is the
number of decimal places in the product.
PRACTICE
Multiply the following.
1) 4.24
X
2) 7.84
6) 9.236
X
4) 8.29
5) 5.17
7) 2.743
12
3) 3.65
54
9) 6.893
10)5.398
28
12) 7.9503
13) 1.4187
16) 6.3702
17) 5.5921
52
91
14) 3.2346
X
18) 2.9150
34
8) 1.615
11) 8.0165
6
15) 4.7860
19) 8.0703
14
72
20) 4.1658
37
69
MORE PRACTICE
Multiply the following.
1) 1.38
2) 2.67
3) 35.08
4) 35.08
5) 14.34
18
6) 45.379
7) 41.327
23
8) 35.633
71
9) 83.621
45
35
10) 52.674
85
96
11) 6.3278
12) 2.3975
16) 351.2895
X
27
13) 6.8214
17) 234.1822
X
18) 159.3164
X
33
94
14) 5.2288
15) 6.3419
19) 321.2854
X
20) 217.3189
X
HOMEWORK
Multiply the following.
1) 2.89
2) 3.84
3) 6.33
4) 7.28
5) 9.31
6) 12.567
7) 34.285
8) 51.611
9) 57.319
10) 61.379
11) 3.2894
12) 6.8279
13) 5.2377
14) 9.1622
15) 3.7755
21
34
52
97
68
Lesson No. 3
## Multiplying Mixed Decimals by Mixed Decimals
Study the example below.
46
23.68
( 2 decimal places )
X 3.1
( 1 decimal place )
23 68
71 04
73.408
( 3 decimal places )
PRACTICE
Multiply the following.
1) 2.7
2) 3.8
3) 9.4
4) 3.1
5) 5.7
X 2.4
X 6.1
X 1.1
X 6.7
X 9.2
6) 3.65
7) 6.28
8) 9.27
9) 3.71
10) 5.61
X 7.1
X 8.4
X 5.4
5.2
11) 32.891
X
5.21
16) 63.2189
X
6.2
12) 52.894
X
6.21
17) 63.2877
X
1.55
6.3
13) 63.821
14) 54.375
5.37
15) 61.975
8.74
6.25
18)95.2118
19)58.3122
20) 14.2975
3.218
MORE PRACTICE
Multiply the following.
47
7.9
6.34
1) 3.8
2) 9.8
3) 4.7
4) 6.7
5) 3.1
X 2.8
X 6.7
X 5.9
X 8.1
X 7.2
6) 6.34
7) 8.19
8) 9.27
9) 8.63
10) 9.34
6.4
11) 8.52
7.5
9.4
4.2
6.5
12) 6.34
13) 7.88
14) 4.22
15) 9.12
X 2.54
X 1.94
X 7.54
X 9.35
16) 7.254
17) 6.934
18) 4.213
19) 9.212
20) 2.643
6.21
5.62
3.21
X 3.96
6.24
21) 3.5218
22) 6.3124
23) 8.2941
3.8
5.21
5.11
## 24) 6.3227 25) 6.2174
5.46
6.315
6.41
HOMEWORK
Multiply the following.
1) 6.9
2) 9.234
3) 6.2491
4) 57.82
X 2.8
2.5
1.56
8.7
5) 22.143
X
6) 6.9724
7) 5.61
8) 6.41
9) 621.34
3.21
3.1
9.2
6.94
6.4
10) 521.8216
X
7.9
Lesson No. 4
## Multiplying Decimals by 10, 100 and 1 000
Study the following examples:
A
B
48
0.8 X 10 = 8
0.34 X 10 = 3.4
0.215 X 10 = 2.15
0.5432 X 10 = 5.432
0.8 X 100 = 80
0.34 X 100 = 34
0.215 X 100 = 21.5
0.5432 X 100 = 54.32
## 0.8 X 1 000 = 800
0.34 X 1 000 = 340
0.215 X 1 000 = 215
0.5432 X 1 000 = 543.2
Now, can you give a short cut for multiplying a decimal number by number like
10, 100 and 1 000 ?
Multiplying decimals by 10 , 100 and 1 000 makes a greater number, so move
the decimal point to the right.
When multiplying decimals by powers of 10, move the decimal point to the right
as many places as there are zeros in the multiplier. Add zero or zeros when needed.
PRACTICE
Complete the table.
Decimal
X 10
X 100
1) 0.934
2) 0.54
3) 0.076
4) 3.248
5) 73.6489
6) 676.5498
7) 835.967
8) 1245.38675
9) 2106.2776
10) 6790.28345
Complete the table below by supplying the product of the given factors.
X 1 000
Decimal
1) 213.049
2) 235.135
3) 841.209
4) 732.143
5) 123.4561
6) 234.5674
7) 892.0203
8) 412.0532
9) 925.2145
10) 373.8214
X 1 000
X 10
X 100
MORE PRACTICE
Multiply the following.
1) 0.386
2) 0.86
3) 0.36
4) 0.473
X 100
X 1000
X 1000
10
49
5) 0.496
X
10
6) 0.85
7) 0.7
8) 0.512
9) 0.93
X 1000
X 1000
X 100
11) 0.289
12) 0.009
13) 0.23
X 1000
X 1000
17) 0.871
18) 0.004
19) 0.70
X 1000
X 1000
X 100
1000
16) 0.39
X
1000
100
14) 0.956
X
100
10) 0.51
X
10
15) 0.365
X
10
20) 0.603
X
10
## Multiply these orally.
1) 0.826 x 10 = _____
## 0.826 X 1000 = _____
2) 0.632 X 10 = _____
## 0.632 X 1000 = _____
3) 0.068 X 10 = _____
## 0.068 X 1000 = _____
4) 0.9 X 10 = _____
## 0.9 X 1000 = _____
5) 0.43 X 10 = _____
## 0.43 X 1000 = _____
HOMEWORK
Complete the table.
Decimal
1) 2.5672
2) 32.18
3) 0.9423
4) 3.8816
5) 21.67211
6) 63.21894
7) 0.2194
8) 6.28545
9) 364.28164
10) 6.38411
X 10
X 100
X 1 000
Lesson No. 5
## Multiplying Decimals by 0.1, 0.01 and 0.001
We move the decimal point 1 place to the left when the decimal number is
multiplied by 0.1. Example:
50
## 18.2 X 0.1 = 1 8 . 2 = 1.82
We move the decimal point 2 places to the left when the decimal number is
multiplied by 0.01. Example:
## 18.2 X 0.01 = 1 8 . 2 = 0.182
We move the decimal point 3 places to the left when the decimal number is
multiplied by 0.001.Example:
## 18.2 X 0.01 = 01 8 . 2 = 0.0182
PRACTICE
Multiply these numbers by:
Factors
1) 38.4
2) 397.8
3) 4 128.76
4) 178.73
5) 839.25
6) 1 473.63
7) 4 472.198
8) 81 813.418
9) 21 934.1473
10) 58 254.0345
11) 788.0854
12) 1 666.765
13) 5 795.275
14) 7 351.379
15) 75 582.145
16) 7 851.98
17) 95.33
18) 5 678.432
19) 859.596
20) 6 785.034
0.1
0.01
MORE PRACTICE
Give the missing numbers in the table.
51
0.001
Factors
1) 8.314
2)
3)
4) 7 638.2
5)
6) 3 468.4
7)
8)
9)
10) 49 875.42
0.1
831.4
0.01
0.001
478.2
47.82
93.54
9 354
76.382
987.635
34.684
9 876.35
975.13
9.7513
635.427
1 025.631
10.25631
498.7542
## Multiply the following.
1) 3256.1978
2) 63.2194
3) 0.52
4) 1.375
5) 621.32
X 0.001
0.001
0.01
0.1
0.01
6) 0.342
7) 12.3
8) 1457.36
9) 921.3
10) 8245.3214
X 0.001
X 0.01
0.1
0.01
0.001
HOMEWORK
Complete the table by multiplying the decimals with 0.1, 0.01 and 0.001.
Decimals
1) 324.12
2) 9 625.3
3) 634.057
4) 0.218
5) 2.31
6) 361.211
7) 348.96
8) 6325.12
9) 85.32
10) 6.3
X 0.1
X 0.01
X 0.001
Lesson No. 6
## Problem Solving Involving Multiplication of
Decimals
Analyze the problem below.
52
Rosemarie sent two packages, each weighing 1.85 kg, to her brothers in Hong Kong. If
she paid P 175.00 per kg of the package, how much did she pay in all ?
What is asked ?
What are given ?
What are the operations to be used ?
Write the equation to solve the problem ?
## Answer more problems !
PRACTICE
Answer the problems below .
1) Edna needs 0.95 m of ribbon to trim a shirt. If a ribbon costs P 0.75 metre how much
will the ribbon cost ?
Given : ______________________________________________________________
Operations: __________________________________________________________
Equation: ____________________________________________________________
2) Carmen can swim 12 meters in one minute. How far can she swim in 30 minutes ?
Given : ______________________________________________________________
Operations: __________________________________________________________
Equation: ____________________________________________________________
3) Milagros bought 6 siopao worth P 9.50 each. How much did she pay in all ?
Given : ______________________________________________________________
Operations: __________________________________________________________
Equation: ____________________________________________________________
53
4) If she added 13 large soft drinks for her friends which cost P 15.50 each, what is the
total cost of the softdrinks ?
Given : ______________________________________________________________
Operations: __________________________________________________________
Equation: ____________________________________________________________
5) What is the area of Aidas backyard if it is 15.935 m long and 6.45 m wide ?
Given : ______________________________________________________________
Operations: __________________________________________________________
Equation: ____________________________________________________________
6) Dante measured their dining table. Its length is 2.643 m while its width is 1.45 m.
What is the area of the dining table ?
Given : ______________________________________________________________
Operations: __________________________________________________________
Equation: ____________________________________________________________
7) In Marikina Elementary, 100 pupil in Grade IV, V, VI joined the field trip in Tagaytay
City. If each pupil paid P 182.25 , how much is the total collection ?
Given : ______________________________________________________________
54
Operations: __________________________________________________________
Equation: ____________________________________________________________
8) A soft drinks company gave out 1 000 free t-shirts to those who patronage their
products. Each t-shirt costs P 35.83. How much did the soft drinks company pay for
the t-shirts?
Given : ______________________________________________________________
Operations: __________________________________________________________
Equation: ____________________________________________________________
9) A coach buys 19 pairs of socks at P 68.75 per pair. What is the total cost of the
socks?
Given : ______________________________________________________________
Operations: __________________________________________________________
Equation: ____________________________________________________________
10) Ellen buys 53 gallons of gasoline at P 79.09 per gallon. What is the total cost of the
gasoline ?
Given : ______________________________________________________________
Operations: __________________________________________________________
Equation: ____________________________________________________________
MORE PRACTICE
Analyze and solve the following problems.
55
## 1) Kenny made a down payment of P 1 750 on a camcorder and agreed to make
payments of P 520.85 per month for one year. Find the total cost of the camcorder.
2) Joel earns P 8 355.35 per month. If Karen earns twice as much as Joel, how much
do they earn together ?
3) The cost of a photocopy in store A is P 0.35 while in store B it costs P 0.53. How
much do I save if I had 317 pages photocopied at store A instead of store B ?
4) The Avis Car Rental Company charges P 1 243.45 per day and P 20 per km to rent
their cars. The ABC Car Rental Company charges P 1 048.25 per day and P 21.25
per km. Which company will you choose if you rent a car for two days and travel 325
km ?
5) If a meter of cloth cost P 72.95 , how much would 6.4 meters cost ?
HOMEWORK
Solve the following problems .
1) Mang Elo, a balut vendor, bought 100 new duck eggs at P 3.65 each. How much did
he pay for all the eggs ?
2) Mrs. Pasana baked 1 000 pineapple pie for a party. If a pie costs P 17.85, how much
did the 1 000 pie cost ?
3) A cone of ice cream cost P 16.25, how much in all did the 6 children spend on ice
56
cream ?
4) Mang Juan sold 46 cotton candies at P 2.15 each. How much altogether is the cost of
the cotton candies ?
## 5) A rope measures 4.63 m. How long is it in centimeters ?
6) The peso dollar exchange rate is P 39.25. If Justin had \$ 27.50, how much would
this amount in pesos ?
7) A factory needs 0.085 ton of glass and 0.012 of rubber to build a truck. How many
tons to glass and rubber are needed for 295 trucks ?
8) A car designed for use as a police car is priced at P 502 200. For municipalities
buying 10 or more units, the price is P 475 500.25. What would be the total cost of 25
cars ?
## 9) If a car travels 55.6 km an hour, how far will it travel in 8 hours ?
10) Rosemarie sent two packages, each weighing 1.85 kg, to her brothers in Hong
Kong. If she paid P 175.00 per kg of the package, how much did she pay in all ?
57
Chapter Five
Division of Decimals
Lesson No. 1
Kinds of Decimals
Lesson No. 2
Lesson No. 3
Lesson No. 4
Numbers
Lesson No. 5
Lesson No. 6
Lesson No. 7
Mentally
Lesson No. 8
## Solve Word Problems Involving Division of
Decimals
Lesson No. 1
KINDS OF DECIMALS
Decimals can be classified as terminating or non terminating.
58
1) TERMINATING Decimal is one that has a definite ending decimal place value.
Example:
_ 51
5 / 255
25
5
5
0
## 2) NON TERMINATING Decimal have an infinite number of decimal places
a) Repeating example
0.333333333333
## b) Non repeating example
0.325841562
Lesson No. 2
Divide Whole Numbers by Decimals and Vice Versa
Check whether you can divide correctly all the numbers given below.
1) 4 / 6 484
2) 6 / 7 044
3) 7 / 9 185
4) 22 / 4850
How about when you divide decimals. Study the example below.
59
1)
0.9 / 63
9 / 630
070
9 /630
0
63
63
0
0
X
____
2) 2/0.24
.12
2/0.24
2
4
4
X
## - move the decimal one place to the right
- add one zero to the dividend
- divide
## - write a decimal directly above the dividend
- divide
PRACTICE
A. Divide the following.
1) 72 0.8 = _____
## 8) 564 0.25 = _____
4) 49 0.7 = _____
60
## B. Divide the following.
1) 0.756 9 = _____
6) 0.846 6 = _____
2) 0.966 6 = _____
7) 0.434 7 = _____
3) 0.686 2 = _____
8) 0.775 5 = _____
4) 0.714 7 =_____
9) 0.872 8 = _____
5) 0.305 5 = _____
## 10) 0.316 2 = _____
MORE PRACTICE
Divide the following.
1) 696 0.09 = _____
6) 0.936 78 = _____
## 2) 888 0.37 = _____
7) 0.924 66 = _____
## 3) 975 0.25 = _____
8) 0.939 69 = _____
## 4) 595 0.85 = _____
9) 0.832 34 = _____
## 10) 0.833 49 = _____
HOMEWORK
Divide the following.
1) 183 0.61 = _____
6) 0.241 2 = _____
## 2) 900 0.45 = _____
7) 0.2564 80 = _____
## 3) 598 0.22 = _____
8) 0.98324 16 = _____
## 4) 775 0.15 = _____
9) 0.925 5 = _____
Lesson No. 3
## Divide Decimals by Decimals
When dividing decimals, move the decimal point of the divisor until it becomes a
whole number. Move also the decimal of the dividend the same number of places the
decimal point of the divisor has been moved. When the divisor is already a whole
number, you can proceed to division. Make sure the decimal point of the quotient is
directly above the newly placed decimal point of the dividend.
Example:
61
_______
0.52 / 0.85241
_______
0.238 / 0.25419
______
0.5 / 0.351
______
52 / 85.241
______
238 / 254.19
_____
5 / 3.51
2 moves
3 moves
1 move
PRACTICE
Divide the following.
1) 0.72 0.3 = _____
## 10) 0.61284 0.04 = _____
MORE PRACTICE
Divide the following.
1) 0.96 0.6 = _____
## 3) 0.212 0.08 = _____
HOMEWORK
Divide the following.
1) 0.24 0.012 = _____
Lesson No. 4
## Division of Mixed Decimals and Whole
Numbers
Study the example below.
Example 1:
_______
3.2 / 160
- move the decimal of the divisor one place to the right and add
62
50
32 / 1600
160
0
0
0
____
- divide
Example 2:
____
4 / 2.4
0.6
4 / 2.4
0
24
24
0
## - place a decimal point in the quotient directly above the decimal
point in the dividend
- divide
PRACTICE
Divide the following.
___
1) 7.4 / 15
____
2) 9.6 / 541
___
3) 8.5 / 22
___
4) 1.5 / 49
____
5) 2.5 / 475
____
6) 3 / 9.6
_____
7) 9 / 1.89
_____
8) 5 / 36.5
_____
9) 16 / 28.80
MORE PRACTICE
Divide the following.
____
1) 1.84 / 273
_____
6) 9 / 1.42
____
2) 1.4 / 24
____
7) 42 / 4.31
_____
3) 2.9 / 4 365
____
8) 19 / 14.2
____
4) 2.1 / 342
____
9) 48 / 16.3
63
_____
10) 8 / 5.60
______
5) 7.8 / 3 568
_____
10) 15 / 28.5
HOMEWORK
Divide the following.
______
1) 3.5 / 241.5
_____
6) 92 / 579.6
_____
2) 4.3 / 51.6
____
7) 18 / 85.3
______
3) 5.16 / 206.41
_____
8) 51 / 36.51
______
4) 10.9 / 62.981
____
9) 13 / 58.5
_______
5) 6.5 / 374.21
____
10) 21 / 75.6
Lesson No. 5
## Division of Mixed Decimals and Mixed
Decimals
Mixed decimals are divided in the same way as whole numbers. In dividing mixed
decimals by mixed decimals, the decimal point in the divisor is moved to the right to
make it a whole number. You only have to move the decimal place of the dividend the
same number the decimal point was moved in the divisor, unless if the dividend in a
64
whole number then you add zero. Align the decimal point of the quotient to that of the
dividend.
PRACTICE
## Divide. Classify the quotient as to terminating or non terminating. If it is non terminating,
classify if it is repeating or non repeating. Write only the letter on the space before the
number.
A. terminating decimal
## B. non terminating, repeating decimal
decimal
____
_____1) 4.2 / 12.4
______
_____6) 1.52 / 395.2
____
_____2) 4.1 / 16.4
____
_____7) 4.5 / 4.28
______
_____3) 3.12 / 7.3008
____
_____8) 6.3 /55.2
______
_____4) 3.60 / 311.04
______
_____9) 7.42 / 244.86
______
_____5) 25.4 / 171.45
_____
_____10) 2.4 / 873.5
MORE PRACTICE
## Divide. Classify the quotient as to terminating or non terminating. If it is non terminating,
classify if it is repeating or non repeating. Write only the letter on the space before the
number.
A. terminating decimal
decimal
65
## _____10) 14.629 92 4.8 = _____
HOMEWORK
Divide. Classify the quotient as to terminating or non terminating. If it is non terminating,
classify if it is repeating or non repeating. Write only the letter on the space before the
number.
A. terminating decimal
decimal
Lesson No. 6
## Divide Decimals by 10 , 100 1,000 Mentally
When a decimal is to be divided by:
10, move the decimal point 1 place to the left to get the quotient
100, move the decimal point 2 places to the left to get the quotient
1 000, move the decimal point 3 places to the left to get the quotient
EXAMPLES:
148.2 10 =
14.82
66
148.2 100 =
1.482
## 148.2 1 000 = 0.1482
The number of zeros in the divisor will tell you the number of places to the left of
the decimal point that would be skipped.
PRACTICE
Divide these numbers by 10, 100 and 1 000.
DIVIDEND
1) 5 147.4
2) 2 183.6
3) 3 452.2
4) 8 318.18
5) 9 493.37
6) 27 345.49
7) 39 146.32
8) 54 987.18
9) 176 347.45
10) 882 963.74
11) 330 153.1
12) 418 253.18
13) 651 210.54
14) 531 053.49
15) 781 255.31
16) 2.364
17) 8.342
18) 52.6217
19) 614.32841
20) 6.328
10
100
1 000
100
43.812
1 000
MORE PRACTICE
FACTORS
1) 4 381.2
2)
3)
4) 6 768.39
5)
6) 3 562.1
10
963.47
157.293
676.839
1 982.418
9.6347
15.7293
198.2418
43.812
67
7)
8)
9)
10) 8 135.87
5 327.8
91 025.83
6 232.4
532.78
9 102.583
62.324
81.3587
HOMEWORK
## Divide these numbers as fast as you can.
1) 43.8 10 = __________
## 11) 59 378.14 100 = __________
2) 196.47 10 = __________
## 12) 41 821.36 100 = __________
3) 343.18 10 = __________
## 13) 97 415.43 100 = __________
4) 456.39 10 = __________
## 14) 186 244.35 1 000 = __________
5) 1 415.43 10 = __________
## 15) 246 132.89 1 000 = __________
6) 9 823.93 10 = __________
Lesson No. 7
## Divide Decimals by 0.1 , 0.01 , 0.001 Mentally
When dividing decimal by:
0.1, move the decimal point of the dividend to one place to the right to get the
quotient
0.01, move the decimal point of the dividend to two places to the right to get the
quotient
68
0.001, move the decimal point of the dividend to three places to the right to get
the quotient
EXAMPLE:
43.2 0.1 =
432
43.2 0.01 =
4320
43.2 0.001 =
43200
## 5.2371 0.001 = 5237.1
PRACTICE
Divide these mentally.
DIVIDEND
1) 4.8471
2) 15.3221
3) 18.1962
4) 35.4325
5) 95.2147
6) 12.1732
7) 89.2473
8) 381.1925
9) 473.1543
10) 2 135.2184
11) 2 529.685
12) 1 283.912
13) 7 568.405
14) 2 567.834
15) 2 456.5852
16) 632.12
17) 1284.3
18) 82.6
19) 6.2
20) 1285.34
0.1
0.01
0.001
0.01
0.001
MORE PRACTICE
Give the mixing numbers in the table.
DIVIDEND
1) 15.3943
2)
3)
4)
5) 4 192.1576
6)
7)
8)
0.1
417.2153
934.2476
81 529.4
351.84
90.804
3 344.95
69
9) 4 376.37
10)
81 529.4
7 611.663
HOMEWORK
Divide the following mentally.
1) 6.32 0.001 = __________
Lesson No. 8
## PROBLEM SOLVING INVOLVING DIVISION
OF DECIMALS
Analyze the problem below.
A customer of Aling Letty paid Php 217.50 for 3 kilos of chicken. How much
did Aling Letty charge for each kilo of chicken ?
What is asked ?
What are given ?
What are the operations to be used ?
70
## Write the equation to solve the problem ?
PRACTICE
Answer the word problems below.
1) Mrs. Lituanias class has to raise Php 287.50 for a globe in Hekasi class. There are
50 pupils in the class. How much will each pupil contribute ?
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
2) Mrs. Malabanan bought 5 reams of coupon bond worth Php 447.50. How much did
each ream of coupon bond cost ?
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
3) Zaldy cut a bamboo 12 meters long. If he were to cut it into pieces 1.2 m long, how
many pieces of bamboo will there be ?
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
4) How many pieces of cloth each 4.2 m long can be cut from a 24 meter piece ?
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
71
5) Mylene bought 7.5 meters of tetoron at Php 198.75. How much does a meter cost ?
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
MORE PRACTICE
Answer the word problem below.
1) A container full of papayas weigh 22.1 kilograms. If each papaya weighs 1.2
kilograms and the container weighs 0.5 kg, how many pieces were in the container ?
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
2) Mrs. Dela Paz bought eggs for Php 686.40. If eggs cost Php 26.40 per dozen, how
many dozen did she buy ?
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
3) Dianne sold 166.25 kg of grapes. If each grapes weighed about 1.75 kg, how many
packs of grapes did Dianne sell ?
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
4) Ana earns Php 43.25 per hour. If her salary for one week was Php 2 335.50 , how
many hours did she work ?
72
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
5) Three men paid a total of P 1 026 for lunch and the cost of the meal is divided
equally. How much did each men pay ?
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
HOMEWORK
Answer the word problem below.
1) Juans odometer read 38 796.1 km at the beginning of his trip. When the trip ended, it
read 39 365.8 km. If he used 31.5 gallons of gasoline, for the whole travel, how many
kilometers did he travel per gallon of gasoline ?
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
2) A ream of bond paper is 6.21 cm thick. If each sheet of paper is 0.003 cm thick, how
many sheets of paper are there in two reams ?
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
3) Rosa found out that she had read 3 618 words in 4.02 minutes. How many words did
she read per minute ?
73
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________
4) If my watch is late by 4.5 seconds per hour, by how many minutes will it be late in two
days ?
Given: ________________________________________________________________
Operation: _____________________________________________________________
Number Sentence: ______________________________________________________ |
# Into Math Grade 6 Module 10 Answer Key Real-World Relationships Between Variables
We included HMH Into Math Grade 6 Answer Key PDF Module 10 Real-World Relationships Between Variables to make students experts in learning maths.
## HMH Into Math Grade 6 Module 10 Answer Key Real-World Relationships Between Variables
Which Relationship Does NOT Belong?
Describe the pattern shown by each relationship.
A.
– The package of 5 markers costs $3 – The package of 10 markers costs$6
– The package of 15 markers costs $9 In the column of a number of markers, the number is increased by adding 5 to the next number. 5 + 5 = 10; 10 +5 = 15; In the cost column, the cost is increased by adding 3 to the next number. 3 + 3 = 6; 6 + 3 = 9. B. Answer: In the column of a number of markers, the number is increased by multiplying 2. 8 x 2 = 16 In the cost column, the cost is increased by multiplying 2 4 x 2 = 8 C. Answer: It is a perfect positive correlation. – A scatter diagram is used to examine the relationship between both the axes (X and Y) with one variable. In the graph, if the variables are correlated, then the point drops along a curve or line. A scatter diagram or scatter plot gives an idea of the nature of the relationship. – In a scatter correlation diagram, if all the points stretch in one line, then the correlation is perfect and is in unity. D. Answer: The pack of 6 markers costs$3.60
The pack of 12 markers costs \$7.20
In the column of a number of markers, the number is increased by multiplying 2.
6 x 2 = 12
In the cost column, the cost is increased by multiplying 2
3.60 x 2 = 7.20
Turn and Talk
Question 1.
Which relationship does not belong? Explain why.
The scatter diagrams do not belong to the relationship.
The Scatter diagram method is a simple representation that is popularly used in commerce and statistics to find the correlation between two variables. These two variables are plotted along the X and Y axis on a two-dimensional graph and the pattern represents the association between these given variables.
Question 2.
Which relationship represents the best deal? Explain your reasoning.
Complete these problems to review prior concepts and skills you will need for this module.
Generate Patterns and Find Relationships
Describe a pattern you see in each table. Use the pattern to complete each table.
Question 1.
In the X column and Y column, the number is increased by multiplying 6.
0 x 6 = 0; 1 x 6 = 6; 2 x 6 = 12; 3 x 6 = 18; 4 x 6 = 24
Therefore, the filled table can be shown below:
Question 2.
In the x column, the number is increased by adding 1
0 + 1 = 1; 1 +1 = 2; 2 + 1 = 3; 3 + 1 = 4
In the Y column, the number is increased by adding 5
8 + 5 = 13; 13 + 5 =18; 18 + 5 = 23; 23 + 5 = 28
Identify Points on a Coordinate Grid.
Write the ordered pair for each point.
Question 3.
A _______________
Ordered pairs are usually used in coordinate geometry to represent a point on a coordinate plane. Also, they are used to represent elements of a relation.
– An ordered pair is a pair formed by two elements that are separated by a comma and written inside the parentheses. For example, (x, y) represents an ordered pair, where ‘x’ is called the first element and ‘y’ is called the second element of the ordered pair.
The ordered pair of point A is (0, 9)
Question 4.
B _______________
Ordered pairs are usually used in coordinate geometry to represent a point on a coordinate plane. Also, they are used to represent elements of a relation.
– An ordered pair is a pair formed by two elements that are separated by a comma and written inside the parentheses. For example, (x, y) represents an ordered pair, where ‘x’ is called the first element and ‘y’ is called the second element of the ordered pair.
The ordered pair of point B is (2, 7)
Question 5.
C _______________
Ordered pairs are usually used in coordinate geometry to represent a point on a coordinate plane. Also, they are used to represent elements of a relation.
– An ordered pair is a pair formed by two elements that are separated by a comma and written inside the parentheses. For example, (x, y) represents an ordered pair, where ‘x’ is called the first element and ‘y’ is called the second element of the ordered pair.
The ordered pair of C is (1, 3)
Question 6.
D _______________
Ordered pairs are usually used in coordinate geometry to represent a point on a coordinate plane. Also, they are used to represent elements of a relation.
– An ordered pair is a pair formed by two elements that are separated by a comma and written inside the parentheses. For example, (x, y) represents an ordered pair, where ‘x’ is called the first element and ‘y’ is called the second element of the ordered pair.
The ordered pair of D is (6, 1)
Question 7.
E _______________
Ordered pairs are usually used in coordinate geometry to represent a point on a coordinate plane. Also, they are used to represent elements of a relation.
– An ordered pair is a pair formed by two elements that are separated by a comma and written inside the parentheses. For example, (x, y) represents an ordered pair, where ‘x’ is called the first element and ‘y’ is called the second element of the ordered pair.
The ordered pair of E is (7, 4)
Question 8.
F _______________
Ordered pairs are usually used in coordinate geometry to represent a point on a coordinate plane. Also, they are used to represent elements of a relation.
– An ordered pair is a pair formed by two elements that are separated by a comma and written inside the parentheses. For example, (x, y) represents an ordered pair, where ‘x’ is called the first element and ‘y’ is called the second element of the ordered pair.
The ordered pair of F is (9, 6)
Graph and label each ordered pair on the coordinate grid.
Question 9.
G(0, 7)
The following are simple and easy steps to locate a point in a cartesian plane. Check out the guidelines, terms, and conditions to plot ordered pairs.
– Let us take one ordered pair.
– Get the sign of the x coordinate, and y coordinate of the ordered pair.
– Based on those signs, identify which quadrant the ordered pair belongs to.
– Count the x coordinate value on the x-axis starting from the origin.
– Similarly, count the y coordinate value on the y-axis from the origin.
– Mark the obtained point as the ordered pair.
Question 10.
H(3, 8)
The following are simple and easy steps to locate a point in a cartesian plane. Check out the guidelines, terms, and conditions to plot ordered pairs.
– Let us take one ordered pair.
– Get the sign of the x coordinate, and y coordinate of the ordered pair.
– Based on those signs, identify which quadrant the ordered pair belongs to.
– Count the x coordinate value on the x-axis starting from the origin.
– Similarly, count the y coordinate value on the y-axis from the origin.
– Mark the obtained point as the ordered pair.
Question 11.
I(5, 5)
The following are simple and easy steps to locate a point in a cartesian plane. Check out the guidelines, terms, and conditions to plot ordered pairs.
– Let us take one ordered pair.
– Get the sign of the x coordinate, and y coordinate of the ordered pair.
– Based on those signs, identify which quadrant the ordered pair belongs to.
– Count the x coordinate value on the x-axis starting from the origin.
– Similarly, count the y coordinate value on the y-axis from the origin.
– Mark the obtained point as the ordered pair.
Question 12.
J(1, 0) |
# NCERT Solutions Class 9 Maths Chapter 2 Polynomials
### Exercise 2.1
1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer:-
(i) 4x2– 3x + 7
(ii) y2+√2
(iii) 3√t+t√ 2
(iv) y+$\frac{2}{y}$
(v) x10+y3+t50
Sol:- (i) 4x2–3x+7 is a polynomial in one variable. The equation 4x2–3x+7 can be written as 4x2–3x1+7x0 and since x is the only variable in the given equation and the powers of x (i.e., 2, 1 and 0) are whole numbers.
(ii) y2+2 is a polynomial in one variable. The equation y2+√2 can be written as y2+2y0 Since y is the only variable in the given equation and the powers of y (i.e., 2 and 0) are whole numbers.
(iii) 3√t+t√2 is not a polynomial in one variable. The equation 3√t+t√2 can be written as 3t1/2+√2t though, t is the only variable in the given equation, the powers of t (i.e.,1/2) is not a whole number.
(iv) y+$\frac{2}{y}$ is not a polynomial in one variable. The equation y+2/y an be written as y+2y-1 though, is the only variable in the given equation, the powers of y (i.e.,-1) is not a whole number.
(v) x10+y3+t50 is not a polynomial in one variable. Here, in the equation x10+y3+t50 though, the powers, 10, 3, 50, are whole numbers, there are 3 variables used in the expression x10+y3+t50.
2. Write the coefficients of x2 in each of the following:
(i) 2+x2+x
(ii) 2-x2+x3
(iii) $\frac{\pi&space;}{2}$x2+x
(iv) √2x-1
Sol:- (i) The coefficient of xin 2+x2+x is 1
(ii) The coefficient of xin 2–x2+xis -1.
(iii) The coefficient of xin $\frac{\pi&space;}{2}$x2+x is $\frac{\pi&space;}{2}$ .
(iv) The coefficient of xin √2x-1 is 0.
3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Sol:- Binomial of degree 35: A polynomial having two terms and the highest degree 35 is called a binomial of degree 35. Eg., 3x35+5.
Monomial of degree 100: A polynomial having one term and the highest degree 100 is called a monomial of degree 100. Eg., 4x100.
4. Write the degree of each of the following polynomials:
(i) 5x3+4x2+7x
(ii) 4 – y2
(iii) 5t –√7
(iv) 3
Ans:- (i) The degree of 5x3+4x2+7x is 3 as 3 is the highest power of x in the equation.
(ii) The degree of 4–y2 is 2 as 2 is the highest power of y in the equation.
(iii) The degree of 5t–√7 is 1 as 1 is the highest power of y in the equation.
(iv) The degree of 3 is 0.
5. Classify the following as linear, quadratic and cubic polynomials:
(i) x2+x
(ii) x-x3
(iii) y+y2+4
(iv) 1+x
(v) 3t
(vi) r2
(vii) 7x3
Sol:- (i) The highest power of x2+x is 2
So, x2+x is a quadratic polynomial.
(ii) The highest power of x–xis 3
So, x–x3 is a cubic polynomial.
(iii) The highest power of y+y2+4 is 2
So, y+y2+4is a quadratic polynomial.
(iv) The highest power of 1+x is 1
So, 1+x is a linear polynomial.
(v) The highest power of 3t is 1
So, 3t is a linear polynomial.
(vi) The highest power of ris 2
So, r2is a quadratic polynomial.
(vii) The highest power of 7xis 3
So, 7x3 is a cubic polynomial. |
Find the sum of the following series:
Question:
Find the sum of the following series:
(i) 2 + 5 + 8 + ... + 182
(ii) 101 + 99 + 97 + ... + 47
(iii) (a − b)2 + (a2 + b2) + (b)2 + ... + [(a + b)2 + 6ab]
Solution:
(i) 2 + 5 + 8 + ... + 182
Here, the series is an A.P. where we have the following:
$a=2$
$d=(5-2)=3$
$a_{n}=182$
$\Rightarrow 2+(n-1)(3)=182$
$\Rightarrow 2+3 n-3=182$
$\Rightarrow 3 n-1=182$
$\Rightarrow 3 n=183$
$\Rightarrow n=61$
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{61}=\frac{61}{2}[2 \times 2+(61-1) \times 3]$
$=\frac{61}{2}[2 \times 2+60 \times 3]$
$=5612$
(ii) 101 + 99 + 97 + ... + 47
Here, the series is an A.P. where we have the following:
$a=101$
$d=(99-101)=-2$
$a_{n}=47$
$\Rightarrow 101+(n-1)(-2)=47$
$\Rightarrow 101-2 n+2=47$
$\Rightarrow 2 n-2=54$
$\Rightarrow 2 n=56$
$\Rightarrow n=28$
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{28}=\frac{28}{2}[2 \times 101+(28-1) \times(-2)]$
$=\frac{28}{2}[2 \times 101+27 \times(-2$
$=2072$
(iii) (a − b)2 + (a2 + b2) + (b)2 + ... + [(a + b)2 + 6ab]
Here, the series is an A.P. where we have the following:
$a=(a-b)^{2}$
$d=\left(a^{2}+b^{2}-(a-b)^{2}\right)=2 a b$
$a_{n}=\left[(\mathrm{a}+\mathrm{b})^{2}+6 a b\right]$
$\Rightarrow(a-b)^{2}+(n-1)(2 a b)=\left[(\mathrm{a}+\mathrm{b})^{2}+6 a b\right]$
$\Rightarrow a^{2}+b^{2}-2 a b+2 a b n-2 a b=\left[a^{2}+b^{2}+2 a b+6 a b\right]$
$\Rightarrow a^{2}+b^{2}-4 a b+2 a b n=a^{2}+b^{2}+8 a b$
$\Rightarrow 2 a b n=12 a b$
$\Rightarrow n=6$
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{6}=\frac{6}{2}\left[2(a-b)^{2}+(6-1) 2 a b\right]$
$=3\left[2\left(a^{2}+b^{2}-2 a b\right)+10 a b\right]$
$=3\left[2 a^{2}+2 b^{2}-4 a b+10 a b\right]$
$=3\left[2 a^{2}+2 b^{2}+6 a b\right]$
$=6\left[a^{2}+b^{2}+3 a b\right]$ |
# Module 22: Surface Area of Common Solids
You may use a calculator throughout this module.
We will now turn our attention from two-dimensional figures to three-dimensional figures, which we often call solids, even if they are hollow inside.
In this module, we will look the surface areas of some common solids. (We will look at volume in a later module.) Surface area is what it sounds like: it’s the sum of the areas of all of the outer surfaces of the solid. When you are struggling to wrap a present because your sheet of wrapping paper isn’t quite big enough, you are dealing with surface area.
There are two different kinds of surface area that are important: the lateral surface area (LSA) and total surface area (TSA).
To visualize the difference between LSA and TSA, consider a can of soup. The lateral surface area would be used to measure the size of the paper label around the can. The total surface area would be used to measure the amount of sheet metal needed to make the can. In other words, the total surface area includes the top and bottom, whereas the lateral surface area does not.
# Rectangular Solids
A rectangular solid looks like a rectangular box. It has three pairs of equally sized rectangles on the front and back, on the left and right, and on the top and bottom.
A cube is a special rectangular solid with equally-sized squares for all six faces.
