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Mariana [72]
3 years ago
15

drag statements and reasons to each row to show why the slope of the line between r and s is the same as the slope between s and

t, given that triangles a and b are similar.

Mathematics
2 answers:
vesna_86 [32]3 years ago
8 0

Answer:

For statement \frac{5}{3}=slope, the reason is Definition of slope.

and for statement  \frac{5}{3}=\frac{15}{9}, the reason is Triangle A is similar to triangle B.

Step-by-step explanation:

If slope of triangles are equal than Triangle A is similar to triangle B.

so, in statement \frac{5}{3}=\frac{15}{9} when we simplify

we get

\frac{5}{3}=\frac{5}{3}

then the reason for this statement is "If slope of triangles are equal"

And the definition of slope in general is m=\frac{y}{x}

so, for the statement \frac{5}{3}=slope, the reason is Definition of slope.

serg [7]3 years ago
6 0

Sorry this came pretty late but I just answered the question and thought it might help someone else later

Answer:                    

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In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC an
Airida [17]

This is a little long, but it gets you there.

  1. ΔEBH ≅ ΔEBC . . . . HA theorem
  2. EH ≅ EC . . . . . . . . . CPCTC
  3. ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
  4. ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
  5. ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
  6. ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
  7. ΔDAC ≅ ΔDAG . . . HA theorem
  8. DC ≅ DG . . . . . . . . . CPCTC
  9. ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
  10. ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
  11. ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
  12. ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
  13. (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
  14. ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
  15. This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
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3 years ago
Is 8.26 an irrational number
densk [106]

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no it is not

Step-by-step explanation:

6 0
3 years ago
The point P(8, −3) lies on the curve y = 3/(7 − x). (a) If Q is the point (x, 3/(7 − x)), use your calculator to find the slope
yan [13]

Answer:

Slope of the line PQ is -63.434948.

Step-by-step explanation:

Given that,

The point P(8,-3) lies on the curve y=\frac{3}{7-x}.

If Q is the point lies on (x,\frac{3}{7-x} ).

To find:- Find the slope of line PQ.

So,  

The coordinates of point Q when it lies on (x,\frac{3}{7-x} )

        if x=1 then y= \frac{3}{7-1} =\frac{3}{6} =\frac{1}{2}

       So,   Q ≡ (1,\frac{1}{2} ) and many points can be calculated by given Equation.

Using the formula when two points (x_{1} ,y_{1} ) \& (x_{2}, y_{2}  ).

                Slope=Tan\theta = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

Then, substituting the coordinates we get,

             Slope = \frac{1-8}{\frac{1}{2}-(-3) }

             Slope = \frac{-7}{\frac{1}{2}+3 } = \frac{-7}{\frac{7}{2} }

             Slope = \frac{-14}{7}=-2

              tan\theta=-2   ⇒  \theta = tan^{-1} (-2)

               \theta= -63.434948

Therefore,

Slope of the line mPQ is -63.434948.

                     

7 0
3 years ago
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