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frosja888 [35]
3 years ago
11

Anyone know the answer to this algebra problem?

Mathematics
1 answer:
m_a_m_a [10]3 years ago
6 0

here you go, the 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).)
5 0
3 years ago
Can you explain this for me 7-9÷3+6×6×2-4÷2​
baherus [9]

Answer:

The Answer Is 74. Your Welcome

8 0
3 years ago
Find the tangent line equation of the curve at the given point. Y=arcsin(7x) at the point where x=sqrt2/14
Mumz [18]

Answer:

Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})

Step-by-step explanation:

The equation of the curve is

Y = sin^{-1}(7x)

To find the equation of tangent we need to differentiate this equation w.r.t x

So, differentiating we get

Y'=\frac{7}{\sqrt{1-49x^2} }

This would give the slope of the tangent line at any given point of which x coordinate is known. In the present case it is  x = \sqrt{\frac{1}{7} }

Then slope would accordingly be

Y'=\frac{7}{\sqrt{1-49/49} }

= ∞

For, x = \sqrt{\frac{1}{7} }, Y = sin^{-1}(7/7)= \pi/2

Equation of tangent line, in the point slope form, would be Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})

4 0
3 years ago
Jan’s age is 3 years less than twice Tritt’s age. The sum of their ages is 30. Write an equation that could be used to find the
musickatia [10]

Answer:

x=2y-3

Step-by-step explanation:

x=Jan's age

y=Tritt's age

4 0
3 years ago
Select the correct answer.
tankabanditka [31]

Answer:

Saving (Start) = 1000

Add to savings = 300

y = 300x + 1000

Motorcycle = 5000

Payment = 250

y = 5000 - 250x

Answer is C

Step-by-step explanation:

8 0
3 years ago
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