The following requires the submission of the details of the parabola related to the equations.
What is a Parabola?A parabola, simply defined, is a plane curve that is created by a point that is moving such that it's distance from a fixed locus is equal to it's distance from a line that is fixed.
What are the details for equation y = -2(x-5)² + 3?Vertex → (5,3)Axis of Symmetry → x = 5Direction of Opening → Opens downMax or Min → (5, 3)Y-intercept → 25/8What are the details for equation y = (x + 2)²Vertex → (-2, 0)Axis of Symmetry → x = -2Direction of Opening → Opens UpMax or Min - (-2, 0)Y-intercept → (0,-47)Learn more about Parabola at;
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help asap 20 points
Select the correct answer from each drop-down menu.
Answer: (99/13, 150/13)
Step-by-step explanation:
I'm too lazy to explain so here's a screen shot from desmos.
You design a logo for your soccer team. The logo is 3 inches by 5 inches. You decide to dilate the logo to 1.5 inches by 2.5 inches. What is the scale factor of this dilation?
Answer:
1/2
Step-by-step explanation:
The dimensions are cut in half ....scale factor 1/2
change the unit of length 6ft 2in= ___ft?
Answer:
Step-by-step explanation:
Comment
You don't have to do anything to the 6 feet. It already is in feet. It's the 2 inches you have to worry about.
2 inches = 2/12 of a foot.
2/12 = 1/6 = 0.167 feet
Answer
6 feet 2 inches = 6.167 feet
Step-by-step explanation:
The formula for the perimeter of a rectangle is P=2(l+w). Part A. Rewrite the formula for the perimeter of a rectangle in terms of the width, w. In your final answer, include all of your work. Part B. In two or more complete sentences, describe the process you followed while isolating the variable w in the equation P=2(l+w).
The expression that represents the width is w = (P-2l)/2
Subject of formulaGiven the formula for calculating the perimeter of a triangle expressed as:
P = 2(l + w)
where
l is the length
w is the width
Make w the subject of the formula
Given
P = 2(l + w)
Expand
P = 2l + 2w
Subtract 2l from both sides
P - 2l = 2w
Divide both sides by2
2w/2 = (P-2l)/2
w = (P-2l)/2
Hence the expression that represents the width is w = (P-2l)/2
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The equation y = ax describes the graph of a line. If the value of a is negative,
the line:
Answer:
The line has a downward slope
Step-by-step explanation:
If y = ax, it means that a represents the slope of the line. So, if A is negative, the line has a downward slope. In other words, for every increase in the value of x, there is a decrease in the value of y (increase in the negative direction). You can imagine the line y = -x, where the value of a is -1. When x is 1, y is -1; when x is 2, y is -2; when x is 5, y is -5. You can imagine that the graph looks like a diagonal line, much like y = x, except the function values are positive in the second quadrant and negative in the fourth (it slants downward to the right side). I hope that answers your question.
helllppppppppppp please
Answer:
No, it will not pass specifications.
Step-by-step explanation:
It says that the max slope of the wheelchair ramp should be 1/12.
To calculate slope, you do rise over run or y values over x values. In this case, the rise over run is 5/50. 5/50 simplifies to 1/10.
1/10 > 1/12, meaning that it is more than the maximum slope.
how to solve part ii and iii
(i) Given that
[tex]\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(xy) = \dfrac{7\pi}{12}[/tex]
when [tex]x=1[/tex] this reduces to
[tex]\tan^{-1}(1) + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}[/tex]
[tex]\dfrac\pi4 + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}[/tex]
[tex]2 \tan^{-1}(y) = \dfrac\pi3[/tex]
[tex]\tan^{-1}(y) = \dfrac\pi6[/tex]
[tex]\tan\left(\tan^{-1}(y)\right) = \tan\left(\dfrac\pi6\right)[/tex]
[tex]\implies \boxed{y = \dfrac1{\sqrt3}}[/tex]
(ii) Differentiate [tex]\tan^{-1}(xy)[/tex] implicitly with respect to [tex]x[/tex]. By the chain and product rules,
[tex]\dfrac d{dx} \tan^{-1}(xy) = \dfrac1{1+(xy)^2} \times \dfrac d{dx}xy = \boxed{\dfrac{y + x\frac{dy}{dx}}{1 + x^2y^2}}[/tex]
(iii) Differentiating both sides of the given equation leads to
[tex]\dfrac1{1+x^2} + \dfrac1{1+y^2} \dfrac{dy}{dx} + \dfrac{y + x\frac{dy}{dx}}{1+x^2y^2} = 0[/tex]
where we use the result from (ii) for the derivative of [tex]\tan^{-1}(xy)[/tex].
