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ANTONII [103]
3 years ago
9

All sides of the building shown above meet at right angles. If three of the sides measure 2 meters, 7 meters, and 11 meters as s

hown, then what is the perimeter of the building in meters?

Mathematics
1 answer:
Svetach [21]3 years ago
3 0

Answer:

Perimeter= 40 units

Step-by-step explanation:

Ok

We are asked to look for the perimeter.

We have some clue given.

All at right angle and some sides are given it's full length.

We have the bae to be 11 unit

The height to be 7 unit.

What this mean is that taking either the base or the height should sum up to either 11 or 7 respectively.

Let's go for the other side of the height.

Let's take all the vertical height and sum it up to 7 because the right side is equal to 7.

So we have 7+7+11

But it's not complete yet.

We are given a dimension 2.

And the 2 is in two places so it's total 2*2= 4

The two is for a small base .

The base is actually an extra to the 11 of the other base.

So summing up

We have 2*11 + 2*7 + 2*2

Perimeter= 22+14+4

Perimeter= 40 units

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Solve the following 3 × 3 system. Enter the coordinates of the solution below.
love history [14]
The system is:

i)    <span>2x – 3y – 2z = 4
ii)    </span><span>x + 3y + 2z = –7
</span>iii)   <span>–4x – 4y – 2z = 10 

the last equation can be simplified, by dividing by -2, 

thus we have:

</span>i)    2x – 3y – 2z = 4
ii)    x + 3y + 2z = –7
iii)   2x +2y +z = -5 


The procedure to solve the system is as follows:

first use any pairs of 2 equations (for example i and ii, i and iii) and equalize them by using one of the variables:

i)    2x – 3y – 2z = 4   
iii)   2x +2y +z = -5 

2x can be written as 3y+2z+4 from the first equation, and -2y-z-5 from the third equation.

Equalize:  

3y+2z+4=-2y-z-5, group common terms:
5y+3z=-9   

similarly, using i and ii, eliminate x:

i)    2x – 3y – 2z = 4
ii)    x + 3y + 2z = –7

multiply the second equation by 2:


i)    2x – 3y – 2z = 4
ii)    2x + 6y + 4z = –14

thus 2x=3y+2z+4 from i and 2x=-6y-4z-14 from ii:

3y+2z+4=-6y-4z-14
9y+6z=-18

So we get 2 equations with variables y and z:

a)   5y+3z=-9 
b)   9y+6z=-18

now the aim of the method is clear: We eliminate one of the variables, creating a system of 2 linear equations with 2 variables, which we can solve by any of the standard methods.

Let's use elimination method, multiply the equation a by -2:

a)   -10y-6z=18 
b)   9y+6z=-18
------------------------    add the equations:

-10y+9y-6z+6z=18-18
-y=0
y=0,

thus :
9y+6z=-18 
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z=-3

Finally to find x, use any of the equations i, ii or iii:

<span>2x – 3y – 2z = 4 
</span>
<span>2x – 3*0 – 2(-3) = 4

2x+6=4

2x=-2

x=-1

Solution: (x, y, z) = (-1, 0, -3 ) 


Remark: it is always a good attitude to check the answer, because often calculations mistakes can be made:

check by substituting x=-1, y=0, z=-3 in each of the 3 equations and see that for these numbers the equalities hold.</span>
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