What is the defination of focus of a parabola on a coordinate grid

Mathematics

1 answer:

AleksAgata [21]2 years ago

4 0

Answer:

Focus of a Parabola. A parabola is set of all points in a plane which are an equal distance away from a given point and given line. The point is called the focus of the parabola and the line is called the directrix. The focus lies on the axis of symmetry of the parabola.

Step-by-step explanation:

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The best fit line is given by the equation y = 0.404x - 5.18, where y represents the height of the plant above ground level, and

IRISSAK [1]

Step-by-step explanation:

<u>The line of best fit given:</u>

y = 0.404x - 5.18

<u>It is given in the slope-intercept form:</u>

y = mx + b, where m- slope, b - y-intercept

The slope is this situation is m = 0.404 cm and it represents the average growth rate of the plant in one day.

It seems reasonable but without means of control it is difficult to trace the growth while still under the ground, during the first 13-14 days.

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2 years ago

-1/6 + 2/3 ( 9 + - 3/4) + - 1/2

Liono4ka [1.6K]

First simplify the section in the parenthesis.
-1/6 + 2/3(8 1/4) + -1/2
Then multiply 2/3 by 8 1/4.
-1/6 + 5 1/2 + -1/2
Add -1/2 to 5 1/2.
-1/6 + 5
Add 5 to -1/6.
4 5/6 is the fully simplified answer.
Hope this helps!

7 0

2 years ago

Help me pls.........!

erica [24]

If each square represents a square foot half of it would be .5 so then add all of them. The formula for Area is A=l•w and the Area formula for a triangle is A=hbb/2. Hope this helps :)

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2 years ago

Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel

Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e ( $x_o, y_o, z_o$ ) and are parallel to $\dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}$

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

$x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct$

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

$\begin {vmatrix} x = 4+ t \\ \\ \dfrac{x-4}{1} = t \begin {vmatirx} \end {vmatrix}$ $\begin {vmatrix} y = - 5+2t \\ \\ \dfrac{y+5}{2} =t \end {vmatrix}$ $\begin {vmatrix} z =2+t \\ \\ \dfrac{z-2}{1} =t \end {vmatrix}$