{(-2, 5), (-1, 2), (0, 1), (2,5)}
Which points are on the graph of the inverse?
Select each correct answer.
(5,2)
(2, 1)
-
(-5, 2)
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let's recall that the domain of a function is the range of its inverse function, namely, the inverse has a pair that's just the same but flipped sideways, Check the picture below.
Related Questions
Each chef at "Sushi Emperor" prepares 15 regular rolls and 20 vegetarian rolls daily. On Tuesday, each customer ate 2 regular rolls and 3 vegetarian rolls. By the end of the day, 4 regular rolls and 1 vegetarian roll remained uneaten.
How many chefs and how many customers were in "Sushi Emperor" on Tuesday?
Please Help!
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Answer: 13 customers and 2 chefs
Step-by-step explanation:
I need help with this practice problem Having trouble solving it If you can use Desmos to graph it
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The graph of the function:
[tex]f(x)=\cot (x+\frac{\pi}{6})[/tex]is shown below:
By graphing at least one full period of the function, we would take the limit of the function as:
[tex]-\pi\le x\le\pi[/tex]Hence, the graph of at least one full period is:
The original price of a riding lawn mower is $1250. Paul bought his for $1000. What percent was the discount?
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we get that the percentage he paid was
[tex]\frac{1000}{1250}\cdot100=80\text{ \% }[/tex]so the percentage of discount is 20%
Write the inequality statement in x describing the numbers [ 11, ∞)
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The inequality [ 11, ∞) represents that value is more than or equal to 11. The interval can be expressed as,
[tex]x\ge11[/tex]In inequality, x is any variable.
So inequality statement in x is,
[tex]x\ge11[/tex]Find the perimeter and area of a square with side 9 inches.
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The perimeter (P) and area (A) of a square of sides a = 9 in, are given by:
[tex]\begin{gathered} P=4a=4\cdot(9in)=36in, \\ A=a^2=(9in)^2=81in^2. \end{gathered}[/tex]Answer
• Perimeter = 36 in
,• Area =, ,81 in²
Evaluate 2^5.32251016
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32
Explanation:The given expression is:
2⁵
This means the product of 2 in 5 places
That is,
2⁵ = 2 x 2 x 2 x 2 x 2
2⁵ = 32
Simplify the following expression(-2v)^4
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We have
[tex]\mleft(-2v\mright)^4[/tex]In order to simplify this expression, we will use the next rule
[tex]\mleft(ab\mright)^m=a^mb^m[/tex]We use the rule and we simplify
[tex](-2)^4v^4=16v^4[/tex]CAN SOMEONE HELP WITH THIS QUESTION?✨
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408-360=48
Answer: [tex]48^{\circ}[/tex]
Step-by-step explanation:
Coterminal angles differ by integer multiples of [tex]360^{\circ}[/tex].
So, an angle coterminal with an angle of [tex]408^{\circ}[/tex] is [tex]408^{\circ}-360^{\circ}=48^{\circ}[/tex], which lies within the required interval.
I need help with 5 and 6. The exponent for part 5 if you can't see it well 2/3
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5.
Given the equation to solve for x:
[tex]3(x+1)^{\frac{2}{3}}=12[/tex]The steps for the solution are as follows:
[tex]\begin{gathered} 3(x+1)^{\frac{2}{3}}=12 \\ \frac{3(x+1)^{\frac{2}{3}}}{3}=\frac{12}{3} \\ (x+1)^{\frac{2}{3}}=4 \\ \lbrack(x+1)^{\frac{2}{3}}\rbrack^{\frac{1}{2}}=(4)^{\frac{1}{2}} \\ \lbrack(x+1)^{\frac{1}{3}}\rbrack=\pm2 \\ \lbrack(x+1)^{\frac{1}{3}}\rbrack^3=(\pm2)^3 \\ x+1=\pm8 \end{gathered}[/tex]From the above equation, we have x + 1 = 8 and x + 1 = -8. These imply x = 7 and x = -9.
Check for extraneous solutions:
If x = 7, then the left-hand side of the equation is:
[tex]3(x+1)^{\frac{2}{3}}=3(7+1)^{\frac{2}{3}}=3(4)=12[/tex]Thus, the equation holds true at x = 7.
If x = -9, then the right-hand side of the equation is:
[tex]3(x+1)^{\frac{2}{3}}=3(-9+1)^{\frac{2}{3}}=3(4)=12[/tex]Thus, the equation holds true at x = -9.
There is no extraneous solution. The solutions of the given equation are x = 7 and x = -9.
