The perimeter of the rectangle below is 94 units. Find the length of side WX.
Write your answer without variables.
Answers
Answer:
24units
Step-by-step explanation:
2(4y + 3) + 2(5y - 1) = 94
18y + 4 = 94
y = 5
WX = 5y - 1 = 5(5) - 1 = 24units
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First we need to find the value of y .
[tex]Perimeter = 2 \times ( \: (4y + 3) + (5y - 1) \: ) \\ [/tex]
[tex]Perimeter = 2 \times (9y + 2)[/tex]
[tex]Perimeter = 18y + 4[/tex]
Question told us that perimeter is 94 ,
Thus :
[tex]94 = 18y + 4[/tex]
[tex]18y + 4 = 94[/tex]
Subtract sides 4
[tex]18y + 4 - 4 = 94 - 4[/tex]
[tex]18y = 90[/tex]
Divide sides by 18
[tex] \frac{18y}{18} = \frac{90}{18} \\ [/tex]
[tex]y = 5[/tex]
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The face sides in the rectangular are same.
Thus :
[tex]VY = WX[/tex]
So ;
[tex] WX = 5y - 1[/tex]
[tex]WX = 5(5) - 1[/tex]
[tex]WX = 25 - 1[/tex]
[tex]WX = 24[/tex]
Done...
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Answer:
Step-by-step explanation:
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Answer:
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Answer:
I hope this helps!
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b. There are no solutions, because the coefficients of the variables on both sides of the equation are different.
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Answers
Answer:
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Step-by-step explanation:
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~theLocoCoco
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Answers
Answer:
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Step-by-step explanation:
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Answer:
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Answer:
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of
be
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
I will mark branliest please
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Answers
Answer:
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Step-by-step explanation:
Answer:
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Answer:
4
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Answers
Answer:
a= -5/8
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Step-by-step explanation:
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Answer:
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A curve is such that (dy)/(dx) = 12/(2x + 1) ^ 2 and P(1, 5) is a point on the curve . (i) The normal to the curve at P crosses the x-axis at Q. Find the coordinates of Q (ii) Find the equation of the curve. (iii) A point is moving along the curve in such a way that the x-coordinate is increasing at a constant rate of 0.3 units per second. Find the rate of increase of the y-coordinate when x=1
Answers
(i) The coordinates of the point Q is [tex](x, y) = (21, 0)[/tex].
(ii) The equation of the curve is [tex]y = 6\cdot \ln (2\cdot x + 1)-1.592[/tex].
(iii) The rate of increase of the y-coordinate is 1.2 units per second.
(i) By analytical geometry, we know that the slope of a line normal of a curve ([tex]m_{\perp}[/tex]) by the following expression:
[tex]m_{\perp} = - \frac{1}{\frac{dy}{dx} }[/tex] (1)
Where [tex]\frac{dy}{dx}[/tex] is the slope of the line tangent to the curve.
If we know that [tex]x = 1[/tex], then the slope of the normal to the curve is:
[tex]\frac{dy}{dx} = \frac{12}{2\cdot (1)+1}[/tex]
[tex]\frac{dy}{dx} = 4[/tex]
And the slope of a line normal of a curve is:
[tex]m_{\perp} = -\frac{1}{4}[/tex]
By definition of secant curve we calculate the remaining coordinate of the point Q:
[tex]m_{\perp} = \frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}[/tex] (2)
Where:
[tex]x_{P}[/tex], [tex]x_{Q}[/tex] - x-coordinates of points P and Q. [tex]y_{P}, y_{Q}[/tex] - y-coordinates of points P and Q.If we know that [tex](x_{P}, y_{P}) = (1, 5)[/tex], [tex](x_{Q}, y_{Q}) = (x, 0)[/tex] and [tex]m_{\perp} = -\frac{1}{4}[/tex], then the remaining coordinate of point Q is:
[tex]-\frac{1}{4} = \frac{0-5}{x-1}[/tex]
[tex]-\frac{1}{4}\cdot (x-1) = -5[/tex]
[tex]-\frac{1}{4}\cdot x +\frac{1}{4} = -5[/tex]
[tex]\frac{1}{4}\cdot x = \frac{21}{4}[/tex]
[tex]x = 21[/tex]
The coordinates of the point Q is [tex](x, y) = (21, 0)[/tex].
