The time taken by Millie to catch up with Ben is 6 minutes, solved using the linear equation in one variable and the unit rates provided.
We assume the minutes taken by Millie to catch up with Ben to be x minutes.
The number of words Ben has already typed = 180.
The unit rate for Ben for the typing of words = 30 per minute.
Therefore, words typed by Ben in x minutes = 30x words.
Therefore, the total number of words typed by Ben = 180 + 30x words.
The number of words Millie has already typed = 60.
The unit rate for Millie for the typing of words = 50 per minute.
Therefore, words typed by Millie in x minutes = 50x words.
Therefore, the total number of words typed by Millie = 60 + 50x words.
As Millie needs to catch up with Ben after x minutes, the total number of words typed by them needs to be equal, that is, 60 + 50x = 180 + 30x, which is the required linear equation in one variable.
To find the minutes taken by Millie to catch up with Ben, we solve this linear equation in one variable as follows:
60 + 50x = 180 + 30x,
or, 60 + 50x - 30x = 180 + 30x - 30x {Subtracting 30x from both sides},
or, 60 + 20x = 180 {Simplifying},
or, 60 + 20x - 60 = 180 - 60 {Subtracting 60 from both sides},
or, 20x = 120 {Simplifying},
or, 20x/20 = 120/20 {Dividing both sides by 20},
or, x = 6 {Simplifying}.
Thus, the time taken by Millie to catch up with Ben is 6 minutes, solved using the linear equation in one variable and the unit rates provided.
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The number of pollinated flowers as a function of time in days can be represented by the function. f(x)
The average increase in the number of flowers pollinated per day between days 4 and 10 is 39, given that the number of pollinated flowers as a function of time in days can be represented by the function [tex]f(x) = (3)^{\frac{x}{2} }[/tex].
In the question, we are asked for the average increase in the number of flowers pollinated per day between days 4 and 10, given that the number of pollinated flowers as a function of time in days can be represented by the function [tex]f(x) = (3)^{\frac{x}{2} }[/tex].
To find the average increase in the number of flowers pollinated per day between days 4 and 10, we use the formula {f(10) - f(4)}/{10 - 4}.
First, we find the value of the function [tex]f(x) = (3)^{\frac{x}{2} }[/tex], for f(10) and f(4).
[tex]f(x) = (3)^{\frac{x}{2} }\\\Rightarrow f(10) = (3)^{\frac{10}{2} }\\\Rightarrow f(10) = 3^5 = 243[/tex]
[tex]f(x) = (3)^{\frac{x}{2} }\\\Rightarrow f(4) = (3)^{\frac{4}{2} }\\\Rightarrow f(10) = 3^2 = 9[/tex]
Thus, the average increase
= {f(10) - f(4)}/{10 - 4},
= (243 - 9)/(10 - 4),
= 234/6
= 39.
Thus, the average increase in the number of flowers pollinated per day between days 4 and 10 is 39, given that the number of pollinated flowers as a function of time in days can be represented by the function [tex]f(x) = (3)^{\frac{x}{2} }[/tex].
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For complete question, refer to the attachment.
Out of 500 people in a room, 200 of them are wearing red shirts. Of the people wearing red shirts, 15% are wearing red pants. What percentage of the people in the room are wearing both red shirts and red pants?
Light travels at a speed of approximately Light travels at a speed of approximately 1,000,000,000 kilometers per second (621,118,012 miles per second). 80,500 kilometers per hour (50,000 mph). 300,000 kilometers per hour (186,336 mph). 300,000 kilometers per second (186,333 miles per second).
The speed of light is 300,000 kilometers per hour. hence option c is correct.
According to the statement
we have to explain that the by how much speed the light travels.
So, Firstly
Light is the natural agent that stimulates sight and makes things visible.
and speed of light means that the distance covered by the light with respect to time.
So, it means those distance which is covered by the light with respect to time is called the speed of light.
The speed of light is measures in km, meters and in other units also.
