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d1i1m1o1n [39]
3 years ago
5

Please answer correctly !!!!!!!!! Will mark brainliest answer !!!!!!!!!!!!

Mathematics
1 answer:
dexar [7]3 years ago
3 0

Answer:

So it will take the stone 1 (one) second to reach its maximum height.

Step-by-step explanation:

Notice that the mathematical quadratic expression given is that of a parabola with branches pointing down. This means that the maximum value for that parabola is at the vertex. Your expression is also given in vertex form:

f(x)=a\,(x-x_v)^2+y_v

where the pair of coordinates (x_v,y_v) are the x coordinate of the vertex and the y coordinate of the vertex.

Since x in your case is the time the stone is on the air, then the maximum (vertex) is at the point (1, 45) where "1" is the time in seconds and 45 is the height of the stone in meters.

So it will take the stone 1 (one) second to reach its maximum height.

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Write the equation of a line that is perpendicular to x=-6x=−6x, equals, minus, 6 and that passes through the point (-1,-2)(−1,−
Alja [10]

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Answer:

  y = -2

Step-by-step explanation:

The given line is a vertical line, so a perpendicular line will be a horizontal line. It will have an equation of the form ...

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In order for the line to go through the given point, the constant in the equation must be the same as the y-coordinate of the point: -2.

Your perpendicular line is ...

  y = -2

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Answer:

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Step-by-step explanation:

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The heights of some sunflowers are shown below.
deff fn [24]

Answer:

2 sunflowers

Step-by-step explanation:

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Help help help help help
Mkey [24]

Answer:

  x = 42

Step-by-step explanation:

The marked angles are supplementary, so their sum is 180°.

  (2x +8) +(2x +4) = 180

  4x +12 = 180 . . . . . . . . . simplify

  x +3 = 45 . . . . . . . divide by 4 (because we can)

  x = 42 . . . . . . subtract 3

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<em>Additional comment</em>

A "two-step" linear equation like this one is usually solved by subtracting the unwanted constant, then dividing by the coefficient of the variable. Here, we have done those steps in reverse order. This makes the numbers smaller and  eliminates the coefficient of the variable. Sometimes I find it easier to solve the equation this way.

5 0
2 years ago
Is x+y+1=0 a tangent of both y^2=4x and x^2=4y parabolas?
Lubov Fominskaja [6]

Answer:

  yes

Step-by-step explanation:

The line intersects each parabola in one point, so is tangent to both.

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For the first parabola, the point of intersection is ...

  y^2 = 4(-y-1)

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The point of intersection is (1, -2).

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For the second parabola, the equation is the same, but with x and y interchanged:

  x^2 = 4(-x-1)

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  x = -2, y = 1 . . . . . one point of intersection only

___

If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.

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Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.

7 0
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