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Lesechka [4]
3 years ago
12

What is the range of the function y = x2? all real numbers xzo y20

Mathematics
1 answer:
IceJOKER [234]3 years ago
6 0
All real numbers is the correct answer
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Which of these equations would give you the balance on an account with a principal of $3000 and an interest rate of 8%, compound
GrogVix [38]
It would be 3000 x 1.08^5
8 0
3 years ago
100 points , please help. I am not sure if I did this correct if anyone can double-check me thanks!
Nookie1986 [14]

Step-by-step explanation:

\lim_{n \to \infty} \sum\limits_{k=1}^{n}f(x_{k}) \Delta x = \int\limits^a_b {f(x)} \, dx \\where\ \Delta x = \frac{b-a}{n} \ and\ x_{k}=a+\Delta x \times k

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

8 0
3 years ago
Which special version of the Pythagorean Theorem can you use to find the length of any square's diagonal, d, using only the leng
kodGreya [7K]
The square's diagonal is the triangle's hypotenuse.

the original Pythagorean theorem is  a^{2} +  b^{2} = c^{2}   where a and b are the two sides and c is the hypotenuse.

that means the Pythagorean theorem for this question is:

s^{2} + s^{2} = d^{2}  or 2( s^{2} )= d^{2}
8 0
3 years ago
Read 2 more answers
Geometry pls help !!! Find the value of AB.<br> AB = [?]
V125BC [204]

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}}

4 0
3 years ago
Mhanifa can you please help? Look at the picture attached. I will mark brainliest!
Nikitich [7]

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
4 0
3 years ago
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