find the equation of the circle where (-9,4),(-2,5),(-8,-3),(-1,-2) are the vertices of an inscribed square.
Answers
so... hmm the diagonals for the square are the diameter of the circle, and keep in mind that the radius of a circle is half the diameter, so let's find the diameter.
[tex]\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -2}}\quad ,&{{ 5}})\quad % (c,d) &({{ -8}}\quad ,&{{ -3}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-8-(-2)]^2+[-3-5]^2} \\\\\\ d=\sqrt{(-8+2)^2+(-3-5)^2}\implies d=\sqrt{(-6)^2+(-8)^2} \\\\\\ d=\sqrt{36+64}\implies d=\sqrt{100}\implies d=10[/tex]
that means the radius r = 5.
now, what's the center? well, the Midpoint of the diagonals, is really the center of the circle, let's check,
[tex]\bf \textit{middle point of 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -2}}\quad ,&{{ 5}})\quad % (c,d) &({{ -8}}\quad ,&{{ -3}}) \end{array}\qquad \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right) \\\\\\ \left( \cfrac{-8-2}{2}~,~\cfrac{-3+5}{2} \right)\implies (-5~,~1)[/tex]
so, now we know the center coordinates and the radius, let's plug them in,
[tex]\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad \begin{array}{lllll} center\ (&{{ h}},&{{ k}})\qquad radius=&{{ r}}\\ &-5&1&5 \end{array} \\\\\\\ [x-(-5)]^2-[y-1]^2=5^2\implies (x+5)^2-(y-1)^2=25[/tex]
The equation of the circle where the vertices of an inscribed square are (-9,4),(-2,5),(-8,-3),(-1,-2) is (x+5)² + (y-1)² = 25. This is found by getting the average x and y coordinates to find the center of the circle, calculating the radius, and applying it to the standard equation of a circle.
Explanation:To find the equation of the circle where the vertices of an inscribed square are located at (-9,4),(-2,5),(-8,-3),(-1,-2), first, you need to find the center of the circle. In the case of a square inscribed in a circle, the center of the square is the same as the center of the circle. This can be found by averaging the x and y coordinates of the vertices of the square.
The average x-coordinate = [(-9) + (-2) + (-8) + (-1)] / 4 = -5
The average y-coordinate = [4 + 5 - 3 - 2] / 4 = 1
So, the center of the circle is at (-5, 1).
Next, you need to calculate the radius of the circle. The radius in an inscribed square is the half of the square's diagonal. Calculate the distance between two opposite vertices of the square - for example (-9,4) and (-1,-2). Apply the distance formula: √[(-1 - (-9))^2 + (-2 - 4)^2 ]. That's √[64 + 36], giving a distance of √100 = 10. As this is the diagonal of the square, the radius would be half of this, giving us a radius of 5.
Finally, the equation of the circle in its standard form (x-h)² + (y-k)² = r² can be written as (x+5)² + (y-1)² = 5².
Thus, the equation of the circle is (x+5)² + (y-1)² = 25.
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Therefore, the equation of the curve is:
[tex]\[ 6y^2 = -\frac{1}{2}x^2 + 1536 \][/tex]
This can also be written in standard form as:
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The value of y'' will be (- 5x/y² - 25x⁴/2y⁵).
What is implicit differentiation?
In implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variable as a function of other.
Given that;
The equation is;
5x³ + 2y³ = 1
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Write the number 63 in four different ways
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1) 63 ones
2) 6 tens and 3 ones
3) 3 tens and 33 ones
4) 4 tens and 23 ones
Solution:
we have been asked to write the number 63 in four different ways. It can be done as follows
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This one of the many way of writing 63.
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Here we have used the concept of subtraction to write 63.
[tex]63=7\times10-7[/tex]
Concept of subtraction and multiplication to write the number 63.
Hence the four ways have been written.
