Can someone please help me and give me reasoning explanation for this? I’m confused.

Mathematics

1 answer:

sattari [20]1 year ago

7 0

Answer is a. (-13, 12) b. (4, 11)

Step by step

When your rule is
(x,y) ➡️ (x -8, y +2)

a)
You take your input (-5, 10) and add your input rule to that input to get your output.

Input of x is -5, add x rule -8 = -13
Input of y is 10, add y rule +2 = 12

Your output or new coordinate is (-13, 12)

b) You take your input (12, 9) and add your input rule to that input to get your output.

Input of x is 12, add x rule -8 = 4
Input of y is 9, add y rule +2 = 11

Your output or new coordinate is ( 4, 11)

If your rule changes the process is still the same. Remember to watch adding or subtracting negatives. You can use a graph and plot your input and count out your “rule” to double check your work if you want.

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Salsk061 [2.6K]

Answer:

The cells membrane is selectively permeable, meaning only certain substances can easily pass through it. Only small, non polar molecules can easily pass through the small spaces between the hydrophobic heads and phospholipids. Water would have trouble due to it being a polar molecule. Proteins and carbs are quite large, making it difficult for them to pass through. Salt, which ionizes water, would also have difficulty passing through as it is a charged particle. Oxygen, on the other hand, is a non polar molecule that is small, so it can easily pass through the phospholipid bilayer.

Step-by-step explanation:

6 0

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.

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