The lateral surface area is the combined total area of the four vertical faces of the solid, but not the top and bottom. If you were painting the four walls of a room, you would be thinking about the lateral surface area.
The total surface area is the combined total area of all six faces of the solid. If you were painting the four walls, the floor, and the ceiling of a room, you would be thinking about the total surface area.
For a rectangular solid with length , width , and height [1]
For a cube with side length
Note: These dimensions are sometimes called base, depth, and height.
Exercises
1. Find the lateral surface area of this rectangular solid.
2. Find the total surface area of this rectangular solid.
# Cylinders
As mentioned earlier in this module, the lateral surface area of a soup can is the paper label, which is a rectangle. Therefore, the lateral surface area of a cylinder is a rectangle; its width is equal to the circumference of the circle, , and its height is the height of the cylinder.
Since a cylinder has equal-sized circles at the top and bottom, its total surface area is equal to the lateral surface area plus twice the area of one of the circles.
For a cylinder with radius and height
Be aware that if you are given the diameter of the cylinder, you will need to cut it in half before using these formulas.
Exercises
3. Find the lateral surface area of this cylinder.
4. Find the total surface area of this cylinder.
# Spheres
The final solid of this module is the sphere, which can be thought of as a circle in three dimensions: every point on the surface of a sphere is the same distance from the center. Because of this, a sphere has only one important measurement: its radius. Of course, its diameter could be important also, but the idea is that a sphere doesn’t have different dimensions such as length, width, and height. A sphere has the same radius (or diameter) in every direction.
We would need to use calculus to derive the formula for the surface area of a sphere, so we’ll just assume it’s true and get on with the business at hand. Notice that, because a sphere doesn’t have top or bottom faces, we don’t need to worry about finding the lateral surface area. The only surface area is the total surface area.
For a sphere with radius or diameter
or
Coincidentally, the surface area of a sphere is times the area of the cross-sectional circle at the sphere’s widest part. You may find it interesting to try to visualize this, or head to the kitchen for a demonstration: if you cut an orange into four quarters, the peel on one of those quarter oranges has the same area as the circle formed by the first cut.
Exercises
5. Find the surface area of this sphere.
6. Find the surface area of this sphere.
1. You might choose to use capital letters for the variables here because a lowercase letter "l" can easily be mistaken for a number "1". |
# The Variance of 20 Observations is 5. If Each Observation is Multiplied by 2, Find the Variance of the Resulting Observations. - Mathematics
The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting observations.
#### Solution
Let $x_1 , x_2 , x_3 , . . . , x_{20}$ be the 20 given observations.
$\text{ Variance} (X) = 5$
$\text{ Variance } (X) = {\frac{1}{20}} \times \sum \left( {x_i - X} \right)^2 = 5 (\text{ Here , is the mean of the given observations } . )$
Let u1,u2,,u3, ..., u20 be the new observations, such that
$u_i = 2 x_i (\text{ for } i = 1, 2, 3, . . . , 20) . . . (1)$
$\text{ Mean } = \bar{U} = \frac{\sum^{20}_{i = 1} u_i}{n}$
$= \frac{\sum^{20}_{i = 1} 2 x_i}{20} \left[ \text{ substituting} u_i \text{ from eq (1) and taking n as } 20 \right]$
$= 2 \times \frac{\sum^{20}_{i = 1}{ x_i} }{20}$
$= 2 \bar{X}$
$u_i - \bar{U} = 2 x_i - 2 \bar{X} (\text{ for } i = 1, 2, . . . , 20)$
$= 2\left( x_i - \bar{X} \right)$
$\left( u_i - \bar{U} \right)^2 = \left( 2\left( x_i - \bar{X} \right) \right)^2 \left(\text{ squaring both the sides } \right)$
$= 4 \left( x_i - \bar{X} \right)^2$
$\therefore \sum^{20}_{i = 1} \left( u_i - \bar{U} \right)^2 = \sum 4^{20}_{i = 1} \left( x_i - \bar{X} \right)^2$
$\frac{\sum^{20}_{i = 1} \left( u_i - \bar{U} \right)^2}{20} = \frac{\sum 4^{20}_{i = 1} \left( x_i - \bar{X} \right)^2}{20}$
$= 4 \frac{\sum^{20}_{i = 1} \left( x_i - \bar{X} \right)^2}{20}$
$\text{ Variance } (U) = 4 \times \text{ Variance }(X)$
$= 4 \times 5$
$= 20$
Thus, variance of the new observations is 20.
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 32 Statistics
Exercise 32.4 | Q 2 | Page 28 |
# Why does $\sum_{k=1}^\infty (ζ[2k+1]-1)=\frac{1}{4}$
Can someone explain why
$$\sum_{k=1}^\infty (ζ[2k+1]-1)=\frac{1}{4}?$$
• Why, that's simple. Expand every zeta, then swap the order of summation. – Ivan Neretin Sep 12 '15 at 13:45
We can rearrange \begin{align}\sum_{k=1}^\infty(\zeta(2k+1)-1)&=\sum_{k=1}^\infty\sum_{n=2}^\infty\frac1{n^{2k+1}}\\ &=\sum_{n=2}^\infty\sum_{k=1}^\infty\frac1{n^{2k+1}}\\ &=\sum_{n=2}^\infty\frac1{n(n^2-1)}\\ &=\sum_{n=2}^\infty\left(\frac1{2n(n-1)}-\frac1{2(n+1)n}\right)\\&=\frac14\end{align}
Hint. You are allowed to switch the order of summation \begin{align} \sum_{k=1}^\infty (ζ(2k+1)-1)&=\sum_{k=1}^\infty \left(\sum_{n=1}^\infty\frac1{n^{2k+1}}-1\right)\\\\ &=\sum_{k=1}^\infty \sum_{n=2}^\infty\frac1{n^{2k+1}}\\\\ &=\sum_{n=2}^\infty \sum_{k=1}^\infty\frac1{n^{2k+1}}\\\\ &=\sum_{n=2}^\infty \frac1{n(n-1)(n+1)} \end{align} Then you may conclude by a partial fraction decomposition and a telescoping sum.
Another approach. Since: $$\zeta(2k+1) = \int_{0}^{+\infty}\frac{x^{2k}}{(2k)!}\cdot\frac{1}{e^x-1}\,dx,\qquad 1=\int_{0}^{+\infty}\frac{x^{2k}}{(2k)!}\cdot\frac{1}{e^x}\,dx\tag{1}$$ we have: $$\begin{eqnarray*}\sum_{k\geq 1}\left(\zeta(2k+1)-1\right)&=&\int_{0}^{+\infty}\frac{1}{e^{2x}-e^{x}}\sum_{k\geq 1}\frac{x^{2k}}{(2k)!}\,dx \\&=&\int_{0}^{+\infty}\frac{\cosh(x)-1}{e^{2x}-e^x}\,dx\\&=&\frac{1}{2}\int_{0}^{+\infty}\left(e^{-x}-e^{-2x}\right)\,dx\\&=&\frac{1}{2}\cdot\left(1-\frac{1}{2}\right)=\color{red}{\frac{1}{4}}.\tag{2}\end{eqnarray*}$$ |
## Monday 3 September 2012
### CBSE Class 9 - Maths - Ch 7 Triangles (MCQs)
Triangles
credits:openclipart
Q1: In Δ ABC, AB = AC and ∠B = 50°. Then ∠C is equal to
(a) 40°
(b) 50°
(c) 80°
(d) 130°
Q2(CBSE 2010): In ΔABC, ∠C = ∠A and BC = 6 cm and AC = 5 cm. Then the length of AB is:
(a) 6 cm
(b) 5 cm
(c) 3 cm
(d) 2.5 cm
Q3: If one angle of a triangle is equal to the sum of other two angles, then the triangle is
(a) an isosceles triangle
(b) an obtuse triangle
(c) an equilateral triangle
(d) a right triangle
Q4(CBSE 2011): If ΔABC, is right angled at B, then :
(a) AB = AC
(b) AC < AB
(c) AB = BC
(d) AC > AB
Q5(CBSE 2011): In ΔABC if AB = BC, then :
(a) ∠B > ∠C
(b) ∠A = ∠C
(c) ∠A = ∠B
(d) ∠A < ∠B
Q6(CBSE 2010): In ΔPQR, ∠P = 60°, ∠Q = 50°. Which side of the triangle is the longest ?
(a) PQ
(b) QR
(c) PR
(d) none
Q7(CBSE 2011): P is a point on side BC of Δ ABC, such that AP bisects ∠BAC, then :
(a) BP = CP
(b) BA > BP
(c) BP > BA
(d) CP < CA
Q8(CBSE 2011): In the give figure q-8, AD is the median, then ∠BAD is:
(a) 55°
(b) 50°
(c) 100°
(d) 40°
Q9(CBSE 2010): If ΔABC ≅ ΔDEF by SSS congruence rule then :
(a) AB = EF, BC = FD, CA = DE
(b) AB = FD, BC = DE, CA = EF
(c) AB = DE, BC = EF, CA = FD
(d) AB = DE, BC = EF, ∠C = ∠F
Q10(CBSE 2011): If Δ ABC is a right angled at B, then
(a) AB = AC
(b) AC < AB
(c) AB = AC
(d) AC > AB
Q11: In a Δ ABC, ∠A = ∠C. If BC = 3 and AC = 4 then the perimeter of the triangle is:
(a) 7
(b) 10
(c) 12
(d) 14
Q12: From the following which condition is not possible for the congruence of two triangles ?
(a) ASA
(b) AAS
(c) AAA
(d) SSS
Q13: In a right angled triangle, if one acute angle is half the other, then the smallest angle is:
(a) 15°
(b) 25°
(c) 30°
(d) 35°
1: (b) 50° [Hint: Δ ABC, AB = AC is an isosceles triangle, ∠B =∠C ]
2: (a) 6 cm [ Hint: Δ ABC, ∠C = ∠A, an isosceles triangle, BC = AB]
3: (d) a right triangle
[Hint: Δ ABC, ∠A = ∠B + ∠C, ∵ ∠A + ∠B + ∠C = 180° ⇒ 2∠A = 180° ⇒ ∠A = 90° ]
4: (d) AC > AB [ Hint: Δ ABC, ∠B = 90°, AC is hypotenuse]
5: (b) ∠A = ∠C [Hint: Δ ABC, AB = BC is an isosceles triangle, ∠A =∠C ]
6: (a) PQ [Hint: ∠R = 70° is the largest, the side opposite to the largest angle is the longest.]
7: (b) BA > BP
[Hint: Consider the triangle as show in fig-q7,
Given, ∠BAP = ∠CAP = z (say)
∠APC is an exterior angle for ΔBAP
⇒ ∠APC = ∠ABP + ∠BAP
⇒ ∠APC > ∠BAP (greater angle has greater side opposite to it.)
⇒ BA > BP]
8: (b) 50°
[Hint:AB = AC ⇒ ∠ABD = ∠ACD = 40°. ⇒ ∠BAC = 100°
Δ ABD ≅ Δ ACD by SSS congruence
9: (c) AB = DE, BC = EF, CA = FD
[Hint: Remember CPCT for corresponding parts of congruent triangles.]
10: (d) AC > AB
[Hint: ∠B = 90°, AC is hypotenuse the longest side. ]
11: (b) 10 [Hint: ∠A = ∠C, ⇒AB = BC = 3.]
12: (c) AAA
13: (c) 30° [Hint: x + 2x + 90° = 180° ⇒ 3x = 90° ⇒ x = 30° ]
.
1. good selection of mcq's
2. In 7th question (d) can also be right
3. This comment has been removed by the author.
4. This helped me for my MCQ test
5. not so good but a little helpful very little |
# In the given figure, O is a point in the interior of square ABCD such that ΔOAB is an equilateral triangle.
Question:
In the given figure, O is a point in the interior of square ABCD such that ΔOAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle.
Solution:
Given: In square ABCD, ΔOAB is an equilateral triangle.
To prove: ΔOCD is an isosceles triangle.
Proof:
$\because \angle D A B=\angle C B A=90^{\circ} \quad$ (Angles of square $A B C D$ )
And, $\angle O A B=O B A=60^{\circ}$ (Angles of equilateral $\triangle O A B$ )
$\therefore \angle D A B-\angle O A B=\angle C B A-\angle O B A=90^{\circ}-60^{\circ}$
$\Rightarrow \angle O A D=\angle O B C=30^{\circ} \quad \ldots \ldots$ (i)
Now, in ΔDAO and ΔCBO,
AD = BC (Sides of square ABCD)
">DAO = ">CBO [From (i)]
AO = BO (Sides of equilateral ΔOAB)
"> By SAS congruence criteria,
ΔDAO "> ΔCBO
So, OD = OC (CPCT)
Hence, ΔOCD is an isosceles triangle. |
# Science:MATH105 Probability/Lesson 1 DRV/1.07 Grade Distribution Example
Consider the grade distribution example that we explored earlier: in a class of 10 people, grades on a test were 30, 30, 30, 60, 60, 80, 80, 80, 90, 100. Let X be the score of a randomly drawn test from this collection.
1. Calculate the probability that a test drawn at random has a score less than or equal to 80.
2. Calculate the probability that a test drawn at random has a score less than or equal to xn, where xn = 0, 10, 20, 30, ... , 100.
## Solution
### Part 1)
Recall the probability distribution, calculated earlier:
Let pk be the probability that the score of a randomly drawn test is xk = 10k. So, for example:
• p0 is the probability that a randomly drawn test score is 0
• p1 is the probability that a randomly drawn test score is 10
• p2 is the probability that a randomly drawn test score is 20
• p3 is the probability that a randomly drawn test score is 30
and so on. Values for each of these probabilities are given in the above bar graph.
The probability that a test drawn at random has a score of no greater than 80 is exactly the value of the CDF of X at x = 80; i.e.,
{\displaystyle {\begin{aligned}\mathrm {Pr} (X\leq 80)&=F(80)\\&=\sum _{k=0}^{8}p_{k}\\&=p_{0}+p_{1}+p_{2}+{\color {Blue}{p_{3}}}+p_{4}+p_{5}+{\color {Blue}{p_{6}}}+p_{7}+{\color {Blue}{p_{8}}}\\&=0+0+0+\mathbf {\color {Blue}{\frac {3}{10}}} +0+0+{\color {Blue}{\frac {2}{10}}}+0+{\color {Blue}{\frac {3}{10}}}\\&={\frac {8}{10}}\\&={\frac {4}{5}}\end{aligned}}}
The colour blue was used in the above calculation to highlight nonzero probabilities.
Because of the sample space of our experiment, if the randomly selected grade is to be less than or equal to 80, then the this grade can only be 30, 60, or 80. Intuitively, the probability that a randomly selected test has a grade of 30, 60, or 80 is the sum of the probabilities that the score is one of these possibilities, which we note is in agreement with our identity concerning probabilities of disjoint events.
### Part 2)
Now we want to calculate the probability that a test drawn at random has a score less than or equal to xk = 10k for k = 0,1,...,10. Again, we identify this as simply finding the value of the CDF of X at each of these xk values.
${\displaystyle \mathrm {Pr} (X\leq 0)=F(0)=\sum _{k=0}^{0}p_{k}=p_{0}=0}$
Similarly, ${\displaystyle F(0)=F(10)=F(20)=0}$. F(30) is non-zero:
${\displaystyle \mathrm {Pr} (X\leq 30)=F(30)=\sum _{k=0}^{3}p_{k}=0+0+0+{\frac {3}{10}}}$
Notice that F(40) is equal to F(30), since p4 = 0.
Other values of F are calculated in the same way using the definition of the cumulative distribution function. The following table contains the values of the CDF of X for xk = 0, 10, 20, 30, ... 100.
k xk F(xk)
0 0 0
1 10 0
2 20 0
3 30 0.3
4 40 0.3
5 50 0.3
6 60 0.5
7 70 0.5
8 80 0.8
9 90 0.9
10 100 1.0
The cumulative distribution function is graphed in the figure below. |
# A concave mirror (f = 36 \ cm) produces an image whose distance from the mirror is one-third the...
## Question:
A concave mirror {eq}(f = 36 \ cm) {/eq} produces an image whose distance from the mirror is one-third the object distance.
Determine:
a) the object distance and
b) the (positive) image distance.
## Spherical mirror
A spherical mirror can be classified into two categories. First, one whose striking surface within the sphere, known as a concave mirror. The second one, whose striking surface does not lie inside that is outside the sphere, is termed as a convex mirror.
## Answer and Explanation:
Given data
• The focal length of the concave mirror is: {eq}f = - 36\;{\rm{cm}} {/eq}
• The image distance is: {eq}v = \dfrac{u}{3} {/eq}
The expression for the mirror formula is,
{eq}\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} {/eq}
Here, {eq}u {/eq} is the object distance.
Substitute the given value in the above expression.
{eq}\begin{align*} \dfrac{3}{u} + \dfrac{1}{u} &= - \dfrac{1}{{36}}\\ \dfrac{4}{u} &= \dfrac{1}{{36}}\\ u &= - 144\;{\rm{cm}} \end{align*} {/eq}
Substitute the given value of the object distance into the expression of the image distance.
{eq}\begin{align*} v &= - \dfrac{{144\;{\rm{m}}}}{3}\\ &= - 48\;{\rm{cm}} \end{align*} {/eq}
Thus, the object distance is {eq}144\;{\rm{cm}} {/eq} in front of the mirror and image distance is {eq}48\;{\rm{cm}} {/eq} in front of mirror. |
# Standard Form and End Behavior of Polynomials
Polynomials include linear expressions and quadratic expressions, as well as expressions adding multiples of higher exponents of the variable. Rational functions are usually written in proper form, where the numerator is of a smaller degree than the denominator. (The degree of a polynomial is its highest exponent.)
For example, these are polynomials:
\begin{align*} x^5-4x^2+1 \hspace{1.5cm} 3x^4+(x+1)^3 \hspace{1.5cm} x^6+(x^2-1)^3+x+1 \end{align*}
On the other hand, these are not polynomials:
\begin{align*} x^2+\sqrt{x}+1 \hspace{1.5cm} \sin(x)+x^3-2 \hspace{1.5cm} |x|+x^5+x^3-1 \end{align*}
Polynomials are usually written in standard form, in which all terms are fully expanded and variable exponents are arranged from greatest to least.
\begin{align*} \text{Original polynomial} \hspace{.5cm} &\Bigg| \hspace{.5cm} x(5+2x^2)(x-1)+3(4+x^3) \\ \text{Simplify} \hspace{.5cm} &\Bigg| \hspace{.5cm} x(5x-5+2x^3-2x^2)+12+3x^3 \\ \text{ } \hspace{.5cm} &\Bigg| \hspace{.5cm} 5x^2-5x+2x^4-2x^3+12+3x^3 \\ \text{Combine like terms} \hspace{.5cm} &\Bigg| \hspace{.5cm} 5x^2-5x+2x^4+x^3+12 \\ \begin{matrix} \text{Arrange exponents from} \\ \text{greatest to least} \end{matrix} \hspace{.5cm} &\Bigg| \hspace{.5cm} 2x^4+x^3+5x^2-5x+12 \end{align*}
The end behavior of a polynomial refers to how it behaves when we substitute extremely large positive or negative values for . If the polynomial evaluates to a very large positive number, we say it approaches infinity. Otherwise, if the polynomial evaluates to a very large negative number, we say it approaches negative infinity.
For example, consider the polynomial $p(x)=-2x^3+x^2+5x-3$. When we substitute a large positive number, such as $x=100$, the output is a large negative number.
\begin{align*} p(100) &= -2(100)^3+(100)^2+5(100)-3 \\ &= -2(1000000) + 10000 + 500 - 3 \\ &= -2000000 + 10000 + 500 - 3 \\ &= -1989503 \end{align*}
When we substitute a large negative number, such as $x=-100$, the output is a large positive number.
\begin{align*} p(100) &= -2(-100)^3+(-100)^2+5(-100)-3 \\ &= -2(-1000000) + 10000 - 500 - 3 \\ &= 2000000 + 10000 - 500 - 3 \\ &= 2009497 \end{align*}
Putting this together, we say that $p(x)$ goes to negative infinity as $x$ goes to positive infinity, and $p(x)$ goes to positive infinity as $x$ goes to negative infinity. We can write this symbolically: $p(x) \to -\infty$ as $x \to +\infty$, and $p(x) \to +\infty$ as $x \to -\infty$. This is the end behavior of the polynomial $p(x)$.
Graphically, end behavior tells us whether the polynomial curves up or down as we travel away from the origin in the right or left direction. Since $p(x) \to -\infty$ as $x \to +\infty$, we know that the polynomial curves down as we travel to the right, and since $p(x) \to +\infty$ as $x \to -\infty$, we know that the polynomial curves up as we travel to the left.
Do you notice any patterns or shortcuts? It’s possible to determine the end behavior of a polynomial without evaluating the full polynomial. The term with the highest exponent controls the end behavior, because it makes the greatest contribution to the result. All the other terms make much smaller contributions – they’re peanuts in comparison to the highest-exponent term.
\begin{align*} p(100) &\approx -2(100)^3 = -2000000 \\ p(-100) &\approx -2(-100)^3 = 2000000 \end{align*}
But we can do even better – we don’t actually have to evaluate anything at all! Within the term having the highest exponent, we just need to look at the exponent and sign of the coefficient. If the exponent is even, then the result after exponentiation will always be positive. Consequently, the term will evaluate to have the same sign as its coefficient.
For example, to find the end behavior of the polynomial $p(x)=2x^2-3x+4$, we just need to look at the $2x^2$ term. Since the exponent is even, $x^2$ will always be positive – if we substitute $x=100$, then $x^2=10000$, and if we substitute $x=-100$, then $x^2=10000$ again. The coefficient $2$ is also positive, so $2x^2$ is always a positive times a positive, which makes a positive. As a result, we have $p(x) \to +\infty$ as $x \to +\infty$ and $p(x) \to +\infty$ as $x \to -\infty$.
Likewise, to find the end behavior of the polynomial $p(x) = -5x^4+7x^3-x-2$, we just need to look at the $-5x^2$ term. Since the exponent is even, $x^4$ will always be positive – if we substitute $x=100$, then $x^4=100000000$, and if we substitute $x=-100$, then $x^4=100000000$ again. But the coefficient $-5$ is negative, so $-5x^4$ is always a negative times a positive, which makes a negative. As a result, we have $p(x) \to -\infty$ as $x \to +\infty$ and $p(x) \to -\infty$ as $x \to -\infty$.
On the other hand, if the exponent is odd, then the result after exponentiation will always have the same sign as the input $x$. Consequently, the term will evaluate to be positive if the coefficient and the input $x$ have the same sign, and negative if they have opposite signs.
For example, to find the end behavior of the polynomial $p(x) = 4x^3-5x^2-2x+1$, we just need to look at the $4x^3$ term. Since the exponent is odd, exponentiation will not change the sign – if we substitute $x=100$, then $x^3=1000000$, and if we substitute $x=-100$, then $x^3=-100000000$. The coefficient $4$ is positive, and multiplying by a positive doesn’t change the sign either. As a result, we have $p(x) \to +\infty$ as $x \to +\infty$ and $p(x) \to -\infty$ as $x \to -\infty$.
Likewise, to find the end behavior of the polynomial $p(x)=-3x^5 + 7x^4 + 3x^2 - 10x$, we just need to look at the $-3x^5$ term. Since the exponent is odd, exponentiation will not change the sign – if we substitute $x=100$, then $x^5=10000000000$, and if we substitute $x=-100$, then $x^5=-10000000000$. But the coefficient $-3$ is negative, and multiplying by a negative changes the sign – if $x^5=10000000000$, then $-3x^5=-30000000000$, and if $x^5=-10000000000$, then $-3x^5=30000000000$. As a result, we have $p(x) \to -\infty$ as $x \to +\infty$ and $p(x) \to +\infty$ as $x \to -\infty$.
Practice Problems
Convert the following polynomials to standard form. Then, write their end behavior symbolically: $p(x) \to \text{__}$ as $x \to +\infty$, and $p(x) \to \text{__}$ as $x \to -\infty$. (You can view the solution by clicking on the problem.)
$1) \hspace{.5cm} p(x)=3x^4-7x+8x^3$
Solution:
$p(x)=3x^4+8x^3-7x$
$p(x) \to +\infty \text{ as } x \to +\infty$
$p(x) \to +\infty \text{ as } x \to -\infty$
$2) \hspace{.5cm} p(x)=1-2x+x^3$
Solution:
$p(x)=x^3-2x+1$
$p(x) \to +\infty \text{ as } x \to +\infty$
$p(x) \to -\infty \text{ as } x \to -\infty$
$3) \hspace{.5cm} p(x)=2+3x^3-x^6-9x^4$
Solution:
$p(x)=-x^6-9x^4+3x^3+2$
$p(x) \to -\infty \text{ as } x \to +\infty$
$p(x) \to -\infty \text{ as } x \to -\infty$
$4) \hspace{.5cm} p(x)=-10x^9-x^3+5x^4-x^2$
Solution:
$p(x)=-10x^9+5x^4-x^3-x^2$
$p(x) \to -\infty \text{ as } x \to +\infty$
$p(x) \to +\infty \text{ as } x \to -\infty$
$5) \hspace{.5cm} p(x)=6x^3-4x^5+1-2x$
Solution:
$p(x)=-4x^5+6x^3-2x+1$
$p(x) \to -\infty \text{ as } x \to +\infty$
$p(x) \to +\infty \text{ as } x \to -\infty$
$6) \hspace{.5cm} p(x)=1+4x^3-5x+x^4$
Solution:
$p(x)=x^4+4x^3-5x+1$
$p(x) \to +\infty \text{ as } x \to +\infty$
$p(x) \to +\infty \text{ as } x \to -\infty$
$7) \hspace{.5cm} p(x)=8x-1+x^4-5x^6-6x^5$
Solution:
$p(x)=-5x^6-6x^5+x^4+8x-1$
$p(x) \to -\infty \text{ as } x \to +\infty$
$p(x) \to -\infty \text{ as } x \to -\infty$
$8) \hspace{.5cm} p(x)=-x^4+x^{11}-3x^8+2$
Solution:
$p(x)=x^{11}-3x^8-x^4+2$
$p(x) \to +\infty \text{ as } x \to +\infty$
$p(x) \to -\infty \text{ as } x \to -\infty$
$9) \hspace{.5cm} p(x)=(x^2+1)(x+1)$
Solution:
$p(x)=x^3+x^2+x+1$
$p(x) \to +\infty \text{ as } x \to +\infty$
$p(x) \to -\infty \text{ as } x \to -\infty$
$10) \hspace{.5cm} p(x)=-(x-1)(x+1)(x-3)$
Solution:
$p(x)=-x^3+3x^2+x-3$
$p(x) \to -\infty \text{ as } x \to +\infty$
$p(x) \to +\infty \text{ as } x \to -\infty$
$11) \hspace{.5cm} p(x)=(2x^5+1)(x^3-2)$
Solution:
$p(x)=2x^8-4x^5+x^3-2$
$p(x) \to +\infty \text{ as } x \to +\infty$
$p(x) \to +\infty \text{ as } x \to -\infty$
$12) \hspace{.5cm} p(x)=x^2(3-2x^4)(x+2)$
Solution:
$p(x)=-2x^7-4x^6+3x^3+6x^2$
$p(x) \to -\infty \text{ as } x \to +\infty$
$p(x) \to +\infty \text{ as } x \to -\infty$
Tags: |
# At a party each guest shook hands with every other guest exactly once. There were a total of 105 handshakes. How many guest were there?
Jan 22, 2016
There are $15$ guests
#### Explanation:
If there are $n$ guests then each guest shakes hands with $\left(n - 1\right)$ other guests. Note, however, that each time a handshake occurs it counts as $2$ handshakes (one for each person involved).
Therefore with $n$ guests the number of handshakes will be $\frac{n \times \left(n - 1\right)}{2}$
We are told
$\textcolor{w h i t e}{\text{XXX}} \frac{n \times \left(n - 1\right)}{2} = 105$
Therefore
$\textcolor{w h i t e}{\text{XXX}} {n}^{2} - n = 210$
$\textcolor{w h i t e}{\text{XXX}} {n}^{2} - n - 210 = 0$
$\textcolor{w h i t e}{\text{XXX}} \left(n + 14\right) \left(n - 15\right) = 0$
Therefore
$\textcolor{w h i t e}{\text{XXX}} n = - 14$ or $n = 15$
And since the number of guest can not be negative:
$\textcolor{w h i t e}{\text{XXX}} n = 15$ |
INTRODUCTION TO COORDINATE GEOMETRY. COO742ARTC
INTRODUCTION TO COORDINATE GEOMETRY Often the integers are shown as specially-marked points evenly spaced on the line. The line includes all real numbers, continuing forever in each direction. figure S.1 The left side of zero consists of negative integers and the right side of zero consists of positive integers. Zero is not marked as positive or Negative. therefore zero is neither positive nor negative. The number line can be constructed horizontally as the above number line and vertically as shown below: figure S.2 Note that, constructing vertically, positive integers must be above zero and negative integers must be below zero. it is like rotating horizontal number line anticlockwise 90° about the origin (0). The following flash video clip will give you a good picture: figure S.3 When joining vertical and horizontal number line we get something called Cartesian plane or XY plane as shown above. Cartesian plane or xy plane has two lines vertical and horizontal axis, the horizontal axis is called x-axis and vertical axis is called y-axis. Points in Cartesian plane are plotted according to their vertical and horizontal position. Consider point A in the figure below: figure S.4 Travelling along horizontal axis from origin i.e. 0 to reach point A, we travel 3 units, while travelling along vertical axis from origin i.e. 0 to reach point A, we travel 4 units. From that explanation we say: The x – coordinate of A is +3 The y – coordinate of A is +4 Therefore we say that, the coordinates of point A are (3,4) or A is the point (3,4), the x-coordinate is given first then the y-coordinate. Now from above figure S.4 the coordinate of the other points are: B (-4, 1), C(4,1), D(-2,-1), E(0,2) and F(2,0) NOTE: the place where the two axes intersect is called the Origin and the point of origin is (0,0) Try your understanding: 1 . Sketch your own axes and plot the following points: A(3, -1), B(-2, -3), C(-1,4), D(0,0), E(0,-4) and F(4, 2). 2. Study the following Coordinate plane and answer the questions Write down the coordinates of the points M, T, P, Y and Q What is the vertical distance from Point G to point Q? Click What is the horizontal distance from point Y to point M? 3. Connect the points A(0,1), B(4,1), C(5,-2), D(-2,-2) on XY Plane and state the shape you obtain. ********** THANKS FOR VISITING ************* .u-star-rating-16 { list-style:none;margin:0px;padding:0px;width:80px;height:16px;position:relative;background: url('/.s/t/1706/rating.png') top left repeat-x } .u-star-rating-16 li{ padding:0px;margin:0px;float:left } .u-star-rating-16 li a { display:block;width:16px;height: 16px;line-height:16px;text-decoration:none;text-indent:-9000px;z-index:20;position:absolute;padding: 0px;overflow:hidden } .u-star-rating-16 li a:hover { background: url('/.s/t/1706/rating.png') left center;z-index:2;left:0px;border:none } .u-star-rating-16 a.u-one-star { left:0px } .u-star-rating-16 a.u-one-star:hover { width:16px } .u-star-rating-16 a.u-two-stars { left:16px } .u-star-rating-16 a.u-two-stars:hover { width:32px } .u-star-rating-16 a.u-three-stars { left:32px } .u-star-rating-16 a.u-three-stars:hover { width:48px } .u-star-rating-16 a.u-four-stars { left:48px } .u-star-rating-16 a.u-four-stars:hover { width:64px } .u-star-rating-16 a.u-five-stars { left:64px } .u-star-rating-16 a.u-five-stars:hover { width:80px } .u-star-rating-16 li.u-current-rating { top:0 !important; left:0 !important;margin:0 !important;padding:0 !important;outline:none;background: url('/.s/t/1706/rating.png') left bottom;position: absolute;height:16px !important;line-height:16px !important;display:block;text-indent:-9000px;z-index:1 } Category: Upper Primary | Added by: Admin (07/Aug/2016) | Author: Yahya Mohamed E W Views: 492 | | Rating: 0.0/0 |
# How To Find Slope Intercept Form On A Graph
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
How To Find Slope Intercept Form On A Graph – One of the numerous forms employed to represent a linear equation the one most commonly seen is the slope intercept form. It is possible to use the formula for the slope-intercept in order to find a line equation assuming that you have the straight line’s slope , and the y-intercept, which is the point’s y-coordinate where the y-axis intersects the line. Learn more about this specific linear equation form below.
## What Is The Slope Intercept Form?
There are three primary forms of linear equations: the standard slope-intercept, the point-slope, and the standard. Even though they can provide the same results when utilized however, you can get the information line more quickly using the slope-intercept form. As the name implies, this form employs the sloped line and it is the “steepness” of the line is a reflection of its worth.
This formula can be used to determine the slope of straight lines, the y-intercept or x-intercept where you can utilize a variety formulas that are available. The equation for a line using this formula is y = mx + b. The straight line’s slope is signified by “m”, while its y-intercept is signified by “b”. Each point of the straight line is represented with an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” need to remain variables.
## An Example of Applied Slope Intercept Form in Problems
When it comes to the actual world in the real world, the slope intercept form is commonly used to represent how an item or problem evolves over the course of time. The value that is provided by the vertical axis demonstrates how the equation handles the intensity of changes over what is represented by the horizontal axis (typically time).
A simple example of using this formula is to determine how the population grows in a particular area as the years pass by. Using the assumption that the area’s population increases yearly by a certain amount, the amount of the horizontal line will grow by a single point as each year passes, and the value of the vertical axis is increased to represent the growing population by the amount fixed.
You can also note the starting point of a particular problem. The starting point is the y-value of the y-intercept. The Y-intercept is the place at which x equals zero. If we take the example of a previous problem the beginning point could be the time when the reading of population begins or when time tracking starts along with the changes that follow.
The y-intercept, then, is the place where the population starts to be documented in the research. Let’s say that the researcher is beginning to perform the calculation or measure in 1995. Then the year 1995 will represent considered to be the “base” year, and the x 0 points would occur in the year 1995. Therefore, you can say that the population of 1995 corresponds to the y-intercept.
Linear equation problems that use straight-line equations are typically solved this way. The beginning value is expressed by the y-intercept and the rate of change is represented through the slope. The principal issue with this form generally lies in the interpretation of horizontal variables in particular when the variable is associated with an exact year (or any other kind number of units). The first step to solve them is to ensure that you understand the variables’ definitions clearly. |
# Lesson 5
Triangles in Circles
## 5.1: One Perpendicular Bisector (5 minutes)
### Warm-up
In this activity, students recall that points on perpendicular bisectors of segments are the same distance from each endpoint of the segment. In the next activity, students will construct the remaining perpendicular bisectors of this triangle and then construct the triangle’s circumscribed circle.
Making dynamic geometry software available gives students an opportunity to choose appropriate tools strategically (MP5).
### Student Facing
The image shows a triangle.
1. Construct the perpendicular bisector of segment $$AB$$.
2. Imagine a point $$D$$ placed anywhere on the perpendicular bisector you constructed. How would the distance from $$D$$ to $$A$$ compare to the distance from $$D$$ to $$B$$? Explain your reasoning.
### Activity Synthesis
Display this image for all to see. Tell students that the dashed line is the perpendicular bisector of segment $$AB$$.
For each of the points $$E$$, $$F$$, and $$G$$, ask students if the point is closer to $$A$$ or closer to $$B$$, and ask how they know. (Point $$F$$ is closer to $$A$$. Point $$G$$ is closer to $$B$$. Point $$E$$ is the same distance from $$A$$ as it is to $$B$$. All points to the left of the perpendicular bisector are closer to $$A$$, points to the right are closer to $$B$$, and points on the bisector are equidistant from $$A$$ and $$B$$.)
## 5.2: Three Perpendicular Bisectors (20 minutes)
### Activity
Students construct the circumcenter of the triangle from the previous activity, or the point where the perpendicular bisectors of the sides of the triangle intersect. They determine that the circumcenter is equidistant from the vertices of the triangle based on the properties of perpendicular bisectors. Finally, they construct the circumscribed circle, observing that this is possible because of the equal distances already noted.
Making dynamic geometry software available gives students an opportunity to choose appropriate tools strategically (MP5).
### Student Facing
1. Construct the perpendicular bisector of segment $$BC$$ from the earlier activity. Label the point where the 2 perpendicular bisectors intersect as $$P$$.
2. Use a colored pencil to draw segments $$PA,PB,$$ and $$PC$$. How do the lengths of these segments compare? Explain your reasoning.
3. Imagine the perpendicular bisector of segment $$AC$$. Will it pass through point $$P$$? Explain your reasoning.
4. Construct the perpendicular bisector of segment $$AC$$.
5. Construct a circle centered at $$P$$ with radius $$PA$$.
6. Why does the circle also pass through points $$B$$ and $$C$$?
### Student Facing
#### Are you ready for more?
Points $$A,B,$$ and $$C$$ are graphed. Find the coordinates of the circumcenter and the radius of the circumscribed circle for triangle $$ABC$$.
### Anticipated Misconceptions
Due to the precision of the tools, students’ circles may not pass through all 3 vertices. Explain that if we had more precise tools, the distances would be the same and the circle would pass through all the vertices.
### Activity Synthesis
The goal is to conclude that all triangles have a circumscribed circle.
• “What kind of circle did you construct in relation to the triangle?” (It is a circumscribed circle.)
• “We saw that some quadrilaterals have a circumscribed circle, but some do not. Do all triangles have a circumscribed circle, or only some?” (All triangles have a circumscribed circle, because what we did for triangle $$ABC$$ could be done for any triangle. We used properties of perpendicular bisectors, not anything special about this triangle.)
• “What other special lines in triangles have we seen that all meet at a central point?” (In the Coordinate Geometry unit, we saw that the medians and the altitudes in a triangle also have single points of intersection.)
Tell students that the point where all 3 perpendicular bisectors of a triangle intersect is called the triangle’s circumcenter. Ask students to add this theorem to their reference charts as you add it to the class reference chart:
The 3 perpendicular bisectors of the sides of a triangle meet at a single point, called the triangle’s circumcenter. This point is the center of the triangle’s circumscribed circle. (Theorem)
If time permits, ask: “For the triangle, once we drew 2 perpendicular bisectors, we knew the third would pass through the point of intersection of the first 2. Why doesn’t a similar argument work for a quadrilateral? Use a general quadrilateral $$ABCD$$ as an example.” (Draw the perpendicular bisectors of segments $$AB$$ and $$BC$$ and call their point of intersection $$P$$. This point is equidistant from points $$A,B,$$ and $$C$$. In a triangle, there are only 3 vertices, so that covers all of them. However, now we have 4 vertices. We have no way to guarantee that point $$P$$ is equidistant from point $$D$$.)
