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ludmilkaskok [199]

3 years ago

6

Fresh cut flowers need to be in at least 4in of water. A spherical vase is filled until the surface of the water is a circle of

5in in diameter. Is the water deep enough for the flowers? Explain.

1 answer:

Aleks [24]3 years ago

8 0

Answer:

No. The depth of the water is less than the required depth(2.5 inches<4 inches)

Step-by-step explanation:

-The deepest point of the half-filled vase is the point perpendicular to the sphere's midpoint.

-This point is equivalent to the extreme end of the radius.

-Since, the diameter is 5 inches, the radius is:

$D=2r\\\\r=D/2\\\\=5/2\\\\=2.5 \ inches$

2.5inches<4 inches

Hence, the water is not deep enough for the flowers.

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Answer:

See explanation

Step-by-step explanation:

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The diagonals are perpendicular

Using the distance formula; $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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$\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}$

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$\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}$

$|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}$

$\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}$

Using the slope formula; $m=\frac{y_2-y_1}{x_2-x_1}$

The slope of EG is $m_{EG}=\frac{2c-0}{0-0}$

$\implies m_{EG}=\frac{2c}{0}$

The slope of EG is undefined hence it is a vertical line.

The slope of DF is $m_{DF}=\frac{c-c}{a+b-(-a-b)}$

$\implies m_{DF}=\frac{0}{2a+2b)}=0$

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since $|DG|\cong |GF| \cong |EF| \cong |DE|$ and $DF \perp FG$ , DEFG is a rhombus

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The slope of DE is $m_{DE}=\frac{2c-c}{0-(-a-b)}$

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The slope of FG is $m_{FG}=\frac{c-0}{a+b-0}$

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