Step-by-step explanation:

In this case we have:
Δx = 3/n
b − a = 3
a = 1
b = 4
So the integral is:
∫₁⁴ √x dx
To evaluate the integral, we write the radical as an exponent.
∫₁⁴ x^½ dx
= ⅔ x^³/₂ + C |₁⁴
= (⅔ 4^³/₂ + C) − (⅔ 1^³/₂ + C)
= ⅔ (8) + C − ⅔ − C
= 14/3
If ∫₁⁴ f(x) dx = e⁴ − e, then:
∫₁⁴ (2f(x) − 1) dx
= 2 ∫₁⁴ f(x) dx − ∫₁⁴ dx
= 2 (e⁴ − e) − (x + C) |₁⁴
= 2e⁴ − 2e − 3
∫ sec²(x/k) dx
k ∫ 1/k sec²(x/k) dx
k tan(x/k) + C
Evaluating between x=0 and x=π/2:
k tan(π/(2k)) + C − (k tan(0) + C)
k tan(π/(2k))
Setting this equal to k:
k tan(π/(2k)) = k
tan(π/(2k)) = 1
π/(2k) = π/4
1/(2k) = 1/4
2k = 4
k = 2
The square's diagonal is the triangle's hypotenuse.
the original Pythagorean theorem is

where
a and
b are the two sides and
c is the hypotenuse.
that means the Pythagorean theorem for this question is:

or
Answer:
AB = 16 Units
Step-by-step explanation:
In the given figure, CD is the diameter and AB is the chord of the circle.
Since, diameter of the circle bisects the chord at right angle.
Therefore, AE = 1/2 AB
Or AB = 2AE...(1)
Let the center of the circle be given by O. Join OA.
OA = OD = 10 (Radii of same circle)
Triangle OAE is right triangle.
Now, by Pythagoras theorem:
![OA^2 = AE^2 + OE^2 \\10^2 = AE^2 + 6^2 \\100= AE^2 + 36\\100-36 = AE^2 \\64= AE^2 \\AE = \sqrt{64}\\AE = 8 \\\because AB = 2AE..[From \: equation\: (1)] \\\therefore AB = 2\times 8\\\huge \purple {\boxed {AB = 16 \: Units}}](https://tex.z-dn.net/?f=%20OA%5E2%20%3D%20AE%5E2%20%2B%20OE%5E2%20%5C%5C%3C%2Fp%3E%3Cp%3E10%5E2%20%3D%20AE%5E2%20%2B%206%5E2%20%5C%5C%3C%2Fp%3E%3Cp%3E100%3D%20%20AE%5E2%20%2B%2036%5C%5C%3C%2Fp%3E%3Cp%3E100-36%20%3D%20AE%5E2%20%5C%5C%3C%2Fp%3E%3Cp%3E64%3D%20AE%5E2%20%5C%5C%3C%2Fp%3E%3Cp%3EAE%20%3D%20%5Csqrt%7B64%7D%5C%5C%3C%2Fp%3E%3Cp%3EAE%20%3D%208%20%5C%5C%3C%2Fp%3E%3Cp%3E%5Cbecause%20AB%20%3D%202AE..%5BFrom%20%5C%3A%20equation%5C%3A%20%281%29%5D%20%5C%5C%3C%2Fp%3E%3Cp%3E%5Ctherefore%20AB%20%3D%202%5Ctimes%208%5C%5C%3C%2Fp%3E%3Cp%3E%5Chuge%20%5Cpurple%20%7B%5Cboxed%20%7BAB%20%3D%2016%20%5C%3A%20Units%7D%7D%20)
Answer:
<u>Given rhombus ABCD with</u>
- m∠EAD = 67°, CE = 5, DE = 12
<u>Properties of a rhombus:</u>
- All sides are congruent
- Diagonals are perpendicular
- Diagonals are angle bisectors
- Diagonals bisect each other
<u>Solution, considering the above properties</u>
- 1. m∠AED = 90°, as angle between diagonals
- 2. m∠ADE = 90° - 67° = 23° as complementary of ∠EAD
- 3. m∠BAE = 67°, as ∠BAE ≅ ∠EAD
- 4. AE = CE = 5, as E is midpoint of AC
- 5. BE = DE = 12, as E is midpoint of BD