## 5.3: Wandering Centers (10 minutes)
### Activity
In this activity, students compare the circumcenters of acute, obtuse, and right triangles. They find that the circumcenter of a right triangle lies on its hypotenuse, the circumcenter of an acute triangle lies inside the triangle, and the circumcenter of an obtuse triangle lies outside the triangle.
This activity works best when each student has access to devices that can run the GeoGebra applet because students will benefit from seeing the relationship in a dynamic way.
### Launch
Arrange students in groups of 2.
Engagement: Develop Effort and Persistence. Provide background knowledge to help focus on increasing the length of on-task orientation in the face of distractions. For example, provide images and definitions of obtuse, acute, and right triangles.
Supports accessibility for: Attention; Social-emotional skills
### Student Facing
Move the vertices of triangle $$ABC$$ and observe the resulting location of the triangle’s circumcenter, point $$D$$. Determine what seems to be true when the circumcenter is in each of these locations:
1. outside the triangle
2. on one of the triangle’s sides
3. inside the triangle
### Launch
Arrange students in groups of 3–4. Ask students, “Suppose you construct 2 perpendicular bisectors of a triangle. Do you need to construct the third to find the circumcenter? Why or why not?” (We don’t need to construct the third perpendicular bisector. Wherever the first 2 intersect, that’s also where the third line will intersect the first 2.)
Engagement: Develop Effort and Persistence. Provide background knowledge to help focus on increasing the length of on-task orientation in the face of distractions. For example, provide images and definitions of obtuse, acute, and right triangles.
Supports accessibility for: Attention; Social-emotional skills
### Student Facing
Each student in your group should choose 1 triangle. It’s okay for 2 students to choose the same triangle as long as all 3 are chosen by at least 1 student.
1. Construct the circumscribed circle of your triangle.
2. After you finish, compare your results. What do you notice about the location of the circumcenter in each triangle?
### Anticipated Misconceptions
If students struggle to get started, ask them to look back at their work from the previous activity.
### Activity Synthesis
The purpose of the discussion is to make informal observations about the locations of the circumcenters. Display this applet for all to see. Don’t immediately move the slider that controls the measure of one of the triangle’s angles.
Alternatively, display these images for all to see.
Ask students why it makes sense that the right triangle has its circumcenter on one of the triangle’s sides (we know that a right angle is inscribed in a half circle, so when you draw the hypotenuse of the triangle, it’s the diameter of the circle).
If using the applet, move the slider to show how the position of the circumcenter changes as the angle moves between being obtuse and acute. Ask students to observe and explain what happens to the circumcenter. (When the angle measure is greater than 90 degrees, the arc in which the angle is inscribed grows to more than 180 degrees. The triangle therefore takes up less than half the circle, so the circumcenter must be outside the circle. Alternatively, when the angle gets smaller, the circle “shrinks” to fit around the triangle. The arc in which the angle is inscribed measures less than 180 degrees, so the triangle fits more snugly in the circle and the circumcenter is inside the triangle.)
Speaking: MLR8 Discussion Supports. Use this routine to support whole-class discussion. Give groups additional time to make sure that each student can explain what they noticed about the location of the circumcenter in each triangle. Invite groups to rehearse what they will say when they share with the whole class. Rehearsing provides students with additional opportunities to speak and clarify their thinking, and will improve the quality of observations shared during the whole-class discussion.
Design Principle(s): Support sense-making; Cultivate conversation
## Lesson Synthesis
### Lesson Synthesis
The goal is to reinforce these 3 concepts: All triangles have circumscribed circles, the center of the circle is the intersection of the perpendicular bisectors of the triangle’s sides, and the circumcenter is equidistant from the triangle’s vertices.
Display this image for all to see.
Tell students that the points represent locations of 3 research stations in a desert, and that scientists want to build a supply hut that is the same distance from all 3 stations. What should they do? Challenge students to be precise in their use of language, and to explain why their ideas will work. Be sure these points come up in the discussion:
• The scientists should create the perpendicular bisectors of the sides of triangle $$RST$$.
• The perpendicular bisectors will meet in one point.
• This point is equidistant from all the vertices of the triangle, because points on perpendicular bisectors of a segment are equidistant from the segment’s endpoints.
Next, tell students that the scientists want to build another research station the same distance from the supply hut as the other stations. What should they do? Be sure these points come up in the discussion:
• The scientists should construct the circumscribed circle and put the new station anywhere on the circle.
• We know this triangle has a circumscribed circle because all triangles have one.
• If the station goes on the circle, it will be the same distance from the supply hut as the others because all points on a circle are the same distance from the circle’s center.
## 5.4: Cool-down - Fair Placement (5 minutes)
### Cool-Down
Suppose we have triangle $$ABC$$ and we construct the perpendicular bisectors of all 3 sides. These perpendicular bisectors will all meet at a single point called the circumcenter of the triangle (label it $$D$$). This point is on the perpendicular bisector of $$AB$$, so it’s equidistant from $$A$$ and $$B$$. It’s also on the perpendicular bisector of $$BC$$, so it’s equidistant from $$B$$ and $$C$$. So, it is actually the same distance from $$A,B,$$ and $$C$$. We can draw a circle centered at $$D$$ with radius $$AD$$. The circle will pass through $$B$$ and $$C$$ too because the distances $$BD$$ and $$CD$$ are the same as the radius of the circle.
In this case, the circumcenter happened to fall inside triangle $$ABC$$, but that does not need to happen. The images show cases where the circumcenter is inside a triangle, outside a triangle, and on one of the sides of a triangle. |
Inverse of a function
Chapter 1 Class 12 Relation and Functions
Serial order wise
### Transcript
Ex 1.3 , 7 (Method 1) Consider f: R R given by f(x) = 4x+ 3. Show that f is invertible. Find the inverse of f. Checking inverse Step 1 f(x) = 4x + 3 Let f(x) = y y = 4x + 3 y 3 = 4x 4x = y 3 x = 3 4 Let g(y) = 3 4 where g: R R Step 2: gof = g(f(x)) = g(4x + 3) = (4 + 3) 3 4 = 4 + 3 3 4 = 4 4 = x = IR Step 3: fog = f(g(y)) = f 3 4 = 4 3 4 + 3 = y 3 + 3 = y + 0 = y = IR Since gof = IR and fog = IR, f is invertible & Inverse of f = g(y) = Ex 1.3 , 7 (Method 2) Consider f: R R given by f(x) = 4x+ 3. Show that f is invertible. Find the inverse of f. f is invertible if f is one-one and onto Checking one-one f(x1) = 4x1 + 3 f(x2) = 4x2 + 3 Putting f(x1) = f(x2) 4x1 + 3 = 4x2 + 3 4x1 = 4x2 x1 = x2 If f(x1) = f(x2) , then x1 = x2 f is one-one Checking onto f(x) = 4x + 3 Let f(x) = y, where y Y y = 4x + 3 y 3 = 4x 4x = y 3 x = 3 4 Here, y is a real number So, 3 4 is also a real number So, x is a real number Thus, f is onto Since f is one-one and onto f is invertible Finding inverse f(x) = 4x + 3 For finding inverse, we put f(x) = y and find x in terms of y We have done that while proving onto y = f(x) y = 4x + 3 x = 3 4 Let g(y) = 3 4 where g: R R Inverse of f = g(y) = |
# What is 136/25 as a decimal?
## Solution and how to convert 136 / 25 into a decimal
136 / 25 = 5.44
136/25 converted into 5.44 begins with understanding long division and which variation brings more clarity to a situation. Both are used to handle numbers less than one or between whole numbers, known as integers. But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. Now, let's solve for how we convert 136/25 into a decimal.
## 136/25 is 136 divided by 25
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators and Denominators. This creates an equation. To solve the equation, we must divide the numerator (136) by the denominator (25). This is how we look at our fraction as an equation:
### Numerator: 136
• Numerators are the parts to the equation, represented above the fraction bar or vinculum. 136 is one of the largest two-digit numbers you'll have to convert. But having an even numerator makes your mental math a bit easier. Large numerators make converting fractions more complex. So how does our denominator stack up?
### Denominator: 25
• Denominators differ from numerators because they represent the total number of parts which can be found below the vinculum. 25 is one of the largest two-digit numbers to deal with. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Have no fear, large two-digit denominators are all bark no bite. So without a calculator, let's convert 136/25 from a fraction to a decimal.
## Converting 136/25 to 5.44
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 25 \enclose{longdiv}{ 136 }$$
We will be using the left-to-right method of calculation. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Solve for how many whole groups you can divide 25 into 136
$$\require{enclose} 00.5 \\ 25 \enclose{longdiv}{ 136.0 }$$
We can now pull 125 whole groups from the equation. Multiply by the left of our equation (25) to get the first number in our solution.
### Step 3: Subtract the remainder
$$\require{enclose} 00.5 \\ 25 \enclose{longdiv}{ 136.0 } \\ \underline{ 125 \phantom{00} } \\ 1235 \phantom{0}$$
If your remainder is zero, that's it! If there is a remainder, extend 25 again and pull down the zero
### Step 4: Repeat step 3 until you have no remainder
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 136/25 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. And the same is true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But each represent values in everyday life! Here are just a few ways we use 136/25, 5.44 or 544% in our daily world:
### When you should convert 136/25 into a decimal
Pay & Salary - Anything to do with finance or salary will leverage decimal format. If you look at your pay check, you will see your labor is worth $20.544 per hour and not$20 and 136/25.
### When to convert 5.44 to 136/25 as a fraction
Carpentry - To build a table, you must have the right measurements. When you stretch the tape measure across the piece of wood, you won't see 10.6 inches. You'll see a tick mark at 10 and 3/5 inches.
### Practice Decimal Conversion with your Classroom
• If 136/25 = 5.44 what would it be as a percentage?
• What is 1 + 136/25 in decimal form?
• What is 1 - 136/25 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 5.44 + 1/2? |
# Exercises Problem (6.1 #19. Introduction to Linear Algebra: Strang) A... three matrix B is known to have eigenvalues 0, 1...
```Exercises on eigenvalues and eigenvectors
Problem 21.1: (6.1 #19. Introduction to Linear Algebra: Strang) A three by
three matrix B is known to have eigenvalues 0, 1 and 2. This information
is enough to find three of these (give the answers where possible):
a) The rank of B
b) The determinant of B T B
c) The eigenvalues of B T B
d) The eigenvalues of ( B2 + I )−1
Solution:
a) B has 0 as an eigenvalue and is therefore singular (not invertible). Since
B is a three by three matrix, this means that its rank can be at most 2.
Since B has two distinct nonzero eigenvalues, its rank is exactly 2.
b) Since B is singular, det( B) = 0 . Thus det( B T B) = det( B T ) det( B) = 0.
c) There is not enough information to find the eigenvalues of B T B. For
example:
⎡
If B = ⎣
⎤
0
⎡
⎦ then B T B = ⎣
1
⎤
0
1
2
⎡
4
⎤
0 1
⎣
1
If B =
⎦.
⎡
⎦ then B T B = ⎣
2
⎤
0
2
⎦.
4
d) If p(t) is a polynomial and if x is an eigenvector of A with eigenvalue
λ, then
p( A)x = p(λ)x.
We also know that if λ is an eigenvalue of A then 1/λ is an eigenvalue
of A−1 . Hence the eigenvalues of ( B2 + I )−1 are 021+1 , 121+1 and 221+1 , or
1, 1/2 and 1/5.
1
Problem 21.2: (6.1 #29.) Find the eigenvalues of A, B, and C when
⎡
⎤
⎡
⎤
⎡
⎤
1 2 3
0 0 1
2 2 2
A = ⎣ 0 4 5 ⎦ , B = ⎣ 0 2 0 ⎦ and C = ⎣ 2 2 2 ⎦ .
0 0 6
3 0 0
2 2 2
Solution: Since the eigenvalues of a triangular matrix are its diagonal
entries, the eigenvalues of A are 1,4, and 6. For B we have:
det( B − λI ) = (−λ)(2 − λ)(−λ) − 3(2 − λ)
= (λ2 − 3)(2 − λ).
√
Hence the eigenvalues of B are ± 3 and 2. Finally, for C we have:
det(C − λI ) = (2 − λ)[(2 − λ)2 − 4] − 2[2(2 − λ) − 4] + 2[4 − 2(2 − λ)]
= λ3 − 6λ2 = λ2 (λ − 6).
The eigenvalues of C are 6, 0, and 0.
We can quickly check our answers by computing the determinants of
A and B and by noting that C is singular.
2
MIT OpenCourseWare
http://ocw.mit.edu
18.06SC Linear Algebra
Fall 2011 |
# Base method of multiplication
## Base method of multiplication
1 st of all question is, why it is called base method. In Vedic mathematics there is a saying in Sanskrit,
“Nikhilam Navatascaramam Dasatah”. It means “all form 9 and last form 10”
In this method we have to determine the base of the number first. Detailed steps of this method is given
below,
• 1 st of all find the base and the difference with the base.
• Number of digits in the RHS is equal to number zeros in the base.
• Multiply the differences on the RHS.
• Put the cross answer in LHS.
### Type 1: Multiply the two numbers with same base.
#### Example 1: 997×995
Answer: 1 st of all the base of the given two number is same which is 1000.
What we see in the above multiplication is in the RHS we put 015 because (5×3=15). But since zero in
the base number is 3, so we have to put a “0” before 15 in the RHS such that total digit in the RHS is
equal to the zeros in the base number.
For LHS we have to subtract either (3 from 995) or (5 from 997). That‟s why we put 992(995-3 or 997-5) in the LHS.
So, 997×995=992015.
#### Example 2: Find 103×105.
In the above illustration we see that since the base is 100 so we have to put only two digit in the RHS of
the slash(/), so we put just 15(3×5). And in the LHS we put 108(103+5 or 105+3).
So 103×105=10815.
### Type 2: Multiply a number below the base with a number above the base
Earlier we multiply two numbers either above the base or below the base. But in Type 2 we multiply one number below the base with other above the base.
#### Example 1: Find 91×106
Answer: Here actual base for 91 is 100 and 106 is 1000. But we will take 100 as working base since 100 is closer to both 91 and 106.
Here final answer = (97×Base)-54= (97×100)-54=9646.
Here also we put RHS and LHS as we put in Type 1. But since there is a minus sign before 9 we put (-54) in RHS. In the final answer we have to multiply the LHS with the base number then subtract RHS to get the final answer. So final answer is 9646.
#### Example 2: 855×1004
Answer: Here actual working base is 1000.
Here we put -580(-145×4) in RHS. Put 859(855+4 or 1004-145) LHS. Final answer will be like below,
(859×Base)-580= (859×1000)-580=858420.
Example 3: 82×105
### Type 3: When number of digits in RHS exceeds number of zeros in the base
You can understand Type 3 with some example given below,
#### Example 1: Find 961×951
Answer: Here working base is 1000. Number zeros in the base are 3.
Here we get 1911 by multiplying 39 by 49, but we need only 3 digits in the RHS since the base has only 3zeros. So what to do next?
Now we must add „1‟ in the extreme left of the number „1911‟of the RHS to the number „912‟of the LHS.And put only „911‟ in the RHS. Numerically final answer is,
913911, since (912+1=913).
#### Example 2: Find 1223×1227
Answer: Here working base is 1000. Number of zeros in the base is 3.
Here we get 50621(223×227) in the RHS. But we need only 3 digits in the RHS since zeros in the base is3. So we add „50‟ in the extreme left of the number „50621‟ of the RHS to the number „1450‟ in the LHSto get the final answer.
1450+50=1500 in the LHS and 621 in the RHS. |
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# CLASS_11_MATHS_SOLUTIONS_NCERT
## Class XI Chapter 4 –
Class XI Chapter 4 – Principle of Mathematical Induction Maths ______________________________________________________________________________ Question 15: Prove the following by using the principle of mathematical induction for all n N: n n n 1 3 5 ..... 2n 1 3 2 2 2 2 2 1 2 1 Solution 15: Let the given statement be P(n), i.e., n n n P n :1 3 5 ..... 2n 1 3 2 2 2 2 2 1 2 1 For n 1, we have 2 1 2.1 1 2.1 1 1.1.3 P1 1 1 1, 3 3 Let P(k) be true for some positive integer k, i.e., k P k which is true. 2 2 2 2 k 2k 1 2k 1 1 3 5 .... 2 1 ...... 1 3 We shall now prove that Consider P k 1 is true. 2 2 2 k 2 k 2 1 3 5 ..... 2 1 2 1 1 2 12 1 k k k 3 2 12 1 k k k 3 2k 2 1 2 2k 1 2 k k k 2 2 2 1 2 1 3 2 1 3 2k 1 k 2k 1 3 2k 1 2k 12k 2 k 6k 3 3 3 2k 12k 2 5k 3 3 2k 12k 2 2k 3k 3 3 2k 1 2k k 1 3k 1 3 Using 1
Class XI Chapter 4 – Principle of Mathematical Induction Maths ______________________________________________________________________________ 2k 1k 12k 3 3 k 1 2 k 1 1 2 k 1 1 3 Thus, Pk 1 is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N. Question 16: Prove the following by using the principle of mathematical induction for all 1 1 1 1 n .... 1.4 4.7 7.10 3 2 3 1 3 1 n n n Solution 16: Let the given statement be P(n), i.e., 1 1 1 1 n Pn: .... 1.4 4.7 7.10 3n 2 3n 1 3n 1 For n = 1, we have 1 1 1 1 P 1 , 1.4 3.11 4 1.4 which is true. Let P(k) be true for some positive integer k, i.e., 1 1 1 1 k Pk ..... ...... 1 1.4 4.7 7.10 3k 2 3k 1 3k 1 We shall now prove that Pk 1 is true. Consider 1 1 1 1 1 .... 1.4 4.7 7.10 3k23k1 3k1 23 k1 1 k 1 3k 1 3k 1 3k 4 1 k 1 3k1 3k4 1 k 3k4 1 3k1 3k4 [Using (1)] n N:
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# Solving Systems of Equations by Substitution
This Lesson, we will cover how we can solve a System of Equations just by using our equations in the system.
Let’s glance over what we covered in the previous lesson. i.e. showing that a point can solve a system if it makes all equations in the system true.
## Question 1
We have the system, y = 3x + 1 and y + 2x = 11, with the point (2, 7) as a solution.
Firstly, show that the point (2, 7) is a solution to the system.
As we recall, to show a point is a solution to a system, we must show it makes both equations true, this is done by entering that point into both equations.
Remember, a solution to a system is a point where the equations meet, an equation is the description of the points that make up a line, we want the point that the equations are both describing the same point.
Let’s check that the point is a solution to the system (a point on where both lines meet)
Enter (2, 7) into y = 3x + 1
(7) = 3(2) + 1
7 = 6 + 1
7 = 7
It makes the first equation true,
now for y + 2x = 11
(7) + 2(2) = 11
7 + 4 = 11,
11 = 11
Therefore, (2, 7) makes both equations true, meaning it is a point on both lines, and is a solution to the system!
Next, substitute 3x + 1 (from y = 3x + 1) into the other equation in place for y, then rearrange and solve for x.
This will be the new concept we introduce in this lesson. Remember, equations are descriptions of how two points interact with each other and how they change value based on what the other point is.
We can rearrange our equations to give us a representation of what one of the points will be based on the other point. When we have a description of that point, we can think of it as having a value for that point, we can then use that description and sub it into another equation, this will give us the point where the two equations meet.
Here we have a description of the point y based on x, we can use that description of y, and sub it into the other equation, as we will be left with only one variable, x, we can then move onto solving for x. When we solve for x, this will give us the x value where the lines meet.
Subbing in the description of y which is the expression 3x + 1 into the other equation in our system, y + 2x = 11,
(3x + 1) + 2x = 11
[If it helps, think of it like you are subbing in a value for y, like subbing in y = something into the equation]
3x + 1 + 2x = 11
5x + 1 = 11
Subtract 1 from both sides,
5x + 1 - 1 = 11 - 1
5x = 10
Divide both sides by 5,
(5x) / 5 = (10) / 5
x = 2,
This also gives us the x value for where the two lines meet.
Finally, use the x value from above and sub it into both equations (this will determine the y point where the two equations meet and hence the solution to the system) then state the solution to the system and comment on the result from each substitution.
So, we have our x value, x = 2, which is the x value for where the two equations meet, now we need to find the y value for where they meet. Let’s sub in x into each equation.
Let’s start with, x = 2 into y = 3x + 1
y = 3(2) + 1
y = 6 + 1
y = 7
Now let’s sub in x = 2 into y + 2x = 11
y + 2(2) = 11
y + 4 = 11
Rearranging for y, subtract 4 from both sides,
y + 4 - 4 = 11 - 4
y = 7
The solution to the system is x = 2 and y = 7 or (2, 7)
ii) Comment on the results from i)
First, think about why we have the same value for y when we sub x into each equation.
x = 2 is the x-value where both the equations meet,
x = 2 is an x-value on both lines,
If we sub that x into either of the equations, it will give us the same y-value, as that is the point where both the equations are equal.
## Question 2
Solve the following systems by our substitution technique.
y = 2x + 1 and y = -2x - 7
Again, solving a solution means finding the point that makes both equations true, it is the point where both the equations meet.
Here, we can take one of the expressions for y and substitute it into the other equation.
Let's take y = 2x + 1 and put (-2x - 7) in place of the y in the equation, (the -2x - 7 comes from y = -2x -7)
(-2x - 7) = 2x + 1
-2x - 7 + 2x = 2x + 1 + 2x
-7 = 4x + 1
Subtract 1 from both sides,
-7 - 1 = 4x + 1 - 1
-8 = 4x
Divide both sides by 4,
(-8) / 4 = (4x) / 4
-2 = x
So, x = -2 is the x-value for our solution to the system (the point where the equations meet), now we need to find the y-value.
Remember, the solution to the system is the point on both equations, so in order to find our y-value, we can choose either of the equations to sub our x-value into as both are correct.
Let’s sub x = -2 into y = 2x + 1,
y = 2(-2) + 1
y = -4 + 1
y = -3,
So, our solution to the system is x = -2 and y = -3 or (-2, -3)
6x - 3y = 9 and y = -x + 6
Sub in (-x + 6) into 6x - 3y = 9
6x - 3(-x + 6) = 9
6x -3(-x) -3(6) = 9
6x + 3x - 18 = 9
9x - 18 = 9
9x - 18 + 18 = 9 + 18
9x = 27
Divide both sides by 9,
(9x) / 9 = (27) / 9
x = 3,
Now, sub x = 3 into one of our equations, let choose, y = - x + 6
y = -(3) + 6
y = 3
So, our point, x = 3 and y = 3 or (3, 3) is the solution to our system.
We can use this method to solve word problems as well.
## Question 3
Lucy and Sarah are figuring out how old they are. They know that the sum of both their ages is 50. They also know that Sarah's age is the same as 14 more than 3 times Lucy’s age.
1. write two equations that make up a system that represents what we know about their ages, with Sarah's age as s and Lucy's age as l.
2. Use our substitution method to solve these equations.
What do we know?
First, we know the sum of both their ages will be 50, this would look like,
Lucy’s age plus Sarah’s age is 50, or,
l + s = 50,
We also know that Sarah’s age is the same as 3 times Lucy’s age with 14 more, this would look like,
Sarah’s age is 3 times Lucy’s age plus 14, or,
s = 3l + 14,
Remembering our substitution method, we can find the solution to this system but equating some of the terms, we can take (s = 3l + 14) and put it into l + (s) = 50,
Write two equations that make up a system that represents what we know about their ages, with Sarah's age as s and Lucy's age as l.
l + (3l + 14) = 50,
4l + 14 = 50
Subtract 14 from both sides,
4l + 14 - 14 = 50 - 14
4l = 36
Divide both sides by 4, m
4l / 4 = 36 / 4
l = 9,
Now we know l = 9 (Lucy’s age), we need to find Sarah’s, so sub l into one of our equations,
use l + s = 50,
(9) + s = 50
Subtract both sides by 9,
9 + s - 9 = 50 - 9
s = 41,
There we have l = 9 and s = 41, or Lucy’s age is 9 and Sarah’s age is 41.
## Question 4
A red train and a blue train are travelling towards each other on the same tracks. They start 300 miles away from each other. The red train is travelling at 70 miles per hour (70h), the blue train is travelling at 50 miles per hour (50h). How long until the trains meet?
Try and write to equations about what we know from the question.
We can represent the two trains positions by picking one of the trains to be “position zero” and the others trains position to be based off the train at “position zero”, we must also include a representation of how their position has changed over time (their speed).
We know that when the two trains meet, the positions will be equal.
For the red train we have:
Starting at the beginning, when they are 300 miles apart, set red trains position as 0, accounting for speed as well,
position = 0 plus (red trains speed)
p = 0 + 70h
For the blue train we have:
Starting at the beginning, when they are 300 miles apart, subtract the speed as the blue train is moving in the opposite direction, accounting for speed as well,
position = 300 minus (blue trains speed)
p = 300 - 50h
When both trains have met, the positions will be equal, so we can equate them,
300 - 50h = 0 + 70h, |
5:30 pm Uncategorized
# Constructing a Triangle: A Comprehensive Guide
Triangles are fundamental geometric shapes that have fascinated mathematicians, architects, and artists for centuries. Their simplicity and versatility make them a cornerstone of various fields, from engineering and physics to art and design. In this article, we will explore the process of constructing a triangle, discussing different methods, properties, and applications. Whether you are a student, a professional, or simply curious about triangles, this guide will provide valuable insights into this fascinating shape.
## The Basics of Triangle Construction
Before delving into the construction techniques, let’s review the basic elements of a triangle. A triangle is a polygon with three sides, three angles, and three vertices. The sum of the interior angles of a triangle always equals 180 degrees. Triangles can be classified based on their side lengths and angle measures, resulting in various types such as equilateral, isosceles, and scalene triangles.
### Tools Required for Triangle Construction
Constructing a triangle requires a few essential tools. These include:
• A ruler or straightedge: Used to draw straight lines and measure distances.
• A compass: Used to draw circles and arcs of specific radii.
• A protractor: Used to measure and draw angles accurately.
• A pencil: Used to mark points and lines during the construction process.
## Methods of Triangle Construction
There are several methods to construct triangles, each with its own set of rules and procedures. Let’s explore some of the most common methods:
### 1. Constructing a Triangle Given Three Sides
If you are given the lengths of all three sides of a triangle, you can construct it using the following steps:
1. Draw a line segment AB of the given length for the first side.
2. From point A, draw an arc with a radius equal to the length of the second side.
3. From point B, draw another arc with a radius equal to the length of the third side.
4. The intersection of these two arcs will be the third vertex of the triangle, C.
5. Connect points A, B, and C to form the triangle.
This method is based on the fact that the sum of any two sides of a triangle must be greater than the length of the third side, according to the triangle inequality theorem.
### 2. Constructing a Triangle Given Two Sides and an Angle
If you are given the lengths of two sides and the measure of the included angle, you can construct the triangle using the following steps:
1. Draw a line segment AB of the given length for the first side.
2. From point A, draw an arc with a radius equal to the length of the second side.
3. Using a protractor, measure the given angle at point A.
4. From the vertex of the angle, draw an arc intersecting the previous arc.
5. The intersection of these two arcs will be the second vertex of the triangle, C.
6. Connect points A, B, and C to form the triangle.
This method utilizes the fact that the length of the third side and the measures of the other two angles can be determined using trigonometric functions.
### 3. Constructing a Triangle Given Two Angles and a Side
If you are given the measures of two angles and the length of a side, you can construct the triangle using the following steps:
1. Draw a line segment AB of the given length for one side of the triangle.
2. Using a protractor, measure the first given angle at point A.
3. From point A, draw an arc with a radius equal to the length of the second side.
4. Using a protractor, measure the second given angle at the intersection of the arc and line AB.
5. Connect points A, B, and C to form the triangle.
This method relies on the fact that the sum of the three angles in a triangle is always 180 degrees.
## Applications of Triangle Construction
The ability to construct triangles accurately is essential in various fields. Here are a few examples of how triangle construction is applied:
### Architecture and Engineering
In architecture and engineering, triangles play a crucial role in structural stability. Triangular trusses and frameworks are commonly used to distribute loads evenly and provide strength to buildings, bridges, and other structures. The precise construction of triangles ensures the stability and integrity of these structures.
### Surveying and Navigation
Surveyors and navigators often use triangles to determine distances and angles. By measuring the angles of a triangle formed by landmarks or celestial objects, they can calculate distances or plot accurate maps. The principles of triangle construction are fundamental to these calculations.
### Art and Design
Artists and designers frequently use triangles to create visually appealing compositions. The balance and harmony achieved through the arrangement of triangles can evoke a sense of stability or dynamism in a piece of art or design. Understanding the principles of triangle construction allows artists and designers to create aesthetically pleasing works.
## Q&A
### 1. Can all triangles be constructed?
Not all combinations of side lengths and angle measures can form valid triangles. According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Additionally, the sum of the interior angles of a triangle must always be 180 degrees.
### 2. Are there any shortcuts or tricks for triangle construction?
While the basic methods of triangle construction involve precise measurements and calculations, there are some shortcuts and tricks that can simplify the process. For example, using a compass to draw arcs of specific radii can help create congruent angles or sides. Additionally, knowledge of special triangles, such as the 30-60-90 or 45-45-90 triangles, can aid in constructing triangles with specific angle measures.
### 3. Can triangles be constructed without using a compass?
While a compass is a valuable tool for triangle construction, it is not always necessary. In some cases, rulers and protractors alone can be used to construct triangles. However, a compass provides greater accuracy and precision, especially when drawing circles or arcs.
### 4. Are there any real-life examples of triangle construction?
Triangle construction is prevalent in various real-life scenarios. For instance, architects and
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# Angular momentum
### Angular momentum
The quantity of the rotational motion, which is identical to mass (m) in the linear motion, is the moment of inertia (I). The quantity of the rotational motion, which is identical to the velocity (v) in the linear motion, is the angular velocity (ω). Thus, the rotating object has angular momentum that can be calculated using the equation:
L = I ω
L = angular momentum (kg m2/s), I = moment of inertia (kg m2), ω = angular velocity (rad/s)
Sample problems of the angular momentum
Sample problem 1.
A particle with a mass of 0.5 grams moves in a circle with a constant angular velocity of 2 rad/s. Determine the angular momentum of the particle if the radius of the particle’s path is 10 cm.
Solution:
The moment of inertia of the particle:
I = m r2 = (0.5 x 10-3 kg)(1 x 10-1 m)2 = (0.5 x 10-3 kg)(1 x 10-2 m2) = 0.5 x 10-5 kg m2
The angular speed of the particle:
The angular momentum of the particle:
L = (0.5 x 10-5 kg m2)(2 rad/s) = 1 x 10-5 kg m2/s
The law of conservation of angular momentum
The law of conservation of angular momentum states that if the resultant moment of force on a rigid body when rotating is zero, then the angular momentum of the rigid body when rotating is always constant. The law of conservation of angular momentum can be derived mathematically by modifying the equation of Newton’s second law of angular momentum. Here is the equation of Newton‘s second law on the angular momentum:
If the resultant moment of force is zero, then the equation above changes to:
It = the final moment of inertia, Io = the initial moment of inertia, ωt = the final angular speed, ωo = the initial angular speed, Lt = the final angular momentum, Lo = the initial angular momentum.
### Sample problems of the law of conservation of angular momentum
Sample problem 1.
A homogeneous solid cylinder disk is initially rotating on its axis with a speed of 4 rad/s. The mass and radius of the disk are 1 kg and 0.5 m. If above the plate is placed a ring that has a mass and radius of 0.2 kg and 0.1 m and the center of the ring,
just above the center of the disk, then the disc and ring will rotate together with the angular velocity of…..
Solution:
Moment of inertia of solid cylinder : I = 1⁄2 m r2 = 1⁄2 (1 kg)(0.5 m)2 = (0.5)(0.25) = 0.125 kg m2
Moment of inertia of ring : I = m r2 = (0.2 kg)(0.1 m)2 = (0.2)(0.01) = 0.002 kg m2
Initial angular momentum (L1) = Final angular momentum (L2)
I1 ω1 = I2 ω2
(0.125 kg m2)(4 rad/s) = (0.125 kg m2 + 0.002 kg m2)(ω2)
(0.5) = (0.127)(ω2)
ω2 = 0.5 : 0.127
Conceptual questions and asnwer
1. What is angular momentum?
Answer: Angular momentum is a measure of the amount of rotation an object has, considering its mass and shape. It is the rotational analog of linear momentum.
2. How is angular momentum defined mathematically?
Answer: Angular momentum (L) is defined as , where is the position vector from the axis of rotation to the point of application, and is the linear momentum.
3. Is angular momentum a scalar or vector quantity?
Answer: Angular momentum is a vector quantity.
4. What is the unit of angular momentum in the SI system?
Answer: The unit of angular momentum in the SI system is kilogram-meter squared per second (kg·m2/s).
5. How is angular momentum conserved?
Answer: In a closed system with no external torques, the total angular momentum remains constant.
6. What is the principle of conservation of angular momentum?
Answer: The principle states that the total angular momentum of a closed system remains constant unless acted upon by an external torque.
7. How does the spin of a figure skater relate to angular momentum?
Answer: When a figure skater pulls their arms and legs close to their body during a spin, they decrease their moment of inertia, causing them to spin faster. This demonstrates the conservation of angular momentum.
8. What is the relationship between torque and angular momentum?
Answer: Torque () is the rate of change of angular momentum. Mathematically, .
9. Is it possible for an object to have angular momentum without angular velocity?
Answer: No, if an object has angular momentum, it has angular velocity. The magnitude and direction of angular momentum depend on both the moment of inertia and angular velocity.
10. How does the moment of inertia affect angular momentum?
Answer: Angular momentum (L) is the product of the moment of inertia (I) and angular velocity (ω). As , an increase in the moment of inertia for a given angular velocity will increase angular momentum.
11. What is the difference between spin and orbital angular momentum in quantum mechanics?
Answer: Spin angular momentum is an intrinsic property of particles, like electrons, and does not arise from motion in space. Orbital angular momentum, on the other hand, arises from the motion of a particle around a central point.
12. What role does angular momentum play in the formation of planetary systems?
Answer: As a molecular cloud collapses to form stars and planets, the conservation of angular momentum causes the material to flatten into a disk around the newborn star, from which planets can form.
13. How does a gyroscope maintain its orientation in space?
Answer: A spinning gyroscope has angular momentum. When an external torque tries to change its orientation, the gyroscope precesses, or changes its axis of rotation, in a direction perpendicular to the applied torque due to the conservation of angular momentum.
14. What causes precession in a spinning top?
Answer: Precession in a spinning top is caused by the torque due to gravity acting on the top’s mass center, which is offset from its pivot point. This torque results in a change in the direction of the top’s angular momentum, causing it to precess.
15. Why do stars flatten at their poles?
Answer: As stars rotate, they experience a centrifugal force pushing matter outward. This effect is stronger at the equator than at the poles, leading to an oblate, or flattened, shape.
16. How is angular momentum related to Kepler’s second law?
Answer: Kepler’s second law states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. This is a consequence of the conservation of angular momentum.
17. Why does a cat always land on its feet when it falls?
Answer: Cats use the conservation of angular momentum. By twisting their bodies and changing their moments of inertia mid-air, they can reorient themselves to land on their feet without violating the conservation laws.
18. How is the magnetic moment related to angular momentum in electrons?
Answer: The magnetic moment of an electron arises from its spin and orbital angular momentum. The magnetic moment is proportional to the electron’s angular momentum.
19. What is angular momentum in terms of polar coordinates?
Answer: In polar coordinates, angular momentum for a point mass can be expressed as , where is the mass, is the radial distance, and is the tangential velocity.
20. What happens to the angular momentum of a closed system when two objects collide?
Answer: In a closed system where two objects collide, the total angular momentum before the collision is equal to the total angular momentum after the collision, assuming no external torques act on the system.
Problems and Solutions
1. Problem: A point mass m = 2 kg moves with a velocity v = 3 m/s in a circle of radius r = 4 m. Calculate its angular momentum about the center of the circle.
Solution: =2×4×3=24 kg.m2/s
2. Problem: A solid sphere of mass 3 kg and radius 0.5 m is rotating about its diameter with an angular velocity ω = 2 rad/s. What is its angular momentum?
Solution: For a solid sphere,
L=2/5×3×(0.5)2×2=0.6 kg.m2/s
3. Problem: A point mass of 3 kg is moving with a speed of 4 m/s in a straight line. What is its angular momentum about a point 5 m away from its line of motion?
Solution:
L=3×4×5=60 kg.m2/s
4. Problem: A disc of mass 5 kg and radius 3 m is rotating with an angular speed of 2 rad/s. Calculate its angular momentum.
Solution: For a disc,
L=1/2×5×32×2=45 kg.m2/s
5. Problem: A rod of length 4 m and mass 2 kg rotates about one end with an angular velocity of 3 rad/s. Find its angular momentum.
Solution: For a rod about one end,
L=1/3×2×42×3=96 kg.m2/s
6. Problem: What is the change in angular momentum if a rotating body doubles its angular velocity?
Solution: The relation between angular momentum (L) and angular velocity (ω) is . If ω is doubled, the angular momentum will also double.
7. Problem: If the radius of a circular path of a moving particle doubles while keeping velocity constant, by what factor does the angular momentum change?
Solution: Angular momentum, . If r doubles, L will also double.
8. Problem: A child of mass 30 kg is standing at the edge of a merry-go-round which is at rest. The radius of the merry-go-round is 2 m. If the child starts running at a speed of 2m/s with respect to the ground along the edge, find the angular momentum of the child with respect to the center.
Solution: =30×2×2=120 kg.m2/s
9. Problem: A cylinder of mass 6 kg and radius 2 m is rotating about its central axis with an angular velocity of 5 rad/s. Calculate its angular momentum.
Solution: For a cylinder,
=1/2×6×22×5=120 kg.m2/s
10. Problem: What will be the new angular momentum of a body if its moment of inertia is halved and its angular velocity is tripled?
Solution: Given , if is halved and is tripled, the new will be 1/2×3=1.5 × the original.
