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Tamiku [17]
3 years ago
6

Draw a triangle that is both a right triangle and an isosceles triangle. You can use either pencil and paper, or a draw program.

2.Label the angles. What are the measurements of the angles that are not equal to 90 degrees? 3.Describe the two sides opposite the angles that are not equal to 90 degrees. 4.Now draw an equilateral triangle. Label the measurements of the angles. Can an equilateral triangle ever be a right triangle too? Explain your answer. 5.Finally, draw a right triangle. Can this triangle ever have an obtuse angle? Explain your answer.
Mathematics
1 answer:
dem82 [27]3 years ago
6 0
2. measurements can be any to complementary angles.
   for example, 60 & 30
3. the 2 sides are equal. Since it is an isosceles triangle.
4.an equi.triangle can never be right angled. Since, all angles are 60° in an             equi.triangle.
5. right triangle can never have an obtuse angle since one angle is already          90°. therefore the other two agles are supposed to be acute angles.[ since        sum of all angles in a triangle is 180°]
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Here's the question. ​
OleMash [197]

Answer:

The value of T₂₀ - T₁₅ is <u>-20</u>.

Step-by-step explanation:

<u>Given</u> :

  • >> If for an A.P, d = -4

<u>To</u><u> </u><u>Find</u> :

  • >> T₂₀ - T₁₅

<u>Using Formula</u> :

General term of an A.P.

\star{\small{\underline{\boxed{\sf{\red{ T_n = a  + (n - 1)d}}}}}}

  • >> Tₙ = nᵗʰ term
  • >> a = first term
  • >> n = no. of terms
  • >> d = common difference

<u>Solution</u> :

Firstly finding the A.P of T₂₀ by substituting the values in the formula :

{\dashrightarrow{\pmb{\sf{ T_n = a  + (n - 1)d}}}}

{\dashrightarrow{\sf{ T_{20} = a  + (20 - 1) d}}}

{\dashrightarrow{\sf{ T_{20} = a  + (19)d}}}

{\dashrightarrow{\sf{ T_{20} = a  + 19  \times d}}}

{\dashrightarrow{\sf{ T_{20} = a  + 19d}}}

{\star \: {\underline{\boxed{\sf{\pink{ T_{20} = a  + 19d}}}}}}

Hence, the value of T₂₀ is a + 19d.

\rule{190}1

Secondly, finding the A.P of T₁₅ by substituting the values in the formula :

{\dashrightarrow{\pmb{\sf{ T_n = a  + (n - 1)d}}}}

{\dashrightarrow{\sf{ T_{15}= a  + (15 - 1) d}}}

{\dashrightarrow{\sf{ T_{15}= a  + (14) d}}}

{\dashrightarrow{\sf{ T_{15}= a  + 14 \times d}}}

{\dashrightarrow{\sf{ T_{15}= a  + 14d}}}

{\star{\underline{\boxed{\sf \pink{ T_{15}= a  + 14d}}}}}

Hence, the value of T₁₅ is a + 14d

\rule{190}1

Now, finding the difference between T₂₀ - T₁₅ :

{\dashrightarrow{\pmb{\sf{T_{20} -  T_{15}}}}}

{\dashrightarrow{\sf{(a + 19d) -  (a + 14d)}}}

{\dashrightarrow{\sf{a + 19d -  a  -  14d}}}

{\dashrightarrow{\sf{a - a + 19d -  14d}}}

{\dashrightarrow{\sf{0+ 19d -  14d}}}

{\dashrightarrow{\sf{19d -  14d}}}

{\dashrightarrow{\sf{5 \times  - 4}}}

{\dashrightarrow{\sf{ - 20}}}

{\star \: \underline{\boxed{\sf{\pink{T_{20} -  T_{15} =  - 20}}}}}

Hence, the value of T₂₀ - T₁₅ is -20.

\underline{\rule{220pt}{3.5pt}}

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Step-by-step explanation:

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