And we can calculate the speed of light per hour or per minute or in other time measurements.
So, The speed of light is 300,000 kilometers per hour. hence option c is correct.
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d all the time and I like the time you
Answer:
Step-by-step explanation:
1a+1b(a+b-c)+1b+1c(b=c-a)+ 1a+1c(c+a-b)
1a+1b-c+1b +1c-a+1a+1c-b
1a+2b-c+a+1c-b
2a+1b+c this is the answer I think lol
A potato was launched in the air using a potato gun. the function for the situation was h(t) = -16t2 + 100t + 10, where t was the time in seconds and h(t) was the height in feet. (3 points each) a. what was the maximum height of the potato? b. what was the lowest height of the potato? c. what was the lowest time that applies to the actual situation? d. what is the greatest time that applied to the actual situation? e. using your answers to parts a through d, what are the domain and range of the function, as they apply to this situation? use set builder notation.
The answers to all the parts are given below.
To find the answers using a function:The height equation is:
h(t) = - 16t^2 + 100t + 10
A) To find the maximum height, we must find the time where the velocity is 0, so we must derive it with respect to the time.
v(t) = - 2 * 16 * t + 100 = 0
t = 100/(2*16) = 3.125s
Now we put this time in our height equation:
h(3.125s) = -16*3.125^2 + 100*3.125 + 10 = 166.25 ft.
B) The lowest height of the potato will be h = 0ft when the potato hits the ground.
C) the lowest time that works for that function is t = 0s when the potato is fired by the gun.
D) the maximum time can be found when the potato hits the ground, after that point the equation does not work anymore, let's find it,
h(t) = 0 = -16t^2 +100t+10
we can solve it using Bhaskara's equation:
[tex]t=\frac{-100+-\sqrt{100^{2} +4*16*10} }{-2*16} =\frac{-100+-103.2}{-32}[/tex]
So the two solutions are:
t = 6.3s
t = -0.1s
We need to choose the positive time, as we already discussed that the minimum time that works for the equation is t = 0s.
so here the answer is t = 6.3s
E) the domain is 0s ≤ t ≤ 6.3s
The range is 0ft ≤ h ≤ 166.25 ft.
Therefore, the answers to all the parts are shown.
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Please Help me with this problem!!! ASAP
The average rate of change is -5 per month
How to determine the average rate of change?The interval is given as:
July to December
From the graph, we have:
July = 110
December = 30
The average rate of a function over the interval (a, b) is calculated as:
Rate = [f(b) - f(a)]/[b - a]
So, we have:
Rate = (30 - 110)/(December- July)
This gives
Rate = (30 - 110)/(12 - 7)
Evaluate
Rate = -16
Hence, the average rate of change is -5 per month
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The function f(x) = (0.2)x
O increases for x > 0
O increases for all x
O decreases for all x
O decreases for x > 0
Answer:
Option (3)
Step-by-step explanation:
Since the base of the exponential is less than 1, the function is decreasing for all x.
I need help I'm stuck on this
a) The length of p is 21.4 cm
b) The measure of angle P is 45.6°
TrigonometryFrom the question, we are to determine the length of p
From the Pythagorean theorem, we can write that
30² = 21² + p²
p² = 30² - 21²
p² = 900 - 441
p² = 459
p = √459
p = 21.4 cm
∴ The length of p is 21.4 cm
b)
Measure of angle P
Using SOH CAH TOA
[tex]cos P= \frac{21}{30}[/tex]
cos P = 0.7
P = cos⁻¹(0.7)
P = 45.6°
Hence, the measure of angle P is 45.6°
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Tatiana has a special puzzle in which all of the pieces fit together in any way. There is no goal picture. Instead, the goal of the puzzle is to make different patterns and pictures using the pieces. If Tatiana has 50 unique puzzle pieces and she plans to use all of them, how many possible pictures can she create
It is an example of Permutation without repetition.