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quotient is division
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(x)(x + 2) = 6[((x) + (x+2)] -1
x² + 2x = 6(2x + 2) - 1
x² + 2x = 12x +12 - 1
And x² - 10x - 11=0
Solve this quadratic expression:
x' = [+10 +√(10²- 4.(1)(-11)]/2 and x" = [+10 -√(10²- 4.(1)(-11)]/2
x' = [10 + √144]/2 and x" = [10 - √64]/2
x' = (10+12)/2 and x" = (10-12)/2
x = 11 and x = -1
We have 2 solutions that satisfy the problem:
1st for x = 11, the numbers at 11 and 13
2nd for x = - 1 , the numbers are -1 and +1
If you plug each one in the original equation :(x)(x + 2) = 6[((x) + (x+2)] -1
you will find that both generates an equlity
Let a be a rational number and b be an irrational number. Which of the following are true statements?
A. The sum of a and b is never rational
B. The product of a and b is rational
C. b^2 is sometimes rational
D. a^2 is always rational
E. √a is never rational
F. √b is never rational
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is a integer a rational number
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Generate a process that would generate random numbers between –1 and 1 for x and y positions of 10000 points and count how many points fall with the circle, then determine an estimated version of π. (see below to generate floating point random numbers. this generation is not needed to describe the process, but to implement the process)
Answers
import random
import math
count_inside = 0
for count in range(0, 10000): d = math.hypot(random.random(), random.random())
if d < 1: count_inside += 1
count += 1
print 4.0 * count_inside / count
The question is about using the Monte Carlo method to estimate the value of π, by generating random points in a square, counting how many fall inside a unit circle, and finally using this count to calculate π.
Explanation:This question involves the implementation of a famous problem in computational mathematics known as the Monte Carlo method for estimating the value of π:
First, initialize a variable, say 'count', to 0. This variable will keep track of the number of points falling within the circle.Next, carry out a 'for' loop to generate 10000 points. Each point will have x and y coordinates, each a random floating point number between -1 and 1.For each generated point (x, y), calculate the distance from the point to the origin (0,0), which is sqrt(x^2 + y^2). If this distance is less than or equal to 1, it means that the point lies within the unit circle, so increment the 'count' variable by one.After all 10000 points have been generated and tested, 'count' will hold the total number of points lying within the circle. To determine an estimation of π, use the formula 4 * (count / total_number_of_points).This process is based on the mathematical fact that the ratio of the area of a unit circle to the square it is inscribed in is π/4.
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Can the sum of two negative numbers ever be positive?
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two negatives equal positive, and one negative and one positive is i think negative
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Answers
2+√4y=11+3=14
√4y=14-2=12
root 4y=2y
2y=12,divide both side by 2
y=6
Final answer:
To solve the equation 2 + √(4y - 3) = 11 for y, subtract 2 from both sides, square both sides to remove the square root, then solve for y to find that y = 21.
Explanation:
The question asks to solve the equation 2 + √(4y - 3) = 11 for y. To solve for y, follow the steps below:
First, subtract 2 from both sides of the equation to isolate the square root term: √(4y - 3) = 9.
Now, square both sides of the equation to eliminate the square root: (√(4y - 3))^2 = 9^2, which simplifies to 4y - 3 = 81.
Add 3 to both sides of the equation to isolate the term with y: 4y = 84.
Finally, divide both sides by 4 to solve for y: y = 21.
Therefore, the solution to the equation is y = 21.
Lean wants to giftwrap a present she got for her little brother. how many square inches of giftwrap will be needed to cover a box that is 5in × 7in × 3in?
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so.. notice, the box for the gift is really just a rectangular prism, which is really just 6 rectangles stacked up to each other at the edges.
the front and back are two 5x3 rectangles
the left and right are just two 3x7 rectangles
and the top and bottom are just two 5x7 rectangles
so, just get the area of each, sum them up, and that's the area of the rectangular prism and thus of the box, and that's how many in² she will need.
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[tex]\bf \begin{array}{ccll} gallons&mowed\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ \frac{1}{2}&\frac{1}{4}\\\\ g&\frac{4}{4} \end{array}\implies \cfrac{\frac{1}{2}}{g}=\cfrac{\frac{1}{4}}{\frac{4}{4}}\implies \cfrac{\frac{1}{2}}{\frac{g}{1}}=\cfrac{\frac{1}{4}}{\frac{4}{4}}\implies \cfrac{1}{2}\cdot \cfrac{1}{g}=\cfrac{1}{4}\cdot \cfrac{4}{4} \\\\\\ \cfrac{1}{2g}=\cfrac{1}{4}\implies 4=2g\implies \cfrac{4}{2}=g\implies 2=g[/tex]