11. Problem: A hoop of mass 2kg and radius 1m is rotating with an angular speed of 4 rad/s. Calculate its angular momentum.
Solution: For a hoop,
L=2×12×4=8 kg.m2/s
12. Problem: A fan has blades of length 0.5m and total mass 1kg. If it rotates with an angular speed of 10 rad/s, determine its angular momentum.
Solution: Assuming the fan blades are like rods rotating about one end,
=1/3×1×0.52×10=0.833 kg.m2/s
13. Problem: Calculate the angular momentum of the earth about its own axis due to its rotation. Given: Mass of earth =5.972×1024 kg, Radius , .
Solution: Assuming earth as a solid sphere,
=2/5×5.972×1024×(6371×103)2×7.27×10−5
14. Problem: A child moves towards the center of a rotating platform. Does the angular momentum increase, decrease or remain the same?
Solution: Angular momentum is conserved. As the child moves towards the center, the moment of inertia decreases. To conserve angular momentum, the angular velocity increases. Thus, the angular momentum remains the same.
15. Problem: If the kinetic energy of a rotating body is doubled, by what factor does its angular momentum change?
Solution: Kinetic energy and Angular momentum . If kinetic energy is doubled and remains constant, is doubled. This implies that is increased by a factor of . Thus, also increases by a factor of .
16. Problem: A bicycle wheel of radius 0.3 m has a rim of mass 1.5 kg. If it rotates with an angular speed of 5 rad/s, find its angular momentum.
Solution: Assuming the rim as a hoop,
=1.5×0.32×5=6.75 kg.m2/s
17. Problem: An ice skater pulls her arms inward while spinning. What happens to her angular momentum?
Solution: The skater’s angular momentum is conserved. By pulling her arms in, she decreases her moment of inertia, and thus she spins faster (increased angular velocity) to conserve her angular momentum.
18. Problem: A flywheel has a moment of inertia of 0.2 kg.m2 and is rotating with an angular velocity of 10 rad/s. Find its angular momentum.
Solution:
=0.2×10=2 kg.m2/s
19. Problem: A system has a net external torque of zero acting on it. What can you say about its angular momentum?
Solution: If the net external torque is zero, then according to the conservation of angular momentum, the total angular momentum of the system remains constant.
20. Problem: A point mass m = 5 kg moves with a velocity v = 3 m/s perpendicular to the line joining it to a point P. If the distance from the point P is 4 m, find its angular momentum about point P.
Solution:
=5×4×3=60 kg.m2/s |
# Write Scientific Notation From Standard Form Worksheet
Assignment will be available soon
Writing scientific notation from standard form involves converting a number expressed in its full decimal form into a product of a coefficient and a power of ten (e.g., converting 35,000 to 3.5*10^4). This process includes identifying the significant digits, placing the decimal point after the first significant digit, and determining the exponent by counting the number of places the decimal point has moved. The coefficient is then multiplied by 10 raised to the appropriate power.
Algebra 1
Scientific Notation
## How Will This Worksheet on "Write Scientific Notation from Standard Form" Benefit Your Student's Learning?
• Worksheets help in enhancing understanding of converting large or small numbers.
• Building confidence in working with scientific notation.
• Improving accuracy in identifying significant figures and applying powers of ten.
• Preparing students for real-world applications in science and math.
• Reinforcing concepts through consistent practice.
• Encouraging critical thinking and problem-solving skills.
## How to Write Scientific Notation from Standard Form?
• Firstly Identify the first non-zero digit in the number.
• Then count the number of places the decimal point needs to be moved so that only one non-zero digit is to the left of the decimal point.
• If the decimal is moved to the right, the exponent is positive. If moved to the left, the exponent is negative.
• Write the first non-zero digit and any other digits to the right of it as the coefficient.
• Multiply the coefficient by 10 raised to the exponent.
## Solved Example
Q. How do you write $603,000$ in scientific notation?$\newline$$\_\_\_\_\_\_ \times 10^{\_\_\_\_\_\_}$
Solution:
1. Find Coefficient: What will be the coefficient when $603,000$ is written in scientific notation? Move the decimal to the left until the number is greater than or equal to $1$ and less than $10$. Coefficient: $6.03$
2. Sign of Exponent: Would $603,000$ have a positive or negative exponent when written in scientific notation? $603,000$ is a number greater than $10$. It can be represented by a positive exponent.
3. Determine Exponent: We know:
The coefficient is $6.03$.
The exponent will be positive.
What is the exponent when $603,000$ is written in scientific notation?
We moved the decimal to the left $5$ times to make the coefficient $6.03$.
$603,000 = 6.03\times 10^5$
Exponent: $5$
4. Write in Scientific Notation: We know the coefficient is $6.03$ and the exponent is $5$. Fill in the values to write $603,000$ in scientific notation. Substitute $6.03$ for $a$ and $5$ for $b$ in $a \times 10^b$. Scientific notation: $6.03 \times 10^5$
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# Image of a function.Determine the image of the function f(x)=x^2-6x+5 for x>=4.
sciencesolve | Certified Educator
You need to find the image of the function, hence, you need to evaluate the range of the given function, such that:
`x^2 - 6x + 5 = y => x^2 - 6x = y - 5`
You need to complete the square to the left using the following formula, such that:
`a^2 - 2ab + b^2 = (a - b)^2`
Reasoning by analogy, yields:
`x^2 - 6x + 9 = y - 5 + 9 => (x - 3)^2 = y + 4 => x - 3 = +-sqrt(y + 4) => x = 3 +- sqrt(y + 4)`
Since the problem provides the information that `x >= 4` yields:
`3 +- sqrt(y + 4) >= 4 => +- sqrt(y + 4) >= 4 - 3 => +- sqrt(y + 4) >= 1`
`sqrt(y + 4) >= 1 => y + 4 >= 1 => y >= 1 - 4 => y = f(x) >= -3`
`-sqrt(y + 4) >= 1 => sqrt(y + 4) <= -1 ` invalid
Hence, evaluating the range of the given function, under the given condition, yields `f(x) >= -3.`
giorgiana1976 | Student
We'll put y = x^2-6x+5
We'll subtract y both sides:
x^2 - 6x + 5 - y = 0
We'll determine the roots of the equtaion with quadratic formula:
x1 = [6 + sqrt(36 - 4(5-y))]/2
x2 = [6 - sqrt(36 - 4(5-y))]/2
But, from enunciation, we know that x>=4
[6 + sqrt(36 - 4(5-y))]/2 >=4
We'll multiply by 2:
6 + sqrt(36 - 4(5-y) >= 8
We'll subtract 6 both sides:
sqrt(36 - 4(5-y)) >= 2
We'll raise to square both sides:
36 - 4(5-y) >= 4
We'll subtract 36 both sides:
- 4(5-y) >= 4 - 36
- 4(5-y) >= -32
We'll divide by -4:
5-y =< 8
y >= -3
The image of the function belongs to the range [-3 , +infinite), for x >=4. |
## Maths Class 10 Notes for Area Related to Circles
PERIMETER AND AREA OF A CIRCLE
We know that
The area of a circle is the measurement of the region enclosed by its boundary. It is measured in square units. -ie- square centimetres or square metres etc.
Area of the circle = πr2
The perimeter of a circle is the length of its boundary. The unit measurement of perimeter is the unit of length.
Perimeter of circle = 2πr
where r is the radius of the circle.
Perimeter of a circle is known as circumference of a circle.
AREA OF SECTOR OF CIRCLE
The part of the circle inclined between two radii (OA & OB) is called sector of circle.
Area of the sector OAPB =0 /360° x πr2
& length of an arc of sector OAPB = length of arc AB =0 /360° x 2πr
Where 0 is the measure of arc AB.
Perimeter of the sector (minor sector)
=0 /360° x 2πr + 2r
AREA OF SEGMENT OF CIRCLE
Any chord AB divides circle into two parts. The bigger part is known as major segment and smaller one is called minor segment.
Area of minor segment APB — Area of sector OAPB — area of ΔOAB
Area of major segment OAQB = πr2 — area of minor segment APB.
NOTE: Area of Δ OAB with ∠AOB = 0 = 1/2 (OA) (OB) sinθ
IN GENERAL
Area of segment of a circle = Area of the corresponding sector — Area of the corresponding triangle.
AREA ENCLOSED BY THE TWO CIRCLES
If R and r are the radii of two concentric circles such that R > r then area enclosed by the two circles = πR2 – πr2
SOME USEFUL RESULTS
(i) If two circles touch internally, then the distance between their centres is equal to the difference of their radii.
(ii) If two circles touch externally, then the distance between their centres is equal to the sum of their radii.
(iii) Distance moved by a rotating wheel in one revolution is equal to the circumference of the wheel.
(iv) The number of revolutions completed by a rotating wheel in one minute Distance moved in one minute
= Distance moved in one minute / circumference |
# Lesson 14
Situations Involving Factors and Multiples
## Warm-up: Number Talk: Dividing by 7 (10 minutes)
### Narrative
This Number Talk encourages students to compose or decompose multiples of 7 and to rely on properties of operations to mentally solve problems. The ability to compose and decompose numbers will be helpful when students divide multi-digit numbers. It also promotes the reasoning that is useful when finding multiples of a number, or when deciding if a number is a multiple of another number.
### Launch
• Display one expression.
• “Give me a signal when you have an answer and can explain how you got it.”
• 1 minute: quiet think time
### Activity
• Record answers and strategy.
• Keep expressions and work displayed.
• Repeat with each expression.
### Student Facing
Find the value of each expression mentally.
• $$21 \div 7$$
• $$35 \div 7$$
• $$140 \div 7$$
• $$196 \div 7$$
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• “What do the expressions have in common?” (They all involve division by 7. The dividends are all multiples of 7. The results have no remainders.)
• “How did the first three expressions help us find the value of the last expression?”
• Consider asking:
• “Who can restate _______ 's reasoning in a different way?”
• “Did anyone have the same strategy but would explain it differently?”
• “Did anyone approach the expression in a different way?”
• “Does anyone want to add on to____’s strategy?”
## Activity 1: Write Multiples (20 minutes)
### Narrative
This activity prompts students to use the relationship between multiplication and division and their understanding of factors and multiples to solve problems about an unknown factor (MP7). Students recognize such problems as division situations. Here the dividends are three-digit numbers and the divisors are one-digit numbers. Students use the context of factors and multiples to interpret division that results in a remainder.
One approach for solving these problems is by decomposing the dividend into familiar multiples of 10 and then dividing the remaining number (which is now smaller) by the divisor. Another is to find increasingly greater multiples of the divisor until reaching the dividend. Both are productive and appropriate. In the synthesis, help students see the connections between the two paths.
MLR8 Discussion Supports. Synthesis: Some students may benefit from the opportunity to rehearse what they will say with a partner before they share with the whole class.
Advances: Speaking
Engagement: Develop Effort and Persistence. If students don’t recognize this as a division situation at first, they may do so when presented with a more accessible value. Consider offering this situation first: “Han starts writing multiples of a number. When he reaches 12, he has written 3 numbers.” Invite students to identify what number Han is writing multiples of (4), and how they know. Then ask how they might apply that reasoning to the task as presented here.
Supports accessibility for: Conceptual Processing
### Launch
• “Who can remind the class of the meaning of ‘multiple?’”
### Activity
• 4–5 minutes: quiet work time for the first set of questions
• Monitor for students who:
• decompose 104 into a multiple of 8 and another number
• compose 104 from increasingly larger multiples of 8
• use multiplication or division equations to show their reasoning
• Pause for a discussion before the second set of questions.
• Select students who use different decomposition strategies to share responses. Record and display their reasoning.
• “Let’s revisit each question we just answered. What equations could we write to represent them?”
• Display equations and highlight their connection to the questions:
1. $$8 \times {?} = 104$$ or $$104 \div 8 = {?}$$
2. $$15 \times 13 = {?}$$ or $$13 \times 15 = {?}$$
3. $${?} \times 13 = 286$$ or $$286 \div 13 = {?}$$
• 4–5 minutes: quiet work time for the second set of questions
### Student Facing
1. Han starts writing multiples of a number. When he reaches 104, he has written 8 numbers.
For each of the following questions, show your reasoning.
1. What number is Han writing multiples of?
2. What is the 15th multiple of this number?
3. Han gets to 286. How many numbers has he written at that point?
2. Kiran wants to know how many multiples of 7 are between 0 and 150.
1. He thinks he can use division to find out. Do you agree? Explain your reasoning.
2. How many multiples will he find? Show your reasoning.
3. Is 150 a multiple of 7? Show how you know.
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• Invite students to share responses for the last set of questions.
• Highlight that to divide a number by a smaller number—say, divide 150 by 7, we can:
• Use familiar multiples or multiplication facts to help us. For example, if we know $$(20 \times 7) + (1 \times 7) = 147$$, we know the result is 21 with a remainder of 3, or $$150 = 21 \times 7 + 3$$.
• Think of the dividend in smaller chunks. For example: We can see the 150 as $$140 + 10$$ and divide each 140 and 10 by 7 separately, which gives $$20 + 1$$ with a remainder of 3.
## Activity 2: Jada’s Mystery Number (15 minutes)
### Narrative
In this activity, students continue to use the relationship between multiplication and division to reason about situations that involve division. Students divide three-digit numbers by single-digit divisors and find results with and without a remainder.
• Groups of 2
### Activity
• 5 minutes: independent work time on the first question
• 2 minutes: partner discussion
• 23 minutes: partner work time on the second question
• Monitor for students who clearly show how they use partial products or partial quotients to answer the questions.
### Student Facing
Jada is writing multiples of a mystery number. After writing a bunch of numbers, she writes out 126.
• Mai says 6 is the mystery number.
• Priya says 8 is the mystery number.
• Andre says 9 could be the mystery number.
1. Which student do you agree with? Show your reasoning and include equations.
2. Jada gives one more clue: “If I keep writing multiples, I’ll get to 153.”
What is the mystery number? Explain or show your reasoning.
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• “How could we use division to help us find out the mystery number?” (If the result has a remainder, then the divisor could not be the mystery number. 153 divided by 6 has 3 as a remainder. 153 divided by 9 has no remainder, so 9 is the mystery number.)
## Lesson Synthesis
### Lesson Synthesis
“Today we tried to find out if a number is a multiple or a factor of another number. For instance: Is 267 a multiple of 8?”
“Is this question a division problem?” (Yes)
“Why?” (We’re looking for how many 8s are in 267.)
“What are we dividing?” (267 by 8)
“One way to answer the question is by using familiar multiplication facts or by finding partial products. How would you start?” (One way is to start with 10 x 8 or its multiples, build the products up to 267 or close to it, and then try smaller multiples of 8.
$$10 \times 8 = 80$$
$$20 \times 8 = 160$$
$$30 \times 8 = 240$$
$$3 \times 8 = 24$$
$$33 \times 8 = 264$$
264 is 3 away from 267, not enough to make another 8, so 267 is not a multiple of 8.)
“Why might it be helpful to start with multiples of 10?” (They’re easy to find and easy to add.)
“Can we use division to answer the question?” (We can start a number close to 267 that is a multiple of 8, divide it by 8, see what is left, and find a multiple of 8 that is close to that number. For example:
• $$160 \div 8 = 20$$. After taking 160 away, there’s 107 left.
• $$80 \div 8 = 10$$. After taking 80 away, there’s 27 left.
• $$24 \div 8 = 3$$. After taking 24 away, there’s 3 left, which is not enough to make 8.)
“How are the two approaches alike?” (They involve using smaller multiples of a number to see if a larger number is a multiple of that number.)
## Cool-down: Reaching 161 with Multiples (5 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners. |
# Compound C shapes. How to calculate the perimeter and area of a C shape.
Here you will be shown to work out the area and perimeter of a C shape. A C shape is a compound shape made up of 3 rectangles. Make sure you know how to find the area and perimeter of an L shape before you attempt this lesson on C shapes.
To work out the perimeter of the C shape first you will need to work out the sides that are missing around the C shape. Once this is done add up all of the side lengths to give the perimeter of the C shape.
Finding the area of a C shape is a little harder than working out the perimeter. You will have to divide the C shape into 3 rectangles. This can be done in two ways - but both ways will give you the same answer. The next thing to do is to work out the area of the 3 rectangles that you have created. The area of each rectangle can be calculated by multiplying the two side lengths of each rectangle together. Make sure you choose the correct side lengths for each rectangle. Once the area of each rectangle has been found, find the total area of the C shape by summing up the areas of all 3 rectangles.
Let’s take a look at an example.
Work out the area of this C shape:
Let’s find the perimeter first. There is only one side length that is missing which can be found by subtracting the two 3m lengths from the total height of 12m:
12 – 3 – 3 = 6m
All you need to do now is add up all of the side lengths:
12 + 3 + 7 + 6 + 10 + 3 + 15 + 12 = 68m
To calculate the area of the C shape, divide the C shape into 3 rectangles:
The area of rectangle 1 can be found by multiplying the two side lengths together:
12 × 3 = 36 m²
The width of rectangle 2 is missing, this can be found by working out 15 – 10 = 5 m or 12 – 7 = 5m. So the area of rectangle 2 is:
5 × 6 = 30 m²
Finally, work out the area of rectangle 3 by multiplying the two side lengths together:
15 × 3 = 45 m²
So the total area of the compound C shape is:
36 + 30 + 45 = 111 m².
So the perimeter of the C shape is 68m and the area of the C shape is 111m².
If you are finding this difficult take a look at these:
Compound shapes. How to find the area of a L shape.
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# 12.06 Speed, distance and time
Lesson
### Speed, distance and time
The average speed of a vehicle is related to the total distance travelled and the total time taken by the formula:
Speed
$\text{average speed}=\frac{\text{total distance travelled}}{\text{total time taken}}$average speed=total distance travelledtotal time taken
$S=\frac{D}{T}$S=DT
We can rearrange the formula for speed to make either $D$D or $T$T the subject. For example, rearranging to make $T$T the subject:
$S$S $=$= $\frac{D}{T}$DT $S\times T$S×T $=$= $\frac{D}{T}\times T$DT×T multiply both sides by $T$T $ST$ST $=$= $D$D $T$T $=$= $\frac{D}{S}$DS divide both sides by $S$S
Similarly we can rearrange to make $D$D the subject and the formulas are below.
Time or distance
Finding unknown time or distance:
Unknown time Unknown distance
$T=\frac{D}{S}$T=DS $D=S\times T$D=S×T
#### Worked example
##### Example 1
Jim drives his car from Sydney to Canberra, a distance of $279$279 km. The journey takes him $3$3 hours and $45$45 minutes, including lunch and fuel stops along the way.
(a) What is Jim's average speed for the journey, correct to the nearest km/h?
Think: Whenever we use a formula, we must keep units consistent. In this case, speed is expressed in kilometres per hour and distance is in kilometres, so time must be expressed in hours.
Our first step is to convert the time of $3$3 hours $45$45 minutes into hours. Because there are $60$60 minutes in an hour, $45$45 minutes is equivalent to $\frac{45}{60}$4560 or $0.75$0.75 hours. Therefore $3$3 hours $45$45 minutes is the same as $3.75$3.75 hours.
Do: Substituting into the speed-distance-time formula:
$S$S $=$= $\frac{D}{T}$DT $=$= $\frac{279}{3.75}$2793.75 substitute $D=279$D=279 and $T=3.75$T=3.75 $=$= $74.4$74.4 $=$= $74$74 km/h (nearest km/h)
(b) How far could Jim travel at this same average speed in $40$40 minutes. Round your answer to the nearest kilometre.
Think: A time of $40$40 minutes is equivalent to $\frac{40}{60}$4060 hours, which simplifies to $\frac{2}{3}$23 of an hour. In this case it is easier to leave the time as a fraction because the decimal is recurring. It is also easier to substitute our values into the formula when $D$D is the subject.
Do:
$D$D $=$= $ST$ST $=$= $74\times\frac{2}{3}$74×23 substitute $S=74$S=74 and $T=\frac{2}{3}$T=23 $=$= $\frac{148}{3}$1483 $=$= $49.\overline{3}$49.3 $=$= $49$49 km (nearest km)
(c) If Jim travelled at this same average speed, how long would it take him to drive from Sydney to Melbourne, a distance of $875$875 km. Give your answer in hours and minutes.
Think: This time, it will be easier to substitute our values into the formula when $T$T is the subject.
Do:
$T$T $=$= $\frac{D}{S}$DS $T$T $=$= $\frac{875}{74}$87574 substitute $D=875$D=875 and $S=74$S=74 $T$T $=$= $11.824\ldots$11.824… $T$T $=$= $11$11 hours $49$49 minutes (nearest minute)
Reflect: In the final steps of the calculation, notice that our value for time was equal to $11.82432\ldots$11.82432 hours. To convert this time into hours and minutes, we simply multiplied the decimal portion by $60$60. To isolate the decimal portion, subtract $11$11 first, then multiply by $60$60. This gives $49.4594\ldots$49.4594 minutes, or $49$49 minutes, rounded to the nearest minute.
#### Practice questions
##### Question 1
A man drives a truck $144$144 km in $2$2 hours, then stops to refuel and eat lunch at a petrol station for an hour. He then drives the truck for another $5$5 hours, covering $96$96 km.
What was the truck’s average speed throughout the whole journey (in kilometres per hour)?
##### Question 2
If a galapagos turtle travels $0.306$0.306 km at a speed of $0.09$0.09 km/hr, how long will it take the animal to cross the whole distance?
##### Question 3
Two animals were tracked and their movements were measured over different time periods. The cat travelled $10$10 km in $0.1$0.1 hours, while the rabbit travelled $34$34 km in $0.3$0.3 hours.
1. Continuing at the rate measured, how far would the cat travel in $0.3$0.3 hours?
##### Question 4
Sally states that her trip between Brisbane and Townsville had an average speed of $91$91 km/h. The table shown displays the distances between two towns (in kilometres). For example, the distance between Adelaide and Brisbane is $2063$2063 km.
How long did it take Sally to travel from Brisbane to Townsville? Give your answer correct to two decimal places.
Adelaide 1542 Alice Springs 2063 3012 Brisbane 3143 2324 1717 Cairns 3053 1511 3415 2727 Darwin 728 2270 1674 3054 3781 Melbourne 2724 3630 4384 5954 4045 3452 Perth 1420 2644 996 2546 4000 868 4144 Sydney 2525 2096 1467 374 2556 2857 5728 2494 Townsville
### Outcomes
#### ACMEM082
compare the time taken to travel a specific distance with various modes of transport
#### ACMEM086
calculate speed, distance or time using the formula speed = distance/time
#### ACMEM089
calculate and interpret average speed; for example, a 4-hour trip covering 250 km |
How do you find the domain and range of f(x)= (3x-1)/(sqrt(x^2+x-2))?
May 7, 2017
Domain : $x : \left(- \infty , - 2\right) \cup \left(1 , \infty\right)$
Range : $f \left(x\right) : \left(- 3 , - \infty\right) \cup \left(3 , \infty\right)$
Explanation:
$f \left(x\right) = \frac{3 x - 1}{\sqrt{{x}^{2} + x - 2}} = \frac{3 x - 1}{\sqrt{\left(x + 2\right) \left(x - 1\right)}}$
Domain: denominator must not be $0$ and under root must not be $< 0$ So $x + 2 \ne 0 \therefore x \ne - 2$ and $x - 1 \ne 0 \therefore x \ne 1$
For under root calculation: critical points are $x = - 2 \mathmr{and} x = 1$
when $x < - 2 , \left(x + 2\right) \cdot \left(x - 1\right) = \left(- \cdot -\right) = +$
when $- 2 < x < 1 , \left(x + 2\right) \cdot \left(x - 1\right) = \left(+ \cdot -\right) = -$
when $x > 1 , \left(x + 2\right) \cdot \left(x - 1\right) = \left(= \cdot +\right) = +$
So in domain : $x < - 2 \mathmr{and} x > 1 \mathmr{and} \left(- \infty , - 2\right) \cup \left(1 , \infty\right)$
Horizontal asymptote is at $y = \frac{3}{\pm 1} = \pm 3$
So range : $f \left(x\right) : \left(- 3 , - \infty\right) \cup \left(3 , \infty\right)$ graph{(3x-1)/(x^2+x-2)^0.5 [-20, 20, -10, 10]} [Ans] |
# Tangents Of Circles
Here we discuss the various symmetry and angle properties of tangents to circles.
In these lessons we will learn about
• Tangent To A Circle And The Point Of Tangency,
• Tangent To A Circle Theorem,
• Secant,
• Two-Tangent Theorem,
• Common Internal And External Tangents.
The following diagrams show the Radius Tangent Theorem and the Two-Tangent Theorem. Scroll down the page for more examples and solutions.
### Tangent To A Circle
A tangent to a circle is a straight line, in the plane of the circle, which touches the circle at only one point. The point is called the point of tangency or the point of contact.
Tangent to a Circle Theorem: A tangent to a circle is perpendicular to the radius drawn to the point of tangency.
### What Is The Tangent Of A Circle?
A tangent is a line in the plane of a circle that intersects the circle at one point. The point where it intersects is called the point of tangency.
### How To Prove The Tangent To A Circle Theorem?
The Tangent to a Circle Theorem states that a line is tangent to a circle if and only if the line is perpendicular to the radius drawn to the point of tangency.
### Secant
A straight line that cuts the circle at two distinct points is called a secant.
Example:
In the following diagram
a) state all the tangents to the circle and the point of tangency of each tangent.
b) state all the secants.
Solution:
AB is a tangent to the circle and the point of tangency is G.
CD is a secant to the circle because it has two points of contact.
EF is a tangent to the circle and the point of tangency is H.
### Tangents From The Same External Point
Two-Tangent Theorem: When two segments are drawn tangent to a circle from the same point outside the circle, the segments are equal in length.
In the following diagram:
If AB and AC are two tangents to a circle centered at O, then:
• the tangents to the circle from the external point A are equal,
• OA bisects the angle BAC between the two tangents,
• OA bisects the angle BOC between the two radii to the points of contact,
• triangle AOB and triangle AOC are congruent right triangles.
The two-tangent theorem is also called the "hat" or "ice-cream cone" theorem because it looks like a hat on the circle or an ice-cream cone.
### How To Prove The Two-Tangent Theorem?
The Two-Tangent Theorem states that when two segments are drawn tangent to a circle from the same point outside the circle, the segments are congruent. (uses Two-Column Proof and CPCTC).
### When Two Tangent Lines Emanate From The Same External Point
How to find an unknown angle using the two-tangent theorem?
### Common Internal And External Tangents
A common tangent is a line that is a tangent to each of two circles.
A common external tangent does not intersect the segment that joins the centers of the circles.
A common internal tangent intersects the segment that joins the centers of the circles.
### Common Internal And External Tangents: Finding Lengths
A lesson on finding the length of common internal and external tangents.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
Trying to find out just how to convert 12/5 into a combined number or fraction? have I acquired the answer because that you! In this guide, we"ll walk you v the step-by-step procedure of converting an not correct fraction, in this case 12/5, come a mixed number. Check out on!
Want to conveniently learn or present students how to transform 12/5 to a blended number? beat this really quick and also fun video now!
Before we begin, let"s revisit part basic portion terms so friend understand precisely what we"re dealing with here:
Numerator. This is the number above the fraction line. For 12/5, the numerator is 12.Denominator. This is the number below the portion line. For 12/5, the denominator is 5.Improper fraction. This is a fraction where the numerator is better than the denominator.Mixed number. This is a way of express an improper portion by simplifying it to totality units and also a smaller in its entirety fraction. It"s an essence (whole number) and also a suitable fraction.
You are watching: What is 12 5 as a mixed number
Now let"s go v the steps needed to convert 12/5 to a combined number.
Step 1: find the totality number
We first want to discover the entirety number, and to carry out this we division the molecule by the denominator. Due to the fact that we are just interested in whole numbers, us ignore any type of numbers to the right of the decimal point.
12/5= 2.4 = 2
Now the we have our whole number for the combined fraction, we need to discover our brand-new numerator for the fraction part that the combined number.
Step 2: get the new numerator
To occupational this the end we"ll usage the entirety number we calculated in step one (2) and multiply that by the initial denominator (5). The an outcome of the multiplication is climate subtracted from the initial numerator:
12 - (5 x 2) = 2
Step 3: Our mixed fraction
We"ve currently simplified 12/5 to a mixed number. To check out it, we simply need to placed the whole number together with our new numerator and original denominator:
2 2/5
Step 4: simplifying our fraction
In this case, our portion (2/5) deserve to be simplified down further. In stimulate to execute that, we need to calculate the GCF (greatest usual factor) the those 2 numbers. You deserve to use our handy GCF calculator to occupational this out yourself if you want to. We currently did that, and the GCF of 2 and also 5 is 1.
We deserve to now division both the brand-new numerator and also the denominator by 1 to simplify this fraction down to its lowest terms.
2/1 = 2
5/1 = 5
When we put that together, we can see that our finish answer is:
2 2/5
Hopefully this tutorial has helped you come understand how to convert any improper fraction you have into a combined fraction, complete with a whole number and also a ideal fraction. You"re free to usage our calculator below to work out more, but do shot and learn just how to carry out it yourself. It"s much more fun than it seems, ns promise!
If you found this content advantageous in your research, please do us a great favor and also use the tool listed below to make certain you correctly reference united state wherever you usage it. We really evaluate your support!
"What is 12/5 as a blended number?". moment-g.com. Accessed top top November 4, 2021. Https://moment-g.com/calculator/improper-to-mixed/what-is-12-5-as-a-mixed-number/.
"What is 12/5 as a blended number?". moment-g.com, https://moment-g.com/calculator/improper-to-mixed/what-is-12-5-as-a-mixed-number/. Accessed 4 November, 2021.
See more: Is A Mushroom A Producer Consumer Or Decomposer, Are Fungi Decomposers Or Producers
What is 12/5 together a combined number?. moment-g.com. Retrieved native https://moment-g.com/calculator/improper-to-mixed/what-is-12-5-as-a-mixed-number/.
Improper fraction to blended Number
Enter one improper portion numerator and also denominator |
# Triangle Inequalities
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Category: Education
## Presentation Description
No description available.
By: rfantster (62 month(s) ago)
By: edz_b_caj (62 month(s) ago)
## Presentation Transcript
### Slide1:
Triangle Inequalities § 7.1 Segments, Angles, and Inequalities § 7.4 Triangle Inequality Theorem § 7.3 Inequalities Within a Triangle § 7.2 Exterior Angle Theorem
### Slide2:
Segments, Angles, and Inequalities You will learn to apply inequalities to segment and angle measures. 1) Inequality
### Slide3:
Segments, Angles, and Inequalities The Comparison Property of Numbers is used to compare two line segments of unequal measures. The property states that given two unequal numbers a and b, either: a < b or a > b The same property is also used to compare angles of unequal measures.
### Slide4:
Segments, Angles, and Inequalities The measure of J is greater than the measure of K. The statements TU > VW and J > K are called __________ because they contain the symbol < or >. inequalities a < b a = b a > b
### Slide5:
Segments, Angles, and Inequalities SN DN 6 – (- 1) 6 – 2 7 4 > >
### Slide6:
Segments, Angles, and Inequalities AB > AC AB > CB A similar theorem for comparing angle measures is stated below. This theorem is based on the Angle Addition Postulate.
### Slide7:
Segments, Angles, and Inequalities A similar theorem for comparing angle measures is stated below. This theorem is based on the Angle Addition Postulate.
### Slide8:
Segments, Angles, and Inequalities mBDA mCDA 45° 40° + 45° < < Use theorem 7 – 2 to solve the following problem. 45° 85°
### Slide9:
Segments, Angles, and Inequalities For any numbers a, b, and c, 1) if a < b and b < c, then a < c. 2) if a > b and b > c, then a > c. if 5 < 8 and 8 < 9, then 5 < 9. if 7 > 6 and 6 > 3, then 7 > 3.
### Slide10:
Segments, Angles, and Inequalities For any numbers a, b, and c, For any numbers a, b, and c, 1) if a < b, then a + c < b + c and a – c < b – c. 2) if a > b, then a + c > b + c and a – c > b – c. 1 < 3 1 + 5 < 3 + 5 6 < 8
### Slide12:
Exterior Angle Theorem You will learn to identify exterior angles and remote interior angles of a triangle and use the Exterior Angle Theorem. 1) Interior angle 2) Exterior angle 3) Remote interior angle
### Slide13:
Exterior Angle Theorem In the triangle below, recall that 1, 2, and 3 are _______ angles of ΔPQR. interior Angle 4 is called an _______ angle of ΔPQR. exterior An exterior angle of a triangle is an angle that forms a _________ with one of the angles of the triangle. linear pair In ΔPQR, 4 is an exterior angle at R because it forms a linear pair with 3. ____________________ of a triangle are the two angles that do not form a linear pair with the exterior angle. Remote interior angles In ΔPQR, 1, and 2 are the remote interior angles with respect to 4.
### Slide14:
Exterior Angle Theorem In the figure below, 2 and 3 are remote interior angles with respect to what angle? 5
### Slide15:
Exterior Angle Theorem remote interior angles m4 = m1 + m2
### Slide16:
Exterior Angle Theorem
### Slide17:
Exterior Angle Theorem remote interior angles m4 > m1 m4 > m2
### Slide18:
Exterior Angle Theorem 1 and 3 Name two angles in the triangle below that have measures less than 74°. acute
### Slide19:
Exterior Angle Theorem
### Slide20:
Exterior Angle Theorem The feather–shaped leaf is called a pinnatifid. In the figure, does x = y? Explain. __ + 81 = 32 + 78 28 28° 109 = 110 No! x does not equal y
### Slide22:
Inequalities Within a Triangle You will learn to identify the relationships between the _____ and _____ of a triangle. sides angles Nothing New!
### Slide23:
Inequalities Within a Triangle in the same order LP < PM < ML mM < mP mL <
### Slide24:
Inequalities Within a Triangle in the same order JK < KW < WJ mW < mK mJ <
### Slide25:
Inequalities Within a Triangle greatest measure WY > XW 3 5 4 WY > XY
### Slide26:
Inequalities Within a Triangle The longest side is So, the largest angle is The largest angle is So, the longest side is
### Slide28:
Triangle Inequality Theorem You will learn to identify and use the Triangle Inequality Theorem. Nothing New!
### Slide29:
Triangle Inequality Theorem greater a + b > c a + c > b b + c > a
### Slide30:
Triangle Inequality Theorem Can 16, 10, and 5 be the measures of the sides of a triangle? No! 16 + 10 > 5 16 + 5 > 10 |
# Which is greater?(i) $0.5$ or $0.05$(ii) $0.7$ or $0.5$(iii) $7$ or $0.7$(iv) $1.37$ or $1.49$(v) $2.03$ or $2.30$(vi) $0.8$ or $0.88$
Given:
Given numbers are:
(i) $0.5$ or $0.05$
(ii) $0.7$ or $0.5$
(iii) $7$ or $0.7$
(iv) $1.37$ or $1.49$
(v) $2.03$ or $2.30$
(vi) $0.8$ or $0.88$
To do:
We have to find the greater number in each of the given numbers.
Solution:
(i) $0.5$ or $0.05$
$0.5$ or $0.05$
$0.5=\frac{5}{10}$ or $0.05=\frac{5}{100}$
On converting decimals them into like fractions, we get
$0.5=\frac{5}{10}\times\frac{10}{10}$ or $0.05=\frac{5}{100}\times\frac{1}{1}$
$\frac{50}{100}$ or $\frac{5}{100}$
$50>5$
This implies,
$0.5>0.05$
Therefore,0.5 is greater than 0.05.
(ii) $0.7$ or $0.5$
$0.7=\frac{7}{10}$ or $0.5=\frac{5}{10}$
$7>5$
Hence, $\frac{7}{10}>\frac{5}{10}$
Therefore, $0.7$ is greater than $0.5$.
(iii) $7$ or $0.7$
$7$ or $\frac{7}{10}$
$7=\frac{7}{1}\times\frac{10}{10}$ or $\frac{7}{10}$
$\frac{70}{10}$ or $\frac{7}{10}$
$70>7$
This implies,
$7 > 0.7$
Therefore, 7 is greater.
(iv) $1.37$ or $1.49$
$1.37=\frac{137}{100}$ or $1.49=\frac{149}{100}$
$\frac{137}{100}$ or $\frac{149}{100}$
$137<149$
This implies,
$1.37$ < $1.49$
Hence, $1.49$ is greater.
(v) $2.03$ or $2.30$
$2.03=\frac{203}{100}$ or $2.30=\frac{230}{100}$
$\frac{203}{100}$ or $\frac{230}{100}$
$203<230$
Therefore,
$2.03 < 2.30$
Hence, $2.30$ is greater.
(vi) $0.8$ or $0.88$
$0.8=\frac{8}{10}$ or $0.88=\frac{88}{100}$
On converting them into like fractions, we get
$\frac{8}{10}\times\frac{10}{10}$ or $\frac{88}{100}$
$\frac{80}{100}$ or $\frac{88}{100}$
$80<88$
Therefore, $0.8$ < $0.88$.
Hence, $0.88$ is greater.
Tutorialspoint
Simply Easy Learning |
# Calendar Math Trick
This calendar math trick is yet another amazing math activity that lets kids (and adults) see just how cool math is! Just grab a calendar from off the wall or wherever you can get one.
The Effect:
• Have a friend circle a group of nine numbers. Make sure the numbers are in a 3 by 3 rectangle like the picture shows here.
• Tell your friend that you have the special ability to add all nine of the numbers circled in your head in a matter of seconds!
• You give the correct answer and have them check it with a calculator.
The Secret: All you have to do to get the sum of all nine numbers is to multiply the number in the center of the circled numbers by nine! So for our example, the center number is 11.
So you just need to multiply in your head, “9 x 11”.
For those of you who’ve memorized your multiplication facts, you know right away that the answer is 99.
HANDY TIP: To multiply by 9 quickly, just multiply by 10 then subtract your number. So to get 22 x 9 you multiply 22 x 10 =220 (easy!) and then subtract 22. With a bit of practice you can do this quickly in your head.
After you perform the trick once, perform it again, this time make it a little more interesting. Challenge them to a race. Say to your friend, “I will add the nine numbers in my head while you add them with the calculator, and I will get the answer first”. You just perform it the same way as before – just multiply the middle number by 9 to get the answer.
They should be pretty impressed and if they are like most, will wonder how it was done. Usually no one will figure out the secret.
Now, how about a grand finale to impress them even more. This time tell them to circle 20 numbers in a 5 by 4 rectangle – and you will add up all the circled numbers in your head in a couple of seconds!
