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MrRissso [65]
2 years ago
11

ory has designed a special table with a circular cutout in the center to be filled with glass. The radius of the cutout is 7 inc

hes. What is the area of the tabletop after he cuts out the inner circle? Use 3.14 for Pi and round the answer to the nearest whole number. 154 square inches
Mathematics
1 answer:
irakobra [83]2 years ago
3 0

Based on the area of the tabletop with the inner circle, the area after the inner circle has been cut out is 12,309 inch².

<h3>How do we find the new area of the tabletop?</h3>

We are told that the area of the tabletop is 12,463 inch².

We then need to find the area of the circlular cutout as:

= π x radius x radius

= 3.14 x 7 x 7

= 154 inch²

The new area after the cutout is removed is therefore:

= Current area - area of circular cutout

= 12,463 - 154

= 12,309 inch ²

Rest of the question:

Area of table including the cutout is 12,463 square inches.

Find out more on area at brainly.com/question/13183446.

#SPJ1

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Answer:

64

Step-by-step explanation:

So here you're just completing the square. the equation you gave is simply: n^2+16n+c where c is the unknown value we're solving for. Whenever you complete the square, you add (b/2)^2

The reason for this, is because whenever you write a binomial as a perfect square it's in the form: (x+b)^2 and this binomial expands out to become: x^2+2bx+b^2

If we write the second term of the binomial as b/2 we get:

(x+\frac{b}{2})^2=x^2+2(\frac{b}{2})x+(\frac{b}{2})^2

which simplifies to:

(x+\frac{b}{2})^2=x^2+bx+(\frac{b}{2})^2

and as you can see the last term is (b/2)^2, which is why we need to add that part for it to be a perfect square.

So we would need to add (16/2)^2 = 8^2 = 64

This way, we can express it as a perfect square binomial: (n+8)^2 which expands out to: n^2+2(8)(n)+8^2 = n^2+16n+64

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1 year ago
8. If3:8=c:56, the value of c will be
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The scale for the second values is 56/8 = 7

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3 years ago
Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and
Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates (x_A,y_A)

Vectors AB and AB' are perpendicular, then

\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0

Vectors AC and AC' are perpendicular, then

\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0

Now, solve the system of two equations:

\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.

Subtract these two equations:

5x_A-y_A-8=0\Rightarrow y_A=5x_A-8

Substitute it into the first equation:

x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2

Then

y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

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the answer would be C

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