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3 years ago

5

Construct equilateral triangle with sides, ABC such that AB = AC = CB = 10.5 cm. Measure and write down the size of angle < A

BC.

1 answer:

tester [92]3 years ago

5 0

Answer:

∠ABC = 60°

Step-by-step explanation:

All angles in equilateral triangles are equal.

all angles in triangles added together = 180°

180°/3 = 60°

All angles in equilateral triangles are equal to 60°.

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Which expression can be used to model the phrase “the sum of three and a number”? 3+x x-3

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The class midpoint of a class is

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Answer:

The center of that class, which is the sum of its largest and smallest values divided by 2 ⇒ E

Step-by-step explanation:

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3 years ago

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

$A=324\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

so

$324\pi=\pi r^{2}$

Solve for r

$r^{2}=324$

$r=18\ units$

Find the volume of the sphere

The volume of the sphere is

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

For $r=18\ units$

substitute

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

$V=7,776\pi\ units^{3}$

Part 2) we have

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

The area of the circle is equal to

$A=\pi r^{2}$

so

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

Solve for r

$r^{2}=36$

$r=6\ units$

Find the volume of the sphere

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

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

$V=972\pi\ units^{3}$

Part 4) we have

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

The area of the circle is equal to

$A=\pi r^{2}$

so

$144\pi=\pi r^{2}$

Solve for r

$r^{2}=144$

$r=12\ units$

Find the volume of the sphere

The volume of the sphere is

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

For $r=12\ units$

substitute

$V=\frac{4}{3}\pi (12)^{3}$

$V=2,304\pi\ units^{3}$

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