Circle R has a radius of line segment of QR and QP is a tangent to circle R at point Q.(Picture attached)
(a) What is the measure of RQP? Explain your answer.
(b) What is the value of x? Explain your answer with work.
(c) What is the measure of QRP? Explain your answer with work.
(d) What is the measure of RPQ? Explain your answer with work.
Answers
Answer:
(a) 90°
(b) 8.75
(c) 63.75°
(d) 26.25°
Step-by-step explanation:
(a) A radius to a point of tangency is always perpendicular to the tangent line there. Q is the point of tangency of line PQ, so the segment RQ from the center of the circle, R, to that point makes a 90° angle with PQ. Angle RQP is 90°.
(b) The sum of the acute angles of a right triangle is 90°, so ...
(5x +20)° + (3x)° = 90° . . . . . the sum of the acute angles is 90°
8x + 20 = 90 . . . . . . . . . . . . simplify, divide by °
8x = 70 . . . . . . . . . . . . . . . . . subtract 20
70/8 = x = 8.75 . . . . . . . . . . . divide by the coefficient of x
(c) ∠QRP = (5x+20)° = (5·8.75 +20)° = 63.75° . . . . . use the value of x in the expression for the angle measure
(d) ∠RPQ = (3x)° = (3·8.75)° = 26.25° . . . . . use the value of x in the expression for the angle measure
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2. A unique solution exists in the entire xy-plane.
3. A unique solution exists in the region y ≤ x.
4. A unique solution exists in the region consisting of all points in the xy-plane except the origin.
5. A unique solution exists in the region x2 + y2 < 1.
Answers
A unique solution exists in the region consisting of all points in the xy-plane except the origin.
The correct option is 4.
The given differential equation is:
(x² + y²)y' = y²
The equation can be rewritten as:
We need to determine a region of the xy-plane for which the differential equation would have a unique solution whose graph passes through a point (x₀, y₀) in the region.
To determine the region, we can use the existence and uniqueness theorem for first-order differential equations.
According to the theorem, a unique solution exists in a region if the differential equation is continuous and satisfies the Lipschitz condition in that region.
To check if the differential equation satisfies the Lipschitz condition, we can take the partial derivative of the equation with respect to y:
dy/dx = y / (x² + y²)
The partial derivative is continuous and bounded in the entire xy-plane except at the origin (x=0, y=0).
Therefore, the differential equation satisfies the Lipschitz condition in the entire xy-plane except at the origin.
Since the differential equation is continuous in the entire xy-plane, a unique solution exists in any region that does not contain the origin. Therefore, the correct answer is:
A unique solution exists in the region consisting of all points in the xy-plane except the origin.
To learn more about the Lipschitz condition;
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Final answer:
The differential equation will have a unique solution in the entire xy-plane except at the origin, as both the function and its partial derivatives are continuous and well-defined everywhere except at that point.
Explanation:
To determine a region of the xy-plane where the differential equation (x2 + y2)y' = y2 has a unique solution passing through a point (x0, y0), we need to consider where the function and its derivative are continuous and well-defined. According to the existence and uniqueness theorem for differential equations, a necessary condition for a unique solution to exist is that the functions of x and y in the equation, as well as their partial derivatives with respect to y, should be continuous in the region around the point (x0, y0).
We note that both the function (x2 + y2)y' and its partial derivative with respect to y, which is 2y, are continuous and well-defined everywhere except at the origin where x = 0 and y = 0. Therefore, a unique solution exists in the region consisting of all points in the xy-plane except the origin.
From the given options, the correct answer is:
4. A unique solution exists in the region consisting of all points in the xy-plane except the origin.
Learn more about Differential Equations here:
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Answers
Answer: Two-way frequency tables are especially important because they are often used to analyze survey results. Two-way frequency tables are also called contingency tables. Two-way frequency tables are a visual representation of the possible relationships between two sets of categorical data.
Step-by-step explanation:
x|y
-2|2
1|5
3|7
---- ----
A. y = x + 4
B. y = -x
C. y = x - 4
D. y = 4 - x
Answers
Answer:
y = x + 4
Step-by-step explanation:
y = mx + b (m = slope; b = y-intercept)
I used 2 points (-2 , 2) and (1 , 5) to calculate the slope of the line
m= (y₂ - y₁)/(x₂ - x₁)
(5-2)/(1--2) = 3/3= 1
y = 1x + b
using point (1, 5)
5 = 1(1) + b
5 = 1 + b
b = 5-1
b=4
y= 1x + 4 or simply y= x + 4
Answer:
A y = x + 4
yeh i did the test
Step-by-step explanation:
Answers
Answer:8
Step-by-step explanation:
Answers
Answer:
Step-by-step explanation:
n+150+7+100=200+50+27
n+257 = 277
n = 277 - 257
n = 20
Answer:
20
Step-by-step explanation:
hope it will help goodbye.if there is any other question feel free to ask
1
2
(x - 3) = 9
Answers
( also your bottom formula is wrong it should be 1/2(x-3)=9)