6.
Given an equation to solve for x:
[tex]\sqrt[]{3x+2}-2\sqrt[]{x}=0[/tex]The steps of the solution are as follows:
[tex]\begin{gathered} \sqrt[]{3x+2}-2\sqrt[]{x}=0 \\ \sqrt[]{3x+2}=2\sqrt[]{x} \\ (\sqrt[]{3x+2})^2=(2\sqrt[]{x})^2 \\ 3x+2=4x \\ 2=4x-3x \\ 2=x \end{gathered}[/tex]Thus, the solution of the equation is x = 2.
I really need help I can’t seem to understand this at all
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Given the sequence below
[tex]8,12,18,27[/tex]The sequence above is a geometric series, therefore the formula for the common ratio(r) is
[tex]r=\frac{2ndterm}{First\text{ term}}=\frac{Thirdterm}{2nd\text{ term}}[/tex]Therefore,
[tex]\begin{gathered} r=\frac{12}{8}=\frac{18}{12} \\ r=\frac{3}{2}=\frac{3}{2} \end{gathered}[/tex]Hence, the answer is
[tex]\frac{3}{2}\text{ \lparen Option 3\rparen}[/tex]a quadrilateral has vertices S (0,0) T (4,0) U (5,4) V (1,4). what is the most precise name for the quadrilateral
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First, plot the points on a coordinate plane:
Notice that ST is parallel to VU, as well as SV is parallel to TU.
A quadrilateral whose opposite sides are parallel, is called a parallelogram.
Therefore, the most precise name for the quadrilateral is: parallelogram.
2) sin X Z 45 36 X 27 Y A) B) no+ D)
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Explanation
For the angle α, the sine function gives the ratio of the length of the opposite side to the length of the hypotenuse.
[tex]\sin \alpha=\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{y}{z}[/tex]then, Let
[tex]\begin{gathered} \text{opposite side= 36} \\ \text{hypotenuse =45} \\ \text{angle}=\angle x \end{gathered}[/tex]Now, replace
[tex]\begin{gathered} \sin \alpha=\frac{\text{opposite side}}{\text{hypotenuse}} \\ \sin \angle x=\frac{36}{45}=\frac{12}{15}=\frac{4}{5} \\ \sin \angle x=\frac{4}{5} \end{gathered}[/tex]so, the answer is
[tex]B)\frac{4}{5}[/tex]I hope this helps you
Write the polynomial function in standard form that has complex roots -2+i and -2-i
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ANSWER
[tex]\text{ x}^2\text{ - 4x + 5}[/tex]EXPLANATION
Given information
The root of the polynomial function are -2 + i and -2- i
To find the standard form of the polynomial function, follow the steps below
Step 1: Express the root of the polynomial in terms of the factor
[tex]\begin{gathered} \text{ Given that the roots of the polynomial function are -2+i and -2 - i} \\ \text{ The factors of the above roots can be expressed as} \\ \text{ \lbrack x + \lparen-2 + i\rparen\rbrack and \lbrack x + \lparen-2 - i\rparen\rbrack} \end{gathered}[/tex]Step 2: Expand the factors of the polynomial in step 1
[tex]\begin{gathered} \text{ \lbrack x + \lparen-2 + i\rparen\rbrack \lbrack x +\lparen-2 -i\rparen\rbrack} \\ [x\text{ -2\rparen + i\rparen\rbrack \lbrack x -2\rparen - i\rparen\rbrack} \\ (x\text{ - 2\rparen}^2\text{ - i}^2 \\ (x\text{ - 2\rparen\lparen x - 2\rparen- i}^2 \\ x^2\text{ - 2x - 2x + 4 - i}^2 \\ x^2\text{ - 4x + 4 - i}^2 \\ \text{ Recall, that i}^2\text{ = -1} \\ \text{ x}^2\text{ - 4x + 4 - \lparen-1\rparen} \\ \text{ x}^2\text{ - 4x + 4 + 1} \\ \text{ x}^2\text{ - 4x + 5} \end{gathered}[/tex][tex]\text{ Hence, the polynomial function in standard form is x}^2\text{ - 4x + 5}[/tex]Plot the image of point C under a reflection across line n.Click to add points
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We can find the image of point C reflected across line n by finding the distance d (perpendicular) from point C to line n, and then placing point C', the image, at an equal and perpendicular distance d on the other side of the line.
