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Wewaii [24]
2 years ago
11

Geometry Pythagorean theorem

Mathematics
1 answer:
lara31 [8.8K]2 years ago
5 0

Answer:

Below!

Step-by-step explanation:

Using Pythagoras theorem, I will solve all of the problems.

<h3>________________________________________________</h3>

<u>Question 9:</u>

  • 10² = 6² + x²
  • => 100 = 36 + x²
  • => 100 - 36 = x²
  • => 64 = x²
  • => x = 8
<h3>________________________________________________</h3>

<u>Question 10:</u>

  • 26² = 24² + x²
  • => 676 = 576 + x²
  • => 676 - 576 = x²
  • => 100 = x²
  • => x = 10
<h3>________________________________________________</h3>

<u>Question 11:</u>

  • 15² = 12² + x²
  • => 225 = 144 + x²
  • => 225 - 144 = x²
  • => 81 = x²
  • => x = 9
<h3>________________________________________________</h3>

<u>Question 12:</u>

  • x² = 8² + 12²
  • => x² = 64 + 144
  • => x² = 208
  • => x = √208
  • => x = 14.2 (Rounded)
<h3>________________________________________________</h3>

<u>Question 13:</u>

  • 7² = 2² + x²
  • => 49 = 4 + x²
  • => 49 - 4 = x²
  • => 45 = x²
  • => x = √45
  • => x = 6.7 (Rounded)
<h3>________________________________________________</h3>

<u>Question 14</u>

First, let's solve for the variable x using Pythagoras theorem.

  • => 5² = 3² + x²
  • => 25 = 9 + x²
  • => 16 = x²
  • => x = 4 units

Now, let's solve for the variable y using Pythagoras theorem.

  • (3 + 6)² = 5² + y²
  • => (9)² = 25 + y²
  • => 81 = 25 + y²
  • => y² = 56
  • => y = √56
  • => y = 7.5 (Rounded) units

Answers (Nearest tenth):

  • x = 4 units
  • y = 7.5 units
<h3>________________________________________________</h3>

<u>Question 15:</u>

First, let's find the value of the variable y using Pythagoras theorem.

  • 8² = 6² + y²
  • => 64 = 36 + y²
  • => 28 = y²
  • => y = √28
  • => y = 5.3 (Rounded) units

Now, let's find the value of the variable x using multiplication.

  • x = 2y
  • => x = 2(5.3)
  • => x = 10.6 units

Answer (Nearest tenth)

  • x = 10.6 units
  • y = 5.3 units
<h3>________________________________________________</h3>
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WILL GIVE BRAINLY FOR ANSWER!! Please help with this question!!! Given the piecewise function: f(x) = 1/2x + 5, x &gt; 2 6, x =
Katarina [22]

Answer:

A)

f'(x) = \left\{        \begin{array}{lIl}            \frac{1}{2} & \quad x >2 \\            0& \quad x =2\\1&\quad x

B) Continuous but not differentiable.

Step-by-step explanation:

So we have the piecewise function:

f(x) = \left\{        \begin{array}{lIl}            \frac{1}{2}x+5 & \quad x >2 \\            6& \quad x =2\\x +4&\quad x

A)

To write the differentiated piecewise function, let's differentiate each equation separately. Thus:

1)

\frac{d}{dx}[\frac{1}{2}x+5}]

Expand:

\frac{d}{dx}[\frac{1}{2}x]+\frac{d}{dx}[5]

The derivative of a linear equation is just the slope. The derivative of a constant is 0. Thus:

\frac{d}{dx}[\frac{1}{2}x+5}]=\frac{1}{2}

2)

\frac{d}{dx}[6]

Again, the derivative of a constant is 0. Thus:

\frac{d}{dx}[6]=0

3)

We have:

\frac{d}{dx}[x+4]

Expand:

\frac{d}{dx}[x]+\frac{d}{dx}[4]

Simplify:

=1

Now, let's substitute our original equations for the differentiated equations. The inequalities will stay the same. Therefore:

f'(x) = \left\{        \begin{array}{lIl}            \frac{1}{2} & \quad x >2 \\            0& \quad x =2\\1&\quad x

B)

For a function to be differentiable at a point, the function <em>must </em>be a) continuous at that point, and b) the left and right hand derivatives must be equivalent.

Let's first determine if the function is continuous at the point. Remember that a function is continuous at a point if and only if:

\lim_{x \to n^-} f(n)= \lim_{x \to n^+}f(n)=f(n)

Let's find the left hand limit of f(x) at it approaches 2.

\lim_{x \to 2^-}f(x)

Since it's coming from the left, let's use the third equation:

\lim_{x \to 2^-}f(x)\\=\lim_{x \to 2^-}(x+4)

Direct substitution:

=(2+4)=6

So:

\lim_{x \to 2^-}f(x)=6

Now, let's find the right-hand limit:

\lim_{x \to 2^+}f(x)

Since we're coming from the right, let's use the first equation:

\lim_{x \to 2^+}(\frac{1}{2}x+5)

Direct substitution:

(\frac{1}{2}(2)+5)

Multiply and add:

=6

So, both the left and right hand limits are equivalent. Now, find the limit at x=2.  

From the piecewise function, we can see that the value of f(2) is 6.

Therefore, the function is continuous at x=2.

Now, let's determine differentiability at x=2.

For a function to be differentiable at a point, both the right hand and left hand derivatives must be equivalent.

So, let's find the derivative of the function as x approahces 2 from the left and from the right.

From the differentiated piecewise function, we can see that as x approaches 2 from the left, the derivative is 1.

As x approaches 2 from the right, the derivative is 1/2.

Therefore, the right and left hand derivatives are <em>not</em> the same.

Thus, the function is continuous but <em>not</em> differentiable.

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3 years ago
How do you do this I have no idea
sveta [45]

Answer:

A

Step-by-step explanation:

Eliminate B C and D

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3 years ago
Graph any third-degree polynomial (a cubic graph) that has exactly one x-intercept, a relative/global minimum at (-2.1), and a r
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Step-by-step explanation:

A notable ______ of the symposium was the speech by a famous entrepreneur.

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2 years ago
Okay some one help meh plz
djverab [1.8K]

Answer:

Point a is (-1/2,-1/4)

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Find the smallest value of $x$ such that $x^2 + 10x + 25 = 8$.
Mashutka [201]

Answer:

-5 -√8  or ≈ -7.83

Step-by-step explanation:

<u>Solving the quadratic equation, to find its roots:</u>

  • x^2 + 10x + 25 = 8           ⇒ The left side is perfect square
  • (x+5)^2=8                         ⇒ Getting square root of both sides
  • x+5 = ± √8                       ⇒ There are 2 roots
  • x = - 5 ± √8

<u>The smallest root is:</u>

  • x= -5 - √8 ≈ -7.83
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4 years ago
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