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3 years ago

5

Work out the diffeence beteeen 77.5 and 2.27

2 answers:

dimaraw [331]3 years ago

8 0

75.37 if you need more help

Stella [2.4K]3 years ago

6 0

77.5
- 2.27
______
75.37

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Simplify the expression -4-6(7-3)

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1) Distribute the -6 to the parentheses

-4 - (6*7) - (-6*3)

2) Simplify

-4 - 42 + 18

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F(x) = –2x – 15<br> g(x) = 2x² + 3x + 3<br> Find: (fxg)(x)

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Match each spherical volume to the largest cross sectional area of that sphere

zlopas [31]

Answer:

Part 1) $324\pi\ units^{2}$ ------> $7,776\pi\ units^{3}$

Part 2) $36\pi\ units^{2}$ ------> $288\pi\ units^{3}$

Part 3) $81\pi\ units^{2}$ ------> $972\pi\ units^{3}$

Part 4) $144\pi\ units^{2}$ ------> $2,304\pi\ units^{3}$

Step-by-step explanation:

we know that

The largest cross sectional area of that sphere is equal to the area of a circle with the same radius of the sphere

Part 1) we have

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$A=\pi r^{2}$

so

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$V=\frac{4}{3}\pi r^{3}$

For $r=18\ units$

substitute

$V=\frac{4}{3}\pi (18)^{3}$

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Part 2) we have

$A=36\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

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$36\pi=\pi r^{2}$

Solve for r

$r^{2}=36$

$r=6\ units$

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The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=6\ units$

substitute

$V=\frac{4}{3}\pi (6)^{3}$

$V=288\pi\ units^{3}$

Part 3) we have

$A=81\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

so

$81\pi=\pi r^{2}$

Solve for r

$r^{2}=81$

$r=9\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=9\ units$

substitute

$V=\frac{4}{3}\pi (9)^{3}$

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Part 4) we have

$A=144\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

so

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Solve for r

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Use pythagoras
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