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3 years ago

What is 2/3 of 460? Just a little easy one for points

Mathematics

2 answers:

In-s [12.5K]3 years ago

8 0

Multiply 460 by 2/3 by first multiplying 460 by 2, then divide that by 3:

460 x 2 = 920

920 /3 = 306 2/3

The answer is 306 2/3

Vikentia [17]3 years ago

4 0

Answer:

2/3 * 460 = 306 and 2/3

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The intersection of any two disjoint sets is a null set. Justify your answer.

Ne4ueva [31]

Answer:

Yes The intersection of any two disjoint sets is a null set

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Step-by-step explanation:

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3 years ago

“An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and wor

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Answer:

Part 1) The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Part 2) The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

Step-by-step explanation:

Step 1

Find the radius of each sphere

we know that

The circumference of a circle is equal to

$C=2\pi r$

Find the radius of the sphere with a 5.50-meter circumference

For $C=5.50\ m$

assume

$\pi =3.14$

substitute and solve for r

$5.50=2(3.14)r$

$r=5.50/[2(3.14)]=0.88\ m$

Find the radius of the sphere with a 7.85-meter circumference

For $C=7.85\ m$

assume

$\pi =3.14$

substitute and solve for r

$7.85=2(3.14)r$

$r=7.85/[2(3.14)]=1.25\ m$

step 2

Find the surface area of each sphere

The surface area of sphere is equal to

$SA=4\pi r^{2}$

Find the surface area of sphere with a 5.50-meter circumference

For $r=0.88\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(0.88)^{2}$

$SA=9.73\ m^{2}$

Find the surface area of sphere with a 7.85-meter circumference

For $r=1.25\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(1.25)^{2}$

$SA=19.63\ m^{2}$

step 3

Find the cost of finishing each sphere

we know that

To find out the cost , multiply the surface area by $92 per square meter

Find the cost of sphere with a 5.50-meter circumference

$9.73*(92)=\$895.16$

therefore

The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Find the cost of sphere with a 7.85-meter circumference

$19.63*(92)=\$1,805.96$

therefore

The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

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3 years ago

What is 2/3 of 460? Just a little easy one for points​

What is 2/3 of 460? Just a little easy one for points