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Setler79 [48]
3 years ago
5

What is the area of the square adjacent to the third side of the triangle?

Mathematics
1 answer:
Naddik [55]3 years ago
3 0

Answer:

85 units^2 = Area of Blue Square

and

x = 5 units

Step-by-step explanation:

To do this we use the Pythagorean theorem (a^2 + b^2 = c^2). a and b represent the legs of the triangle whereas c represents the longest side of the triangle, or the hypotenuse.

Since we know the area of a square is the side length multiplied by itself (or the side length squared), \sqrt{35} is the side length of the pink square and \sqrt{50} is the side length of the green square.

That means a = \sqrt{35} and b = \sqrt{50} , so...

(\sqrt{35} )^{2} + (\sqrt{50})^{2}  = c^{2}

35 + 50 = c^2

85 = c^2

\sqrt{85} = c

Now we need to square the square root of 85 to find the area of the blue square.

(\sqrt{85})^{2}  = Area of blue square

85 units^2 = Area of Blue Square

To solve the other question we use the same formula again.

x^2 + 12^2 = 13^2\\x^2 +144 = 169\\-144\\x^2 =  25\\x=\sqrt{25} \\x=5

x = 5 units

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What is the largest possible integral value in the domain of the real-valued function
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Answer:

Max Value: x = 400

General Formulas and Concepts:

<u>Algebra I</u>

  • Domain is the set of x-values that can be inputted into function f(x)

<u>Calculus</u>

  • Antiderivatives
  • Integral Property: \int {cf(x)} \, dx = c\int {f(x)} \, dx
  • Integration Method: U-Substitution
  • [Integration] Reverse Power Rule: \int {x^n} \, dx = \frac{x^{n+1}}{n+1} + C

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = \frac{1}{\sqrt{800-2x} }

<u>Step 2: Identify Variables</u>

<em>Using U-Substitution, we set variables in order to integrate.</em>

u = 800-2x\\du = -2dx

<u>Step 3: Integrate</u>

  1. Define:                                                                                                            \int {f(x)} \, dx
  2. Substitute:                                                                                         \int {\frac{1}{\sqrt{800-2x} } } \, dx
  3. [Integral] Int Property:                                                                                     -\frac{1}{2} \int {\frac{-2}{\sqrt{800-2x} } } \, dx
  4. [Integral] U-Sub:                                                                                           -\frac{1}{2} \int {\frac{1}{\sqrt{u} } } \, du
  5. [Integral] Rewrite:                                                                                          -\frac{1}{2} \int {u^{-\frac{1}{2} }} \, du
  6. [Integral - Evaluate] Reverse Power Rule:                                                 -\frac{1}{2}(2\sqrt{u}) + C
  7. Simplify:                                                                                                         -\sqrt{u} + C
  8. Back-Substitute:                                                                                            -\sqrt{800-2x} + C
  9. Factor:                                                                                                           -\sqrt{-2(x - 400)} + C

<u>Step 4: Identify Domain</u>

We know from a real number line that we cannot have imaginary numbers. Therefore, we cannot have any negatives under the square root.

Our domain for our integrated function would then have to be (-∞, 400]. Anything past 400 would give us an imaginary number.

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