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Kazeer [188]
3 years ago
7

What is 37.726 rounded to the nearest tenth

Mathematics
2 answers:
pochemuha3 years ago
7 0
If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up (+1).

If the number you are rounding is followed by 0, 1, 2, 3, or 4, round the number down (no change).

37.7\fbox26\approx\huge\boxed{37.7}
marin [14]3 years ago
4 0
37.726\approx37.7
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Find the midsegment of triangle ABC so that it is parallel to side BC and label it as segment EG on the graph coordinate of E: _
Sholpan [36]

Answer:

Coordinate of E: (-2,3) Coordinate of G: (2.5, 1.5)

Step-by-step explanation:

You first have to find the midpoints of AC and AB.

AC X coordinate:

X1+X2/2

Substitute the x points of AC in

1+(-5)/2=-2

AC Y coordinate:

Y1+Y2/2=3

Substitute the y points of AC in

5+1/2=3

You do the same thing with the AB line. I am pretty sure this works and I hope this helps!




5 0
2 years ago
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 
0+6z=-18
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>
3 0
2 years ago
Read 2 more answers
Solve<br> 2<br> X = 6.<br> x =<br> Help would be greatly appreciated
Murrr4er [49]

Answer:

x=3

Step-by-step explanation:

6 0
2 years ago
How does a digit in the ten thousands place compare to a digit in the thousands place?
andre [41]
Um I think it's like because there's a zero difference like ten thousands have more zeros
3 0
2 years ago
Help me please! If you DO decide to help, please explain it to me since I'm extremely confused! THANK YOU!
den301095 [7]

\bf \cfrac{(-4x^2)(2x^{-2}y)^3}{(16x^5)(4y^3)^2}\implies \cfrac{(-4x^2)(2^3x^{-2\cdot 3}y^3)}{(16x^5)(4^2y^{3\cdot 2})}\implies \cfrac{(-4x^2)(8x^{-6}y^3)}{(16x^5)(16y^6)} \\\\\\ \cfrac{-32x^{2-6}y^3}{256x^5y^6}\implies -\cfrac{x^{-4} y^3}{8x^5y^6}\implies -\cfrac{1}{8x^5x^{4}y^6y^{-3}}\implies -\cfrac{1}{8x^{5+4}y^{6-3}} \\\\\\ -\cfrac{1}{8x^9y^3}

8 0
3 years ago
Read 2 more answers
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