Take a look a the picture below.
The Secret: Add the smallest and largest numbers that are circled and multiply their sum by 10. And that's it. Pretty cool, huh.
So for the above example;
6 + 31 = 37 and then,
37 x 10 = 370
So the sum of all 20 numbers is 370!
Have fun with this trick. No one will be able to figure out how you get the answer so fast! If you liked this calendar math trick, we will be adding another real soon that is also really cool.
Go to main Math Tricks page
Return from Calendar Math Trick to Learn With Math Games Home |
# 1470 Can You Find Factor Pairs That Make Sum-Difference?
Contents
### Today’s Puzzles:
I bet you can find a factor pair of 30 that adds up to 13 as well as another factor pair of 30 that subtracts to give you 13.
If you can solve that simple puzzle, then you will be able to solve the puzzle next to it. Even though 1470 has 12 different factor pairs, you don’t have to worry too much about them: All of the answers to the second puzzle are just _____ times the answers to the first puzzle! (And 1470 is _____² times 30.)
Likewise, don’t get scared off with this next set of puzzles one of which wants you to find the factor pairs of 518616 that add up or subtract to 1470. Crazy, right? Again, if you can solve the first puzzle in the set, and if you can multiply a 3-digit number by a 1-digit number, you can easily solve the second puzzle because the answers are just _________ times the answers to the first puzzle in the set! (And 518616 is merely _________² times 6.)
If you need more help than what I’ve already said, scroll down to the factor trees for 1470. I selected those particular trees for a reason!
### Factors of 1470:
• 1470 is a composite number.
• Prime factorization: 1470 = 2 × 3 × 5 × 7 × 7, which can be written 1470 = 2 × 3 × 5 × 7²
• 1470 has at least one exponent greater than 1 in its prime factorization so √1470 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1470 = (√49)(√30) = 7√30
• The exponents in the prime factorization are 1, 1, 1, and 2. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1)(1 + 1)(2 + 1) = 2 × 2 × 2 × 3 = 24. Therefore 1470 has exactly 24 factors.
• The factors of 1470 are outlined with their factor pair partners in the graphic below.
### More Facts about the Number 1470:
1470 is the average of 14² and 14³. That simple fact makes 1470 the 14th Pentagonal Pyramidal Number.
1470 is the hypotenuse of a Pythagorean triple:
882-1176-1470 which is (3-4-5) times 294
Hmm… that same factor pair showed up again!
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# What is 1/305 as a decimal?
## Solution and how to convert 1 / 305 into a decimal
1 / 305 = 0.003
Fraction conversions explained:
• 1 divided by 305
• Numerator: 1
• Denominator: 305
• Decimal: 0.003
• Percentage: 0.003%
1/305 or 0.003 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
1 / 305 as a percentage 1 / 305 as a fraction 1 / 305 as a decimal
0.003% - Convert percentages 1 / 305 1 / 305 = 0.003
## 1/305 is 1 divided by 305
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 1 is being divided into 305. Think of this as our directions and now we just need to be able to assemble the project! Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 1 divided by 305. Now we divide 1 (the numerator) into 305 (the denominator) to discover how many whole parts we have. Here's 1/305 as our equation:
### Numerator: 1
• Numerators are the portion of total parts, showed at the top of the fraction. Small values like 1 means there are less parts to divide into the denominator. 1 is an odd number so it might be harder to convert without a calculator. Ultimately, having a small value may not make your fraction easier to convert. Now let's explore the denominator of the fraction.
### Denominator: 305
• Denominators represent the total parts, located at the bottom of the fraction. Larger values over fifty like 305 makes conversion to decimals tougher. But 305 is an odd number. Having an odd denominator like 305 could sometimes be more difficult. Have no fear, large two-digit denominators are all bark no bite. Now it's time to learn how to convert 1/305 to a decimal.
## How to convert 1/305 to 0.003
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 305 \enclose{longdiv}{ 1 }$$
Use long division to solve step one. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 305 \enclose{longdiv}{ 1.0 }$$
Uh oh. 305 cannot be divided into 1. Place a decimal point in your answer and add a zero. This doesn't add any issues to our denominator but now we can divide 305 into 10.
### Step 3: Solve for how many whole groups you can divide 305 into 10
$$\require{enclose} 00.0 \\ 305 \enclose{longdiv}{ 1.0 }$$
How many whole groups of 305 can you pull from 10? 0 Multiple this number by our furthest left number, 305, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.0 \\ 305 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$
If there is no remainder, you’re done! If you have a remainder over 305, go back. Your solution will need a bit of adjustment. If you have a number less than 305, continue!
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. They each bring clarity to numbers and values of every day life. And the same is true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 1/305 and 0.003 bring clarity and value to numbers in every day life. Here are examples of when we should use each.
### When you should convert 1/305 into a decimal
Dollars & Cents - It would be silly to use 1/305 of a dollar, but it makes sense to have $0.0. USD is exclusively decimal format and not fractions. (Yes, yes, there was a 'half dollar' but the value is still$0.50)
### When to convert 0.003 to 1/305 as a fraction
Carpentry - To build a table, you must have the right measurements. When you stretch the tape measure across the piece of wood, you won't see 10.6 inches. You'll see a tick mark at 10 and 3/5 inches.
### Practice Decimal Conversion with your Classroom
• If 1/305 = 0.003 what would it be as a percentage?
• What is 1 + 1/305 in decimal form?
• What is 1 - 1/305 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.003 + 1/2?
### Convert more fractions to decimals
From 1 Numerator From 305 Denominator What is 1/306 as a decimal? What is 2/305 as a decimal? What is 1/307 as a decimal? What is 3/305 as a decimal? What is 1/308 as a decimal? What is 4/305 as a decimal? What is 1/309 as a decimal? What is 5/305 as a decimal? What is 1/310 as a decimal? What is 6/305 as a decimal? What is 1/311 as a decimal? What is 7/305 as a decimal? What is 1/312 as a decimal? What is 8/305 as a decimal? What is 1/313 as a decimal? What is 9/305 as a decimal? What is 1/314 as a decimal? What is 10/305 as a decimal? What is 1/315 as a decimal? What is 11/305 as a decimal? What is 1/316 as a decimal? What is 12/305 as a decimal? What is 1/317 as a decimal? What is 13/305 as a decimal? What is 1/318 as a decimal? What is 14/305 as a decimal? What is 1/319 as a decimal? What is 15/305 as a decimal? What is 1/320 as a decimal? What is 16/305 as a decimal? What is 1/321 as a decimal? What is 17/305 as a decimal? What is 1/322 as a decimal? What is 18/305 as a decimal? What is 1/323 as a decimal? What is 19/305 as a decimal? What is 1/324 as a decimal? What is 20/305 as a decimal? What is 1/325 as a decimal? What is 21/305 as a decimal?
### Convert similar fractions to percentages
From 1 Numerator From 305 Denominator 2/305 as a percentage 1/306 as a percentage 3/305 as a percentage 1/307 as a percentage 4/305 as a percentage 1/308 as a percentage 5/305 as a percentage 1/309 as a percentage 6/305 as a percentage 1/310 as a percentage 7/305 as a percentage 1/311 as a percentage 8/305 as a percentage 1/312 as a percentage 9/305 as a percentage 1/313 as a percentage 10/305 as a percentage 1/314 as a percentage 11/305 as a percentage 1/315 as a percentage |
# Solving trinomials
When Solving trinomials, there are often multiple ways to approach it. Math can be a challenging subject for many students.
## Solve trinomials
We will also give you a few tips on how to choose the right app for Solving trinomials. How to quickly and effectively let students solve equations? Through summary and analysis, I summarized the skills of solving various equations and compiled a pithy formula to help students remember In the past, although equations were also taught in primary school mathematics, they were simple equations, similar to 5-x = 2. They mainly used the method of strongly regulating equations, and did not emphasize the background of establishing equations. Now primary school mathematics has strengthened the nature, relationship and law behind the letters to represent numbers. Is it true that the sooner you master algebra and equations, the better? no, it isn't.
The better the problem is solved, the smoother the child's growth and learning will be. The ability to solve problems is also the basis for judging a person's ability to learn and do things. Therefore, the stronger your ability to solve problems, the more problems you can solve, the less resistance to growth, and the smoother your growth. Problem solving skills are very important to cultivate children's psychological resilience, and problem-solving ability can be taught.
Here I will analyze why this method can solve these two problems separately. Therefore, the correct way to recite words must be a surprise way, a surprise memory, and a gradual review. In this way, the problem of forgetting the front behind the back can be solved and the efficiency of reciting words can be improved. It is recommended that parents let their children recite 100 words every day, so that it only takes 16 days to recite 1600 words in junior high school.
But you won't nag or remind about homework, then, shut up, leave, and only appear when the child asks for help. Homework is the responsibility of children. When you keep nagging, homework becomes your responsibility. 4. Tell the child that if he needs help with his homework, you will be there.
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# Solving Direct Variation Problems
As the numbers for a problem can be scored is fixed, it is a constant = K = $$\frac$$ = 10For solving 5 problems total numbers scored T = KN = 10 x 5 = 50.From the above example we can understand that the ratio of two variables is a constant K and T, N are the variables which changes in proportion with value of constant.When two variables change in proportion it is called as direct variation.
Tags: Read An Essay Concerning Human UnderstandingCapstone Projects IdeasArt History Dissertation QuestionsData Analysis For DissertationPaper EssayResearch Paper On Recycling
Solution: As P varies directly with Q, ratio of P and Q is constant for any value of P and Q.
So constant K = $$\frac$$ = $$\frac$$ = $$\frac$$So the equation that describes the direct variation of P and Q is P = $$\frac$$Q.
Graphically, we have a line that passes through the origin with the slope of k. If the distance is 10 cm when the mass is kg, what is the distance when the mass is 5 kg?
We're told that the total cost of filling up your car with gas varies directly with the number of gallons of gasoline you are purchasing. So if x is equal to 1-- this statement up here, a gallon of gas-- that tells us if we get 1 gallon, if x is equal to 1, then y is $2.25, right? They tell us 1 gallon costs$2.25, so you could write it right here, \$2.25 is equal to k times x, times 1. So the equation, how y varies with x, is y is equal to 2.25x, where x is the number of gallons we purchase. Now if we want to solve for x, we can divide both sides by 2.25, so let's do that. So first of all, I just like to think of it as a fraction.
Direct variation is the simplest type of variation and in practical life we can find many situations which can be co-related with direct variation.
If two variables A and B are so related that when A increases ( or decreases ) in a given ratio, B also increases ( or, decreases ) in the same ratio, then A is said to vary directly as B ( or, A is said to vary as B ).Direct variation word problems may be solved using the following steps: 1) Recognizing the word problem consists of direct variation as exemplified by presence of verbiage listed above; 2) Writing an equation containing known values for both variables resulting in an expression containing only the unknown representing the constant of proportionality; 3) Solving the equation for the constant of proportionality; 4) Using the calculated constant of proportionality to determine the value of one of the variables given the other.The amount of money raised at a charity fundraiser is directly proportional to the number of attendees.This is symbolically written as, A ∝ B (read as, ‘A varies as B’ ).Like in a math examination if for one problem solving we can score 10 numbers, so five problems solving we can get 50 numbers.The speed of a car and the distance traveled in a certain amount of time.The following statements are equivalent * y varies directly as x * y is directly proportional to x * y = kx for some constant k What is the direct variation formula? If a gallon of gas costs .25, how many gallons could you purchase for ? Now, they give us more information, and this will help us figure out what k is.A direct variation is a linear equation that can be written in the form y = kx , where k is a nonzero constant. Determine the direct variation equation and then determine y when x = 3.5 2. The number k is called the constant of proportionality or constant of variation. Hooke's Law states that the displacement, d, that a spring is stretched by a hanging object varies directly as the mass of the object.
## Comments Solving Direct Variation Problems
• ###### Solve direct variation problems College Algebra
Solve direct variation problems. In the example above, Nicole's earnings can be found by multiplying her sales by her commission. The formula e = 0.16s tells us.…
• ###### Direct variation word problem filling gas video Khan.
Worked example Model a context about filling gas with a direct variation equation. Inverse variation word problem string vibration · Proportionality constant for. Now if we want to solve for x, we can divide both sides by 2.25, so let's do that.…
• ###### Direct Variation - How to Solve Variation Problems Part 1 fbt.
Jul 19, 2016. This video by Fort Bend Tutoring shows the process of solving direct variation problems. Six 6 examples are shown in this FBT video.…
• ###### Direct Variation Solving Direct Variation Word Problem
When two variables change in proportion it is called as direct variation. In direct variation one variable is constant times of other. If one variable increases other.…
• ###### Direct Variation - - Algebra Help - YouTube
Apr 24, 2008. For a complete lesson on direct variation, go to. are examples of direct variation, as well as solve direct variation word problems.…
• ###### Solving Direct Variation Problems - Practice Problems
Solving Direct Variation Problems – Practice Problems Move your mouse over the "Answer" to reveal the answer or click on the "Complete Solution" link to.…
• ###### Write and solve direct variation equations Algebra 1. - IXL
Improve your math knowledge with free questions in "Write and solve direct variation equations" and thousands of other math skills.… |
# SAT Math : Outcomes
## Example Questions
### Example Question #11 : How To Find The Probability Of An Outcome
Michelle is randomly drawing cards from a deck of of 52 cards. What is the chance she will draw a black queen followed by a 5 of any color, without replacing the cards?
4/169
2/169
2/663
8/663
2/2652
2/663
Explanation:
There are 2 black queens in the deck, one of spades and one of clubs, so there is a 2/52 chance a black Queen will be drawn and 4/51 chance of drawing a 5 of any color, since the queen has already been removed from the deck. Thus: 2/52 * 4/51 = 8/2652 → 2/663.
1
### Example Question #12 : How To Find The Probability Of An Outcome
Zack has 10 green, 14 red, 2 blue, and 6 black marbles in a bag. What is the probability that Zack will not randomly pick a red or blue marble from the bag?
3/16
1/2
1/3
15/32
5/16
1/2
Explanation:
To NOT choose a red or blue, leaves 6 black and 10 green to choose from. That leaves 16 marbles out of a total of 32 marbles, or a 1/2 chance.
### Example Question #13 : How To Find The Probability Of An Outcome
A bag contains four blue marbles, four red marbles, and two green marbles in a bag. If one marble is drawn, and then a second is drawn without replacement, what is the probability that at least one of the two marbles will be red?
11/15
2/5
26/45
2/3
4/9
Explanation:
We have blue, red, and green marbles in a bag. We need to consider all of the ways that we could draw at least one red. This means we can draw a red the first time, a red the second, or a red both times. These are the possible ways we could at least draw one red marble:
We can draw a red and then a blue.
We can draw a red and then a green.
We can draw a blue and then a red.
We can draw a green and then a red.
We could draw a red and then another red.
So we need to find the probability of each of these five scenarios. Then, we need to add these probabilities.
Let's look at the probability of the first scenario (red, then blue). The probability of drawing a red on the first time would be 4/10, because there are 10 marbles, and four are red. On the second draw, we don't put this marble back. This means we only have 9 marbles in the bag, and four of them are blue. Thus, the probability of the second draw being blue would be 4/9.
The probability of drawing a red and then a blue is equal to the product of these two events. Whenever we want to find the probability of one event AND another, we need to multiply. Thus, the probability of drawing the red AND then a blue would be (4/10)(4/9) = 16/90.
We can calculate the probability of the other four possibilities in a similar fashion.
The probability of drawing a red and then a green is (4/10)(2/9) = 8/90
The probability of drawing a blue then a red is (4/10)(4/9) = 16/90
The probability of drawing a green then a red is (2/10)(4/9) = 8/90
The probabilty of drawing a red then a red is (4/10)(3/9) = 12/90
To find the total probabilty, we need to add up the probabilities of the five different scenarios. Whenever we want to find the probability of one event OR another, we add. So the final probablity is
16/90 + 8/90 + 16/90 + 8/90 + 12/90 = 60/90 = 2/3.
The probability of drawing at least one red is 2/3.
### Example Question #14 : How To Find The Probability Of An Outcome
There are 5 girls and 4 boys in Homeroom A, and 3 boys and 7 girls in Homeroom B. If each homeroom selects one captain at random from among its students, what is the probability that a boy is selected in at least one of the homerooms?
12/19
11/18
11/10
67/90
7/18
11/18
Explanation:
Note that the choosing of a captain from each homeroom is independent of the other, such that the events are independent. This means that the probabilities of each event would be multiplied.
Find the probability that girls are selected for both homerooms. As this is the only possibility in which a boy is NOT selected, the answer is 1 minus this probability.
There are 9 students in Homeroom A. Of them, 5 are girls.
There are 10 students in Homeroom B. Of them, 7 are girls.
Probability of 2 girls = P(girl for A) x P(girl for B)
= (5/9) x (7/10) = 35/90 = 7/18
Probability of at least one boy = 1 - 7/18 = 11/18
### Example Question #15 : How To Find The Probability Of An Outcome
How many positive four-digit integers are multiples of 5 and less than 7,000?
1200
1850
1600
800
1000
1200
Explanation:
First, observe that multiples of 5 are all those numbers that have a 0 or a 5 as their final digit (5, 10, 15, 20, 25, etc.), so the question is asking how many 4-digit integers under 7000 end in a 0 or a 5. Second, notice that the smallest 4-digit integer is 1,000. So, to rephrase the question, we want to find how many integers that are between 1000 and 6999 end in a 0 or a 5.
Writing the 4-digit integer as WXYZ, we see that there are six possibilties for W (1, 2, 3, 4, 5, and 6), ten possibilities for X (0–9), ten possibilities for Y (0–9), and two possibilities for Z (0 and 5). So there are 6 * 10 * 10 * 2 = 1200 total integers between 1000 and 6999 that end in a 0 or a 5. Therefore, there are 1200 four-digit integers that are multiples of 5 and less than 7,000.
### Example Question #16 : How To Find The Probability Of An Outcome
A street light in Anytown, USA, completes a cycle from red to green to yellow, and back to red, in 50 seconds. During this time, the light will be red for 20 seconds, green for 25 seconds, and yellow for the remaining time. Over a period of 100 seconds, what is the probability that the light will be yellow at a randomly chosen time?
1/10
1/25
1/50
1/5
1/20
1/10
Explanation:
First, find the amount of time that the light is yellow, which is 50 – 20 – 25 = 5 seconds out of a 50 second cycle. The period of 100 seconds is irrelevant because the probability of getting a yellow light would be the same for any complete period or number of complete periods. So the probability of getting a yellow light would be 5/50 = 1/10.
### Example Question #17 : How To Find The Probability Of An Outcome
A fair die is rolled 30 times. How many times would you expect a 2 or a 6 to be rolled?
10
12
5
15
10
Explanation:
There is a 1/3 chance that a 2 or 6 will come up for any given roll. The probability that one will come up in 30 rolls is (1/3) * 30 = 10
A 2 or 6 will, on average, appear 10 times in 30 rolls.
### Example Question #11 : How To Find The Probability Of An Outcome
A coin is flipped 4 times. What is the probability of getting 4 heads in a row?
1/16
1/100
1/4
1/2
Explanation:
The probability a heads will show on any flip is 1/2. To find the probability of 4 consecutive heads, multiply the probability of each individual flip together.
1/2 * 1/2 * 1/2 * 1/2 = 1/16
### Example Question #12 : How To Find The Probability Of An Outcome
A restaurant offers two different entrees, four different side dishes, and three desserts. If a customer wants to order a meal special that consists of one entree, two different side dishes, and one dessert, how many different meals are possible?
36
72
24
144
288
Explanation:
Our strategy is as follows: We must determine the number of possible entrees, the number of possible combinations of two side dishes, and the number of desserts. Then we must multiply these four numbers together to get the number of all of the possible meals.
Let's label the two different entrees E1 and E2 The customer can choose one of these, so he has two different options for his entree.
Let's label the four different side dishes S1, S2, S3, and S4. The customer can only choose two of these four dishes, and the order in which he chooses doesn't matter. For example, if he were to choose mashed potatoes and broccoili, he would end up with the same meal as if he were to choose broccoli and then mashed potatoes. So we need to consider all of the different pairs that can be selected from these four side dishes. Here are all of the possible combinations:
S1 and S2
Sand S3
Sand S4
S2 and S3
S2 and S4
S3 and S4
This means that the customer has six different combinations of two side dishes.
Finally, the customer can choose one dessert from three different choices. This means there are three possibilities for dessert.
So, the customer has two different options for entrees, six different options for the two side dishes, and three different choices for dessert. To find the total number of meals possible, we can multiply these three numbers together.
Number of meals = 2 x 6 x 3 = 36.
### Example Question #13 : How To Find The Probability Of An Outcome
How many positive four-digit integers have the thousands digit equal to 8 and the units digit (ones digit) equal to 5?
99
200
100
20
10 |
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# 6 is not the case lets see what we are doing on a
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Unformatted text preview: 39 What determines test results? This time lets start with the Critical Value Approach: After stating the hypotheses and calculating the z-statistic, the decision rule is going to be: x −µ ¯ Reject H0 if | σ/√0 | > zα/2 n In other words, reject the null if the test statistic is outside the interval (−zα/2 , zα/2 ) where critical value zα/2 is the z-value for upper tail probability α/2. For our example, we have z = 2.5−2.6 √ 1/ 25 = 0.1/0.3 = −0.33. Let’s pick α = 0.05, then α/2 = 0.025 and zα/2 = 1.96. Because z = 0.33 lies inside the interval (−1.96, 1.96), we do not reject the null hypothesis. We don’t have enough evidence to assert that µ = 2.6 is not the case. Let’s see what we are doing on a picture. Utku Suleymanoglu (UMich) Hypothesis Testing 24 / 39 What determines test results? Graphical Explanation 0 Utku Suleymanoglu (UMich) Hypothesis Testing z 25 / 39 What determines test results? p-value Approach for Two-Tailed Tests We can also calculate a p-value for the test statistic z , to compare with different α’s to get a conclusion. The p-value can be calculated as the area outside the interval (−z , z ), or simply (due to symmetricity) as 2 × P (Z > z = 0.33) In our example we get a p-value 2 × P (Z > z = 0.33) = 2 × 0.3707 = 0.7414. This value is bigger than any reasonable α so reach at the same conclusion as before. We fail to reject the null hypothesis. Let’s go back one slide and see this on a picture. Utku Suleymanoglu (UMich) Hypothesis Testing 26 / 39 What determines test results? Hypothesis Testing Fundamentals Recap Before we go on to different cases, let’s repeat the general idea of hypothesis testing: We have an hypothetical value for a population parameter (µ = µ0 ) as a claim and we want to test this. We have a sample and a point estimate x = 2, let’s say. ¯ We know the sampling distribution of x assuming the claim is true from chapter 6. ¯ Then we can evaluate the probability of x or a similar draw from this sampling distribution. ¯ To do that we need to transform our normal random variable...
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## This note was uploaded on 03/17/2014 for the course ECON 404 taught by Professor Staff during the Spring '08 term at University of Michigan.
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## Areas of Parallelograms and Triangles
#### 9.1 Introduction
In Chapter 5, you have seen that the study of Geometry, originated with the measurement of earth (lands) in the process of recasting boundaries of the fields and dividing them into appropriate parts. For example, a farmer Budhia had a triangular field and she wanted to divide it equally among her two daughters and one son. Without actually calculating the area of the field, she just divided one side of the triangular field into three equal parts and joined the two points of division to the opposite vertex. In this way, the field was divided into three parts and she gave one part to each of her children. Do you think that all the three parts so obtained by her were, in fact, equal in area? To get answers to this type of questions and other related problems, there is a need to have a relook at areas of plane figures, which you have already studied in earlier classes.
You may recall that the part of the plane enclosed by a simple closed figure is called a planar region corresponding to that figure. The magnitude or measure of this planar region is called its area. This magnitude or measure is always expressed with the help of a number (in some unit) such as 5 cm2, 8 m2, 3 hectares etc. So, we can say that area of a figure is a number (in some unit) associated with the part of the plane enclosed by the figure.
We are also familiar with the concept of congruent figures from earlier classes and from Chapter 7. Two figures are called congruent, if they have the same shape and the same size.
Fig. 9.1
In other words, if two figures A and B are congruent (see Fig. 9.1) ,then using a tracing paper, you can superpose one figure over the other such that it will cover the other completely. So if two figures A and B are congruent, they must have equal areas. However, the converse of this statement is not true. In other words, two figures having equal areas need not be congruent. For example, in Fig. 9.2, rectangles ABCD and EFGH have equal areas (9 × 4 cm 2 and 6 × 6 cm 2) but clearly they are not congruent. (Why?)
Fig. 9.2
Now let us look at Fig. 9.3 given below:
Fig. 9.3
You may observe that planar region formed by figure T is made up of two planar regions formed by figures P and Q. You can easily see that
Area of figure T = Area of figure P + Area of figure Q.
You may denote the area of figure A as ar(A), area of figure B as ar(B), area of figure T as ar(T), and so on. Now you can say that area of a figure is a number
(in some unit) associated with the part of the plane enclosed by the figure with the following two properties:
(1) If A and B are two congruent figures, then ar(A) = ar(B);
and (2) if a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then ar(T) = ar(P) + ar(Q).
You are also aware of some formulae for finding the areas of different figures such as rectangle, square, parallelogram, triangle etc., from your earlier classes. In this chapter, attempt shall be made to consolidate the knowledge about these formulae by studying some relationship between the areas of these geometric figures under the condition when they lie on the same base and between the same parallels. This study will also be useful in the understanding of some results on ‘similarity of triangles’.
9.2 Figures on the Same Base and Between the Same Parallels
Look at the following figures:
Fig. 9.4
In Fig. 9.4(i), trapezium ABCD and parallelogram EFCD have a common side DC. We say that trapezium ABCD and parallelogram EFCD are on the same base DC. Similarly, in Fig. 9.4 (ii), parallelograms PQRS and MNRS are on the same base SR; in Fig. 9.4(iii), triangles ABC and DBC are on the same base BC and in
Fig. 9.4(iv), parallelogram ABCD and triangle PDC are on the same base DC.
Now look at the following figures:
Fig. 9.5
In Fig. 9.5(i), clearly trapezium ABCD and parallelogram EFCD are on the same base DC. In addition to the above, the vertices A and B (of trapezium ABCD) opposite to base DC and the vertices E and F (of parallelogram EFCD) opposite to base DC lie on a line AF parallel to DC. We say that trapezium ABCD and parallelogram EFCD are on the same base DC and between the same parallels AF and DC. Similarly, parallelograms PQRS and MNRS are on the same base SR and between the same parallels PN and SR [see Fig.9.5 (ii)] as vertices P and Q of PQRS and vertices M and N of MNRS lie on a line PN parallel to base SR.In the same way, triangles ABC and DBC lie on the same base BC and between the same parallels AD and BC [see Fig. 9.5 (iii)] and parallelogram ABCD and triangle PCD lie on the same base DC and between the same parallels AP and DC [see Fig. 9.5(iv)].
So, two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.
Keeping in view the above statement, you cannot say that PQR and DQR of Fig. 9.6(i) lie between the same parallels l and QR. Similarly, you cannot say that
Fig. 9.6
parallelograms EFGH and MNGH of Fig. 9.6(ii) lie between the same parallels EF and HG and that parallelograms ABCD and EFCD of Fig. 9.6(iii) lie between the same parallels AB and DC (even though they have a common base DC and lie between the parallels AD and BC). So, it should clearly be noted that out of the two parallels, one must be the line containing the common base.Note that ∆ABC and ∆DBE of Fig. 9.7(i) are not on the common base. Similarly, ∆ABC and parallelogram PQRS of Fig. 9.7(ii)are also not on the same base.
Fig. 9.7
Exercise 9.1
1. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.
Fig. 9.8
9.3 Parallelograms on the same Base and Between the same Parallels
Now let us try to find a relation, if any, between the areas of two parallelograms on the same base and between the same parallels. For this, let us perform the following activities:
Activity 1 : Let us take a graph sheet and draw two parallelograms ABCD and PQCD on it as shown in Fig. 9.9.
Fig. 9.9
The above two parallelograms are on the same base DC and between the same parallels PB and DC. You may recall the method of finding the areas of these two parallelograms by counting the squares.
In this method, the area is found by counting the number of complete squares enclosed by the figure, the number of squares a having more than half their parts enclosed by the figure and the number of squares having half their parts enclosed by the figure. The squares whose less than half parts are enclosed by the figure are ignored. You will find that areas of both the parallelograms are (approximately) 15 cm2. Repeat this activity* by drawing some more pairs of parallelograms on the graph sheet. What do you observe? Are the areas of the two parallelograms different or equal? If fact, they are equal. So, this may lead you to conclude that parallelograms on the same base and between the same parallels are equal in area. However, remember that this is just a verification.
Activity 2 : Draw a parallelogram ABCD on a thick sheet of paper or on a cardboard sheet. Now, draw a line-segment DE as shown in Fig. 9.10.
Fig. 9.10
Next, cut a triangle A′ D′ E′ congruent to triangle ADE on a separate sheet with the help of a tracing paper and place A′ D′ E′ in such a way that A′ D′ coincides with BC as shown in Fig 9.11.Note that there are two parallelograms ABCD and EE′ CD on the same base DC and between the same parallels AE′ and DC. What can you say about their areas?
Fig. 9.11
As ∆ ADE ≅ ∆ A′ D′ E′
Therefore ar (ADE) = ar (A′ D′ E′ )
Also ar (ABCD) = ar (ADE) + ar (EBCD)
= ar (A′ D′ E′ ) + ar (EBCD)
= ar (EE′ CD)
So, the two parallelograms are equal in area.
Let us now try to prove this relation between the two such parallelograms.
*This activity can also be performed by using a Geoboard.
Theorem 9.1 : Parallelograms on the same base and between the same parallels are equal in area.
Proof : Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF and DC are given (see Fig.9.12).
Fig. 9.12
We need to prove that ar (ABCD) = ar (EFCD).
∠ DAE =∠ CBF (Corresponding angles from AD || BC and transversal AF) (1)
∠ AED =∠ BFC (Corresponding angles from ED || FC and transversal AF) (2)
Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3)
Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)
So, ∆ADE≅ ∆BCF [By ASA rule, using (1), (3), and (4)]
Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)
Now, ar (ABCD) = ar (ADE) + ar (EDCB)
= ar (BCF) + ar (EDCB) [From(5)]
= ar (EFCD)
So, parallelograms ABCD and EFCD are equal in area.
Let us now take some examples to illustrate the use of the above theorem.
Example 1 : In Fig. 9.13, ABCD is a parallelogram and EFCD is a rectangle.
Also, AL DC. Prove that
(i) ar (ABCD) = ar (EFCD)
(ii) ar (ABCD) = DC × AL
Fig. 9.13
Solution : (i) As a rectangle is also a parallelogram,
therefore, ar (ABCD) = ar (EFCD) (Theorem 9.1)
(ii) From above result,
ar (ABCD) = DC × FC (Area of the rectangle = length × breadth) (1)
As AL DC, therefore, AFCL is also a rectangle
So, AL = FC (2)
Therefore, ar (ABCD) = DC × AL [From (1) and (2)]
Can you see from the Result (ii) above that area of a parallelogram is the product of its any side and the coresponding altitude. Do you remember that you have studied this formula for area of a parallelogram in Class VII. On the basis of this formula, Theorem 9.1 can be rewritten as parallelograms on the same base or equal bases and between the same parallels are equal in area.
Can you write the converse of the above statement? It is as follows: Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels. Is the converse true? Prove the converse using the formula for area of the parallelogram.
Example 2 : If a triangle and a parallelogram are on the same base and between the same parallels, then prove that the area of the triangle is equal to half the area of the parallelogram.
Solution : Let ∆ABP and parallelogram ABCD be on the same base AB and between the same parallels AB and PC (see Fig. 9.14).
Fig9.14
You wish to prove that ar (PAB) = ar(ABCD)
Draw BQ || AP to obtain another parallelogram ABQP. Now parallelograms ABQP and ABCD are on the same base AB and between the same parallels AB and PC.
Therefore, ar (ABQP) = ar (ABCD) (By Theorem 9.1) (1)
But PAB ≅ BQP (Diagonal PB divides parallelogram ABQP into two congruent triangles.)
So, ar (PAB) = (BQP) (2)
Therefore, ar (PAB) = ar (ABQP) [From (2)] (3)
This gives ar (PAB) =ar(ABCD) [From (1) and (3)]
ExerCise 9.2
1. In Fig. 9.15, ABCD is a parallelogram, AE DC and CF AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Fig. 9.15
2. If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that
ar (EFGH) = ar(ABCD).
3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
Fig. 9.16
(i) ar (APB) + ar (PCD) = ar(ABCD)
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
[Hint : Through P, draw a line parallel to AB.]
5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
Fig. 9.17
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AX S) = ar(PQRS)
6. A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
9.4 Triangles on the same Base and between the same Parallels
Let us look at Fig. 9.18. In it, you have two triangles ABC and PBC on the same base BC and between the same parallels BC and AP. What can you say about the areas of such triangles? To answer this question, you may perform the activity of drawing several pairs of triangles on the same base and between the same parallels on the graph sheet and find their areas by the method of counting the
squares. Each time, you will find that the areas of the two triangles are (approximately) equal. This activity can be performed using a geoboard also. You will again find that the two areas are (approximately) equal.
Fig. 9.18
To obtain a logical answer to the above question, you may proceed as follows:
In Fig. 9.18, draw CD || BA and CR || BP such that D and R lie on line AP(see Fig.9.19).
Fig. 9.19
From this, you obtain two parallelograms PBCR and ABCD on the same base BC and between the same parallels BC and AR.
Therefore, ar (ABCD) = ar (PBCR) (Why?)
Now ∆ABC ≅ ∆CDA and ∆PBC ∆ ≅ CRP (Why?)
So, ar (ABC) = ar(ABCD)and ar (PBC) = ar(PBCR) (Why?)
Therefore, ar (ABC) = ar (PBC)
In this way, you have arrived at the following theorem:
Theorem 9.2 : Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Now, suppose ABCD is a parallelogram whose one of the diagonals is AC
(see Fig. 9.20).Let AN | DC. Note that
Fig. 9.20
So, ar (ADC) = ar (CBA) (Why?)
= (DC x AN) (Why?)
So, area of ∆ADC = × base DC × corresponding altitude AN
In other words, area of a triangle is half the product of its base (or any side) and the corresponding altitude (or height). Do you remember that you have learnt this formula for area of a triangle in Class VII ? From this formula, you can see that two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.
For having equal corresponding altitudes, the triangles must lie between the same parallels. From this, you arrive at the following converse of Theorem 9.2 .
Theorem 9.3 : Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.
Let us now take some examples to illustrate the use of the above results.
Example 3 : Show that a median of a triangle divides it into two triangles of equal areas.
Solution : Let ABC be a triangle and let AD be one of its medians (see Fig. 9.21).
Fig. 9.21
You wish to show that
ar (ABD) = ar (ACD).
Since the formula for area involves altitude, let us draw AN | BC.
Now ar(ABD) = × base × altitude (of ∆ABD)
= x BD xAN
= x CD x AN (As BD = CD)
= × base × altitude (of ∆ACD)
= ar(ACD)
Example 4 : In Fig. 9.22, ABCD is a quadrilateral and BE || AC and also BE meets DC produced at E. Show that area of ∆ADE is equal to the area of the quadrilateral ABCD.
Fig. 9.22
Solution : Observe the figure carefully .
∆BAC and ∆EAC lie on the same base AC and between the same parallels AC and BE.
Therefore, ar(BAC) = ar(EAC) (By Theorem 9.2)
Exercise 9.3
1. In Fig.9.23, E is any point on median AD of a ∆ABC. Show that ar (ABE) = ar (ACE).
Fig. 9.23
2. In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = .
3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.
4. In Fig. 9.24, ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD).
Fig. 9.24
5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ABC. Show that
(i) BDEF is a parallelogram. (ii) ar (DEF) =
(iii) ar (BDEF) = ar (ABC)
6. In Fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.
Fig. 9.25
If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint : From D and B, draw perpendiculars to AC.]
7. D and E are points on sides AB and AC respectively of ∆ABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
8. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that
ar (ABE) = ar (ACF)
9. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. 9.26). Show that
ar (ABCD) = ar (PBQR).
Fig. 9.26
[Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]
10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
11. In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)
Fig. 9.27
12. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC
at Y. Prove that ar (ADX) = ar (ACY).
[Hint : Join CX.]
14. In Fig.9.28, AP || BQ || CR. Prove that
ar (AQC) = ar (PBR).
Fig. 9.28
15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that
ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
16. In Fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
Fig. 9.29
Exercise 9.4 (Optional)*
1. Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
2. In Fig. 9.30, D and E are two points on BC
such that BD = DE = EC. Show that
ar (ABD) = ar (ADE) = ar (AEC).
Fig. 9.30
Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ∆ABC into n triangles of equal areas.]
3. In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that
4. In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that
AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).
[Hint : Join AC.]
5. In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
Fig. 9.33
(i) ar (BDE) =
(ii) ar (BDE) = ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint : From A and C, draw perpendicular to BD.]
7. P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar (PRQ) = ar (ARC) (ii) ar (RQC) = ar (ABC)
(iii) ar (PBQ) = ar (ARC)
8. In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AXDE meets BC at Y. Show that:
Fig. 9.34
(i) ∆MBC ≅ ∆ABD (ii) ar (BYXD) = 2 ar (MBC)
(iii) ar (BYXD) = ar (ABMN) (iv) ∆FCB ≅ ∆ACE
(v) ar (CYXE) = 2 ar (FCB) (vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
*These exercises are not from examination point of view.
9.5 Summary
In this chapter, you have studied the following points :
1. Area of a figure is a number (in some unit) associated with the part of the plane enclosed by that figure.
2. Two congruent figures have equal areas but the converse need not be true.
3. If a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then ar (T) = ar (P) + ar (Q), where ar (X) denotes the area of figure X.
4. Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices, (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.
5. Parallelograms on the same base (or equal bases) and between the same parallels are equal in area.
6. Area of a parallelogram is the product of its base and the corresponding altitude.
7. Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels.
