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pashok25 [27]
2 years ago
11

Yo can someone help me out

Mathematics
1 answer:
Gnesinka [82]2 years ago
6 0

Answer:12

Step-by-step explanation:

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Find the area of the figure below
alex41 [277]

Answer: 29

Step-by-step explanatio?

3 0
2 years ago
Which of the following describes the square root of 41. 5,6 6,7 20,21 40,42
cestrela7 [59]

Answer:

6,7

Step-by-step explanation:

the squre root of 41 is 6.403

3 0
3 years ago
An altitude is drawn from the vertex of an isosceles triangle, forming a right angle
Hatshy [7]

Answer:56.8 inches

Step-by-step explanation:

For perimeter we must find the lengths of the 2 diagonal sides of the triangle which are the same because it is isosceles.

the triangle is cut directly in half so both halves of the base are 17/2

17/2 = 8.5

solve through pythagoren theorum

altitude^2 + (half of base)^2 = (diagonal length)^2

18^2 + (8.5)^2 = (diagonal length)^2

324 + 72.25 = (diagonal length)^2

396.3 = (diagonal length)^2

\sqrt{396.3} = diagonal length

diagonal length = 19.9

diagonal length + diagonal length + base = perimeter

19.9 + 19.9 + 17 = perimeter

56.8 inches = perimeter

6 0
2 years ago
HELP I NEED TO GET MY HOMEWORK DONE ASAP PLEASE
sertanlavr [38]
R=0.8 might be wrong tho
8 0
2 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>
<span><span>|d+1|</span>=2⋅3</span> => <span><span>|d+1|</span>=6</span>First solution:
<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>
8 0
2 years ago
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