6x + y = -40 is the correct answer, please brainliest if you can. Hope that helps :)
Answer:
The proof is below
Step-by-step explanation:
Given a parallelogram ABCD. Diagonals AC and BD intersect at E. We have to prove that AE is congruent to CE and BE is congruent to DE i.e diagonals of a parallelogram bisect each other.
In ΔACD and ΔBEC
AD=BC (∵Opposite sides of a parallelogram are equal)
∠DAC=∠BCE (∵Alternate angles)
∠ADC=∠CBE (∵Alternate angles)
By ASA rule, ΔACD≅ΔBEC
By CPCT(Corresponding Parts of Congruent triangles)
AE=EC and DE=EB
Hence, AE is conruent to CE and BE is congruent to DE
Answer:
Degree = 1
Step-by-step explanation:
Given:
The differential equation is given as:

The given differential equation is of the order 2 as the derivative is done 2 times as evident from the first term of the differential equation.
The degree of a differential equation is the exponent of the term which is the order of the differential equation. The terms which represents the differential equation must satisfy the following points:
- They must be free from fractional terms.
- Shouldn't have derivatives in any fraction.
- The highest order term shouldn't be exponential, logarithmic or trigonometric function.
The above differential equation doesn't involve any of the above conditions. The exponent to which the first term is raised is 1.
Therefore, the degree of the given differential equation is 1.
Answer: A. u = ⟨3, 4⟩ and v = ⟨4, –3⟩
Step-by-step explanation:
edge2020
Answer:
Step-by-step explanation:
17 units