To find the number of pictures can create.
what is permutation and combination?An arrangement of things in a certain sequence is what we refer to as a permutation. The constituents or components of sets are presented in this section in a sequential or chronological fashion. A coming together or merging of several parts or characteristics in which each of the constituent parts or traits retains its unique identity.
Given that:
50 unique pieces of puzzle
Use them all to solve puzzle
No repetition is allowed therefore
Since no repetition is allowed because all puzzle pieces are required to be used.
Therefore each of the 50 pieces can be used max one time
Therefore if we start with piece like this " 1 2 3 .. up to ... 50 " it is one combination
Another could be "2 1 3 .. up to .. 50" or "3 1 2 .. up to 50" and so on
So we see that we can see that we can select the first piece in 50 ways 49 left which could be selected for second piece then 48, 47 respectively
Therefore 50X49X48X47X46X....X1 = 50!
Hence, It is an example of Permutation without repetition.
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Option is missing
A. Combination with repetition
B. Permutation with repetition
C. Combination without repetition
D. Permutation without repetition
A scale drawing of a house addition shows a scale factor of 1 in. = 3.3 ft. Josh decides to make the house addition smaller, and he changes the scale of the drawing to 1 in. = 1.1 ft.
What is the change in the scale factor from the old scale to the new scale?
The change in the scale factor from the old scale to the new scale is 3.
What is the change in the scale factor?A scale drawing is a reduced form in terms of dimensions of an original image / building / object. The scale drawing is usually reduced at a constant dimension.
The change in the scale factor = larger scale / smaller scale
3.3 / 1.1 = 3
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Solve for B.
R = x(A + B)
PLS HELP!!!!!!!!!!!!! math related
Answer: B
Step-by-step explanation:
When you evaluate f(-1)=2x^4+x^2-5 we will get -2 in basic notation (-1,-2). We need to find the root of the function to find the other roots or factors, where y=0.
What is the solution to |x – 5| + 2 < 20?
Answer:
Step-by-step explanation:
hello :
|x – 5| + 2 < 20 means : |x – 5| + 2-2 < 20-2
|x – 5| < 18
-18< x - 5 < 18
-18+5< x - 5+5 < 18+5
-13< x< 23 ( solutions)
I legit forgot how to do this I need help with this
Answer: 3/4
Step-by-step explanation:
Tangent is opp/adj
Therefore tan (x) = 18/24, which reduces to 3/4
Proof : tan10 x tan20 x tan…70 x tan80=1
This follows from the identity
[tex]\cot(x) = \dfrac1{\tan(x)} = \tan(90^\circ - x)[/tex]
The overall product is 1 because
[tex]\tan(10^\circ) \tan(80^\circ) = \cot(80^\circ) \tan(80^\circ) = 1[/tex]
[tex]\tan(20^\circ) \tan(70^\circ) = \cot(70^\circ) \tan(70^\circ) = 1[/tex]
[tex]\tan(30^\circ) \tan(60^\circ) = \cot(60^\circ) \tan(60^\circ) = 1[/tex]
[tex]\tan(40^\circ) \tan(50^\circ) = \cot(50^\circ) \tan(50^\circ) = 1[/tex]
Find a linear inequality with the following solution set. Each grid line represents one unit. [asy] size(200); fill((-2,-5)--(5,-5)--(5,5)--(3,5)--cycle,yellow); real ticklen=3; real tickspace=2; real ticklength=0.1cm; real axisarrowsize=0.14cm; pen axispen=black+1.3bp; real vectorarrowsize=0.2cm; real tickdown=-0.5; real tickdownlength=-0.15inch; real tickdownbase=0.3; real wholetickdown=tickdown; void rr_cartesian_axes(real xleft, real xright, real ybottom, real ytop, real xstep=1, real ystep=1, bool useticks=false, bool complexplane=false, bool usegrid=true) { import graph; real i; if(complexplane) { label("$\textnormal{Re}$",(xright,0),SE); label("$\textnormal{Im}$",(0,ytop),NW); } else { label("$x$",(xright+0.4,-0.5)); label("$y$",(-0.5,ytop+0.2)); } ylimits(ybottom,ytop); xlimits( xleft, xright); real[] TicksArrx,TicksArry; for(i=xleft+xstep; i 0.1) { TicksArrx.push(i); } } for(i=ybottom+ystep; i 0.1) { TicksArry.push(i); } } if(usegrid) { xaxis(BottomTop(extend=false), Ticks("%", TicksArrx ,pTick=gray(0.1),extend=true),p=invisible);//,above=true); yaxis(LeftRight(extend=false),Ticks("%", TicksArry ,pTick=gray(0.1),extend=true), p=invisible);//,Arrows); } if(useticks) { xequals(0, ymin=ybottom, ymax=ytop, p=black, Ticks("%",TicksArry , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize)); yequals(0, xmin=xleft, xmax=xright, p=black, Ticks("%",TicksArrx , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize)); } else { xequals(0, ymin=ybottom, ymax=ytop, p=axispen, above=true, Arrows(size=axisarrowsize)); yequals(0, xmin=xleft, xmax=xright, p=axispen, above=true, Arrows(size=axisarrowsize)); } }; draw((-2,-5)--(3,5),dashed+red, Arrows(size=axisarrowsize)); rr_cartesian_axes(-5,5,-5,5); f
The linear inequality of the graph is: -x + 2y + 1 > 0
How to determine the linear inequality?First, we calculate the slope of the dashed line using:
[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
Two points on the graph are:
(1, 0) and (3, 1)
The slope (m) is:
[tex]m = \frac{1 - 0}{3 - 1}[/tex]
This gives
m = 0.5
The equation of the line is calculated as:
[tex]y = m(x -x_1) + y_1[/tex]
So, we have;
[tex]y = 0.5(x -1) + 0[/tex]
This gives
[tex]y = 0.5x -0.5[/tex]
Multiply through by 2
[tex]2y = x - 1[/tex]
Now, we convert the equation to an inequality.
The line on the graph is a dashed line. This means that the inequality is either > or <.
Also, the upper region of the graph that is shaded means that the inequality is >.
So, the equation becomes
2y > x - 1
Rewrite as:
-x + 2y + 1 > 0
So, the linear inequality is: -x + 2y + 1 > 0
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Complete question
Find a linear inequality with the following solution set. Each grid line represents one unit. (Give your answer in the form ax+by+c>0 or ax+by+c [tex]\geq[/tex] 0 where a, b, and c are integers with no common factor greater than 1.)
100 POINTS PLEASE HELP!!!
The coordinate plane below represents a community. Points A through F are houses in the community.
graph of coordinate plane. Point A is at negative 5, 5. Point B is at negative 4, negative 2. Point C is at 2, 1. Point D is at negative 2, 4. Point E is at 2, 4. Point F is at 3, negative 4.
Part A: Using the graph above, create a system of inequalities that only contains points C and F in the overlapping shaded regions. Explain how the lines will be graphed and shaded on the coordinate grid above. (7 points)
Part B: Explain how to verify that the points C and F are solutions to the system of inequalities created in Part A. (5 points)
Part C: Erica wants to live in the area defined by y < 7x − 4. Explain how you can identify the houses in which Erica is interested in living. (2 points)
Answer:
[tex]\sf A) \quad \begin{cases}\sf y > -5x+5\\\sf y < -5x+12\end{cases}[/tex]
B) see below
C) points C, E and F
Step-by-step explanation:
Given points:
A = (-5, 5)B = (-4, -2)C = (2, 1)D = (-2, 4)E = (2, 4)F = (3, -4)Part AA system of inequalities is a set of two or more inequalities in one or more variables.
To create a system of inequalities that only contains C and F in the overlapping shaded region, create a linear equation where points C, F and E are to the right of the line and a linear equation where points C, F, A, B and D are to the left of the line.
The easiest way to do this is to find the slope of the line that passes through points C and F, then add values to move the lines either side of the points.