Solve for [tex]\frac{dy}{dx}[/tex] :
[tex]\dfrac1{1+x^2} + \left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} + \dfrac y{1+x^2y^2} = 0[/tex]
[tex]\left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} = -\left(\dfrac1{1+x^2} + \dfrac y{1+x^2y^2}\right)[/tex]
[tex]\dfrac{1+x^2y^2 + x(1+y^2)}{(1+y^2)(1+x^2y^2)} \dfrac{dy}{dx} = - \dfrac{1+x^2y^2 + y(1+x^2)}{(1+x^2)(1+x^2y^2)}[/tex]
[tex]\implies \dfrac{dy}{dx} = - \dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2) (1 + x^2y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2) (1+x^2y^2)}[/tex]
[tex]\implies \dfrac{dy}{dx} = -\dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2)}[/tex]
From part (i), we have [tex]x=1[/tex] and [tex]y=\frac1{\sqrt3}[/tex], and substituting these leads to
[tex]\dfrac{dy}{dx} = -\dfrac{\left(1 + \frac13 + \frac1{\sqrt3} + \frac1{\sqrt3}\right) \left(1 + \frac13\right)}{\left(1 + \frac13 + 1 + \frac13\right) \left(1 + 1\right)}[/tex]
[tex]\dfrac{dy}{dx} = -\dfrac{\left(\frac43 + \frac2{\sqrt3}\right) \times \frac43}{\frac83 \times 2}[/tex]
[tex]\dfrac{dy}{dx} = -\dfrac13 - \dfrac1{2\sqrt3}[/tex]
as required.
Kirsten is knitting and hates to donate to a local charity. Each hat takes 3 hours to make. If Kirsten knitted 3 hats and a third of another hat. how many hours did she spend knitting?
A group of summer campers went on a trip.
22 campers rode in cars with their parents
while the rest filled four buses. How many
campers were on each bus if a total of 150
campers went on the trip?
Answer:
Answer is 32
Step-by-step explanation:
150-22=128
128÷4=32
Consider the following scenario describing the residents of Charlestown: The list pairs resident's ages with their zip codes.
Does this scenario represent a function?
Answer:
Step-by-step explanation:
If B=3n-10B=3n−10 and C=n^{2}-6n-6,C=n
2
−6n−6, find an expression that equals 2B-3C2B−3C in standard form.
The expression of 2B - 3C in standard form is -3n^2 + 24n - 2
Functions and valuesGiven the following expressions
B = 3n - 10
C = n^2-6n-6
Required
2B - 3C
Substitute
2B - 3C = 2(3n -10) - 3(n^2-6n-6)
2B - 3C = 6n - 20 -3n^2+18n+18
2B - 3C = -3n^2 + 24n - 2
Hence expression of 2B - 3C in standard form is -3n^2 + 24n - 2
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−9x+2>18 OR 13x+15≤−4 solve for x please
Inequalities help us to compare two unequal expressions. The value of x must be less than -19/13.
What are inequalities?Inequalities help us to compare two unequal expressions. Also, it helps us to compare the non-equal expressions so that an equation can be formed. It is mostly denoted by the symbol <, >, ≤, and ≥.
The given inequalities can be simplified as,
1.)
−9x+2>18
−9x>18-2
−9x>16
x < -16/9
Thus, the value of x must be less than -16/9.
2.)
13x+15≤−4
13x ≤ -19
x ≤ -19/13
Thus, the value of x must be less than -19/13.
For the given condition −9x+2>18 OR 13x+15≤−4, the value of x must follow any one of the inequality. Hence, the value of x must be less than -19/13.