8. If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.
9. Triangles on the same base (or equal bases) and between the same parallels are equal in area.
10. Area of a triangle is half the product of its base and the corresponding altitude.
11. Triangles on the same base (or equal bases) and having equal areas lie between the same parallels.
12. A median of a triangle divides it into two triangles of equal areas. |
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# Modulus and Phase
Complex numbers are very abstract in themselves, and to better understand their properties, we need to draw parallels between them and other less abstract quantities. One way of doing this is treating a complex number as a vector and finding similar properties, such as the magnitude (also known as the modulus) of a complex number, its direction, and phase. It is suggested that you go over the article on Complex Numbers here before continuing with this article!
## A Quick Review of Complex Numbers - the Formula and Symbol
It is a very fruitful idea to represent complex numbers as points in the XY plane. But this plane is not a regular Cartesian plane with normal cartesian coordinates, but we shall call it the complex plane. Also sometimes known as the Argand plane, after the mathematician Jean-Robert Argand.
Complex numbers are plotted very naturally in this plane. Let there be a complex plane of the form where . Here a is called the real part, and b is called the imaginary part. The coordinates of z on the complex plane will be (a, b). The y-axis is called the imaginary axis, and the x-axis is the real axis. The line 'r' is referred to as the terminal arm in this article.
Complex numbers are made up of a real and an imaginary part, StudySmarter Originals
## The Modulus of a Complex Number
Let us denote the distance of the point from the origin as r, and we can use the Pythagorean theorem to calculate its distance from the origin, which is
This distance r is known as the Modulus of z hence we get,
where denotes the modulus of z.
The modulus of a complex number is the distance of that point from the origin on a complex plane. Alternatively, the square of the modulus of a complex number is equal to the sum of the squares of its real and imaginary parts. ()
Calculate the modulus of the complex number z=3+4i
Solution:
Step 1: From the definition of the modulus of a complex number,
Step 2: The number beside the is always the imaginary part. From the question, 3 and 4 are the real and imaginary parts, respectively:
Thus, the modulus of z=3+4i is 5
### A relation between a complex number, its conjugate, and its modulus
Let be a complex number, then the conjugate of it is (recall the Complex Number article section about conjugates)!
Step 1: If we multiply the conjugates together, like so:
Step 2: And when we expand the brackets, we get
And as , therefore
Step 3: Substituting for into the equation in step 2, we get
The right-hand side is just the square of the modulus! So, we have
In other words,
## Modulus and Phase Angle
Let us go back to the geometric view of the complex number,
The phase of a complex number is the angle between the positive x-axis and the terminal arm, StudySmarter Originals
Let be the angle made by z with the positive x-axis. We call this angle the phase angle of z, also known as the argument of z. In polar form, z is expressed as,
and from the graph, we can use trigonometric ratios to say that,
Solving for we get
IMPORTANT! For the purposes of this article, the -1 on tan is not an exponent (i.e., it is not the reciprocal of tangent - cotangent). This is used because it represents the inverse tangent function, arctangent, on most calculators.
An alternative way to calculate the phase angle is using sine and cosine.
Subbing in variables from figure 2 above, we get:
We just learned that r is the modulus, . Substituting this expression for r in the equation above, we get
Using Pythagoras theorem, we can determine r in terms of a and b:
Substituting for r into cosine, we get (as we did for sine):
### Principal Arguments of Complex Numbers
It can be noticed that if the angle is changed by and so on, the complex number remains the same. So there will not be a unique value of but infinitely many. Hence, we call the first value of the principal argument of z, Arg(z).
Calculate the principal argument of the complex number .
Solution:
Step 1: We know that . From the question we have . Thus, we have the value of tangent as,
Step 2: Solving for , we get
Thus, the principal argument, Arg(z), of is . This means that we can add or subtract multiples of 360 degrees to 45 and get the same complex number!
### De Moivre’s Theorem
De Moivre’s Theorem is fundamental in complex numbers. The theorem is stated as follows:
Theorem: Let be two distinct complex numbers and their polar forms are
and
Then the product will be a complex number with the phase as .
A significant consequence of De Moivre’s Theorem: We have defined as the phase angle of a complex number z. We shall now see how the phase angle changes according to the power raised to z. Let us multiply z with itself n times. We have a new complex number, , with a different phase angle. A consequence of the above theorem is that
simplifies to
which has a phase angle of . This is true for all integer values of n i.e.
This consequence helps us find the argument of a complex number raised to a power much more efficiently rather than multiplying all the terms.
## Modulus and Phase Examples
Without expanding the complex number , find its argument.
Solution:
Step1: Convert the complex number into its polar form:
Step2: Comparing with the general form, we deduce that
Giving us the principle argument as follows:
Step3: Plug in the value of n , which is 6:
Hence the principle argument is radians.
## Applications of Complex numbers
The concept of complex numbers is not entirely pure and abstract but has applications everywhere.
One of the most straightforward applications of complex numbers is finding the roots of a quadratic equation. It is also one of the places where the concept of a complex number arose.
Find the root of the quadratic equation .
Solution:
Step 1: Using the quadratic formula to find the roots of this equation:
Step 2: Upon simplification, we get:
But how do we simplify the part because we do not know the square root of a negative number?
This is where we employ our knowledge of using which simplifies the given form into:
Which can now be treated according to the laws of complex numbers.
A whole branch of mathematics known as Complex Analysis is based on studying complex numbers and applying their properties to other fields of mathematics.
A very popular equation in physics also employs the use of complex numbers, the Schrodinger Equation:
Which is a wave equation used to model the orbits of electrons around the nucleus of an atom.
The applications are plenty, from the differential equation of an inductor-capacitor (LC) circuit to modeling the signals received.
## Modulus and Phase - Key takeaways
• The modulus of a complex number is its distance from the origin on a complex plane.
• A complex number is related to its modulus and conjugate by the equation
• The phase angle or argument of a complex number is the angle r(line segment connecting the origin to the complex number) makes with the positive x-axis.
• The phase angle of a complex number z=a+ib is .
• De Moivre’s theorem describes how a complex number raised to an integral power can be calculated easily without using the binomial theorem.
• A significant consequence of De Moivre’s theorem states that (the argument of a complex number raised to power n is n times the argument of the original complex
## Frequently Asked Questions about Modulus and Phase
The phase of a complex number is the angle it makes with the positive x axis in a complex plane.
Modulus essentially means the magnitude of a complex number i.e. the distance of a complex point from the origin.
The phase angle of a complex number z=a+ib can be found using the formula "inverse tan of (y/x) = phase"
## Final Modulus and Phase Quiz
Question
What do you get when you multiply a complex number $$z$$ and its complex conjugate $$\bar{z}$$ together?
$$|z|^2$$.
Show question
Question
What is the plane in which complex numbers are plotted known as?
Complex plane or Argand plane
Show question
Question
What is Modulus of a complex number?
The distance of a complex point from the origin on a complex plane is called the modulus of a complex number.
Show question
Question
How will the phase of a complex number z change when it is raised to a power of n?
The phase will now be n times the original phase.
Show question
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# Find the coordinates of the minimum point of the curve y = 3x^(2) + 9x + 10
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(Note: in this answer I will use the notation y' to represent the differential of y, and by the same reasoning y'' denotes the differential of y', or the second derivative of y.)
A curve's maximum or minimum will be found at the stationary point of the curve (for a continuous function), where the gradient of the curve is flat and equal to zero.
We can find the gradient of a curve by differentiation, where the first derivative is found by the rule of 'bring the power down and minus 1 from the power.' Formally, this can be written as if y = Axn, then the differential of y with respect to x is y' = nAxn-1. Consider the curve y = 3x2 + 9x + 10 to see how this works in practice, first including the equation's hidden powers for more clarity:
y = 3x+ 9x+ 10x0
y' = (2)(3)x2-1 + (1)(9)x1-1 + (0)(10)x0-1
= y' = 6x + 9, as x= 1, and the last term disappears as it is multiplied by zero, constants (terms with no 'x' included) will always drop out once differentiated.
As stated earlier, the maximum or minimum is found where the curve is stationary, and the gradient is equal to zero, and so we set our newly derived gradient equal to zero, and rearrange the equation to find the value of x at this point:
6x + 9 = 0 (minus 9 from both sides)
6x = -9 (divide both sides by 6)
x = -9/6 = -3/2, and so the curve is stationary at the point where x is = to -3/2. What is the value of y at this point? Simply substitute this newly found x value into the original equation of the curve:
y = 3(-3/2)+ 9(-3/2) + 10 = (27/4) - (27/2) + 10 = 13/4. So there is a stationary point at (-3/2, 13/4).
Finally, in order to ensure that this point is a minimum point, we need to perform the second derivative test. This test states that we must differentiate y again (differentiate y'), if:
y'' > 0, then the curve is convex and minimised at that point.
y'' < 0, then the curve is concave and maximised at that point.
Therefore, lets differentiate y' = 6x + 9 again:
y'' = 6 which is > 0 and so the curve is in fact minimised, thus we can conclude that the minimum point of the curve is found at the point (-3/2, 13/4)
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Home > Preschool Ideas > Regular Arrangement of Objects Number Pattern Recognition Activity For Preschoolers Inspired by Mouse Count
# Regular Arrangement of Objects Number Pattern Recognition Activity For Preschoolers Inspired by Mouse Count
In the book Mouse Count by Ellen Stoll Walsh, the snake places the mice into a jar as he counts them in and they escape. Inspired by this we have decided for our Number Week for the Virtual Book Club for Kids we are going to learn some Regularly Number Pattern Arrangements inside some jars to help with number recognition and counting. So here goes a simple to do Number Pattern Recognition for Preschoolers inspired by Mouse Count.
Although we’ve done this activity indoors I’ve provided an alternative suggestion for doing the same activity outside and on a larger scale great for getting those little bodies moving around and learning at the same time.
## Recognising Number Patterns
This is a really useful skill for kids to learn. It can help them to instantly see a number and count a regular arrangement of objects quickly.
You can help your child to learn to do this by playing games with dice like Snakes and Ladders but also doing this activity together.
We did this and read some of our numbers and counting books and sang some counting rhymes as part of our Virtual Book Club for Kids this week.
## Materials Needed for this Preschool Maths Activity tO HELP COUNTING OF OBJECTS IN REGULAR ARRANGEMENTS
• Wooden Hoops – curtains rings work ours are from a Spielgaben set
• Wooden Beads or Dots again the ones used in this activity are from the set above.
• Large sheet of Paper
• Marker
## Setting up this Simple Maths Centre Activity for Preschoolers learning numbers in regular Arrangements.
Write the numbers 1 to 5 on the large sheet of paper giving lots of space between them
Set up each of the circles with 1, 2, 3, 4, and 5 dots or beads in as you would see on a dice
Place the circles with dots in randomly around the paper
## How Many? 0 to 5 Regular Arrangement of OBjects Counting Activity for Preschoolers
We started off with looking at the numbers that I had written on the paper – going through the numbers in random order first and then in numerical order.
Once I was certain that she knew the placement of each of the numbers we then looked at the patterns.
I said “Find me the circle with 1 in?”, “Which circle has 2 dots in?”, then asked her to point to another circle and I asked her “How many dots is there – without counting!”.
It was interesting 1, 2 and 3 she was great on 4 and 5 caused some problems. We then looked just at those two, talking about the similarities – the dots made a square shape, and the differences one had a dot in the centre, different coloured dots.
We then moved each of the circle in numerical order over to the correct number. Dragging it across the sheet of paper and positioning it on top.
## Adapting this Number Pattern Recognition Activity for outdoors
Yes as promised this is my plan – materials needed a Hula Hoop, some Ball Pit Balls and the numbers 1 to 5 from a Numbers Mat.
Position the numbers around the garden similar to how we did on our number activity here and then inside the hula hoop place the balls in the same arrangement as we did with the circles. You can use 1 hula hoop and after the hoop has moved, swap them out and change up the number pattern. Or if you have 5 hula hoops and at least 15 balls then set them up and get the children to drag them to the correct numbers as well.
## Mouse Count Activities for Preschoolers
We loved the book Mouse Count by Ellen Stoll Walsh so much that we also created this Mice in the Jar Mouse Count Activity for Preschoolers.
## More Number and Counting Activities on Rainy Day Mum
Inspired by the Book Jack and the Beanstalk we worked on counting to 10 with a DIY ten frame and number line in this Jack and the Beanstalk Counting to 10 Activity.
Feed the Hungry Lion in this Zoo Themed Counting Activity for Preschoolers. Count out how many animals it eats and with older children add and subtract too.
Get outside and create a giant Number Line with your Preschoolers and Toddlers in this simple activity to work on number recognition and ordering.
## More counting activities from Our Preschool Blogging Friends.
Author
##### Cerys Parker
Cerys is a marine biologist, environmental educator, teacher, mum, and home educator from the UK. She loves getting creative, whether it is with simple and easy crafts and ideas, activities to make learning fun, or delicious recipes that you and your kids can cook together you'll find them all shared here on Rainy Day Mum. |
# If area of triangle is $35\; sq\; units$ with vertices $(2, -6), (5, 4)$ and $(k, 4)$. Then $k$ is
$(A) 12\qquad (B) -2\qquad (C)-12,-2\qquad (D) 12,-2$
Toolbox:
• The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
• $\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
• On expanding along row $R_1$
• $|\bigtriangleup|=\frac{1}{2}\begin{bmatrix}x_1\begin{vmatrix}y_2 & 1\\y_3 & 1\end{vmatrix}-y_1\begin{vmatrix}x_2 & 1\\x_3 & 1\end{vmatrix}+1\begin{vmatrix}x_2 & y_2\\x_3 & y_3\end{vmatrix}\end{bmatrix}$
If area of a triangle is 35sq,units with vertices (2,-6),(5,4) and (k,4).Then k is$(A)\;12\quad(B)\;-2\quad(C)\;-12,-2\quad(D)\;12.-2$
The area of the triangle is given by
$\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
Let (2,-6),(5,4) and (k,4) be ($x_1,y_1),(x_2,y_2),(x_3,y_3)$
Now substituting the respective values,we get
$0=\frac{1}{2}\begin{vmatrix}2 & -6 & 1\\5 & 4 & 1\\k & 4 & 1\end{vmatrix}$
Now expanding along the row $R_1$ we get
$35=\frac{1}{2}\begin{bmatrix}2\begin{vmatrix}4 & 1\\4 & 1\end{vmatrix}-(-6)\begin{vmatrix}5 & 1\\k & 1\end{vmatrix}+1\begin{vmatrix}5 & 4\\k& 4\end{vmatrix}\end{bmatrix}$
70=[2(4-4)+6(5-k)+1(20-4k)]
70=0+30-6k+20-4k
70=-10k+50
20=-10k$\Rightarrow k=-2.$
Hence B is the correct answer. |
# Ultimate Guide to Writing Numbers in Expanded Form
0
1496
Writing numbers in expanded form is a way of writing a number by writing the place value of its digit. The different place values are like Ones, Tens, Hundreds, Thousands, and so on.
For easy and convenient expansion, refer to the place value chart.
2,311 is in standard form.
We can use different ways to express numbers in expanded form.
Read More- The Interpretation of Angel Number 222
### #Method 1
To expand the number 2,311, we separate a number by their place value and expand it to show the value of each digit.
• The value of digit 2 at Thousands place =2000,
• The value of digit 3 at Hundreds place =300,
• The value of digit 1 at Tens place = 10,
• The value of digit 1 at Ones place =1,
If you add the place value of each digit then you will get a total of 2,311
2000+300+ 10+ 1= 2,311
So, the expanded form of 2,311 is 2000 + 300 + 10 + 1
Another method that can be used for writing numbers in expanded form is to write the digits one by one and convert all the c9ming numbers to zero.
### Method #2
This method works by multiplying the digit with its place value.
Example: 2,311
• 2 x 1000
• 3 x 100
• 1 x 10
• 1 x 1
(2 x 1000) + (3 x100) + (1 x10) + (1 x 1) = 2,311 is the expanded form of 2,311.
### Method #3
In this method, the first digit is written and then its place value is written adjacent to it.
Example: 2,311
2 Thousands + 3 Hundreds + 1 Tens + 1 ones = 2,311.
In this way you can write the expanded form of any number. We hope that you liked this article. Make sure to tell us how you liked these mathematical tips and tricks.
Read More- What is 911 Angel Number Meaning and How it Impacts Your Life? |
# GMAT Math : DSQ: Calculating the length of the side of a rectangle
## Example Questions
### Example Question #1 : Dsq: Calculating The Length Of The Side Of A Rectangle
Ronald is making a bookshelf with a rectangular base that will be two yards tall. What is the area of the base?
I) The distance around the base will be yards.
II) The smaller sides of the base are half the length of the longer sides.
Either statement alone is sufficient to answer the question.
Statement I is sufficient to answer the question, but Statement II is not sufficient to answer the question.
Both statements together are needed to answer the question.
Statement II is sufficient to answer the question, but Statement I is not sufficient to answer the question.
Both statements together are needed to answer the question.
Explanation:
To find the area we need the length and width of the rectangle. We can use II together with I to make an equation for perimeter with only one unknown.
So we need both to solve.
Solve for and then go back to find and then with that you can find the area of the base and you are finished.
### Example Question #2 : Dsq: Calculating The Length Of The Side Of A Rectangle
Find a possible width of rectangle .
I) has a perimeter of fathoms.
II) has a diagonal length of fathoms.
Either statement is sufficient to answer the question.
Statement II is sufficient to answer the question, but statement I is not sufficient to answer the question.
Both statements are needed to answer the question.
Statement I is sufficient to answer the question, but statement II is not sufficient to answer the question.
Both statements are needed to answer the question.
Explanation:
When asked to find the width of a rectangle we will need to use both statemests together.
For Statement I) we can use the perimeter formula.
Now, for Statement II) we will use the length of the diagonal along with the Pythagorean Theorem.
From here you can solve the perimeter equation in terms of either l or w. Then you can use substitution into the Pythagorean Theorem to solve for a possible width.
### Example Question #221 : Data Sufficiency Questions
Find the length of the side of a rectangle with a width three times the length.
1. The area of the rectangle is .
2. The perimeter of the rectangle is
Statement 2 alone is sufficient, but statement 1 alone is not sufficient to answer the question.
Statement 1 alone is sufficient, but statement 2 alone is not sufficient to answer the question.
Both statements taken together are sufficient to answer the question, but neither statement alone is sufficient.
Each statement alone is sufficient to answer the question.
Statements 1 and 2 are not sufficient, and additional data is needed to answer the question.
Each statement alone is sufficient to answer the question.
Explanation:
Statement 1:
Statement 1 is sufficient to answer the question
Statement 2:
Statement 2 is also sufficient to answer the question
### Example Question #222 : Data Sufficiency Questions
A rectangle has a width measuring twice the length. Find the length.
1. The rectangle has a perimeter of .
2. The rectangle's area is .
Both statements taken together are sufficient to answer the question, but neither statement alone is sufficient.
Statement 1 alone is sufficient, but statement 2 alone is not sufficient to answer the question.
Statement 2 alone is sufficient, but statement 1 alone is not sufficient to answer the question.
Statements 1 and 2 are not sufficient, and additional data is needed to answer the question.
Each statement alone is sufficient to answer the question.
Each statement alone is sufficient to answer the question.
Explanation:
Statement 1:
Recall the formula to find the perimeter of a rectangle. Substitute in the given information and solve.
Statement 2:
Recall the formula for the area of a rectangle. Substitute in the given information and solve.
Each statement alone is sufficient to answer the question.
### Example Question #223 : Data Sufficiency Questions
Given parallelogram with diagonal . Is this parallelogram a rectangle?
1)
2)
BOTH statements TOGETHER are insufficient to answer the question.
BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.
Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.
Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.
EITHER statement ALONE is sufficient to answer the question.
BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.
Explanation:
The length of one diagonal alone does not prove the parallelogram to be a rectangle, nor do the lengths of the sides.
Suppose we know all of these lengths, though. Since is a parallelogram, if , then
The sides and diagonal form a triangle with sidelengths 25, 60, and 65. The parallelogram is a rectangle if and only if is a right angle; therefore, we must determine whether the conditions of the Pythagorean Theorem hold:
This is true; is a right angle and is a rectangle.
Therefore, both statements together are sufficient to answer the question, but neither statement alone is sufficient to answer the question.
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# How do you write an equation of a line passing through (4, -2), perpendicular to y = 4x + 2?
Sep 5, 2017
See a solution process below:
#### Explanation:
The equation in the problem is in slope-intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$
Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.
$y = \textcolor{red}{4} x + \textcolor{b l u e}{2}$
The slope of this line is: $\textcolor{red}{m = 4}$
Let's call the slope of the line perpendicular to the line in the problem: ${m}_{p}$
The formula for the slope of a perpendicular line is: ${m}_{p} = - \frac{1}{m}$
Substituting gives the slope of the line we are looking for as:
${m}_{p} = - \frac{1}{4}$
We can substitute this into the slope-intercept formula giving:
$y = \textcolor{red}{- \frac{1}{4}} x + \textcolor{b l u e}{b}$
We can now substitute the values from the point in the problem for $x$ and $y$ in the formula and solve for $\textcolor{b l u e}{b}$ giving:
$- 2 = \left(\textcolor{red}{- \frac{1}{4}} \times 4\right) + \textcolor{b l u e}{b}$
$- 2 = \textcolor{red}{- 1} + \textcolor{b l u e}{b}$
$1 - 2 = 1 \textcolor{red}{- 1} + \textcolor{b l u e}{b}$
$- 1 = 0 + \textcolor{b l u e}{b}$
$- 1 = \textcolor{b l u e}{b}$
Substituting this back into the formula gives:
$y = \textcolor{red}{- \frac{1}{4}} x + \textcolor{b l u e}{- 1}$
$y = \textcolor{red}{- \frac{1}{4}} x - \textcolor{b l u e}{1}$ |
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# Jasmine allows 4% discount on the marked price of her goods and still earns a profit of 20%. What is the cost price of a shirt for her marked at Rs. 850.
Last updated date: 21st Apr 2024
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Hint: We will use the fact that gain is equal to selling price subtracted from cost price, i.e. gain= S.P -C.P. Using the unitary method we will find the gain percentage. We will use $gain%=\dfrac{gain}{C.P}\times 100$ to get the gain%. Also, the formula for calculating the discount % is that,
$discount\%=\dfrac{discount}{M.P}\times 100$
(Where M.P. is the marked price of an article)
One more important thing that needs to be kept in mind is that, Discount=M.P.-S.P.
Let the cost price of the shirt be x.
Now, using the formula for finding the discount %, we can write the following
\begin{align} & \Rightarrow discount\%=\dfrac{discount}{M.P}\times 100 \\ & \Rightarrow 4=\dfrac{discount}{850}\times 100 \\ & \Rightarrow discount=Rs.34 \\ \end{align}
Hence, the discount on the shirt is of Rs. 34.
Now, for finding the selling price of the product, we can write and use the following formula which is
\begin{align} & \Rightarrow Discount=M.P.-S.P. \\ & \Rightarrow 34=850-S.P. \\ & \Rightarrow Rs.\ 816=S.P. \\ \end{align}
Now, we know that the selling price of the shirt is Rs. 816.
Now, as we know the gain as well as the the gain%, we can write the following
\begin{align} & \Rightarrow gain\%=\dfrac{gain}{C.P}\times 100 \\ & \Rightarrow 20=\dfrac{gain}{x}\times 100 \\ & \Rightarrow Gain=\dfrac{x}{5} \\ \end{align}
Now, again, as we know the C.P. as well as the S.P., we can write the following
\begin{align} & \Rightarrow S.P.-C.P.=gain \\ & \Rightarrow 816-x=\dfrac{x}{5} \\ & \Rightarrow \dfrac{6x}{5}=816 \\ & \Rightarrow x=Rs.680 \\ \end{align}
Hence, the cost price of the shirt is Rs. 680.
Note: In this question, it is very important to know all the formulae beforehand and also, the calculations should also be done very carefully as there are very high chances that the students might commit a mistake while solving the question resulting in a wrong solution. Some students may write the formula of discount as SP - MP, but this wrong and the correct formula is discount = MP - SP. |
Vector Geometry
Many ways problem
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Solution
Each of the following situations, A-D, shows a system of forces acting on a particle of given mass and the initial velocity of the particle. Work out the single extra constant force that can be added to this system, so that the particle would be brought instantaneously to rest at the specified time.
Mass and forces Initial velocity Time when at rest
A $\quantity{4}{s}$
There are many ways to solve these problems. We’ll approach the first couple of examples using different mechanical principles and different notation. Both approaches could be used for all four problems, but a particular example may tempt you to choose one approach rather than another.
Without any extra force, this $\quantity{6}{kg}$ particle is moving with velocity $\quantity{8}{m\,s^{-1}}$ and accelerating perpendicular to this, due to the $\quantity{60}{N}$ force.
How will the force of magnitude $\quantity{F}{N}$ bring the particle to rest?
Can we say anything about the approximate values of $F$ and $\theta$? Thinking about this now may help us to check that our final answers are sensible. For example, $F\sin \theta = 60$, so the magnitude of the force is greater than $\quantity{60}{N}.$ Is $\theta$ likely to be greater than or less than $45^{\circ}$?
To work out the actual value of $F$ and the angle $\theta$, let’s consider perpendicular components along the dotted lines in the diagram.
Firstly, we know that $F\sin \theta = 60$ because we want to balance out the $\quantity{60}{N}$ force.
Why do we want to put the particle in equilibrium in this direction?
Secondly, $F\cos \theta$ needs to provide the acceleration to bring the $\quantity{6}{kg}$ particle to rest in $4$ seconds. Using $v=u+at$ and taking left to be the positive direction, we have $0=-8+4a.$ So the acceleration we need is $\quantity{2}{m\,s^{-2}}$ in the opposite direction to the initial velocity. It follows from Newton’s Second Law that $F\cos \theta=12.$
We now know the perpendicular components of the force, $F\cos \theta$ and $F\sin \theta$, so we can find $F$ and $\theta$. $F=\sqrt{60^2+12^2}=\sqrt{3744}=61.18...$ so the force has magnitude $\quantity{61.2}{N}\,\,\,(3\,s.f.)$ and we now need to work out its direction.
As $\tan \theta =\dfrac{60}{12}=5$, we have $\theta=\arctan{5}=78.69...$ so $\theta=79^{\circ}$ to the nearest degree.
Does this combination of magnitude and direction seem reasonable?
Describe how the particle behaves over time if this extra force is applied, assuming all the given forces acting on the particle remain unchanged. Compare this with how the particle would have behaved if the extra force had not been applied.
With the extra force added, the resultant force on the particle is acting in the opposite direction to the initial velocity. Therefore the particle will be restricted to moving in a straight line, reversing direction after $4$ seconds. Without this force, the particle would have maintained the $\quantity{8}{m\,s^{-1}}$ component of its initial velocity, but also accelerated in the direction of the $\quantity{60}{N}$ force at a rate of $\quantity{10}{m\,s^{-2}}$, so the component of velocity in this direction after time $t$ seconds would have been $\quantity{10t}{m\,s^{-1}}.$ Try to sketch this path of the particle.
Mass and forces Initial velocity Time when at rest
B $\quantity{2}{s}$
For this example we’ll use two approaches.
Compare these approaches with each other and our approach to the first example.
• How did using momentum and impulse compare with using acceleration?
• How did using a triangle of forces compare with using components?
• How did using $\mathbf{i}$ and $\mathbf{j}$ affect the type of calculations we had to do? Note that we chose to define $\mathbf{i}$ and $\mathbf{j}$ in the directions given, but we could have chosen different directions for these. For example, we could have chosen $\mathbf{i}$ to be in the direction of the initial velocity.
Describe how the particle behaves over time if this extra force is applied, assuming all the given forces acting on the particle remain unchanged. Compare this with how the particle would have behaved if the extra force had not been applied.
With the extra force in place the impulse is directed in the opposite direction to the initial momentum, so the particle will move in a straight line, coming to rest after $2$ seconds as it reverses direction along this line. Without this extra force, the component of momentum in the direction of $\mathbf{i}$ would have been maintained, but the particle would have received an impulse in the $-\mathbf{j}$ direction, first bringing the momentum in this direction to zero, but after this the velocity and momentum would have had components in the positive $\mathbf{i}$ and negative $\mathbf{j}$ directions.
Before tackling the next two examples, it would be worth considering how they are similar to or different from the first two and each other. How will this affect
• the approach you take?
• your choice of notation?
Mass and forces Initial velocity Time when at rest
C $\quantity{3}{s}$
Why will the extra force shown bring this particle to rest in $\quantity{3}{s}$?
Describe how the particle behaves over time if this extra force is applied, assuming all the given forces acting on the particle remain unchanged. Compare this with how the particle would have behaved if the extra force had not been applied.
If the extra force is applied, the particle will continue to move in the direction of its initial velocity, travelling in a straight line but reversing direction after $5$ seconds. Without this force, the component of the particle’s velocity parallel to its initial velocity would have decreased to zero and then reversed direction, whereas the component of velocity perpendicular to this would have continued to increase from zero, so at time $\quantity{3}{s}$ the particle would have been moving perpendicular to its initial velocity. Can you sketch the two paths of the particle?
Mass and forces Initial velocity Time when at rest
D $\quantity{2}{s}$
Why will the extra force shown bring this particle to rest in $\quantity{2}{s}$?
Describe how the particle behaves over time if this extra force is applied, assuming all the given forces acting on the particle remain unchanged. Compare this with how the particle would have behaved if the extra force had not been applied.
If the extra force is applied, the particle will move in a straight line, reversing direction after $2$ seconds. If the extra force had not been applied, the particle would have continued to move along the same straight line, but as it would be in equilibrium, it would have maintained its speed of $\quantity{8}{m\,s^{-1}}.$ |
# Basic Math Vectors and Scalars Addition/Subtraction of Vectors Unit Vectors Dot Product.
## Presentation on theme: "Basic Math Vectors and Scalars Addition/Subtraction of Vectors Unit Vectors Dot Product."— Presentation transcript:
Basic Math Vectors and Scalars Addition/Subtraction of Vectors Unit Vectors Dot Product
Scalars and Vectors (1) Scalar – physical quantity that is specified in terms of a single real number, or magnitude Ex. Length, temperature, mass, speed Vector – physical quantity that is specified by both magnitude and direction Ex. Force, velocity, displacement, acceleration We represent vectors graphically or quantitatively: Graphically: through arrows with the orientation representing the direction and length representing the magnitude Quantitatively: A vector r in the Cartesian plane is an ordered pair of real numbers that has the form. We write r= where a and b are the components of vector v. Note: Both and r represent vectors, and will be used interchangeably.
Scalars and Vectors (2) The components a and b are both scalar quantities. The position vector, or directed line segment from the origin to point P(a,b), is r=. The magnitude of a vector (length) is found by using the Pythagorean theorem: Note: When finding the magnitude of a vector fixed in space, use the distance formula.
Operations with Vectors (1) Vector Addition/Subtraction The sum of two vectors, u= and v= is the vector u+v =. Ex. If u= and v=, then u+v= Similarly, u-v= =
Operations with Vectors (2) Multiplication of a Vector by Scalar If u= and c is a real number, the scalar multiple cu is the vector cu=. Ex. If u= and c=2, then cu= cu=
Unit Vectors (1) A unit vector is a vector of length 1. They are used to specify a direction. By convention, we usually use i, j and k to represent the unit vectors in the x, y and z directions, respectively (in 3 dimensions). i= points along the positive x-axis j= points along the positive y-axis k= points along the positive z-axis Unit vectors for various coordinate systems: Cartesian: i, j, and k Cartesian: we may choose a different set of unit vectors, e.g. we can rotate i, j, and k
Unit Vectors (2) To find a unit vector, u, in an arbitrary direction, for example, in the direction of vector a, where a=, divide the vector by its magnitude (this process is called normalization). Ex. If a=, then is a unit vector in the same direction as a.
Dot Product (1) The dot product of two vectors is the sum of the products of their corresponding components. If a= and b=, then a·b= a 1 b 1 +a 2 b 2. Ex. If a= and b=, then a·b=3+32=35 If θ is the angle between vectors a and b, then Note: these are just two ways of expressing the dot product Note that the dot product of two vectors produces a scalar. Therefore it is sometimes called a scalar product.
Dot Product (2) Convince yourself of the following: Conclusion: After you define the direction of an arbitrary vector in terms of the Cartesian system, you can find the projection of a different vector onto the arbitrary direction. By dividing the above equation by the magnitude of b, you can find the projection of a in the b direction (and vice versa). |
Chapter 13 Probability and Data Analysis
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# Chapter 13 Probability and Data Analysis - PowerPoint PPT Presentation
Chapter 13 Probability and Data Analysis. 13.1 Probabilities and Odds. Vocabulary: outcome, event, sample space, probability, odds What is probability???. 13.1 Continued. You flip 3 coins. How many possible outcomes are in the sample space? List the possible outcomes.
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## PowerPoint Slideshow about ' Chapter 13 Probability and Data Analysis' - anila
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### Chapter 13 Probability and Data Analysis
13.1 Probabilities and Odds
• Vocabulary: outcome, event, sample space, probability, odds
• What is probability???
13.1 Continued
• You flip 3 coins. How many possible outcomes are in the sample space? List the possible outcomes.
• You roll a dice three times. How many possible outcomes are in the sample space?
Odds versus probability
• Odds = favorable/ unfavorable
• Probability = favorable/ total outcomes
• A study indicates that out of every 60 telephone calls, 6 result in busy signals and 12 result in no answer. What are the odds in favor of someone answering?
• What are the odds in favor of a busy signal?
13.2 Permutations
• Vocabulary: permutation, factorial
• How many permutations can be formed using the letters in the word DOG?
• In how many ways can you arrange the letters in the word BEAR?
• A garage door has a keypad with 10 different digits. A sequence of 4 digits must be selected to open the door. How many keypad codes are possible?
• There are 12 members on a committee. You want to choose a president, vice-president, and treasurer. In how many ways can you choose the officers?
13.3 Combinations
• Vocabulary: Combination
• You have three seashells. How many different combinations of two seashells can you make?
• You order a pizza at a restaurant. You can choose 3 toppings from a list of 12. How many combinations of toppings are possible?
• A committee must award three students out of 15 students with scholarships. What is the probability that you and your two best friends are awarded the scholarships?
13.4 Compound Events
• Vocabulary: compound event, mutually exclusive events, overlapping events, independent events, dependent events
1) Find the probability of A or B. You roll a number cube. Find the probability that you roll a 4 or a prime number. Rolling a 4 and rolling a prime number are mutually exclusive events because 4 is not a prime number.
13.4 Continued
• You roll a number cube. Find the probability that you roll an even number or a number greater than 3. These are overlapping events.
13.4 continued
• Find the probability of A and B. You roll two number cubes. What is the probability that you roll a 1 first and a 2 second?
• The events are _____________
• You and a friend must each select a golf ball from a bucket to play miniature golf. There are 3 yellow balls, 4 red balls, 5 green balls, and 4 purple balls. You select a golf ball, and then your friend selects a golf ball. What is the probability that both balls are green.
• The events are _______________
13. 5 Analyze Surveys and Samples
• Vocabulary: Survey, population, sample, biased sample, biased question
Sampling Methods:
In a random sample, every member of the population has an equal chance of being selected.
In a stratified random sample, the population is divided into distinct groups. Members are selected at random from each group
In a systematic sample, a rule is used to select members of the population.
In a convenience sample, only members of the population who are easily accessible are selected.
In a self-selected sample, members of the population select themselves by volunteering.
Examples
• A high school is conducting a survey to determine the average number of hours that their students spend doing homework each week. At the school, only the members of the sophomore class are chosen to complete the survey. Identify the population and classify the sampling method.
Biased Questions
Tell whether the question is potentially biased. Explain your answer. If the question is potentially biased, rewrite it so that it is not.
• Do you still support the school basketball team, even though the team is having its worst season in 5 years?
• Don’t you think that dogs are better pets than cats?
13.6 Measures of Central Tendency and Dispersion
• Vocabulary: measure of dispersion, range, mean absolute deviation
• Mean, median, mode= measures of central tendency
• Example: Your last 8 test scores are 81, 87, 91, 93, 95, 98, 100. Find the mean, median, mode.
Measures of dispersion: range, mean absolute deviation
• The 9-hole scores of golfers on two different high school teams are given. Compare the spread of the data sets using the range and the mean absolute deviation.
Team 1: 51, 46, 40, 49, 55, 47
Team 2: 41, 47, 54, 50, 42, 42
13.7 Interpret Steam and Leaf Plots and Histograms
• Vocabulary: stem-and-leaf plot, frequency, frequency table, histogram
• A survey asked people how many miles they commute to work. The results are listed below. Make a stem-and-leaf plot of the data.
5, 10, 18, 15, 9, 27, 10, 35, 12, 4, 8, 14, 23, 2, 20, 5, 15
The birth weight (in ounces) of babies born at a hospital are listed below. Make a histogram of the data.
96, 128, 115, 120, 107, 125, 136, 122, 131, 112, 110
13.8 Interpret Box-and-Whisker Plots
• Vocabulary: box-and-whisker plot, quartile, interquartile range, outlier
• Make a box-and-whisker plot of the heights (in inches) of 7 family members: 34, 67, 70, 62, 46, 75, 54
• Identify an outlier: The average monthly high temperatures (in degrees Fahrenheit) in Atlanta are: 52, 57, 65, 73, 80, 87, 89, 88, 82, 73, 63, 55 |
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# Copy of Assignment 1 Answers-2 - #2 Payment of \$70,000...
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#2 Payment of \$70,000 arrives in period 3. (a) \$62,229.75 (b) \$165,894.32 \$165,894.32 The present value = 70,000/(1.04 3 )= The value in time period 25 = 70,000 x (1.04 22 ) = Alternatively, you can calculate this as \$62,229.75 x (1.07 25 ) =
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#3(a) I. Cost of RV in present value terms if he pays cash = \$42,000.00 II. PV of paying \$45,000 in time 2 = 45,000/(1.07^2) = \$39,304.74 II. PV of paying 72 monthly payments of \$675= =PV(0.07/12,6*12,-675) = \$39,591.75 3(b) To solve this, try different discount rates in part II until the PV reaches just over \$42,000 r= 3.5098% PV= \$42,000.03 If r<3.51%, then option I will be preferred to option II.
