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3 years ago

Plot on graph.

Mathematics

1 answer:

Rus_ich [418]3 years ago

7 0

This equation is in vertex form so finding the vertex is easy.

Vertex: (-1,4)

Once you find the vertex, axis of symmetry is easy.

x=-1

Then find y intercept by finding h(0).

h(0)=(0+1)^2-4=-3

Now find x intercepts by solving for x when h(x)=0.

(x+1)^2-4=0

(x+1)^2=4

x+1=±2

x=±2-1

x=1,-3

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Where would (0.5,2) be potted ? Please answer this !!

NISA [10]

(0.5,2)

0.5 is x value

2 is y value

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x value is on dashed line (0.5)

y value is on more separated line going up and down. (2)

To plot 0.5, you can put it as an equation. 0.5= 1/2. So to plot it, 0.5 is half of 1. So plot it as in between the 1 and 0.

To plot 2 go up to the two on the y axis and then go over 0.5 and plot a point.

That is you're plotted point for (0.5,2)

Hope it helps!

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Step-by-step explanation:

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3 years ago

Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

hram777 [196]

Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

Put $(x,y)=(1,1)$

$\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9$

This is true.

So, the point $(1,1)$ satisfies the equation $\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

So, point H is $(1,1)$ .

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Anon25 [30]

Answer:

7-(3x)=11

Step-by-step explanation:

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