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Which exponential equation is equivalent to this logarithmic equation? log x = 4
Answer:
[tex]x=10^{4}[/tex]
Step-by-step explanation:
Switching from Logarithmic to Exponential form:An exponential base is the inverse of a logarithm with the same base. Given the equation log x = 4, note that the base of the logarithm isn't written in (it has no subscript). By default, the base of the logarithm function is base-10. So, to rewrite the equation using an exponential, we need to undo the logarithm with an exponential of the same base.
[tex]\log(x)=4[/tex]
[tex]10^{\log{(x)}}=10^{4}[/tex]
[tex]x=10^{4}[/tex]
Analogies with Addition/SubtractionThink about the equation x + 7 = 10.
I could tell you this is the equation in "addition form". What if I asked you to write an equation in "subtraction form"? While neither "addition form" nor "subtraction form" have been defined explicitly, one could undo the addition, and get an equation with subtraction in it, which one could argue is a "subtraction form" of the equation.
[tex]x+7=10[/tex]
Subtracting 7 from both sides to undo the addition...
[tex](x+7)-7=(10)-7[/tex]
Simplifying the left side since the "+7" and "-7" cancel...
[tex]x=10-7[/tex]
This is arguably in a "subtraction form" (there's a subtraction in it), whereas the original equation had addition in it.
While we could simplify this particular equation's right-hand side, we might not always be able to (what if the 10 had been a "y"... then the "y" and the "7" aren't like terms, and they would have to remain separate)
Similarly, in the logarithm problem, one could simplify 10^4 (it's 10,000), but one doesn't have to, and one won't always be able to.
For the logarithm problem, x=10^4 is the exponential form, as requested.
A courier travels due North at an average speed of 40km/h for 6 minutes to collect a parcel, before travelling 10km due East to deliver it. He then travels due South at an average speed of 30km/h for 12 minutes to collect another parcel. Find the shortest distance between the courier and his starting point.
Answer:
Step-by-step explanation:
Answer:10.2km
He took the route eastwards then southwards
URGENT PLEASE ANSWER THESE
Answer:
1A) 2 gallons of 20% solution and 3 gallons 15% solution needed
1B) 4 gallons of 20% solution and 1 gallons 15% solution needed
Step-by-step explanation:
1A) adding 20% salt and 15% water making 5 gallons of 17%
20% salt + 15% salt = 5 gallons of 17% salt
0.20x + 0.15(5 - x) = 0.17(5)
0.20x + 0.75 - 0.15x = 0.85
0.20x + 0.75 - 0.75 - 0.15x = 0.85 - 0.75
0.20x - 0.15x = 0.10
0.05x = 0.10
0.05x/ 0.05 = 0.10/0.05
x = 2 gallons (amount of 20% solution needed)
5 - x = 5 - 2 = 3 gallons (amount of 15% solution needed)
1B)
0.20x + 0.15(5 - x) = 0.19(5)
0.20x + 0.75 - 0.15x = 0.95
0.20x + 0.75 - 0.75 - 0.15x = 0.95 - 0.75
0.20x - 0.15x = 0.20
0.05x = 0.20
0.05x / 0.05 = 0.20/0.05
x = 4 gallons (amount of 20% solution needed)
5 - x = 5 - 4 = 1 gallons (amount of 15% solution needed)
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Simplify the expression to a polynomial in
standard form:
(3x-1)(3x^2+3x+8)
Answer:
9x^3+6x^2+21x-8
Step-by-step explanation:
(3x-1) (3x ^ 2 + 3x + 8) =3x(3x ^ 2 + 3x + 8) +(-1)(3x ^ 2 + 3x + 8) =9x^3+9x^2+24x-3x^2-3x-8 = 9x^3+6x^2+21x-8
Answer:
[tex]\huge\boxed{\sf 9x^3+6x^2+21x-8}[/tex]
Step-by-step explanation:
Given expression:= [tex](3x-1)(3x^2+3x+8)[/tex]
[tex]Expand[/tex]
[tex]= 3x(3x^2+3x+8)-1(3x^2+3x+8)\\\\Multiply\\\\=9x^3+9x^2+24x-3x^2-3x-8\\\\Combine \ like \ terms\\\\= 9x^3+9x^2-3x^2+24x-3x-8\\\\= 9x^3+6x^2+21x-8\\\\\rule[225]{225}{2}[/tex]
Which represents where f(x) = g(x)?