4(a) If you will be making equal deposits into a retirement account for 10 years (with each payment at the end of the year), how much must you deposit each year if the account earns 6% compounded annually and you wish to have \$200,000 after 30 years? There are many ways to do this. One is to do all calculations in PV. Step 1: PV of 200,000 in 30 years at 6%: \$34,822.03 Step 2: Annuity with PV = \$34,822.03 and 10 equal annual payments at 6%: (\$4,731.20) So 10 payments of \$4,731.20 is the answer. Alternative method 1: Step 1: Find FV in time 10 of 200,000 in time 30: \$62,360.95 Step 2: Solve for the payment that equals FV10 of 62,360.95 using =PMT()
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Subject: prealgebra Name: nina Who are you: Student Find the amount in an account if 1250 is invested at 6% compounded annually for 3 years Hi Nina, If you have \$1250 invested for 1 year at 6% per year then at the end of the year you have \$1250 plus 6% of \$1250. Thus at the end of the first year you have \$1250 + 6/100 \$1250 = \$1250 + 0.06 \$1250 = 1.06 \$1250 = \$1325 Now, in the second year you have \$1325 invested at 6% for 1 year so how much do you have at the end of the second year? In the third year you have that much invested at 6% for 1 year so how much do you have at the end of the third year? That's the amount that will be in your account at the end of the third year but can you see how to get this number in one calculation rather than three? Penny Hi Nina. First, recall that a percentage is just the amount out of 100. Compount interest is interest that is paid and then is added to the total so that future interest is paid on the sum. Here's an example: Start with \$1000 at 10% compound annual interest. Year 0: You have \$1000 total. Year 1. You have earned interest of 10% x \$1000. That is \$100. This gets added to the \$1000 you had. So now your total is \$1000 + \$100 = \$1100. Year 2: You have earned interest of 10% x \$1100. That is \$110. This gets added to the \$1100 you had. So now your total is \$1100 + \$110 = \$1210. Year 3: 10% of \$1210 is \$121. \$121 + \$1210 = \$1331. and so on. When you multiply something by 1, you get the original number. So here is a shortcut: new total = previous total x (1 + interest rate) For example for year 1 above: new total = \$1000 x (1 + 10%) = \$1000 x (1 + 0.10) = \$1000 x 1.10 = \$1100. In fact, you could keep doing this repeatedly: total for year 3 would have three multiplications: total for year 3 = \$1000 x (1 + 10% ) x (1 + 10%) x (1 + 10%) = \$1000 x 1.10 x 1.10 x 1.10 = \$1331. That's a good shortcut you can use with your question. Stephen La Rocque> |
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# Solving ax + b = cx+d
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Solving linear equations of the form ax + b = cx + d
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### Solving ax + b = cx+d
1. 1. Practice Problem Five more than three times a number is three more than 5 times the number. What is the number??
2. 2. Objective- To solve equations in the form ax + b = cx + d and inequalities in the form ax + b < cx + d Solve. 14k + 20 = 21k - 43 -14k -14k 20 = 7k - 43 + 43 + 43 63 = 7k 7 7 9 = k
3. 3. Solve. x + 1.05x = 2x - 1000 2.05x = 2x - 1000 -2x -2x 0.05x = -1000 0.05 0.05 x = -20,000
4. 4. Solve. - 2y + 11 = - 6y +2y +2y 11 = - 4y - 4 - 4 -2.75 = y
5. 5. Problem In 1990, the population of El Paso was 515,000 and increasing at 9000 people per year. Milwaukee was 628,000 and decreasing at 800 people per year. When will they be the same? Let n = number of years Let P = population El Paso Milwaukee P = 515,000 + 9000n P = 628,000 - 800n
6. 6. Let n = number of years Let P = population El Paso Milwaukee P = 515,000 + 9000n P = 628,000 - 800n 515,000 + 9000n = 628,000 - 800n + 800n +800n 515,000 + 9800n = 628,000 -515,000 -515,000 9800n = 113,000 9800 9800 n 11.53 years
7. 7. Problems Involving Inequalities Solve. -3( x - 5 ) + 7 < 4( 3x + 2) -3x + 15 + 7 < 12x + 8 -3x + 22 < 12x + 8 +3x +3x 22 < 15x + 8 - 8 - 8 14 < 15x 15 15 < x x >
8. 8. Tree Problem A beech tree is 8 ft. tall and grows ft. per year. A maple tree is 3 ft. tall and grows 1 foot per year. When will the maple be taller than the beech? Let x = number of years Let h = height of tree Beech tree Maple tree h = 8 + 0.5x h = 3 + x
9. 9. Let x = number of years Let h = height of tree Beech tree Maple tree h = 8 + 0.5x h = 3 + x When is maple taller than beech? < 8 + 0.5x < 3 + x - 0.5x - 0.5x 8 < 3 + 0.5x 5 < 0.5x 10 < x x > 10 years |
## Monday, March 30, 2020
### Newton's Second Law and Displacement
In this article, the topic is the relationship between Newton's Second Law and Motion, and this time, displacement will be the focus.
How do you find a total displacement when given time (t), initial displacement (di), initial velocity (vi), but instead of acceleration (a), you are given force (F) and mass (m)?
Once again, it is necessary to combine two principles in order to see the full relationship.
The distance formula is:
df = di + (vi )(t)+ 1/2(a)(t2)
However, to find final displacement, you need an acceleration(a), but you have mass and force.
Using Newton's Second Law, fortunately, will allow the calculation of a using the force and mass given.
So, we'll be doing two steps.
Step 1:
Use
F=ma to find a
Step 2:
Then use THAT in the distance formula.
df = di + (vi )(t)+ 1/2(a)(t2)
EXAMPLES
Example 1
Let's take a look at another example (even easier!) and work it out:
A force of 50 N acts on a object with a mass of 10 kg. If it has an initial displacement (di) of 10 m and has an initial velocity (vi) of 20 m/s, what is its final displacement after an elapsed time of 4 seconds?
The question tells us that we need to find final velocity (vf). But there is no acceleration given. So, do step 1 (above):
Find a where:
F = 50 N
m = 10 kg
F = ma
50 N = (10 kg)(a)
50 N/10 kg = a
5 m/s/s = a
Next, use THAT calculated a to find final velocity (step 2 above):
Find df where
di = 10 m
vi = 20 m/s
t = 4 s
a = 5 m/s/s
df = di + (vi )(t)+ 1/2 (a)(t2)
df = 10 m + (20 m/s )(4 s)+ 1/2 (5 m/s/s)(42)
df = 10 m + 80 m + 1/2 (5 m/s/s)(16 s2)
df = 10 m + 80 m + 40 m
df = 130 m
Example 2
How about seeing one worked out?
https://youtu.be/TjxUEwMjmvc
SUMMARY:
You gotta do it in steps!
This process requires doing the work in steps. Depending on what is given, you use the two formulas below:
F = ma
df = di + (vi )(t)+ 1/2(a)(t2)
So, read the problem, write down what is being looked for, write down what is given, THEN... use the two formulas. Two steps! |
# How do you write the standard form of a line given (3,-2) and (7,6)?
Jun 28, 2016
y = 2x - 8
#### Explanation:
The equation of a line in $\textcolor{b l u e}{\text{slope-intercept form}}$ is
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y = m x + b} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where m represents the slope and b, the y-intercept.
We require to find m and b to obtain the equation.
To find m use the $\textcolor{b l u e}{\text{gradient formula}}$
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 points} .$
The 2 coordinate points here are (3 ,-2) and (7 ,6)
let $\left({x}_{1} , {y}_{1}\right) = \left(3 , - 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(7 , 6\right)$
$\Rightarrow m = \frac{6 - \left(- 2\right)}{7 - 3} = \frac{8}{4} = 2$
We now have a partial equation: y= 2x + b and to find b use either of the 2 given coordinate points.
Using (7 ,6) and substituting x = 7 , y = 6 into the partial equation.
$: 6 = \left(2 \times 7\right) + b \Rightarrow 14 + b = 6 \Rightarrow b = 6 - 14 = - 8$
$\Rightarrow y = 2 x - 8 \text{ is the equation}$ |
# Second Derivative
A derivative basically gives you the slope of a function at any point.
The "Second Derivative" is the derivative of the derivative of a function. So:
• Find the derivative of a function
• Then take the derivative of that
A derivative is often shown with a little tick mark: f'(x)
The second derivative is shown with two tick marks like this: f''(x)
### Example: f(x) = x3
• Its derivative is f'(x) = 3x2
• The derivative of 3x2 is 6x, so the second derivative of f(x) is:
f''(x) = 6x
A derivative can also be shown as: dy , and the second derivative shown as: d2y dx dx2
### Example: (continued)
The previous example could be written like this:
y = x3
dy = 3x2 dx
d2y = 6x dx2
## Distance, Speed and Acceleration
A common real world example of this is distance, speed and acceleration:
### Example: A bike race!
You are cruising along in a bike race, going a steady 10 m every second.
Distance: is how far you have moved along your path. It is common to use s for distance (from the Latin "spatium").
So let us use:
• distance (in meters): s
• time (in seconds): t
Speed: is how much your distance s changes over time t ...
... and is actually the first derivative of distance with respect to time: ds dt
And we know you are doing 10 m per second, so ds = 10 m/s dt
Acceleration: Now you start cycling faster! You increase your speed to 14 m every second over the next 2 seconds.
When you are accelerating your speed is changing over time.
So ds is changing over time! dt
We could write it like this:
d ds dt
dt
But it is usually written d2s dt2
Your speed increases by 4 m/s over 2 seconds, so d2s = 4/2 = 2 m/s2 dt2
Your speed changes by 2 meters per second per second.
And yes, "per second" is used twice!
It can be thought of as (m/s)/s but is usually written m/s2
(Note: in the real world your speed and acceleration changes moment to moment, but here we assume you can hold a constant speed or constant acceleration.)
So: Example
Measurement
Distance: s 100 m
First Derivative is Speed:
ds dt
10 m/s
Second Derivative is Acceleration:
d2s dt2
2 m/s2
And the third derivative (how acceleration changes over time) is called "Jolt" ... !
## Play With It
Here you can see the derivative f'(x) and the second derivative f''(x) of some common functions:
Notice how the slope of each of those functions is the derivative plotted below it. |
h�bbdb"���)��"���E.��sA��)df��H� ��i0� Next lesson. Use expansion of cofactors to calculate the determinant of a 4X4 matrix. 3x3 and 4x4 matrix determinants and Cramer rule for 3x3.notebook 1 April 14, 2015 Sect 6.8: Determinants 3x3 Lesson on determinants, inverses, and Create customized worksheets for students to match their abilities, and watch their confidence soar through excellent practice! Formula for the determinant We know that the determinant has the following three properties: 1. det I = 1 2. 0. This method entails three kinds of row operations: Recall the triangular property of the determinant which says that if each element in the matrix above or below the main diagonal is zero, the determinant is equal to the product of the elements in the diagonal. Find more Mathematics widgets in Wolfram|Alpha. The determinant is a scalar value obtained from the elements of the square matrix. Determinant as scaling factor. It would be very time consuming and challenging to find the determinant of 4x4 matrix by using the elements in the first row and breaking the matrix into smaller 3x3 sub-matrices. Rj1. In this determinant worksheet, students find the determinants of each matrix. Linear Algebra: Simpler 4x4 determinant Calculating a 4x4 determinant by putting in in upper triangular form first. An example of a 4x4 determinant. There are 10 problems on this worksheet. We de ne the factor of every row operation as follows: Type assmuption Row Operation Factor I j 6= k Rj $Rk -1 II 6= 0 Rj ! So the Determinant of Minor 2 is (0+0+0)(-1)= 0 Now on to Minor number 3. This is the currently selected item. 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. The determinant of the 2x2 matrix is given below: As we are given the higher order matrices, the calculation of determinants becomes more and more challenging. We will multiply the elements in the diagonal to get the determinant. R w mAtl tl t zrVi1gzhdt Csv jr1e DsHear 0v7eWdd.h T WMlaEdaeB Iw jiRtChm FIzn If1isn WiEt Eey UAClAgle db1r oa4 l2 x.R Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Determinants of 3×3 Matrices Date_____ Period____ Evaluate the determinant … We explain Finding the Determinant of a 4x4 Matrix with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. DETERMINANTS To compute the determinant of a 3 × 3 or n× nmatrix, we need to introduce some notation. For instance, we have included a spreadsheet containing fields Matrix and Determinants. They use diagonals and the expansion by minors method to help. Find the inverse of the Matrix: 41 A 32 ªº «» ¬¼ Method 1: Gauss – Jordan method Step1: Set up the given matrix with the identity matrix as the form of 4 1 1 0 3 2 0 1 ªº «» ¬¼ %%EOF Determinants Worksheets. A series of linear algebra lectures given in videos: 4x4 determinant, Determinant and area of a parallelogram, Determinant as Scaling Factor and Transpose of a Matrix. If you need a refresher, check out my other lesson on how to find the determinant of a 2×2. Rj 1 De nition 1.2. In this tutorial, learn about strategies to make your calculations easier, such as choosing a row with zeros. 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. The Formula of the Determinant of 3×3 Matrix The standard formula to find the determinant of a 3×3 matrix is a break down of smaller 2×2 determinant problems which are very easy to handle. Get the free "3x3 Determinant calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. The matrix determinant is a number derived from the values in array. Suppose we … Determinant of 3×3 Matrix Read More » M3 -->$-1^4 = 1$... Determinant of 4x4 Matrix by Expansion Method. �N˂��� I�P ;LDr��H��r:�d6�l.����Vv�C �_������uH�Qr��&�8w4F��t5J���Qr��FX����S�?ө? determinants of 2x2 matrices worksheet answers with work, As a hint, I'll take the determinant of a very similar two by two matrix. Example 1 It has made evaluating determinant dead-easy for users working with matrices. Launch Excel 2010 spreadsheet on which you want to apply Matrix determinant. The determinant of matrices we define as inductive, that is, the determinant of a square matrix of the$n$-th order we define using the determinant of a square matrix of the$(n-1)$-th order. For row reduction, we apply a series of arithmetic operations on the matrix, so that each element below the main diagonal of a matrix becomes zero. Simpler 4x4 determinant. Determinant 4x4. 0 So, we've reduced our calculation of a 4x4 determinant to a couple of 3x3 determinants and some addition and multiplication. Linear Algebra: Determinants along other rows/cols Finding the determinant by going along other rows or columns �RZ:vM�V����(9�z"�*��Da��Y�UNG�[���AL�f�Zd0;��b٭ � �! Find more Mathematics widgets in Wolfram|Alpha. Use our printable worksheets to help high school students find the determinants of order 2 x 2 or 3 x 3 with ease. Math Worksheets; A series of linear algebra lectures given in videos. About This Quiz & Worksheet Solving for a determinant is a very useful skill to have, particularly if you can do it without the aid of a calculator. Everything above or below the main diagonal is zero. III j 6= k Rj+ Rk ! Definition 4.2. h�bf�gcQfd@ A�rD�8��T����] �0 �S��$|������������ HB1�)~�;l Transpose of a matrix. This is the currently selected item. [4] Compute the determinant of the following 4×4 matrix: 1110 2202 3033 0444 What can you say about the determinant of the n×n matrix with the same pattern? It is basically a series of operations applied to the matrix elements. In this section, we will see how to compute the determinant of a 4x4 matrix using Gaussian elimination and matrix properties. Determinant of a $2\times 2$ block matrix. The determinant of a matrix $\mathbf{A}=[a]$ of order $1$ is the number $a$: \ma… Cramer uses determinant to identify the solutions of systems of equations in two and three variables. The determinant of the matrix is an important concept in linear algebra as it is quite helpful in solving linear equations, altering variables in integrals, and telling us how linear transformations alter area or volume. Let A= [ajk] be an n×nmatrix.Let Mjk be that (n−1)× (n− 1) matrix obtained from Aby deleting its jth row and kth column. The determinant of the matrix is denoted by two vertical lines ||. I like to spend my time reading, gardening, running, learning languages and exploring new places. It will not affect the determinant of the matrix. h��Ymo7�+�eE��D�E �/k���{9�7�e;H\���{�;�ۉ�%@�K2I���D�LtFie�#E�{�B�ީ�Gc��Ã�LN. Enter the coefficients. The determinant of the 1x1 matrix is the number itself. They are also useful in computing the matrix inverse and have some applications in calculus. 150 CHAPTER4. 1786 0 obj <>stream To find the det (B), I multiplied B 14 by det (B 14) and B 24 by det (B 24) and followed the + - + - pattern as showed by the formula here (scroll below for 4x4 formula). The determinant is a real function such that each square matrix $\mathbf{A}$ joins a real number (the determinantof a matrix $\mathbf{A}$). |�� For example, the determinant of the matrix . It does not affect the value of the determinant. 1312 0 obj <> endobj Gaussian elimination is also called as a row reduction. Transpose of a matrix. Interchanging two rows. Online Calculator for Determinant 4x4. A 4x4 matrix has 4 rows and 4 columns in it. Determinant and area of a parallelogram. –32 + 30 + (–42) = –44. Determinant as scaling factor. Determinants and inverses A matrix has an inverse exactly when its determinant is not equal to 0. I am passionate about travelling and currently live and work in Paris. 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. Determinants Worksheet Exercise 1 Prove, without developing, that the following determinants are zero: $A = begin {vmatrix} 1 & a & b + c \ 1 & b & a + c \ 1 & c & a + b \ end {vmatrix}$ $B = begin… 11. After we have converted a matrix into a triangular form, we can simply multiply the elements in the diagonal to get the determinant of a matrix. In the next section, we will see how to compute the determinant of the 4x4 matrix. And let's see if we can figure out its determinant, the determinant of A. det (B) = [-2 (0)] - [-5 (-1)] + [0] - [0] = -5. Determinant and area of a parallelogram. | 4 2 6 −1 −4 5 3 7 2 |→| 4 2 6 −1 −4 5 3 7 2 | 4 2 −1 −4 3 7. ***** *** 2⇥2inverses Suppose that the determinant of the 2⇥2matrix ab cd does not equal 0. For this matrix, you need to break down the larger matrix into smaller 2x2 matrices. The determinant remains unchanged. Oct 6, 2019; 3 min read; Inverse Of 4x4 Matrix Example Pdf Download ⎠.. We are working with a 4x4 matrix, so it has 4 rows and 4 columns. In other words, we can say that while computing the determinant, input is a square matrix and output is a scalar number. endstream endobj 1313 0 obj <> endobj 1314 0 obj <> endobj 1315 0 obj <>stream The determinants of such matrices are the product of the elements in their diagonals. −72 140 −4 −| 4 2 6 1 −4 5 3 7 2 | 4 2 −1 −4 3 7 −32 30 −42. Get the free "4x4 Determinant calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. Step 1: Rewrite the first two columns of the matrix. Step 2: Multiply diagonally downward and diagonally upward. In the previous lecture we gave examples of pairs of nxn matrices whose.. Matrices & Determinants Worksheet Finding the Inverse of a Matrix Answers & Solutions 1. This video shows how to calculate determinants of order higher than 3. Proving generalized form of Laplace expansion along a row - determinant. The determinant of a square matrix$\mathbf{A}$is denoted as$det \mathbf{A}$or$|\mathbf{A}|$. Video transcript. 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## 4x4 determinant worksheet
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##### by Spencer Bowen
In Mind Reading with Math we showed how you could use binary to read someone’s mind.
For that activity, we used a set of cards representing the first four binary place values: 8, 4, 2, 1.
Binary only has two digits: “1” or “0” (which could also be thought of as “yes” or “no”). This made our mind reading trick pretty easy. If a number we were trying to “mind read” appeared on the card corresponding to a particular place value, then that place value was required as an addend of the number we’re trying to “mind read.” So if you’re reading a volunteer’s mind, and they respond “Yes, yes, no, and yes,” then you add together the place values of the cards showing their number and get 13.
This trick can extend to higher numbers by including more cards, with each card doubling the possible numbers someone could chose from. So if you added a card, you would now have a “16s” place value and could choose from 32 possible numbers (0-31).
We can extend this trick to base 3 or ternary, but we will need to rethink how we make our cards. Our binary cards just needed two digits: “on the card” = 1 and “not on the card” = 0. Ternary has three possible digits (0, 1, 2), so we cannot use the simple off/on distinction.
Instead, we can use symbols to represent our digits. For example, let “Circle” = 0, “Square” = 1, and “Diamond” = 2. Using colors to represent place values like before, we get:
Now Mind Reading involves a bit more mental math. For each card, we ask our volunteer what shape their number is on the card, then multiply the corresponding digit by the card’s place value.
For example, let’s say a volunteer chooses “50.” Then they would respond “Square, Diamond, Square, Diamond.”
This method works for all numbers 0 – 80. The only information the mind reader has to memorize is each digit’s shape and each place value’s color.
### Even More Bases
This activity generalizes nicely to any base. You just have to decide how you want to distinguish your place values and digits. Here are base 4 or quaternary cards using our same color scheme for place values and the four playing card suits as our digits.
In fact, it’s surprisingly easy to make mind reading cards for different bases! If you look at the ternary and quaternary cards, you can notice some pretty distinct geometry. The “ones” place value has the digits in a repeating pattern. Looking at the above example, you can see on the “1s” card (which is red) that “hearts” = 0, “spades” = 1, “diamonds” = 2, and “clubs” = 3. And the suits keep repeating in a cycle.
Next, the “fours” place value (which is orange) has a cycling pattern as well (“hearts” , “spades”, “diamonds”, “clubs”); the only difference is that each digit is repeated four times in a row before moving to the next digit in the cycle. This pattern continues with the “sixteens” place value as well, except with each digit being repeated sixteen times before moving to the next digit in the cycle.
This geometry generalizes to any base assuming you lay out the numbers in reading order. For each card corresponding to a particular place value, the symbols will form into sets of increasing numbers sharing the same symbol. The length of these sets will be to the card’s place value. The symbol of these groups will then cycle between your digits’ symbols in order.
### How to make your own base cards
1. Choose your favorite base. Call it b.
2. Decide the largest number you want your volunteers to be able to choose. This number has to be equal to one less than a power of b.
3. Make a number of cards equal to the power you chose in step 2. Make a way to distinguish between your cards. The examples shown above color-code the cards. But you can use different methods as long as you can tell the cards apart.
4. Lay your numbers out on the cards in increasing order from left to right, making new rows as you run out of room (start with 0 in the top left corner and end with your largest number in the bottom right corner).
5. Determine which symbols represent your digits. They should be distinct and easy to see.
6. For your “ones” place value card, label each of your bases’ digits with the correct symbol. These symbols then repeat in a cycle all the way to your largest number.
7. Your next card should represent the b place value. Label the first b (0, …, b-1) numbers with your “0” symbol, then label the next set of b numbers with your “1” symbol. Keep cycling through your digits’ symbols, labeling sets of numbers with a symbol before moving onto the next symbol.
8. For each card representing some place value b^k, repeat step 7 but instead label sets of numbers of length b^k with the same symbol before moving onto the next digit. |
# Multiples of 8
Multiples of 8
• The multiples of 8 are numbers that can be divided exactly by 8, leaving no remainder.
• The first few multiples of 8 are the numbers in the 8 times table.
• For example 2 × 8 = 16 and so, the second multiple of eight is 16.
• For example 100 × 8 = 800 and so, the hundredth multiple of 8 is 800.
• A number is a multiple of 8 if it can be halved, halved and halved again to leave a whole number.
• For example, 24 is a multiple of 8 because half of 24 is 12, half of 12 is 6 and half of 6 is 3.
A multiple of 8 is any number that can be divided by 8 exactly, leaving no remainder.
#### Multiples of 8 Flashcards
Click on the multiples of 8 flashcards below to memorise the 8 times table:
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#### Multiples of 8: Interactive Questions
Multiples of 8: Interactive Questions
# Multiples of 8
## What are Multiples of 8?
Multiples of 8 are numbers that can be divided exactly by 8. Multiples of 8 are found by counting up in eights. The first few multiples of 8 are: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88 and 96.
The first few multiples of 8 are the numbers in the 8 times table. Here is a list of the first twelve multiples of 8:
To teach the multiples of 8, place the numbers on a number grid. We can count on in eights from one number to the next. Here are the multiples of 8 shown on the chart:
To learn the multiples of 8, it can help to look for patterns in the positions of the numbers on the chart.
We can see that the multiples of 8 form diagonal patterns on the chart.
Multiples of 8 always end in 0, 2, 4, 6 and 8. Multiples of 8 are always even numbers.
To test if a number is a multiple of 8, half it three times. If the result is a whole number, the number is a multiple of 8. If the result is not a whole number, the number is not a multiple of 8.
For example, we can halve 88 three times in a row. Half of 88 is 44, half of 44 is 22 and half of 22 is 11. 11 is a whole number and so, 88 is a multiple of 8.
20 is not a multiple of 8. It cannot be halved three times to leave a whole number. For example, half of 20 is 10, half of 10 is 5 but half of 5 is 2.5. 2.5 is not a whole number and so, 20 is not a multiple of 8.
All multiples of 8 are also multiples of 4. This is because 4 divides exactly into 8.
## How do you Find Multiples of 8?
To find multiples of 8, multiply any whole number by 8. For example, the tenth multiple of 8 is 80 because 10 × 8 = 80. Alternatively, multiples of 8 can be found by starting at zero and counting up in eights.
We can start at zero and count up in eights to find the multiples of 8.
0 + 8 = 8, 8 + 8 = 16, 16 + 8 = 24, 24 + 8 = 32, 32 + 8 = 40 and so on.
Here are the multiples of 8 to 100.
There are infinite multiples of 8.
We can continue to add on 8 to find further multiples of 8.
96 + 8 = 104, 104 + 8 = 112 and so on.
## Multiples of 8 up to 100
There are 12 multiples of 8 that are less than 100. They are:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88 and 96.
Here is a longer list of multiples of 8:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, 248, 256, 264, 272, 280, 288, 296, 304, 312, 320, 328, 336, 344, 352, 360, 368, 376, 384, 392, 400.
Now try our lesson on Lattice Multiplication Method where we learn how to use the lattice method for multiplying numbers.
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SECTION 1.3 THE LIMIT OF A FUNCTION.
Presentation on theme: "SECTION 1.3 THE LIMIT OF A FUNCTION."— Presentation transcript:
SECTION 1.3 THE LIMIT OF A FUNCTION
THE LIMIT OF A FUNCTION Our aim in this section is to explore the meaning of the limit of a function. We begin by showing how the idea of a limit arises when we try to find the velocity of a falling ball. 1.3
Find the velocity of the ball after 5 seconds.
Example 1 Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 5 seconds. 1.3
Example 1 SOLUTION Through experiments carried out four centuries ago, Galileo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. Remember, this model neglects air resistance. If the distance fallen after t seconds is denoted by s(t) and measured in meters, then Galileo’s law is expressed by the following equation. s(t) = 4.9t2 1.3
Example 1 SOLUTION The difficulty in finding the velocity after 5 s is that you are dealing with a single instant of time (t = 5). No time interval is involved. However, we can approximate the desired quantity by computing the average velocity over the brief time interval of a tenth of a second (from t = 5 to t = 5.1). 1.3
Example 1 SOLUTION 1.3
Example 1 SOLUTION The table shows the results of similar calculations of the average velocity over successively smaller time periods. It appears that, as we shorten the time period, the average velocity is becoming closer to 49 m/s. 1.3
Example 1 SOLUTION The instantaneous velocity when t = 5 is defined to be the limiting value of these average velocities over shorter and shorter time periods that start at t = 5. Thus, the (instantaneous) velocity after 5 s is: v = 49 m/s 1.3
INTUITIVE DEFINITION OF A LIMIT
Let’s investigate the behavior of the function f defined by f(x) = x2 – x + 2 for values of x near 2. The following table gives values of f(x) for values of x close to 2, but not equal to 2. 1.3
INTUITIVE DEFINITION OF A LIMIT
From the table and the graph of f (a parabola) shown in Figure 1, we see that, when x is close to 2 (on either side of 2), f(x) is close to 4. 1.3
INTUITIVE DEFINITION OF A LIMIT
In fact, it appears that we can make the values of f(x) as close as we like to 4 by taking x sufficiently close to 2. We express this by saying “the limit of the function f(x) = x2 – x + 2 as x approaches 2 is equal to 4.” The notation for this is: 1.3
Definition 1 We write and say “the limit of f(x), as x approaches a, equals L” if we can make the values of f(x) arbitrarily close to L (as close to L as we like) by taking x to be sufficiently close to a (on either side of a) but not equal to a. 1.3
THE LIMIT OF A FUNCTION Roughly speaking, this says that the values of f(x) tend to get closer and closer to the number L as x gets closer and closer to the number a (from either side of a) but x a. An alternative notation for is as which is usually read “f(x) approaches L as x approaches a.” 1.3
Notice the phrase “but x a” in the definition of limit.
THE LIMIT OF A FUNCTION Notice the phrase “but x a” in the definition of limit. This means that, in finding the limit of f(x) as x approaches a, we never consider x = a. In fact, f(x) need not even be defined when x = a. The only thing that matters is how f is defined near a. 1.3
Figure 2 shows the graphs of three functions.
THE LIMIT OF A FUNCTION Figure 2 shows the graphs of three functions. Note that, in the third graph, f(a) is not defined and, in the second graph, However, in each case, regardless of what happens at a, it is true that 1.3
Example 2 Guess the value of . SOLUTION
Notice that the function f(x) = (x – 1)/(x2 – 1) is not defined when x = 1. However, that doesn’t matter—because the definition of says that we consider values of x that are close to a but not equal to a. 1.3
Example 2 SOLUTION The tables give values of f(x) (correct to six decimal places) for values of x that approach 1 (but are not equal to 1). On the basis of the values, we make the guess that 1.3
Example 2 is illustrated by the graph of f in Figure 3.
THE LIMIT OF A FUNCTION Example 2 is illustrated by the graph of f in Figure 3. 1.3
This new function g still has the same limit as x approaches 1.
THE LIMIT OF A FUNCTION Now, let’s change f slightly by giving it the value 2 when x = 1 and calling the resulting function g: This new function g still has the same limit as x approaches 1. See Figure 4. 1.3
Example 3 Estimate the value of . SOLUTION
The table lists values of the function for several values of t near 0. As t approaches 0, the values of the function seem to approach … So, we guess that: 1.3
What would have happened if we had taken even smaller values of t?
THE LIMIT OF A FUNCTION What would have happened if we had taken even smaller values of t? The table shows the results from one calculator. You can see that something strange seems to be happening. If you try these calculations on your own calculator, you might get different values but, eventually, you will get the value 0 if you make t sufficiently small. 1.3
Does this mean that the answer is really 0 instead of 1/6?
THE LIMIT OF A FUNCTION Does this mean that the answer is really 0 instead of 1/6? No, the value of the limit is 1/6, as we will show in the next section. The problem is that the calculator gave false values because is very close to 3 when t is small. In fact, when t is sufficiently small, a calculator’s value for is 3.000… to as many digits as the calculator is capable of carrying. 1.3
THE LIMIT OF A FUNCTION Something very similar happens when we try to graph the function of the example on a graphing calculator or computer. 1.3
THE LIMIT OF A FUNCTION These figures show quite accurate graphs of f and, when we use the trace mode (if available), we can estimate easily that the limit is about 1/6. 1.3
THE LIMIT OF A FUNCTION However, if we zoom in too much, then we get inaccurate graphs—again because of problems with subtraction. 1.3
Example 4 Guess the value of . SOLUTION
The function f(x) = (sin x)/x is not defined when x = 0. Using a calculator (and remembering that, if , sin x means the sine of the angle whose radian measure is x), we construct a table of values correct to eight decimal places. 1.3
From the table at the left and the graph in Figure 6 we guess that
Example 4 SOLUTION From the table at the left and the graph in Figure 6 we guess that This guess is in fact correct, as will be proved in the next section using a geometric argument. 1.3
Example 5 Investigate . SOLUTION
Again, the function of f(x) = sin (p /x) is undefined at 0. 1.3
Example 5 SOLUTION Evaluating the function for some small values of x, we get: Similarly, f(0.001) = f(0.0001) = 0. 1.3
On the basis of this information, we might be tempted to guess that
Example 5 SOLUTION On the basis of this information, we might be tempted to guess that This time, however, our guess is wrong. Although f(1/n) = sin np = 0 for any integer n, it is also true that f(x) = 1 for infinitely many values of x that approach 0. 1.3
The graph of f is given in Figure 7.
Example 5 SOLUTION The graph of f is given in Figure 7. The dashed lines near the y-axis indicate that the values of sin(p/x) oscillate between 1 and –1 infinitely as x approaches 0. 1.3
Example 5 SOLUTION Since the values of f(x) do not approach a fixed number as approaches 0, does not exist. 1.3
THE LIMIT OF A FUNCTION Examples 3 and 5 illustrate some of the pitfalls in guessing the value of a limit. It is easy to guess the wrong value if we use inappropriate values of x, but it is difficult to know when to stop calculating values. As the discussion after Example 3 shows, sometimes, calculators and computers give the wrong values. In the next section, however, we will develop foolproof methods for calculating limits. 1.3
The Heaviside function H is defined by:
Example 6 The Heaviside function H is defined by: The function is named after the electrical engineer Oliver Heaviside (1850–1925). It can be used to describe an electric current that is switched on at time t = 0. 1.3
The graph of the function is shown in Figure 8.
Example 6 The graph of the function is shown in Figure 8. As t approaches 0 from the left, H(t) approaches 0. As t approaches 0 from the right, H(t) approaches 1. There is no single number that H(t) approaches as t approaches 0. So, does not exist. 1.3
ONE-SIDED LIMITS We noticed in Example 6 that H(t) approaches 0 as t approaches 0 from the left and H(t) approaches 1 as t approaches 0 from the right. We indicate this situation symbolically by writing and The symbol ‘‘ ’’ indicates that we consider only values of t that are less than 0. Similarly, ‘‘ ’’ indicates that we consider only values of t that are greater than 0. 1.3
Definition 2 We write and say the left-hand limit of f(x) as x approaches a [or the limit of f(x) as x approaches a from the left] is equal to L if we can make the values of f(x) arbitrarily close to L by taking x to be sufficiently close to a and x less than a. 1.3
ONE-SIDED LIMITS Notice that Definition 2 differs from Definition 1 only in that we require x to be less than a. Similarly, if we require that x be greater than a, we get ‘’the right-hand limit of f(x) as x approaches a is equal to L;’ and we write Thus, the symbol ‘‘ ’’ means that we consider only 1.3
The definitions are illustrated in Figures 9.
ONE-SIDED LIMITS The definitions are illustrated in Figures 9. 1.3
ONE-SIDED LIMITS By comparing Definition 1 with the definition of one-sided limits, we see that the following is true. Definition 3 1.3
Example 7 The graph of a function g is shown in Figure 10. Use it to state the values (if they exist) of: (a) (b) (c) (d) (e) (f) 1.3
Example 7(a) & (b) SOLUTION
From the graph, we see that the values of g(x) approach 3 as x approaches 2 from the left, but they approach 1 as x approaches 2 from the right. Therefore (a) and (b) 1.3
Example 7(c) SOLUTION (c) As the left and right limits are different, we conclude that does not exist. 1.3
Example 7(d) & (e) SOLUTION
The graph also shows that (d) and (e) 1.3
For , the left and right limits are the same.
Example 7(f) SOLUTION For , the left and right limits are the same. So, we have Despite this, notice that 1.3
Example 8 Find if it exists. SOLUTION
As x becomes close to 0, x2 also becomes close to 0, and 1/x2 becomes very large. 1.3
Example 8 SOLUTION In fact, it appears from the graph of the function f(x) = 1/x2 that the values of f(x) can be made arbitrarily large by taking x close enough to 0. Thus, the values of f(x) do not approach a number. So, does not exist. 1.3
PRECISE DEFINITION OF A LIMIT
Definition 1 is appropriate for an intuitive understanding of limits, but for deeper under-standing and rigorous proofs we need to be more precise. 1.3
PRECISE DEFINITION OF A LIMIT
We want to express, in a quantitative manner, that f(x) can be made arbitrarily close to L by taking to x be sufficiently close to a (but x a. This means f(x) that can be made to lie within any preassigned distance from L (traditionally denoted by e, the Greek letter epsilon) by requiring that x be within a specified distance (the Greek letter delta) from a. 1.3
PRECISE DEFINITION OF A LIMIT
That is, | f(x) – L | < e when | x – a | < and x a. Notice that we can stipulate that x a by writing 0 < | x – a | . The resulting precise definition of a limit is as follows. 1.3
Definition 4 Let f be a function defined on some open interval that contains the number , except possibly at a itself. Then we say that the limit of f(x) as x approaches a is L, and we write if for every number e > 0 there is a corresponding number > 0 such that if 0 < | x – a | < then | f(x) – L | < e 1.3
PRECISE DEFINITION OF A LIMIT
If a number e > 0 is given, then we draw the horizontal lines y = L + e and y = L – e and the graph of f. 1.3
PRECISE DEFINITION OF A LIMIT
If , then we can find a number > 0 such that if we restrict x to lie in the interval (a – ) and (a + ) take x a, then the curve y = f(x) lies between the lines y = L – e and y = L + e . 1.3
PRECISE DEFINITION OF A LIMIT
It’s important to realize that the process illustrated in Figures 12 and 13 must work for every positive number e, no matter how small it is chosen. Figure 14 shows that if a smaller e is chosen, then a smaller may be required. 1.3
PRECISE DEFINITION OF A LIMIT
In proving limit statements it may be helpful to think of the definition of limit as a challenge. First it challenges you with a number e. Then you must be able to produce a suitable . You have to be able to do this for every e > 0, not just a particular . 1.3
Example 9 Prove that SOLUTION
Let e be a given positive number. According to Definition 4 with a = 3 and L = 7, we need to find a number such that if 0 < | x – 3 | < then | (4x – 5) – 7 | < e 1.3
Example 9 SOLUTION But |(4x – 5) – 7| = |4x – 12| = |4(x – 3)| = 4|(x – 3)| . Therefore, we want: if 0 < | x – 3 | < then 4|(x – 3)| < e We can choose to be e/4 because if 0 < | x – 3 | < = e/ then 4|(x – 3)| < e Therefore, by the definition of a limit, 1.3
PRECISE DEFINITION OF A LIMIT
For a left-hand limit we restrict x so that x < a, so in Definition 4 we replace 0 < | x – a | < by x – < x < a. Similarly, for a right-hand limit we use a < x < a + . 1.3
Example 10 Prove that SOLUTION
Let e be a given positive number. We want to find a number such that if 0 < x < then 1.3
Example 10 SOLUTION But So if we choose = e2 and 0 < x < = e2, then (See Figure 16.) This shows that 1.3
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## Algebra 2 (Eureka Math/EngageNY)
### Course: Algebra 2 (Eureka Math/EngageNY)>Unit 2
Lesson 7: Topic B: Lessons 15-17: Trigonometric identities
# Proof of the Pythagorean trig identity
The Pythagorean identity tells us that no matter what the value of θ is, sin²θ+cos²θ is equal to 1. We can prove this identity using the Pythagorean theorem in the unit circle with x²+y²=1. Created by Sal Khan.