(ii) The equation of the curve is derived by separating the variables and integrating the resulting expression:
[tex]dy = \frac{12}{2\cdot x + 1}\,dx[/tex]
[tex]\int\, dy = 6\int {\frac{2\,dx}{2\cdot x + 1} }[/tex]
[tex]\int \, dy = 6\int {\frac{du}{u} }[/tex]
[tex]y = 6\cdot \ln u[/tex]
[tex]y = 6\cdot \ln (2\cdot x + 1)+C[/tex]
And the integration constant is found for [tex](x,y) = (1, 5)[/tex]:
[tex]5 = 6\cdot \ln(2\cdot 1 + 1) + C[/tex]
[tex]C = -1.592[/tex]
The equation of the curve is [tex]y = 6\cdot \ln (2\cdot x + 1)-1.592[/tex].
(iii) The rate of increase of the y-coordinate by using the differential equation presented in statement and the definition of rate of change:
[tex]\frac{dy}{dt} = \left(\frac{12}{2\cdot x +1}\right)\cdot \frac{dx}{dt}[/tex] (3)
If we know that [tex]\frac{dx}{dt} = 0.3[/tex] and [tex]x = 1[/tex], then the rate of increase of the y-coordinate is:
[tex]\frac{dy}{dt} = \left[\frac{12}{2\cdot (1)+1} \right]\cdot (0.3)[/tex]
[tex]\frac{dy}{dt} = 1.2[/tex]
The rate of increase of the y-coordinate is 1.2 units per second.
To learn more on differential equations, we kindly invite to check this verified question: https://brainly.com/question/25731911
i). The coordinates of Q is equal to:
[tex](x,y) = (21, 0)[/tex]
ii). The equation of the curve is:
[tex]y = 6. In(2.x + 1) - 1.592[/tex]
iii). The rate of increase for y-coordinate when x = 1:
[tex]1.2 units[/tex] per second
i). Using analytical geometry,
[tex]xp[/tex] and [tex]xq[/tex] are the P and Q's x-coordinates.
[tex]yp[/tex] and [tex]yq[/tex] are the P and Q's y-coordinates.
Since ([tex]xp[/tex], [tex]yp[/tex]) = (1,5) and ( [tex]xq[/tex], [tex]yq[/tex]) = [tex](x, 0)[/tex] and m⊥ = - 1/4, the left coordinate of Q's point:-
-1/4 = (0-5)/(x - 1)
∵[tex]x = 21[/tex]
Thus, the coordinates of y = (21, 0)
ii). Through isolating the variables and employing integration:-
[tex]dy = \frac{12}{(2.x + 1)} dx[/tex]
by integration and integration constant (1,5), the equation is:
[tex]y = 6. In(2.x + 1) - 1.592[/tex]
iii). The rate of increase for y-coordinate:
[tex]dy/dt[/tex] = [tex]\frac{(12)}{(2.1 + 1)} .0.3[/tex]
∵ [tex]dy/dt[/tex] [tex]= 1.2[/tex]
Thus, the rate of increase for y-coordinate when x = 1 is 1.2 units per second.
Learn more about "Co-ordinates" here:
brainly.com/question/21902081
27. Use Structure Use the table at the right. Yolanda is planning a party that will take place in three rooms. a. Write an expression that can be used to represent the total amount Yolanda will need to rent all three rooms and the sound system for t hours. b. How can you use a property to write a simplified equivalent expression?
Answers
Answer:
25t+95 or 5(5t + 19)
Step-by-step explanation:
The total of the sound fee = $25
The total of the rental fee = $95
Write is as an equation and you can get 25t + 95 or 5(5t + 19)