[tex]\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{y_F-y_C}{x_F-x_C}=\dfrac{-4-1}{3-2}=-5[/tex]
Therefore:
[tex]\sf y = -5x + 5[/tex] → points C, F and E are to the right of the line.
[tex]\sf y=-5x+12[/tex] → points C, F, A, B and D are the left of the line.
Therefore, the system of inequalities that only contains points C and F in the overlapping shaded regions is:
[tex]\begin{cases}\sf y > -5x+5\\\sf y < -5x+12\end{cases}[/tex]
To graph the system of inequalities:
Plot 2 points on each of the lines.Draw a dashed line through each pairs of points.Shade the intersected region that is above the line y > -5x + 5 and below the line y < -5x + 12.Part BTo verify that the points C and F are solutions to the system of inequalities created in Part A, substitute the x-values of both points into the system of inequalities. If the y-values satisfy both inequalities, then the points are solutions to the system.
Point C (2, 1)
[tex]\implies \sf x=2 \implies 1 > -5(2)+5 \implies 1 > -5\quad verified[/tex]
[tex]\implies \sf x=2 \implies 1 < -5(2)+12\implies 1 < 2 \quad verified[/tex]
Point F (3, -4)
[tex]\implies \sf x=3 \implies -4 > -5(3)+5 \implies -4 > -10\quad verified[/tex]
[tex]\implies \sf x=3 \implies -4 < -5(3)+12\implies -4 < -3 \quad verified[/tex]
Part CMethod 1
Graph the line y = 7x - 4 (making the line dashed since it is y < 7x - 4).
Shade below the dashed line.
Points that are contained in the shaded region are the houses in which Erica is interested in living: points C, E and F.
Method 2
Substitute the x-value of each point into the given inequality y < 7x - 4.
Any point where the y-value satisfies the inequality is a house that Erica is interested in living.
[tex]\sf Point\: A: \quad x=-5 \implies 5 < 7(-5)-4 \implies -5 < -39 \implies no[/tex]
[tex]\sf Point\: B: \quad x=-4 \implies -2 < 7(-4)-4 \implies -2 < -32 \implies no[/tex]
[tex]\sf Point\: C: \quad x=2 \implies 1 < 7(2)-4 \implies 1 < 10 \implies yes[/tex]
[tex]\sf Point\: D: \quad x=-2 \implies 4 < 7(-2)-4 \implies 4 < -18 \implies no[/tex]
[tex]\sf Point\: E: \quad x=2 \implies 4 < 7(2)-4 \implies 4 < 10 \implies yes[/tex]
[tex]\sf Point\: F: \quad x=3 \implies -4 < 7(3)-4 \implies -4 < 17 \implies yes[/tex]
( will give brainlyest)
Write an expression equivalent to the following problems using the fewest number of terms. ( number 1)
Amira's mother was in her garden, tying tomato plants to wooden stakes: it took her 10 minutes to stake each plant and she placed the plants 2 feet apart. IF she started at one end of the row at 8:00 am, how far from the first plant was the tomato plant she finished tying up at 9:00 am?
Answer:
12ft
Step-by-step explanation:
It takes 10 minutes for each plant and she spent 1 hour in total or 60 minutes: 60/10 = 6 plants
If each of the plants is 2 feet apart and there are 6 plants: 2ft*6 = 12ft
PLEASE HELPPPPP WILL GIVE BRAINLIEST
consider this equation
5/8x = 1/2x + 2
generate a plan to solve for the variable describe the steps you’ll use
Answer:
the value of the variable x is 16
Step-by-step explanation:
[tex]given \: expression \\ \frac{5}{8}x = \frac{1}{2} x + 2 \\ substract \: \frac{1}{2} x from \: both \: sides \\ \frac{5}{8}x - \frac{1}{2} x = \frac{1}{2} x - \frac{1}{2} x + 2 - \frac{1}{2} x \\ factor \: out \: common \: x \\ x( \frac{5}{8} - \frac{1}{2} ) = \frac{1}{8} \\ simplify \\ \frac{1}{2} x + 2 - \frac{1}{2} x \: \frac{1}{2} x - \frac{1}{2} x = 0 = 2 \\ \frac{1}{8} x = 2 \\ multiply \: both \: sides \: by \: 8 \\ 8. \frac{1}{8} x = 2.8 \\ simplify \\ x = 16.[/tex]
The difference in length of a spring on a pogo stick from its non-compressed length when a teenager is jumping on it after θ seconds can be described by the function f of theta equals 2 times cosine theta plus radical 3.
