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zavuch27 [327]
3 years ago
11

What is the midpoint of AB if A = (−2, 2) and B = (3, −1)?

Mathematics
1 answer:
Irina-Kira [14]3 years ago
6 0
Midpoint formula (x1 + x2) / 2, (y1 + y2)/2
(-2,2)...x1 = -2 and y1 = 2
(3,-1)...x2 = 3 and y2 = -1
now we sub
m = (-2 + 3) / 2 , (2 - 1) / 2
m = (1/2, 1/2) <===
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4. Gloria the grasshopper is working on her hops.
aivan3 [116]

The path that Gloria follows when she jumped is a path of parabola.

The equation of the parabola  that describes the path of her jump is \mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}

The given parameters are:

\mathbf{Height = 20}

\mathbf{Length = 28}

<em>Assume she starts from the origin (0,0)</em>

The midpoint would be:

\mathbf{Mid = \frac 12 \times Length}

\mathbf{Mid = \frac 12 \times 28}

\mathbf{Mid = 14}

So, the vertex of the parabola is:

\mathbf{Vertex = (Mid,Height)}

Express properly as:

\mathbf{(h,k) = (14,20)}

A point on the graph would be:

\mathbf{(x,y) = (28,0)}

The equation of a parabola is calculated using:

\mathbf{y = a(x - h)^2 + k}

Substitute \mathbf{(h,k) = (14,20)} in \mathbf{y = a(x - h)^2 + k}

\mathbf{y = a(x - 14)^2 + 20}

Substitute \mathbf{(x,y) = (28,0)} in \mathbf{y = a(x - 14)^2 + 20}

\mathbf{0 = a(28 - 14)^2 + 20}

\mathbf{0 = a(14)^2 + 20}

Collect like terms

\mathbf{a(14)^2 =- 20}

Solve for a

\mathbf{a =- \frac{20}{14^2}}

\mathbf{a =- \frac{20}{196}}

Simplify

\mathbf{a =- \frac{5}{49}}

Substitute \mathbf{a =- \frac{5}{49}} in \mathbf{y = a(x - 14)^2 + 20}

\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}

Hence, the equation of the parabola  that describes the path of her jump is \mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}

See attachment for the graph

Read more about equations of parabola at:

brainly.com/question/4074088

7 0
2 years ago
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Answer:

Step-by-step explanation:

2/5 + x = 1

Multiply each term by 5

2 + 5x = 5

5x = 5 - 2

5x = 3

x = 3/5

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