We can graph this as:
120+m=203d+59=33c-87=-42
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Let's solve the following equation
c - 87 =42
Adding 87 at both sides:
c - 87 + 87 = -42 + 87
c = 45
Consider the line y=4x-5.Find the equation of the line that is perpendicular to this line and passes through the point (6. 4).Find the equation of the line that is parallel to this line and passes through the point (6, 4).Equation of perpendicular line: Equation of parallel line:
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Solution
gradient = 4
Slope for Perpendicular = -1/4
Slope for Parallel = 4
Equation of perpendicular line:
[tex]\begin{gathered} y-4=-\frac{1}{4}(x-6) \\ \\ 4y-16=-x+6 \\ \\ 4y+x=22 \end{gathered}[/tex]Equation of parallel line:
[tex]\begin{gathered} y-4=4(x-6) \\ \\ y-4=4x-24 \\ \\ y=4x-20 \end{gathered}[/tex]Which function has the graph shown?O A. y = secx-1)O B. y = - secxO C. y = csexO D. y = csc(x) +1
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We will have that the graph of the function shown belongs to:
[tex]y=csc(x)+1[/tex]This can be seeing as follows:
Enter the exponential function using t (for time) as the independent variable to model the situation. Then find the value of the function after the given amount of time. The value of a textbook is $65 and decreases at a rate of 14% per year for 13 years. The exponential function that models the situation is y =__After 13 years, the value of the textbook is $__
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Please, give me some minutes to take over your question
_________________________________
For my practice review, I need help to determine if these are functions or not.
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Answer:
1: no
2: no
3: yes
4: no
5: yes
6: yes.
Step-by-step explanation:
Think of a vertical line sweeping across the graph from left to right. If ever this line crosses two points of the graph at the same time, it cannot be a function, since a function can only have max. 1 result per x value.
Not a timed or graded assignment. Need a quick answer showing work. Please DRAW factor tree. Thank you so much.
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ANSWER:
[tex]40\sqrt[]{7}m^3[/tex]STEP-BY-STEP EXPLANATION:
Using the following factor tree, we decompose the number 112 to be able to simplify, just like this:
Therefore, it would be:
[tex]\begin{gathered} 10\sqrt[]{2\cdot2\cdot2\cdot2\cdot7\cdot m^6} \\ 10\sqrt[]{2^4\cdot7\cdot m^{3\cdot2}} \\ 10\cdot2^2\cdot m^3\sqrt[]{7} \\ 10\cdot4\cdot m^3\sqrt[]{7}=40\sqrt[]{7}m^3 \end{gathered}[/tex]what is the answer for this pls answer
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Answer: A
Step-by-step explanation: You merge the equations, -2x and 2x cancel out, 4y + 1y = 5y, and 12 + (-7) = 5
You'll be left with 5y = 5
Dividing both sides by 5 to isolate the y results in y = 1
The mean annual salary at the company where Samuel works is $37,000, with standard deviation $4,000. Samuel's salary is $32,500. Based on the mean and standard deviation, is Samuel's salary abnormal compared to other salaries at this company? When choosing your answer, be careful to select the answer with the correct explanation. A. Samuel's salary falls within the standard deviation, so his salary is not abnormal compared to other salaries at this company. B. Samuel's salary falls outside the standard deviation, so his salary is abnormal compared to other salaries at this company. C. Samuel's salary falls within the standard deviation, so his salary is abnormal compared to other salaries at this company. D. Samuel's salary falls outside the standard deviation, so his salary is not abnormal compared to other salaries at this company?
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Answer : Samuel salary falls within the standard deviation and his salary is not abnormal
The mean annual salary at the company where samuel works is $37, 000
The standard deviation is given as $4, 000
Samule's annual salary is $32, 500
Using the Z- score formula
[tex]\begin{gathered} z\text{ = }\frac{x\text{ - }\mu}{\sigma} \\ \text{Where x = sample score} \\ \mu\text{ = mean} \\ \sigma\text{ = standard deviation} \end{gathered}[/tex][tex]\begin{gathered} x\text{ = \$32, 500} \\ \mu\text{ = \$37, 000} \\ \sigma=\text{ \$ 4000} \\ z\text{ = }\frac{32,\text{ 500 - 37000}}{4000} \\ z\text{ = }\frac{-4500}{4000} \\ z\text{ = -1.125} \end{gathered}[/tex]Since, the value of Z- score is -1. 125, then, the salary is 1 standard deviation below the mean.