Part A: Determine all values where the pogo stick's spring will be equal to its non-compressed length.
Part B: If the angle was doubled, that is θ became 2θ, what are the solutions in the interval [0, 2π)? How do these compare to the original function?
Part C: A toddler is jumping on another pogo stick whose length of their spring can be represented by the function g of theta equals 1 minus sine squared theta plus radical 3 period At what times are the springs from the original pogo stick and the toddler's pogo stick lengths equal?
The given function for the difference in length is presented as follows;
[tex]f( \theta) = 2 \cdot cos( \theta) + \sqrt{3} [/tex]
When the pogo stick will be equal to its non compressed length, the difference is zero, therefore;
[tex]f( \theta) = 2 \cdot cos( \theta) + \sqrt{3} = 0[/tex]
[tex] 2 \cdot cos( \theta) = - \sqrt{3} [/tex]
[tex]\theta= arccos \left( \frac{ - \sqrt{3} }{2} \right) [/tex]
Which gives;
[tex] \theta = \frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{6} [/tex]
[tex] \theta = -\frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{6} [/tex]
Part B; If the angle was doubled, we have;
[tex]f( \theta) = 2 \cdot cos(2 \cdot \theta) + \sqrt{3} = 0[/tex]
Therefore;
[tex] 2 \cdot cos(2 \cdot \theta) = - \sqrt{3} [/tex]
Which gives;
[tex] \theta = \frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{12} [/tex]
[tex] \theta = -\frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{12} [/tex]
Between 0 and 2•π, we have;
[tex] \theta = \frac{5\cdot \pi}{12} [/tex]
[tex] \theta = \frac{ 17\cdot \pi}{12} [/tex]
Part C;
The toddler's pogo stick is presented as follows;
[tex]g( \theta) = 1- son^2( \theta) + \sqrt{3} [/tex]
Integrating the original function between 0 and theta gives;
[tex]2 \cdot sin( \theta) + \sqrt{3}\cdot \theta [/tex]
The original length =
Therefore, when the lengths are equal, we have;
[tex]1- son^2( \theta) + \sqrt{3} = 2 \cdot sin( \theta) + \sqrt{3}\cdot \theta [/tex]
A group of 18 patrons each owe $15 at a restaurant. What is the total amount owed by all 18 customers?
Answer:
$270
Step-by-step explanation:
18x15=$270
HOPE THAT HELPED:)
Help me solve this question!
Answer:
1st quartile is 136
The second quartile is also the median 142
The third quartile is 162
The interquartile range is the difference between the 3rd and first quartile or 26
Step-by-step explanation:
Question 2: If you were to take a cross-section parallel to the base for one of your items, what shape would
you see? Can a cross-section be a sphere? Explain in two to three sentences.
The cross-section would be the same shape as the base. A cross-section cannot be a sphere because a cross-section must be two-dimensional, but a sphere is three-dimensional.