The equation represents the intersections. The correct option is A:
f(2) = g(2) = 0f(0) = g(0) = 4.Which represents where f(x) = g(x)?
If we have two functions:
y = f(x) and y = g(x), the equation:
f(x) = g(x) gives the value of x such that the two functions have the same output. So, that equation gives the intersection points between the two graphs.
By looking at the graph, we can see that we have two intersections, one at:
(2, 0) and other at (0, 4).
This means that:
f(2) = g(2) = 0f(0) = g(0) = 4.So the correct option is the first one.
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Please help? I’ve been stuck on this question
Answer:
y-intercept = -3
another point on the line = (6,-1)
Step-by-step explanation:
1. Underline important parts of the question (slope of 1/3, point (3, -2))
2. mark the point that is given (3, -2)
3. slope form is rise over run which basically means that the number one is how much the line rises, and the number three is how much the line goes to the right (since the slope is positive then it runs to the right, if it was negative it would go to the left.) Basically 1/3 = rise over run
4. put your pencil on the point (3, -2) and go up one point and over three points. continue doing so from each new point (two or three points is good enough)
5. take a ruler and draw a line that connects all three and keep it going until it crosses the y- axis (which is the vertical line)
6. the point that crosses through the y -axis is the y-intercept.
7. now you can go and look at your points that you have made and choose one of them to answer the question with
(I hope this cleared it up a bit I'm not very good at explaining things this is the best I can do)
Determine the number of possible solutions for a triangle with b = 37, a = 32, and b = 27.
There are two(2) possible solutions for the given parameters of a triangle. Using the law of cosines, the parameters of the triangle are calculated.
What is the law of cosines?The law of cosines is used to relate the angles and side lengths of a triangle.
They are given as follows:
a² = b² + c² -2bc cos(A)
b² = a² + c² - 2ac cos(B)
c² = a² + b² - 2ab cos(C)
Calculation:The given triangle has B = 37°, a = 32 and b = 27
So, using the law b² = a² + c² - 2ac cos(B)
On substituting,
27² = 32² + c² - 2(32)c cos(37°)
⇒ c² - 50.56 c + 1024 = 729
⇒ c² - 50.56 c + 295 = 0
Since the obtained equation is quadratic, it has 2 solutions.
Using quadratic formula the solutions are calculated.
c = [50.56 ± [tex]\sqrt{50.56^2-4(1)(295)}[/tex]]/2(1)
On simplifying,
c = 43.8 and c = 6.7
Therefore, there are 2 possible solutions for the given triangle.
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Explain weather this equation is a linear equation
4. y = 1 - x
Answer:
Yes, it is a linear equation.
Step-by-step explanation:
Linear equation:An equation of the form ax + by +c = 0, where x and y are two variables and a, b,c are the non-zero real numbers is called a linear equation and the degree of the equation is 1.
y = 1 - x
x + y - 1 = 0
Where a = 1 ; b=1 and c = -1
So, y = 1 -x is a linear equation.
HELP!!
[tex]\sf If \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 47, \: and \: x > 0, \: then \: what \: is \: value \: of \: x + \dfrac{1}{x} [/tex]
DON'T SPAM.
Answer:
[tex]x+\cfrac{1}{x} =7[/tex]================
Given:[tex]x^2+\cfrac{1}{x^2} =47[/tex]Add 2 to both sides of equation:
[tex]x^2+\cfrac{1}{x^2}+2 =47+2[/tex]Then follow the steps:
[tex]x^2+\cfrac{1}{x^2}+2*x*\cfrac{1}{x} =49[/tex][tex](x+\cfrac{1}{x})^2=7^2[/tex]Take square root of both sides to get:
[tex]x+\cfrac{1}{x} =\pm\ 7[/tex]We are taking the positive value as we are told x > 0, hence the answer is 7.
