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taurus [48]
3 years ago
15

Without drawing the graph of the equation, determine how many points the given equation have in common with the x-axis and where

is the vertex in relation to the x-axis?
y = -2x^2 + x + 3

Mathematics
1 answer:
Gala2k [10]3 years ago
4 0
Remark
It's asking you to complete the square to find the vertex. 

Solve
y = -2x^2 + x + 3 Put brackets around the first 2 terms.
y = (-2x^2 + x) + 3    Pull out the 2.
y = -2(x^2 - 1/2 x) + 3 Be especially observant of the sign on 1/2x Divide the middle term's number by 2 and square it. 
y = -2(x^2 - 1/2 x + [(1/2) / 2]^2 ) + 3 [1/2 divided by 2 is 1/4. (1/4)^2 = 1/16; 2*(1/16) = 1/8
y = - 2(x^2 - 1/2 x + (1/4)^2 ) + 3 + 1/8
y = -2(x - 1/4)^2 + 3 1/8

Comment
Answer to your question. The maximum value (3 1/8) is above the y axis. That means the graph of the equation crosses the x axis twice. I'm going to include the graph because in answering the question, there is no harm in seeing the graph.

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So it’s going to be d
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4 years ago
Suppose a change of coordinates T:R2→R2 from the uv-plane to the xy-plane is given by x=e−2ucos(5v), y=e−2usin(5v). Find the abs
anzhelika [568]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The solution is  \frac{\delta  (x,y)}{\delta (u, v)} | = 10e^{-4u}

Step-by-step explanation:

From the question we are told that

        x =  e^{-2a} cos (5v)

and  y  =  e^{-2a} sin(5v)

Generally the absolute value of the determinant of the Jacobian for this change of coordinates is mathematically evaluated as

     | \frac{\delta  (x,y)}{\delta (u, v)} | =  | \ det \left[\begin{array}{ccc}{\frac{\delta x}{\delta u} }&{\frac{\delta x}{\delta v} }\\\frac{\delta y}{\delta u}&\frac{\delta y}{\delta v}\end{array}\right] |

        = |\ det\ \left[\begin{array}{ccc}{-2e^{-2u} cos(5v)}&{-5e^{-2u} sin(5v)}\\{-2e^{-2u} sin(5v)}&{-2e^{-2u} cos(5v)}\end{array}\right]  |

Let \   a =  -2e^{-2u} cos(5v),  \\ b=-2e^{-2u} sin(5v),\\c =-2e^{-2u} sin(5v),\\d=-2e^{-2u} cos(5v)

So

     \frac{\delta  (x,y)}{\delta (u, v)} | = |det  \left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] |

=>    \frac{\delta  (x,y)}{\delta (u, v)} | = | a *  b  - c* d |

substituting for a, b, c,d

=>    \frac{\delta  (x,y)}{\delta (u, v)} | =  | -10 (e^{-2u})^2 cos^2 (5v) - 10 e^{-4u} sin^2(5v)|

=>   \frac{\delta  (x,y)}{\delta (u, v)} | =  | -10 e^{-4u} (cos^2 (5v)   + sin^2 (5v))|

=>  \frac{\delta  (x,y)}{\delta (u, v)} | = 10e^{-4u}

7 0
3 years ago
Triangle ABC has vertices at A(−3, 4), B(4, −2), C(8, 3). The triangle is translated 4 units down and 3 units left. Which rule r
crimeas [40]

Option C: (x, y) \rightarrow(x-3, y-4) ; (5,-1) is the coordinates of C after translation.

Explanation:

The triangle ABC has vertices at A(−3, 4), B(4, −2), C(8, 3).

Also, given that the triangle is translated 4 units down and 3 units left.

We need to determine the coordinates of the vertex C after translation.

The triangle is shifted 4 units down means subtracting 4 units from the y - coordinate of the graph.

Thus, it can be written as y-4

Also, the triangle is shifted 3 units left means subtracting 3 units from the x - coordinate of the graph.

Thus, it can be written as x-3

Thus, the translation is given by

(x, y) \rightarrow(x-3, y-4)

The vertices of C after translation can be determined by

(8, 3) \rightarrow(8-3, 3-4)\implies (5,-1)

Thus, (x, y) \rightarrow(x-3, y-4) ; (5,-1) is the coordinates of C after translation.

Hence, Option C is the correct answer.

3 0
3 years ago
Read 2 more answers
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Marrrta [24]
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2. 2x + 5x = 8x

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Definition of vertical line test
Serhud [2]

the vertical line test is a graphical way to determine whether or not the graph is a function of x

if you can draw a vertical line on a graph and cross the graph at least 2 times, the graph is not a function

a function will have 1 y value for every x value, therefore, if we can draw a vertical line (x=c, a constant) that intersects the curve at more than 1 point, then that point where it intersects at more than 1 point is where 1 value of x coresponds to more than 1 value of y and is therefore not a function

4 0
3 years ago
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