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Lapatulllka [165]
3 years ago
10

First letter of word best fits a mark left after injured tissue has healed

Mathematics
2 answers:
GuDViN [60]3 years ago
6 0

The mark left after injured tissue has healed is called a Scar.

So, the first letter of the word is 'S'

Schach [20]3 years ago
5 0
<h2>Answer:</h2>

<u>The mark left after injured tissue has healed is called a </u><u>scar.</u>

<h2>Step-by-step explanation:</h2>

Scar is a mark which is left on the skin where a wound, burn, or sore has not healed completely r just after it is healed. The human body produces a tough fiber in the body which gives the skin strength and flexibility. The fiber is known as collagen and it helps the body to reconnect the tissues broken apart by the injury. Since the question has been asked to tell the first letter of the word so the first letter of the word SCAR is "S".

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Solve this inequality: 8z + 3 - 2z &lt; 51
slava [35]

Answer:

9x+4<m

Step-by-step explanation:

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3 years ago
Write the equation of the line that goes through each set of points------ (-1.6),(3-,2)
Maksim231197 [3]

Answer:

The equation that goes through this set of points is y = -x + 5

Step-by-step explanation:

In order to find this, we need to start by finding the slope. For that we use the slope formula.

m(slope) = (y2 - y1)/(x2 - x1)

m = (6 - -2)/(-1 - 3)

m = 4/-4

m = -1

Now that we have this, we can use the slope and a point in point-slope form to get the equation.

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y - 6 = -1(x - -1)

y - 6 = -1(x + 1)

y - 6 = -x - 1

y = -x + 5

3 0
3 years ago
Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?
zhuklara [117]

Lets Find It Out..

First we'll find the equation of ALL planes parallel to the original one.

As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
<span><span>→n</span>=<1,2−2></span>

The equation of the plane parallel to the original one passing through <span>P<span>(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span></span>is:

<span><span>→n</span>⋅< x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span><1,2,−2>⋅<x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span>x−<span>x0</span>+2y−2<span>y0</span>−2z+2<span>z0</span>=0</span>
<span>x+2y−2z−<span>x0</span>−2<span>y0</span>+2<span>z0</span>=0</span>

Or

<span>x+2y−2z+d=0</span> [1]
where <span>a=1</span>, <span>b=2</span>, <span>c=−2</span> and <span>d=−<span>x0</span>−2<span>y0</span>+2<span>z0</span></span>

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when <span>x=0</span> and <span>y=0</span>:
<span>x+2y−2z=1</span> => <span>0+2⋅0−2z=1</span> => <span>z=−<span>12</span></span>
<span>→<span>P1</span><span>(0,0,−<span>12</span>)</span></span>

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping <span>D=2</span>, and d as d itself, we get:

<span><span>D=<span><span>|a<span>x1</span>+b<span>y1</span>+c<span>z1</span>+d|</span><span>√<span><span>a2</span>+<span>b2</span>+<span>c2</span></span></span></span></span>
<span>2=<span><span><span>∣∣</span>1⋅0+2⋅0+<span>(−2)</span>⋅<span>(−<span>12</span>)</span>+d<span>∣∣</span></span><span>√<span>1+4+4</span></span></span></span>
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<span>d+1=6</span> => <span>d=5</span>
<span>→x+2y−2z+5=0</span>Second solution:
<span>d+1=−6</span> => <span>d=−7</span>
<span>→x+2y−2z−7=<span>0</span></span></span>
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Anastaziya [24]
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