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Luba_88 [7]
3 years ago
6

What is 1 over 2 x + 13 = 9

Mathematics
1 answer:
vekshin13 years ago
6 0
1/2x+13=9

2x1/2x+2x13=2x9

x+26=18

x=18-26

x=-8
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rewona [7]

Answer:

C and H

Explanation:

All others are not the same shape, they tend to be pointing in different directions.

7 0
2 years ago
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Proportional relationships between x and y help anyone
mojhsa [17]

Answer:

The table on the right, the green table, is the one that shows a proportional relationship

Step-by-step explanation:

6 0
3 years ago
In the Exercises below, find the scale factor. Then list all pairs of congruent angles and write the ratios of the corresponding
nadezda [96]

Answer:

8) scale factor: 3

angles Q & L, S & N, r & M

sides 10 x 3 = 30, 6 x 3 = 18, 13 x 3 = 39

9) scale factor: 0.4

angles B & F, C & G, D & H, A & E

sides 9 x 0.4 = 3.6, 7.5 x 0.4 = 3, 10 x 0.4 = 4, 12 x 0.4 = 4.8

4 0
3 years ago
Please help me:
Tasya [4]

Answer:

6 units

Step-by-step explanation:

This coordinate is a three dimensional coordinate, which involves positive and negative x,y, and z axis.

The y axis is from left to right, i.e from negative to positive.

So 6 unit left is = -6 but it explains moving left .

Thank you

3 0
3 years ago
A line is tangent to the circle x^2+y^2=41, at the point (-4,-5)
VashaNatasha [74]
If the equation of the circle is x^2+ y^2 = 41, we must first understand the parts of the equation.
 A general circle's equation is (x-h)^2+(y-k)^2= r^2  
(h.k) is the radius of the circle
r is the radius of the circle
Another useful fact to know is that tangent lines touch the circle at one point (4,5)

Since in our original equation there are no h or k values, we can assume that the center of the circle is (0,0). 

The formula for slope is <u>Y1-Y2</u>
                                   X1-X2
We can break this down with our two points (center and tangent)
  (0,0)    and   (-4,-5)
(X1,Y1) and (X2,Y2)

therefore, we will put the equation as such
<u>0-(-5)= 5</u>  = <em> </em><u><em>5</em></u>
0-(-4)= 4     <em> 4</em>

<em>this is our slope from the center to the point of tangency.</em>

We know that tangent lines are perpendicular to the radius, which we've already found the slope of. Perpendicular lines are opposite reciprocals of the line they are perpendicular to. 

Therefore, we take our slope from center to the tangent, and make it opposite and then take the reciprocal of that slope, which will give us the slope of the tangent line itself. (note: reciprocal means flip the numerator and denominator)
<u>5</u> =  <u>-5</u>  =  <u>-4</u><u>
</u>4     4        5   

Now, we have a point on the line, and the line's slope. We can use slope-intercept equation to find the equation of the line.

Slope-int      y=mx+b
(x,y) is a point,
m is the slope
b is the y intercept  ( the point where x=0, or where its on the y axis)

now we plug things in
(-4,-5) is our point,
<u>-4</u>  is our slope
5

-5=<u>-4</u>(-4)+b       After we plug things in, solve for b
     5

-5= 3.2+b

-1.8= b  or  b=  <u />1 <u>4</u>
                         5

Now we just need to rewrite our equation with all our components.
(-4.-5) = point
<u>-4</u>  = slope<u>
</u>5
1 <u>4</u> = y-intercept<u>
</u>   5

<em>y=</em><u><em>-4</em></u><em> x+ 1 </em><u><em>4</em></u><em>     This is the equation of the tangent line</em><u>
</u><em>    5         5</em>


Hope that helped
7 0
4 years ago
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