[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]
x + 1/x = 7[tex]\textsf{ \underline{\underline{Steps to solve the problem} }:}[/tex]
[tex] \qquad❖ \: \sf \: {x}^{2} + \cfrac{1}{ {x}^{2} } = 47[/tex]
[tex] \qquad❖ \: \sf \: {x}^{2} + \cfrac{1}{ {x}^{2} } + 2 - 2 = 47[/tex]
( adding and subtracting 2, doesn't change the value )
[tex] \qquad❖ \: \sf \: {x}^{2} + \cfrac{1}{ {x}^{2} } + 2 = 47 + 2[/tex]
[tex] \qquad❖ \: \sf \: {x}^{2} + \cfrac{1}{ {x}^{2} } + \bigg(2 \sdot x \sdot \cfrac{1}{x} \bigg) = 49[/tex]
( x cancel out, so no change in value )
[tex] \qquad❖ \: \sf \: \bigg(x + \cfrac{1}{x} \bigg) {}^{2} = 49[/tex]
( use identity a² + 2ab + b² = (a + b)² )
[tex] \qquad❖ \: \sf \: \bigg(x + \cfrac{1}{x} \bigg) {}^{} = \sqrt{49} [/tex]
[tex] \qquad❖ \: \sf \: x + \cfrac{1}{x} {}^{} = \pm7[/tex]
since we have to take positive value, i.e greater than 0
[tex] \qquad❖ \: \sf \: x + \cfrac{1}{x} {}^{} = 7[/tex]
[tex] \qquad \large \sf {Conclusion} : [/tex]
Therefore, the required value is 7
from two points one on each leg of an isosceles triangle perpendicular are drawn to the base prove that the triangles formed are similar
The description below proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.
How to prove an Isosceles Triangle?Let ABC be an isosceles triangle such that AB = AC.
Let AD be the bisector of ∠A.
We want to prove that BD=DC
In △ABD & △ACD
AB = AC(Thus, △ABC is an isosceles triangle)
∠BAD =∠CAD(Because AD is the bisector of ∠A)
AD = AD(Common sides)
By SAS Congruency, we have;
△ABD ≅ △ACD
By corresponding parts of congruent triangles, we can say that; BD=DC
Thus, this proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.
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Area=
Help me please thanks
The area of the shaded region shown in the figure is 80π unit²
What is an equation?An equation is an expression that shows the relationship between two or more variables and numbers.
Area of the shaded region = area of semicircle IL - area of semicircle JK + area of semicircle IJ + area of semicircle KL
Area of the shaded region = area of semicircle IL + area of semicircle IJ = (0.5 * π * 12²) + (0.5 * π * (12/3)²) = 80π
The area of the shaded region shown in the figure is 80π unit²
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Researchers at a food company are interested in how a new ketchup made from green tomatoes (and green in color) will compare to their traditional red ketchup. They randomly assign participants to either the green or red ketchup condition. Participants indicate the tastiness of the sauce on a 20-point scale. Tastiness scores tended to be skewed (not normal). What statistical test should be used to analyze that data
The statistical test which can be used to analyze the data is chi square test.
Given two categories of tomatoes , one is red and other one is green.
When we want to compare more than two categorical variables, we can use the chi-square test. The main objective of chi square test is to compare the distribution of responses or the proportions of participants in each response category. This test does not require mean, standard deviation or anything. In the given problem we have not given sample mean, population mean, population standard deviation, etc. So we cannot use z test, t test, f test because they require sample mean, population mean , standard deviation in their calculation. The formula to be use in chi square test is as under:
[tex]X^{2}=[/tex]∑[tex](f_{0} -f_{e}) ^{2} /f_{e}[/tex]
where [tex]f_{0}[/tex]= observed frequency of each of response category.
[tex]f_{1}[/tex]=expected frequency in each of the response categories.
We can find the critical value in a table of probabilities for the chi square distribution with degree of freedom df=K-1.
Hence we can use chi square test to analyze the data.
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True or False? If ab<0, then a<0, b>0 or a>0, b<0
Answer:
True
Step-by-step explanation:
If ab < 0, then ab = negative #.
