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Oxana [17]
3 years ago
13

What is the approximate area of the shaded region?

Mathematics
2 answers:
Gwar [14]3 years ago
8 0

Answer:

21.5^2

Step-by-step explanation:

tensa zangetsu [6.8K]3 years ago
3 0
That 10 is the diameter of the circle.

First let's find the area of the square.

A = s^2

Where 's' is the side.

A = 10^2

A = 100

Now let's find the area of the circle.

The '10' is also the diameter of the circle.

A = pi * r^2

A = 3.14 * 5^2

A = 3.14 * 25

A = 78.5

Now subtract this from the area of the square.

100 - 78.5 = 21.5

So the approximate area of the shaded region is 21.5cm^2
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Step-by-step explanation:

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that is the answer

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I have these factoring questions I need help with.
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f(x)=4x^2+4x-3

1. Determine y-intercept.

<em>The y-intercept is (0,-3). Substitute x = 0 in the equation as we get f(x) = -3 as the y-intercept.</em>

2. Determine the zeros.

<em>Factor the polynomials first. </em>(2x-1)(2x+3) <em>This is the factored form. The zeros are the roots of equation. Therefore, the zeros are 1/2 and -3/2</em>

3. Determine the Axis of Symmetry

<em>We can solve this by using the formula of </em>x=-\frac{b}{2a}  <em>However, I'll be solving the Axis of Symmetry with Calculus instead.</em>

<em />f'(x)=2(4x^{2-1})+1(4x^{1-1})-0\\f'(x)=8x+4<em />

<em>Then let f'(x) = 0 to find the Axis of Symmetry.</em>

<em />8x+4=0\\8x=-4\\x=-\frac{4}{8}\\x=-\frac{1}{2}<em />

<em>Therefore, the Axis of Symmetry is -1/2</em>

4. Determine the vertex.

<em>Substitute the value of Axis of Symmetry in f(x).</em>

<em />f(x)=4(-\frac{1}{2})^2+4(-\frac{1}{2})-3\\f(x)=4(\frac{1}{4})-2-3\\f(x)=1-2-3\\f(x)=-4<em />

<em>Therefore the vertex is at (-1/2, -4)</em>

5. Does the vertex representing max-pont or min-point?

<em>The vertex represents the minimum point. The graph is upward, meaning the minimum point is the point that gives the LOWEST Y-VALUE. </em>

7. Graph f(x)

<em>Unfortunately I won't be able to graph. But I can tell you to graph a parabola that has the most curve at the vertex and intercepts y-axis at (0,-3).</em>

<em>--------------------------- </em>End of Part 1 --------------------------------

2. Write the general form of Quadratic Function in standard form.

y=ax^2+bx+c

<em>The graph can be easily determined about the y-intercept and being an upward or downward parabola along with how narrow or wide the parabola is. </em>

<em>For example, c value is the y-intercept as defined. When a > 0, the parabola is upward and when a < 0, the parabola is downward. The more value of | a | is, the more narrow it will be and the less value of | a | it is, the wider it will be.</em>

3. Write in Factored Form

y=(x+a)(x+b)\\y=(x-a)(x-b)\\y=(x+a)(x-b)\\y=(x-a)(x+b)

<em>These are factored forms with different types of operators.</em>

<em>The equation can be easily determined about the roots of equation. For example, if the function is in f(x) = (x+2)(x-1) Then the roots would be x = -2 and 1 as the graph will intercept x-axis at (-2,0) and (1,0)</em>

4. Write in Vertex Form.

y=a(x-h)^2+k

<em>The equation can be easily determined for the vertex, axis of symmetry and the same narrow/wide/upward/downward parabola again.</em>

<em>The vertex is at (h,k) and the axis of symmetry is at x = h.</em>

<em>For example, </em>y=(x-2)^2+3<em />

<em>The vertex would be at (2,3) and the axis of symmetry is x = 2.</em>

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