Therefore, Samuel salary falls within the standard deviation and his salary is not abnormal
Lynette is covering shapes with wrapping paper to make a design for the school carnival how much paper and square feet will Lynette need to cover the figure shown below
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The area of paper needed is;
[tex]7\frac{1}{2}ft^2[/tex]Here, we want to get the square feet of paper needed
What we have to do here is to get the area of the parallelogarm
Mathematically, that would be the product of the base of the parallelogram and its height
We have the base as 3 3/4 ft which is same 15/4 ft and the height as 2 ft
Thus, we have the area calculated as follows;
[tex]\frac{15}{4}\times\text{ 2 = }\frac{30}{4}\text{ = 7}\frac{1}{2}ft^2[/tex]The graph shows the distance ofa remote control drone above theground as it flies west to east. Thex-axis represents the distance from acentral point and the y-axis representsthe distance above the ground, in m.411-21021. What is the range of the functionand what does it represent?
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Part 1
For this question we need to remember that the range is defined as:
[tex]\text{Range}=\text{Max}-Mi[/tex]And if we look at the function we see that Min =0 and Max= 5 so then we have:
[tex]\text{Range}=5-0=5[/tex]And the range represent the lenght of the codomain of a function
Part 2
The domain for this case is given by:
[tex]\text{Domain}=\left\lbrack -4,4\rbrack\right?[/tex]And it represent all te possible values of x that the function can assume
Part 3
For this case we identify two intervals where the height is increasing:
[-4,-2] and [0,4]
But the longest interval is :[0,4]
Part 4
The x intercept represent the values when the function satisfy that y=0 and we have:
x intercepts: x=-4, x=0
Part 5
The average rate of change between [-4,4] is given by:
[tex]m=\frac{3-0}{4-(-4)}=\frac{3}{8}[/tex]And then the answer for this case would be 3/8
What is the value of x in the figure at the right? 60° (2x)°
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The angle whose measure is 60° and the angle (2x)° are vertical angles, if two angles are vertical angles, then they are congruent, then we can express:
[tex]2x=60[/tex]From this expression, we can solve for x to get:
[tex]\begin{gathered} \frac{2x}{2}=\frac{60}{2} \\ x=30 \end{gathered}[/tex]Then, x equals 30
What is the slope of the line created by this equation? Round your answer out to two decimal places. 10x+5y=3
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Given the Linear Equation:
[tex]10x+5y=3[/tex]You can write it in Slope-Intercept Form, in order to identify the slope of the line.
By definition, the Slope-Intercept Form of the equation of a line is:
[tex]y=mx+b[/tex]Where "m" is the slope of the line and "b" is the y-intercept.
Therefore, you can rewrite the given equation in Slope-Intercept Form by solving for "y":
[tex]\begin{gathered} 5y=-10x+3 \\ \\ y=\frac{-10x}{5}+\frac{3}{5} \end{gathered}[/tex][tex]y=-2x+\frac{3}{5}[/tex]You can identify that:
[tex]\begin{gathered} m=-2 \\ \\ b=\frac{3}{5} \end{gathered}[/tex]Hence, the answer is:
[tex]m=-2[/tex]Use the formula d = vt + 1672, where d is the distance in feet, v is the initial velocity in feet per second, and t is the time in seconds.An object is released from the top of a building 320 ft high. The initial velocity is 16 ft/s. How many seconds later will the object hit the ground?
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We got to use the given formula:
[tex]d=v\cdot t+16t^2[/tex]The distance, d, given is 320 ft and the initial velocity, v, 16 ft/s. We want the time, t. So:
[tex]\begin{gathered} d=v\cdot t+16t^2 \\ 320=16t+16t^2 \\ 16t^2+16t-320=0 \\ \frac{16t^2}{16}+\frac{16t}{16}-\frac{320}{16}=\frac{0}{16} \\ t^2+t-20=0 \end{gathered}[/tex]Now, we have a quadratic equation, so we can use Bhaskara formula:
[tex]\begin{gathered} t=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-20)}}{2\cdot1}=\frac{-1\pm\sqrt[]{1+80}}{2}=\frac{-1\pm\sqrt[]{81}}{2}=\frac{-1\pm9}{2} \\ t_1=\frac{-1-9}{2}=-\frac{10}{2}=-5 \\ t_2=\frac{-1+9}{2}=\frac{8}{2}=4 \end{gathered}[/tex]Because we can't have a negative time, we consider only the second one, which it t = 4s.
Diagram 3 shows a piece of rectangularcardboard and an open box that is made from the cardboard.The box is made by cutting out four squares of equal size from the cornersof the cardboard then folding up the sides. Finda) the length in cm of sides of the squares to be cut out in order to get a box with largest volume.b) the minimum number of the boxes needed to fill with 5645 cm³ of pudding
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SOLUTION:
Step 1:
In this question, we are given the following:
Diagram 3 shows a piece of rectangular cardboard and an open box that is made from the cardboard.