Please help me it would mean a lot!! due right now. Please show work thank youuuu
Answer:
area of A'B'C'D' = 18.4 units²
Step-by-step explanation:
given 2 similar figures with
ratio of corresponding sides = a : b , then
ratio of areas = a² : b²
here ratio of sides = 5 : 2
ratio of areas = 5² : 2² = 25 : 4
let x be the area of A'B'C'D' , then by proportion
[tex]\frac{25}{115}[/tex] = [tex]\frac{4}{x}[/tex] ( cross- multiply )
25x = 460 ( divide both sides by 25 )
x = 18.4
area of A'B'C'D' = 18.4 units²
Help checking my answer (solving inequalities)
(-4x + 2) / 4 >= 6
-4x + 2 >= 24
-4x >= 22
x <= 22/4 OR x <= 5 1/2 OR x <= 5.5
Hope this helps!
what does a irrational number look like
A irrational number is a number that can't be expressed as a ratio of two whole numbers. That's it.
For examples (in increasing order of difficulty)
1 is a rational number because it is 1/1
0.75 is a rational number because it is equal to 3/4
2.333... (infinite number of digits, all equal to three) is rational because it is equal to 7/3.
sqrt(2) is not a rational number. This is not completely trivial to show but there are some relatively simple proofs of this fact. It's been known since the greek.
pi is irrational. This is much more complicated and is a result from 19th century.
As you see, there is absolutely no mention of the digits in the definition or in the proofs I presented.
Now the result that you probably hear about and wanted to remember (slightly incorrectly) is that a number is rational if and only if its decimal expansion is eventually periodic. What does it mean ?
Take, 5/700 and write it in decimal expansion. It is 0.0057142857142857.. As you can see the pattern "571428" is repeating in the the digits. That's what it means to have an eventually periodic decimal expansion. The length of the pattern can be anything, but as long as there is a repeating pattern, the number is rational and vice versa.
As a consequence, sqrt(2) does not have a periodic decimal expansion. So it has an infinite number of digits but moreover, the digits do not form any easy repeating pattern.
Quick algebra 1 question for 50 points!
Only answer if you know the answer, quick shout-out to Yeony2202, tysm for the help!
Answer:
35449 people
Explanation:
Given equation: y = 1100.74x -1976.47
Here year '2005' represents x = 005 = 5
So, the year '2034' will represent x = 034 = 34
Insert this into equation:
y = 1100.74(34) -1976.47
y = 35448.69
y ≈ 35449 (rounded to nearest whole number)
The magnitude of vector λa is 5. Find the value of λ, if: a = (−6, 8)
The value of λ in the vector is 0.2
What are vectors?Vectors are the opposite of scalar quantities and they are quantities that have directions and magnitude
How to determine the value of λ?The vector a is given as:
a = (−6, 8)
The magnitude of vector a is calculated using
|a| = √(x^2 + y^2)
So, the equation becomes
|a| = √((-6)^2 + 8^2)
Evaluate the exponent
|a| = √(36 + 64)
Evaluate the sum
|a| = √100
Take the square root of both sides
|a| = 10
Given that
λa = 5
This means that
λ * 10 = 5
Divide both sides by 10
λ = 0.2
Hence, the value of λ is 0.2
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Write [tex]sec^2x[/tex] in two equivalent forms
The two equivalent forms of [tex]sec^2x[/tex] are [tex]\frac{1}{cos^2x}[/tex] and [tex]1+tan^2x[/tex]
Equivalent forms forms of trigonometric expressionsThe given trigonometric expression is:
[tex]sec^2x[/tex]
Note that:
[tex]sec(x)=\frac{1}{cos(x)}[/tex]
Substitute this equivalence to the given expression
[tex]sec^2(x)=\frac{1}{cos^2(x)}[/tex]
Also from [tex]cos^2(x)+sin^2(x)=1[/tex]
Divide through by [tex]cos^2x[/tex]
[tex]\frac{cos^2x}{cos^2x} +\frac{sin^2x}{cos^2x}=\frac{1}{sin^2x}[/tex]
Simplifying the resulting expression:
[tex]1+tan^2x=sec^2x\\\\sec^2x=1+tan^2x[/tex]
Therefore, the two equivalent forms of [tex]sec^2x[/tex] are [tex]\frac{1}{cos^2x}[/tex] and [tex]1+tan^2x[/tex]
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