## Want to join the conversation?
• How is tan theta= cosine theta over sin theta?
• Think back algebra. Slope of a line is y/x. Tan represents that slope. Since sin θ is y and cos θ is x. So sin θ/cos θ = y/x = tan θ.
• Is there an alternative way to prove the Pythagorean trig identities (there are three of them)?
• We can use the definitions of the functions and the following property: x^2 + y^2 = r^2.
sinθ = y/r
cosθ = x/r
sin²θ + cos²θ
= (y/r)² + (x/r)²
= y²/r² + x²/r²
= (x² + y²)/r²
= r²/r²
= 1
We have proven sin²θ + cos²θ = 1. To get to the other identities, all we have to do is divide by sin²θ or cos²θ.
(sin²θ + cos²θ = 1)/cos²θ
tan²θ + 1 = sec²θ
(sin²θ + cos²θ = 1)/sin²θ
1 + cot²θ = csc²θ
• Doesn't this proof. only prove the Pythagorean Identity in the unit circle? If the radius was not one, wouldn't this not work.
• I had the same question. After a while I figured that it would still work because of how ratios work. Basically to have any other circle you would have to multiply by the same factor:
sin²Θ + cos²Θ = 1
(sin²Θ + cos²Θ)*factor = 1*factor(for different radius)
If you divide each side by the factor, you're back where you started.
I know this answer is super late, but I hope someone else can learn from it...I hope this is correct.
• At , why does he put absolute value around the sin and cos?
• Good question! Actually, I don't see why Sal would use the absolute value of sin and cos as the square of a negative integer is equal to the square of the absolute value of the same integer.
-a^2 = a^2
• What if the radius is not 1?
cos^2(theta) + sin^2(theta) = 4
Do we just divide both sides of the equation by 4?
• For the pythagorean trig identity? It works for any radius circle. Try going throught he video again but instead of x^2 + y^2 = 1, pretend that it is x^2 + y^2 = a^2 where we can insert any number for a, so it would be applicable for any circle.
The one twist is that where he has a point's coordinates be cos(theta), sin(theta), it would be a*cos(theta), a*sin(theta) since each point is found by manipulating cos(theta) = x/a and sin(theta) = y/a. let me know if you do not understand why this is.
You would not wind up with cos^2(theta) + sin^2(theta) = 4. Keep in mind x^2 + y^2 = 1 is the same as x^2 + y^2 = 1^2, because int he equation for a circle the number at the end is the radius squared.
• wait, I understand that the hypotenuse and the x and y are 1, but how does 1+1=1?
• The unit circle does not go through (1,1). At a 45 degree angle where sin and cos are equal, the point is (√2/2,√2/2).
• Why are the exponents on sine and cosine and not theta?
• We do this to remove some ambiguity. If we had sinx^2 instead of sin^2x when we wrote it, you could either regard it as (sin(x))^2 or sin(x^2), which would give you different values. When the exponent is on the trig function, it means to raise the output of the function to the power. However, this solution can also lead to confusion with the inverse trig functions.
We write inverse sine as sin^-1(x). However, raising anything to the -1 is the same as dividing 1 by it. So, we can write sin^-1(x) as 1 / sin(x), which is equal to the cosecant of x. The cosecant is in no way the same thing as the inverse sine, so we have yet more confusion.
Hope this sheds light on the issue.
• Does this only work for 90-degree triangles?
• No, whatever the value of θ is, sin^2(θ) + cos^2(θ) = 1
• , I don't know if I am misunderstanding or making my own mistake in thought, but shouldn't the cos of theta be the green dotted line and sin of theta be the pink dotted line? Isn't cos of theta the base of the triangle in reference to the triangle the x-axis and the ray make?
• This was my big doubt too... but cos is the x value and sin the y value. The pink line connects the point you want to find with the x axis (it's a kind of projection of that point on the line), while the green line connects the point with the y axis... hope this helps!
• I know the trig identities, but am having trouble applying them. Can someone help me by explaining the following equation to me please?
(secθ+tanθ)(secθ-tanθ)=1 |
16
Q:
# A drink vendor has 80 liters of Mazza, 144 liters of Pepsi, and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required?
A) 35 B) 36 C) 37 D) 38
Explanation:
If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can.
So, the number of liters in each can = HCF of 80,144 and 368 = 16 liters.
Now, number of cans of Maaza = 80/16 = 5
Number of cans of Pepsi = 144/16 = 9
Number of cans of Sprite = 368/16 = 23
Thus, the total number of cans required = 5 + 9 + 23 = 37
Q:
Three different containers contain 1365 litres, 1560 litres and 1755 litres of mixtures of milk and water respectively. What biggest measure that can measure all the different quantities exactly ?
A) 37 lit B) 97 lit C) 129 lit D) 195 lit
Explanation:
From the given data,
containers contain mixtures of milk and water as 1365 lit, 1560 lit and 1755 lit.
Biggest measure that can measure all these different quantities exactly can be given by
HCF of (1365, 1560, 1755)
That can be found as
HCF of 1365, 1755 = 195
Now, HCF of 195 and 1560 = 195
Hence, biggest measure that can measure all the given different quantities exactly is 195 lit.
1 160
Q:
What is a Common Multiple?
If the Multiples for two or more numbers which are Common, then those Multiples are said to be Common Multiples.
For Example :
Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30,...
Multiples of 5 are 5, 10, 15, 20, 25, 30,...
Here common multiples of 3 & 5 upto 30 are 15, 30.
1148
Q:
LCD of 12 and 18
A) 36 B) 42 C) 12 D) 6
Explanation:
LCD is nothing but Lowest or Least Common Denominator
Here LCD of 12 and 18 means LCD of two fractions with denominators 12 and 18 respectively.
Therefore, LCM of 12 & 18 = 6 x 3 x 2 = 36
• How to calculate LCD ::
The lowest common denominator or least common denominator (LCD) is the least common multiple (LCM) of the denominators of a set of fractions.
17 1873
Q:
54, 60 LCM is
A) 600 B) 540 C) 60 D) 54
Explanation:
Here the LCM of 54, 60 is
=> 2 x 3 x 3 x 3 x 10
= 6 x 9 x 10
= 54 x 10
540.
21 2352
Q:
If the HCF of 210 and 55 is expressible in the form of 210 x 5 + 55P, then value of P = ?
A) -23 B) 27 C) 16 D) -19
Explanation:
HCF of 210 and 55 is 5
Now, 210x5 + 55P = 5
=> 1050 + 55P = 5
=> 55P = -1045
=> P = -1045/55
=> P = -19.
20 3263
Q:
A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is ?
A) 63 B) 31 C) 16 D) 27
Explanation:
To get the least number of coconuts :
LCM = 30 => 30 + 1 = 31
23 4258
Q:
A drink vendor has 368 liters of Maaza, 80 liters of Pepsi and 144 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required ?
A) 47 B) 46 C) 37 D) 35
Explanation:
The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = 368/16 = 23
Number of cans of Pepsi = 80/16 = 5
Number of cans of Sprite = 144/16 = 9
The total number of cans required = 23 + 5 + 9 = 37 cans.
24 3362
Q:
H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 and 16 x 81 x 5 x 11 x 49 is :
A) 360 B) 180 C) 90 D) 120
Explanation:
4 x 27 x 3125 = ;
8 x 9 x 25 x 7 =
16 x 81 x 5 x 11 x 49 =
H.C.F = = 180. |
# Factoring Trinomials of the Form ax^2 + bx + c; Perfect Squares
## The polynomial $ax^2 + bx + c$ can be factored using a variety of methods, including trial and error.
#### Key Points
• Some trinomials, known as perfect square trinomials, can be factored into two equal binomials.
• We can factor $a^2 − b^2$, the difference of two squares, by finding the terms that produce the perfect squares and substituting these quantities into the factorization form. When using real numbers, there is no factored form for the sum of two squares.
• Perfect square trinomials factor as the square of a binomial. To recognize them, look for whether (1) the first and last terms are perfect squares, and (2) the middle term is divisible by 2, and when halved, equals the product of the terms that when squared produce the first and last terms.
#### Terms
• A polynomial expression consisting of three terms, or monomials, separated by addition and/or subtraction symbols.
• A polynomial consisting of two terms, or monomials, separated by addition or subtraction symbols.
#### Figures
1. ##### Factorization
The polynomial $x^2+cx+d$ is factored to the form $(x+a)(x+b)$. This is based on the relationships that $a+b=c$ and $a*b=d$.
2. ##### Interactive Graph: Polynomial
Graph of a trinomial with the example equation $y=x^2-x-2$. Notice what happens when the signs are flipped from negative to positive.
## Factoring Trinomials
The polynomial $ax^2+bx+c$ can be factored using a variety of methods. One such method is trial and error (Figure 2).
Ultimately, the trinomial should be factored in the form $(px+q)(rx+s)$, where p, q, r, and s are constants, and x is a variable. Using trial and error, we can find values for each of the constants, using the FOIL method to determine whether the constants used produce the trinomial $ax^2+bx+c$. As seen in (Figure 1), FOIL stands for the order in which one multiplies the four terms - first, outside, inside, last. The order isn't important; however, the acronym is useful to prevent missing a multiplication term or accidentally multiplying the same terms twice. We know that the product of px and rx must equal ax2. Additionally, the sum of products $px \ast s$ and $q \ast rx$ must equal bx. Finally, the product of q and s must equal c.
## Perfect Squares
Some trinomials, known as perfect square trinomials, can be factored into two equal binomials. For example:
$a^2+2ab+b^2=(a+b)^2$
and
$a^2-2ab+b^2=(a-b)^2$
Perfect square trinomials always factor as the square of a binomial.
To recognize a perfect square trinomial, look for the following features:
1. The first and last terms are perfect squares.
2. The middle term is divisible by 2.
For example, $x^2-10x+25$ can be identified as a perfect square because x2 is the square of x, and 25 is the square of 5. The middle term (-10x) is divisible by 2 (equalling -5x).
Given that the coefficient of x2 is 1, we know that the factored form will be $(x+a)(x+b)$, where a and b are to-be-determined coefficients. We need $x \ast b+a \ast x$ to equal -10x, and a*b to equal 25. Filling in -5 for a and b, we find a plausible solution that reads (x-5)(x-5), or (x-5)2. This is a perfect square.
#### Key Term Glossary
binomial
A polynomial consisting of two terms, or monomials, separated by addition or subtraction symbols.
##### Appears in these related concepts:
coefficient
a constant by which an algebraic term is multiplied.
##### Appears in these related concepts:
constant
An identifier that is bound to an invariant value.
##### Appears in these related concepts:
factor
To find all the factors of (a number or other mathematical object) (the objects that divide it evenly).
##### Appears in these related concepts:
polynomial
an expression consisting of a sum of a finite number of terms, each term being the product of a constant coefficient and one or more variables raised to a non-negative integer power, such as $a_n x^n + a_{n-1}x^{n-1} + ... + a_0 x^0$. Importantly, because all exponents are positive, it is impossible to divide by x.
##### Appears in these related concepts:
square
The second power of a number, value, term or expression.
##### Appears in these related concepts:
term
any value (variable or constant) or expression separated from another term by a space or an appropriate character, in an overall expression or table.
##### Appears in these related concepts:
trinomial
A polynomial expression consisting of three terms, or monomials, separated by addition and/or subtraction symbols.
##### Appears in these related concepts:
variable
A symbol that represents a quantity in a mathematical expression, as used in many sciences |
Repeated Prime Factors
We know if the factors of a number are prime then it is called prime factor. Now if the same prime factor is repeated more than once then it is called repeated prime factor.
For Example:
Find the repeated prime factors of 64
Solution:
The factors of 64 are:
2 × 2 × 2 ×2 × 2 × 2
Therefore, the repeated factors of 64 are:
2 × 2 × 2 ×2 × 2 × 2
Here are few examples illustrated on repeated factors:
1. List the repeated prime factors of 128
Solution:
The repeated prime factors of 128 are:
2 × 2 × 2 ×2 × 2 × 2 × 2
2. List the repeated prime factors of 256
Solution:
The repeated prime factors of 256 are:
2 × 2 × 2 ×2 × 2 × 2 × 2 × 2
3. Find the repeated prime factors of 81
Solution:
The repeated prime factors of 81 are:
3 × 3 × 3 ×3
4. Find the repeated prime factors of 125
Solution:
The repeated prime factors of 125 are:
5 × 5 × 5
5. List the repeated prime factors of 72
Solution:
The repeated prime factors of 72 are:
2 × 2 × 2 × 3 × 3
6. List the repeated factors of 225
Solution:
The repeated prime factors of 225 are:
3 × 3 × 5 × 5
7. List the repeated factors of 108
Solution:
The repeated prime factors of 108 are:
3 × 3 × 3 × 2 × 2
8. List the repeated factors of 144
Solution:
The repeated prime factors of 144 are:
3 × 3 × 2 × 2 × 2 × 2
9. List the repeated factors of 27
Solution:
The repeated prime factors of 27 are:
3 × 3 × 3
10. List the repeated factors of 625
Solution:
The repeated prime factors of 625 are:
5 × 5 × 5 × 5
11. List the repeated factors of 4096
Solution:
The repeated prime factors of 4096 are:
2 × 2 × 2 ×2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
12. List the repeated factors of 1024
Solution:
The repeated prime factors of 1024 are:
2 × 2 × 2 ×2 × 2 × 2 × 2 × 2 ×2 × 2
13. Find the repeated prime factors of 529
Solution:
The repeated prime factors of 529 are:
23 × 23
14. Find the repeated prime factors of 784
Solution:
The repeated prime factors of 784 are:
2 × 2 × 2 ×2 × 7 × 7
15. Find the repeated prime factors of 2048
Solution:
The repeated prime factors of 2048 are:
2 × 2 × 2 ×2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
16. Find the repeated prime factors of 121
Solution:
The repeated prime factors of 121 are:
11 × 11
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# Evaluate $\displaystyle \Large \lim_{x \,\to\, 0} \normalsize \dfrac{1-\sqrt{1-\tan x}}{\sin x}$
$x$ is a literal number and represents an angle of the right angled triangle as well. It formed a function in ratio form with trigonometric functions sine and tangent. The value of the function is required to evaluate when the value of $x$ tends to $0$.
$\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{1-\sqrt{1-\tan x}}{\sin x}$
01
### Substitute x is equal to 0
$= \,\,\,$ $\dfrac{1-\sqrt{1-\tan (0)}}{\sin (0)}$
$= \,\,\,$ $\dfrac{1-\sqrt{1-0}}{0}$
$= \,\,\,$ $\dfrac{1-\sqrt{1}}{0}$
$= \,\,\,$ $\dfrac{1-1}{0}$
$= \,\,\,$ $\dfrac{0}{0}$
The function become indeterminate when the value of $x$ approaches zero. So, the limit problem must be solved in another method.
02
### Apply Rationalizing method
Use rationalising method by multiplying both numerator and denominator of the function by the rationalising factor of the numerator.
$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{1-\sqrt{1-\tan x}}{\sin x}$ $\times$ $\dfrac{1+\sqrt{1-\tan x}}{1+\sqrt{1-\tan x}}$
$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{ (1-\sqrt{1-\tan x}) \times (1+\sqrt{1-\tan x}) }{ \sin x (1+\sqrt{1-\tan x})}$
$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{ {(1)}^2 -{(\sqrt{1-\tan x})}^2}{\sin x (1+\sqrt{1-\tan x})}$
$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{1 -(1-\tan x)}{\sin x (1+\sqrt{1-\tan x})}$
$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{1-1+\tan x}{\sin x (1+\sqrt{1-\tan x})}$
$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\require{cancel} \dfrac{\cancel{1}-\cancel{1}+\tan x}{\sin x (1+\sqrt{1-\tan x})}$
$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{\tan x}{\sin x (1+\sqrt{1-\tan x})}$
03
### Use Reciprocal identity of sine and cosine
There is a $\sin x$ term in denominator and it can be cancelled by expressing $\tan x$ in terms of ratio of $\sin x$ to $\cos x$ by the reciprocal identity of sine and cosine functions.
$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{\dfrac{\sin x}{\cos x}}{\sin x (1+\sqrt{1-\tan x})}$
$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{\sin x}{\sin x \times \cos x \times (1+\sqrt{1-\tan x})}$
$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\require{cancel} \dfrac{\cancel{\sin x}}{\cancel{\sin x} \times \cos x \times (1+\sqrt{1-\tan x})}$
$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{1}{\cos x (1+\sqrt{1-\tan x})}$
04
### Substitute x is equal to 0
Now substitute $x$ is equal to zero to evaluate the function when $x$ tends to zero.
$=\,\,\,$ $\dfrac{1}{\cos (0) (1+\sqrt{1-\tan (0)})}$
$=\,\,\,$ $\dfrac{1}{1 \times (1+\sqrt{1-0})}$
$=\,\,\,$ $\dfrac{1}{1 \times (1+\sqrt{1})}$
$=\,\,\,$ $\dfrac{1}{1 \times (1+1)}$
$=\,\,\,$ $\dfrac{1}{1 \times 2}$
$=\,\,\,$ $\dfrac{1}{2}$
It is the required solution for this limit problems of the calculus mathematics.
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# How to find an equivalent angle between $-\pi$ and $\pi$? [closed]
Let's say we have an angle such as $$270$$ degrees or $$-5892$$ degrees, or similarly in radians.
How do we convert it to its equivalent value between $$-\pi$$ and $$\pi$$?
• $-5892 = (-16 \times 360) + (-132).$ Then $~\displaystyle -132 \times \frac{\pi}{180} = \frac{-11}{15}\pi ~$ radians. May 16, 2022 at 12:01
Given $$\theta$$, you can find an equivalent angle $$\alpha = \theta + 2k\pi, \; k \in \mathbb{Z} \;$$ s.t. $$\;-\pi \le \alpha \le \pi$$.
$$k = \left\lfloor \frac{\pi - \theta}{2\pi} \right\rfloor$$ where $$\lfloor \cdot \rfloor$$ is the floor function.
Note: If $$\theta$$ is in degrees, it helps to convert it to radians first.
• @azerila I don't think so $-$ it's really just a $360$ degree rotation.
It helps to express the angle in radians first. $$1$$ radian is equal to $$\frac{180}{\pi}$$ degrees. Equivalently, 1 degree is equal to $$\frac{\pi}{180}$$ radians. Use this and express the given angle in radians. Next, recognise that the complete angle around a point is equal to $$2\pi$$ radians. Therefore, $$2\pi+1$$ radians is equivalent to $$1$$ radian. If the angle you so get is greater than or equal to $$\pi$$ radians, then you can express it as a negative angle in $$[-\pi,0]$$. Say you get $$x$$ radians as the answer. The same angle expressed as a negative angle will be $$-(2\pi-x)$$radians. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Newton's Third Law
## For every action, there is an equal and opposite reaction.
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Newton's Third Law
Students will learn that Newton's 3rd law holds that for every force there is an equal but opposite reaction force. Important to note that the force and reaction force act on different objects.
### Key Equations
F=F\begin{align*} \vec{F} = - \vec{F'} \end{align*}
Guidance
Newton’s 3rd\begin{align*}3^{rd}\end{align*} Law states for every force there is an equal but opposite reaction force. To distinguish a third law pair from merely oppositely directed pairs is difficult, but very important. Third law pairs must obey three rules: (1) Third law force pairs must be of the same type of force. (2) Third law force pairs are exerted on two different objects. (3) Third law force pairs are equal in magnitude and oppositely directed. Example: A block sits on a table. The Earth’s gravity on the block and the force of the table on the block are equal and opposite. But these are not third law pairs, because they are both on the same object and the forces are of different types. The proper third law pairs are: (1) earth’s gravity on block/block’s gravity on earth and (2) table pushes on block/ block pushes on table.
#### Example 1
Question: Tom and Mary are standing on identical skateboards. Tom and Mary push off of each other and travel in opposite directions.
a) If Tom (M)\begin{align*}(M)\end{align*} and Mary (m)\begin{align*}(m)\end{align*} have identical masses, who travels farther?
b) If Tom has a bigger mass than Mary, who goes farther?
c) If Tom and Mary have identical masses and Tom pushes twice as hard as Mary, who goes farther?
Solution
a) Neither. Both Tom and Mary will travel the same distance. The forced applied to each person is the same (Newton's Third Law). So Ma=ma\begin{align*} \cancel{M}a=\cancel{m}{a} \end{align*} which cancels to a=a\begin{align*} a=a \end{align*} Therefore both people will travel the same distance because the acceleration controls how far someone will travel and Tom and Mary have equal acceleration.
b) Mary will go farther. Again, the same force is applied to both Mary and Tom so Ma=ma\begin{align*} Ma=ma \end{align*} Since Tom has the larger mass, his acceleration must be smaller (acceleration and mass are inversely proportional). Finally, because Mary's acceleration is greater, she will travel farther.
c) Neither. Newton's Third Law states that for every action there is an equal and opposite reaction. Therefore if Tom pushes twice as hard as Mary, Mary will essentially be pushing back with the same strength. They will therefore travel the same distance.
### Time for Practice
Refer to the Newtons Law problems section of this chapter for practice.
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By: mkkosariya (111 month(s) ago)
very good sir i want to save this ppt sir
## Presentation Transcript
### Connecting addition to multiplication :
Connecting addition to multiplication Instructional Activity By: Melissa Merritt
### Objective :
Objective To make connections between addition and multiplication I know you all know how to ADD! Multiplication is just REPEATED addition!
### Lets try a problem together! :
Lets try a problem together! How many birds are there in all?
### You got it! There are 12 birds! :
You got it! There are 12 birds! 6 + 6 = 12
### Now lets try a different way! :
Now lets try a different way! There are 2 groups and each group has 6 birds! How many in each group? How many groups? 6 2
### It’s EASY :
It’s EASY How many groups? How many in each group? _______ x ______ 2 6 2 x 6 = 12
### Review time! :
Review time! AdditionMultiplication 6 + 6 = 12 2 x 6 = 12 6 twos equals 12 2 times 6 equals 12
### Lets try a new problem! :
Lets try a new problem! How many apples are in all?
### You guessed it! There are 12 apples! :
You guessed it! There are 12 apples!
### Now lets try a different way! :
Now lets try a different way! There are 3 groups and each group has 4 apples! How many in each group? How many groups? 4 3
### It’s EASY :
It’s EASY How many groups? How many in each group? _______ x ______ 4 3 4 x 3 = 12
### Review time! :
Review time! AdditionMultiplication 4 + 4 + 4 = 12 4 x 3 = 12 3 fours equals 12 3 times 4 equals 12
### Remember that… :
Remember that… Multiplication is just repeated addition! Don’t forget: To solve the problems both ways! Make groups! Check your work! |
# If y varies inversely with x, and y= 6 when x= 18, how do you find y when x= 8?
Jul 9, 2016
$y = 13.5$
#### Explanation:
y prop 1/x or y = k*1/x or x*y=k; y=6 ; x=18 :. 6*18=k or k=108 k= constant of variation. So the inverse variation equation is $x \cdot y = 108$
Now $x = 8 \therefore 8 \cdot y = 108 \mathmr{and} y = \frac{108}{8} = 13.5$[Ans]
Jul 9, 2016
$y = \frac{13}{5}$
#### Explanation:
Inverse variations look like this: $y = \frac{k}{x}$
We need to make our own inverse variation given $y = 6$ and $x = 18$. We can plug this into the standard inverse variation with variables.
$y = \frac{k}{x}$
$6 = \frac{k}{18}$
$108 = k$
We've determined that $k = 108$. Now we have what it takes to create an inverse variation. This is our final equation:
$y = \frac{108}{x}$
We still need to find what $y$ equals when $x$ is $8$. Let's plug in $8$ for $x$.
$y = \frac{108}{x}$
$y = \frac{108}{8}$
$y = 13.5$ |
# You asked: What is the probability of rolling a 7 with a pair of fair dice?
Contents
Total Number of combinations Probability
7 6 16.67%
8 5 13.89%
9 4 11.11%
10 3 8.33%
## What is the probability of getting a 7 on the first throw of two dice and an 11 on the second throw?
2 Answers. The probability is 25% .
## What is the probability of not rolling a sum of 7 with two fair dice?
Explanation: When rolling 2 dice there are 36 possible outcomes. [to see this imagine one die is red and the other green; there are 6 possible outcomes for the red die and for each of these red outcomes there are 6 possible green outcomes]. That is 30 out of 36 outcomes will not be a 7 total.
## What is the probability of getting a 7 or 11 if a pair of dice is tossed at once?
So, the total is 8 pairs that would give either a sum of 7 or 11. 8/36 or 2/9, 22.22%. Nearly a 1 in 4 chance of getting 7 or 11.
## What is the probability of rolling a 7?
Probabilities for the two dice
THIS IS IMPORTANT: Is the lottery a sacrifice?
Total Number of combinations Probability
4 3 8.33%
5 4 11.11%
6 5 13.89%
7 6 16.67%
## What is the probability of rolling a 7 or 11?
The probability of winning on the first roll is the probability of rolling 7 or 11, which is 1/6 plus 1/18, which equals to 2/9. Suppose we roll 4 on the first roll (the probability of rolling 4 is 1/12).
## What is the probability of not rolling a 7?
The probability of not rolling a 7 on any one roll is 5/6.
## What’s the probability of rolling a 2?
Two (6-sided) dice roll probability table
Roll a… Probability
2 1/36 (2.778%)
3 2/36 (5.556%)
4 3/36 (8.333%)
5 4/36 (11.111%)
## How many ways can you roll 7?
Since there are 6 ways to get 7 and two ways to get 11, the answer is 6+2=8.
## How many ways can you roll a 7 with 3 dice?
This gives 15 distinct ways of obtaining a total of 7. There are 216 possible outcomes when rolling three dice. Originally Answered: If we roll 3 dice, what is the probability that the sum is 7?
## How many ways can you roll 6 with 2 dice?
How many total combinations are possible from rolling two dice? Since each die has 6 values, there are 6∗6=36 6 ∗ 6 = 36 total combinations we could get. |
# 2022 AMC 10B Problems/Problem 23
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The following problem is from both the 2022 AMC 10B #23 and 2022 AMC 12B #22, so both problems redirect to this page.
## Problem
Ant Amelia starts on the number line at $0$ and crawls in the following manner. For $n=1,2,3,$ Amelia chooses a time duration $t_n$ and an increment $x_n$ independently and uniformly at random from the interval $(0,1).$ During the $n$th step of the process, Amelia moves $x_n$ units in the positive direction, using up $t_n$ minutes. If the total elapsed time has exceeded $1$ minute during the $n$th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most $3$ steps in all. What is the probability that Amelia’s position when she stops will be greater than $1$?
$\textbf{(A) }\frac{1}{3} \qquad \textbf{(B) }\frac{1}{2} \qquad \textbf{(C) }\frac{2}{3} \qquad \textbf{(D) }\frac{3}{4} \qquad \textbf{(E) }\frac{5}{6}$
## Solution 1 (Geometric Probability)
Let $x$ and $y$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that $$P(x+y\leq 1)=\frac{\frac12\cdot1^2}{1^2}=\frac12,$$ as shown below: $[asy] /* Made by MRENTHUSIASM */ size(200); real xMin = -0.25; real xMax = 1.25; real yMin = -0.25; real yMax = 1.25; //Draws the horizontal ticks void horizontalTicks() { for (real i = 1; i < yMax; ++i) { draw((-1/32,i)--(1/32,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = 1; i < xMax; ++i) { draw((i,-1/32)--(i,1/32), black+linewidth(1)); } } horizontalTicks(); verticalTicks(); label("0",(0,0),2*SW); label("1",(1,0),2*S); label("1",(0,1),2*W); fill((0,0)--(1,0)--(0,1)--cycle,yellow); draw((0,1)--(1,1)^^(1,0)--(1,1),dashed); draw((0,1)--(1,0)); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(8)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(8)); label("x",(xMax,0),(2,0)); label("y",(0,yMax),(0,2)); [/asy]$ Let $x,y,$ and $z$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that $$P(x+y+z\leq 1)=\frac{\frac13\cdot\left(\frac12\cdot1^2\right)\cdot1}{1^3}=\frac16,$$ as shown below: $[asy] /* Made by MRENTHUSIASM */ size(200); import graph3; import solids; currentprojection=orthographic((0.3,0.1,0.1)); draw(surface((1,0,0)--(0,1,0)--(0,0,1)--cycle),yellow); draw(surface((1,0,0)--(0,1,0)--(0,0,0)--cycle),yellow); draw(surface((1,0,0)--(0,0,1)--(0,0,0)--cycle),yellow); draw(surface((0,1,0)--(0,0,1)--(0,0,0)--cycle),yellow); draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--cycle,dashed); draw((0,1,0)--(1,1,0)--(1,0,0),dashed); draw((0,1,1)--(0,1,0)^^(1,1,1)--(1,1,0)^^(1,0,1)--(1,0,0),dashed); draw((-0.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10)); draw((0,-0.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10)); draw((0,0,-0.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10)); draw((-0.1,0,1)--(0.1,0,1),linewidth(1)); draw((0,1,-0.1)--(0,1,0.1),linewidth(1)); draw((1,-0.1,0)--(1,0.1,0),linewidth(1)); label("x",(1.5,0,0),4*dir((1.5,0,0))); label("y",(0,1.5,0),2*dir((0,1.5,0))); label("z",(0,0,1.5),2*dir((0,0,1.5))); label("0",(0,0,0),2*dir((0,0.5,-0.5))); label("1",(1,0,0),4*dir((0,-1,0))); label("1",(0,1,0),4*dir((0,0,-1))); label("1",(0,0,1),5*dir((-1,0,0))); draw((1,0,0)--(0,1,0)--(0,0,1)--cycle); [/asy]$ We have two cases:
1. Amelia takes exactly $2$ steps.
2. We need $x_1+x_2>1$ and $t_1+t_2>1.$ So, the probability is $$P(x_1+x_2>1)\cdot P(t_1+t_2>1)=\left(1-\frac12\right)\cdot\left(1-\frac12\right)=\frac14.$$
3. Amelia takes exactly $3$ steps.
4. We need $x_1+x_2+x_3>1$ and $t_1+t_2\leq1.$ So, the probability is $$P(x_1+x_2+x_3>1)\cdot P(t_1+t_2\leq1)=\left(1-\frac16\right)\cdot\frac12=\frac{5}{12}.$$
Together, the answer is $\frac14 + \frac{5}{12} = \boxed{\textbf{(C) }\frac{2}{3}}.$
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~MRENTHUSIASM
## Solution 2 (Generalization and Induction)
We can in fact find the probability that any number of randomly distributed numbers on the interval $[0, 1]$ sum to more than $1$ using geometric probability, as shown in the video below.
If we graph the points that satisfy $x + y < 1$, $0 < x, y < 1$, we get the triangle with points $(0, 0)$, $(1, 0)$, and $(0, 1)$. If we graph the points that satisfy $x + y + z < 1$, $0 < x, y, z < 1$, we get the tetrahedron with points $(0, 0, 0)$, $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$.
Of course, the probability of either of these cases happening is simply the area/volume of the points we graphed divided by the total area of the graph, which is always $1$ (this would be much simpler than my calculus proof above).
Thus, we can now solve for the probability that the sum is less than one for $n$ numbers using induction.
$\textbf{Claim:}$ The probability that the sum is less than one is $\frac{1}{n!}$.
$\textbf{Base Case:}$ For just $1$ number, the probability is $1$.
$\textbf{Induction step:}$ Suppose that the probability for $n$ numbers is $\frac{1}{n!}$. We will prove that the probability for $n+1$ numbers is $\frac{1}{(n+1)!}$. To prove this, we consider that the area of an $n+1$-dimensional tetrahedron is simply the area/volume of the base times the height divided by $n+1$.
Of course, the area of the base is $\frac{1}{n!}$, and the height is $1$, and thus, we obtain $\frac{1}{n! \cdot (n+1)} = \frac{1}{(n+1)!}$ as our volume (this may be hard to visualize for higher dimensions). The induction step is complete.
The probability of the sum being less than $1$ is $\frac{1}{n!}$, and the probability of the sum being more than $1$ is $\frac{n!-1}{n!}$. This trivializes the problem. The answer is $$\frac{1}{2} \cdot \frac{2! - 1}{2!} + \frac{1}{2} \cdot \frac{3! - 1}{3!} = \boxed{\textbf{(C) }\frac{2}{3}}.$$
~mathboy100
## Solution 3 (Observations)
There are two cases: Amelia takes two steps or three steps.
The former case has a probability of $\frac{1}{2}$, as stated in Solution 1, and thus the latter also has a probability of $\frac{1}{2}$.
The probability that Amelia passes $1$ after two steps is also $\frac{1}{2}$, as it is symmetric to the probability above.
Thus, if the probability that Amelia passes $1$ after three steps is $x$, our total probability is $\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot x$. We know that $0 < x < 1$, and it is relatively obvious that $x > \frac{1}{2}$ (because the probability that $x > \frac{3}{2}$ is $\frac{1}{2}$). This means that our total probability is between $\frac{1}{2}$ and $\frac{3}{4}$, non-inclusive, so the only answer choice that fits is $\boxed{\textbf{(C) }\frac{2}{3}}$.
~mathboy100
## Remark (Calculus)
It is not immediately clear why three random numbers between $0$ and $1$ have a probability of $\frac{5}{6}$ of summing to more than $1$. Here is a proof:
Let us start by finding the probability that two random numbers between $0$ and $1$ have a sum of more than $x$, where $0 \leq x \leq 1$.
Suppose that our two numbers are $y$ and $z$. Then, the probability that $y > x$ (which means that $y + z > x$) is $1 - x$, and the probability that $y < x$ is $x$.
If $y < x$, the probability that $y + z > x$ is $1 - x + y$. This is because the probability that $y + z < x$ is equal to the probability that $z < x - y$, which is $x - y$, so our total probability is $1 - (x - y) = 1 - x + y$.
Let us now find the average of the probability that $y + z > x$ when $y < x$. Since $y$ is a random number between $0$ and $x$, its average is $\frac{x}{2}$. Thus, our average is $1 - x + \frac{1}{2} = 1 - \frac{x}{2}$.
Hence, our total probability is equal to $$1(1-x) + \left(1 - \frac{x}{2}\right)(x) = 1 - \frac{1}{2}x^2.$$ Now, let us find the probability that three numbers uniformly distributed between $0$ and $1$ sum to more than $1$.
Let our three numbers be $a$, $b$, and $c$. Then, the probability that $a + b + c > 1$ is equal to the probability that $b + c$ is greater than $1 - a$, which is equal to $1 - \frac{1}{2}(1 - a)^2$.
To find the total probability, we must average over all values of $a$. This average is simply equal to the area under the curve $1 - \frac{1}{2}(1-x)^2$ from $0$ to $1$, all divided by $1$. We can compute this value using integrals: \begin{align*} \frac{\int_0^1 \! 1 - \frac{1}{2}(1 - x)^2 \mathrm{d}x}{1} &= \int_0^1 \! 1 - \frac{1}{2}(1 - x)^2 \mathrm{d}x \\ &= 1 - \frac{1}{2}\int_0^1 \! (1 - x)^2 \mathrm{d}x \\ &= 1 - \frac{1}{2}\int_0^1 \! x^2 \mathrm{d}x \\ &= 1 - \frac{1}{2}\left(\frac{1}{3}\right) \\ &= \frac{5}{6}. \end{align*} For those who don't know calculus, $\int_m^n \! f(x) \mathrm{d}x$ is the area under the curve $f(x)$ from $m$ to $n$.
~mathboy100
~ pi_is_3.14
## Video Solution
~ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~Interstigation |
# Difference between revisions of "The Mathematical Papyri"
## The Rhind Mathematical Papyrus
This papyrus is sometimes called the Ahmes papyrys. Written by a scribe named Ahmes (Ahmose) in ca. 1550 BCE and bought in modern times by Charles Henry Rhind. The papyrus lists many problems covering topics in algebra, geometry and other areas in mathematics.
### Problems from the Ahmes (or Rhind) Papyrus
The Rhind mathematical papyrus
Some of the following examples of the problems presented come from [Egyptian Mathematics] a website by G. Donald Allen, Professor at Texas A&M University, College Station. Others come from Corinna Rossi's book [1]
Definition: The unknown, x , is called the heep.
Problem 24. Find the heep if the heep and a seventh of the heep is 19.
Problem 48 This problem shows how the formula from problem 50 is arrived at.
Problem 50. A circular field of diameter 9 has the same area as a square of side 8.
Problem 56 This problem indicates an understanding of the idea of geometric similarity. This problem discusses the ratio $\frac{rise}{run}$ The problem essentially asks to compute the cotangent for some angle . Such a formula would be need for building pyramids.
Problem 57. The height of a pyramid is calculated from the base length and the seked (egyptian for slope).
Problems 58, 59 and 60. These problems, together with 56 and 57, contain calculations involving the seked of pyramids.
Problem 63. 700 loaves are to be divided among recipients where the amounts they are to receive are in the continued proportion $\frac{2}{3} : \frac{1}{2} : \frac{1}{3} : \frac{1}{4}$
Definition: The strength is given by the following formula $strength = \frac{1}{grain \ density}$
Problem 72. How many loaves of strength 45 are equivalent to 100 loaves of strength 10?
Problem 79. This problem cites only seven houses, 49 cats, 343 mice, 2401 ears of spelt, 16,807 hekats."
## The Moscow Papyrus
The Moscow mathematical papyrus
The Moscow Mathematical Papyrus is also called the Golenischev Mathematical Papyrus, after its first owner, Egyptologist Vladimir Goleniščev. The Papyrus dates to the Middle Kingdom, ca. 1700 BCE. The Moscow papyrus is smaller than the Rhind papyrus. The most often quoted problem is problem 14. In this problem the volume of a truncated pyramid is computed.
Problem 14. The base is a square of side 4 cubits, the top is a square of side 2 cubits and the height of the truncated pyramid is 6 cubits.
## Other Mathematical sources
• Berlin Papyrus
• Egyptian Mathematical Leather Roll
• Kahun Papyri
## References
1. C. Rossi Architecture and Mathematics in Ancient Egypt Cambridge University Press 2004 |
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