The box is made by cutting out four squares of equal size from the corners
of the cardboard then folding up the sides.
Find
a) the length in cm of sides of the squares to be cut out in order to get a box with the largest volume.
[tex]\begin{gathered} The\text{ volume of the rectangle would be expressed as:} \\ \text{V = ( 30-2x )(16-2x) ( x)} \\ Multiply\in g\text{ out, we have that:} \\ V=480x-92x^2+4x^3 \\ \text{Differentiating V with respect to x, we have that:} \\ \frac{dV}{dx}=480-184x+12x^2=0 \\ \text{Factorizing the quadratic equation, we have that:} \\ \text{x = 12 or x =}\frac{10}{3} \end{gathered}[/tex][tex]\begin{gathered} \text{Differentiating again, we have that:} \\ \frac{d^2V}{dx^2\text{ }}\text{ = -184 + 24 x} \end{gathered}[/tex]To get the maximum, we need to substitute the values of :
[tex]\begin{gathered} x\text{ = 12, we have that:} \\ \frac{d^2V^{}}{dx^2\text{ }}\text{ = -184 + 24( 12) = }-184\text{ +288 = 104} \\ x=\frac{10}{3},\text{ we have that:} \\ \frac{d^2V}{dx^2}\text{ = -184 + 24 (}\frac{10}{3})\text{ = -184 +}\frac{240}{3}\text{ = - 184 + 80 = -104 }<0 \end{gathered}[/tex]At this stage, we can see that:
[tex]\begin{gathered} x\text{ =}\frac{10\text{ }}{3}cm\text{ is the length of the squares to be cut in order to get a box with } \\ \text{largest volume} \end{gathered}[/tex]b) Find the minimum number of the boxes needed to fill with 5645 cm³ of pudding
[tex]\begin{gathered} \text{From the equation,} \\ V=(30-2\text{x )(16-2x)(x)} \\ \text{put x =}\frac{10}{3}\text{ in the equation, we have that:} \\ V\text{ = \lbrack}30\text{ -2(}\frac{10}{3})\rbrack\text{ \lbrack 16-2(}\frac{10}{3}\rbrack\lbrack\frac{10}{3}\rbrack \\ V\text{ = ( 30 -}\frac{20}{3})\text{ ( 16 - }\frac{20}{3})(\frac{10}{3}) \\ V=725.93cm^3 \\ Now\text{, we asked to find the minimum number of boxes ne}eded^{} \\ to^{} \\ \text{fill with 5645cm}^{3\text{ }}\text{ of pudding.} \\ \text{Then, we ne}ed\text{ to do the following:} \end{gathered}[/tex]
Minimum number of boxes =
[tex]\begin{gathered} \frac{5645}{725.93} \\ =\text{ 7.78} \\ \approx\text{ 8} \end{gathered}[/tex]CONCLUSION:
A minimum of 8 boxes will be needed to fill with 5645 cm³ of pudding
Want to check if I got the correct answer, thank you
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To find:
The division of the polynomial.
Solution:
The division in given in the image below:
Thus, the result is:
[tex]x^3+3x^2-1x-5-\frac{11}{x+3}[/tex]Option D is correct.
Can you please help me out with a question
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To determine the green rectangle, each side of the blue rectangle was multiplied by a determined scale factor k.
To determine the measure of x, the first step is to determine the scale factor.
The information that you have to use is the areas of both rectangles.
After dilation, the area of the resulting shape is equal to the area of the original shape multiplied by the square of the scale factor:
[tex]A_{\text{green}}=k^2A_{\text{blue}}[/tex]A.green=50 m²
A.blue= 72m²
[tex]50=72k^2[/tex]-Divide both sides by 72
[tex]\begin{gathered} \frac{50}{72}=\frac{72k^2}{72} \\ \frac{25}{36}=k^2 \end{gathered}[/tex]-Apply the square root to both sides of the equal sign:
[tex]\begin{gathered} \sqrt[]{\frac{25}{36}}=\sqrt[]{k^2} \\ \frac{5}{6}=k \end{gathered}[/tex]Now, to determine the value of x, multiply the length of the corresponding side on the blue rectangle by the scale factor:
[tex]\begin{gathered} x=\frac{5}{6}\cdot12 \\ x=10 \end{gathered}[/tex]The length of the side on the green triangle is 10m