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andrew11 [14]
3 years ago
11

Help?

Mathematics
2 answers:
posledela3 years ago
4 0

Answer:

B

Step-by-step explanation:

Rashid [163]3 years ago
3 0
*Hint: When a figure is reflective about the y-axis, the rule would be (-x, y)

So as you can see, all the points of the figure changed from a negative x to a positive one but kept the same y value. And then after the translation up by 6 units, the final rule would be (-x, y + 6) and if you were to plug in the points, you would see they would end up as the image.

The answer to this question would be:

B.) Reflection about the y-axis followed by translation up by 6 units.
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Rearrange the equation so r is the independent variable.<br><br> 10q - 5r = 30
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Answer:

q = 3 + 1/2r

Step-by-step explanation:

10q - 5r = 30

We want to solve for q

Add 5r to each side

10q - 5r+5r = 30+5r

10q = 30+5r

Divide by 10

10q/10 = 30/10 +5r/10

q = 3 + 1/2r

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Pls answer the question attached
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\huge\mathfrak{\underline{answer:}}

\large\bf{\angle ACD = 105°}

__________________________________________

\large\bf{\underline{Here:}}

  • BCD is an isosceles right triangle , right angled at D
  • ABC is an equilateral triangle

\large\bf{\underline{To\:find:}}

  • ∠ ACD

\large\bf{In\: triangle\:ABC:}

❒ Sum of all angles is 180° , since it is an equilateral triangle all the three angles would be same

\large\bf{\underline{So}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle ACB = \frac{180}{3}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle ACB = 60°}

\large\bf{In\: triangle\:BDC}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle D = 90°}

\bf{⟼\angle DBC =\angle DCB }(Isosceles triangle)

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle DBC + \angle BDC + \angle DCB = 180° }

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼2\angle DCB + 90° = 180°}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼2\angle DCB = 180 -90}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle DCB = \frac{90}{2}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle DCB = 45°}

\large\bf{\underline{Therefore,}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟹\angle ACD = \angle ACB + \angle DCB}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟹\angle ACD = 60+45}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\angle ACD = 105°}

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