In order for ab to be a negative #, one of them has to be negative while the other one needs to be positive.
Example:
a = -2, b = 1
ab < 0
(-2)(1) < 0
-2 < 0, TRUE
a < 0
-2 < 0, TRUE
b > 0
1 > 0, TRUE
If I switch a = -2 to 1 and b = 1 to -2, a > 0 and b < 0 is true too.
Can someone help me with question 3 please. Please make sure you actually answer the question and not type something that isn't regarding the question.
Answer:
Option C
Step-by-step explanation:
We can see that these two triangles are similar triangles. And based on the sides we also know triangle ZXY and triangle FDE have a ratio of 3:2. We can find side x by the equation:
3:2=9:x
3x=18
x=6 or option C
Answer:
C.6
Step-by-step explanation:
12x=8
x=8/12=2/3
2/3*9=18/3=6
what is the decimal of 1/2
Answer:
0.5
one divide by two is zero point five
[tex]\frak{Hi!}[/tex]
[tex]\orange\hspace{300pt}\above2[/tex]
We need to find the decimal form of [tex]\boldsymbol{\sf{\displaystyle\frac{1}{2}}}[/tex].
But first we need to convert it into a fraction whose
denominator is 10, or 100, or 1,000, etc.
The closest number that's divisible by 2 is 10.
So that's the new denominator of the fraction.
Now we should also multiply the top times the number
that we multiplied the bottom by. See how this works?
[tex]\boldsymbol{\sf{\displaystyle\frac{1\times5}{2\times5}}}[/tex]. This yields
[tex]\boldsymbol{\sf{\displaystyle\frac{5}{10}}}[/tex]. And 5/10 in decimal-form yields
[tex]\boldsymbol{\sf{0.5}}[/tex].
[tex]\orange\hspace{300pt}\above3[/tex]
[tex]\LARGE\boldsymbol{\sf{calligraphy}}[/tex]
Solve the questions by factoring! Please help asap!
Answer:
Step-by-step explanation:
A tennis tournament has 2n contestants. We want to pair them up for the first round of singles matches. Show that the number of different ways in which first round matches can be conducted is
There are
[tex]\dbinom{2n}2 = \dfrac{(2n)!}{2! (2n-2)!}[/tex]
ways of pairing up any 2 members from the pool of [tex]2n[/tex] contestants. Note that
[tex](2n)! = 1\times2\times3\times4\times\cdots\times(2n-2)\times(2n-1)\times(2n) = (2n-2)! \times(2n-1) \times(2n)[/tex]
so that
[tex]\dbinom{2n}2 = \dfrac{(2n)\times(2n-1)\times(2n-2)!}{2! (2n-2)!} = \boxed{n(2n-1)}[/tex]
The number of different ways in which first-round matches can be conducted is n (2n - 1).
We have,
Number of contestant = 2n
Number of contestants in each match = 2
Now,
The number of different ways in which first-round matches can be conducted.
A combination formula is used.
= [tex]^{2n}C_2[/tex]
= 2n! / 2! (2n - 2)!
= 2n (2n - 1)(2n - 2)! / [2 x (2n - 2)!]
= 2n (2n - 1) / 2
= n (2n - 1)
Thus,
The number of different ways in which first-round matches can be conducted is n (2n - 1).
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2. A map is drawn so that every 1 inch on the map represents 50 actual miles. The table below can be used to record the actual distances between towns on the map. Actual Distance (mi.) Map Distance (in.) Actual Distances and Map Distances 50 1 Complete the table. What is the actual distance between two towns that are 4 inches apart on the map? 2. A map is drawn so that every 1 inch on the map represents 50 actual miles . The table below can be used to record the actual distances between towns on the map . Actual Distance ( mi . ) Map Distance ( in . ) Actual Distances and Map Distances 50 1 Complete the table . What is the actual distance between two towns that are 4 inches apart on the map ?
Answer:
200miles
Step-by-step explanation
4 x 50 = 200
As for the table, just multiply each inch by 50.
For example; 7in = 350 miles because 7x50 = 350