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natita [175]
1 year ago
13

According to the graph, what is(are) the solution(s) of -2x2 + 4x−2 = 0?

Mathematics
1 answer:
Ymorist [56]1 year ago
4 0

Answer:

At X=1

Step-by-step explanation:

You just look for the point that has Y=0 on the graph of the equation

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A pair of parallel lines is cut by another pair of parallel lines as shown in the figure. Which angles are congruent to angle K?
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B. angles M, E, and C.

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Angles are congruent by being corresponding angles, or vertical angles, or alternate interior or exterior angles.

Answer:   B. angles M, E, and C.

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What equation represents a circle whose center is located at (-6,2) and whose radius is 10 units?
Anna [14]
One of the equations for a circle is (x - h)² + (y - k)² = r² where (h, k) are the center and r is the radius.
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3 0
3 years ago
Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea
Aleks [24]

Recall that variation of parameters is used to solve second-order ODEs of the form

<em>y''(t)</em> + <em>p(t)</em> <em>y'(t)</em> + <em>q(t)</em> <em>y(t)</em> = <em>f(t)</em>

so the first thing you need to do is divide both sides of your equation by <em>t</em> :

<em>y''</em> + (2<em>t</em> - 1)/<em>t</em> <em>y'</em> - 2/<em>t</em> <em>y</em> = 7<em>t</em>

<em />

You're looking for a solution of the form

y=y_1u_1+y_2u_2

where

u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt

u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt

and <em>W</em> denotes the Wronskian determinant.

Compute the Wronskian:

W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}

Then

u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t

u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}

The general solution to the ODE is

y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}

which simplifies somewhat to

\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}

4 0
3 years ago
Let log base aU=X and log base aV=Y, then a to the x power =? and a to y power =?
Andrews [41]

Answer:

{ \bf{ log_{a}(U)  = x}} \\ { \boxed{ \tt{ {a}^{x}  = U}}} \\  \\ { \bf{ log_{a}(V)  = y}} \\ { \boxed{ \tt{ {a}^{y} = V }}}

3 0
3 years ago
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