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2 years ago

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Jack's mother gave him 505050 chocolates to give to his friends at his birthday party. He gave 333 chocolates to each of his fri

ends and still had 222 chocolates left.

1 answer:

Margarita [4]2 years ago

4 0

Answer:

Is Jack's mom trying to give him diabetes?

Step-by-step explanation:

You might be interested in

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

Artist 52 [7]

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

$C=2\pi r$

<u><em>Find the radius of the sphere with a 5.50-meter circumference</em></u>

For $C=5.50\ m$

assume

$\pi =3.14$

substitute and solve for r

$5.50=2(3.14)r$

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

<u><em>Find the radius of the sphere with a 7.85-meter circumference</em></u>

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

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

<u><em>Find the surface area of sphere with a 5.50-meter circumference</em></u>

For $r=0.88\ m$

assume

$\pi =3.14$

substitute

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

$SA=9.73\ m^{2}$

<u><em>Find the surface area of sphere with a 7.85-meter circumference</em></u>

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

<u><em>Find the cost of sphere with a 5.50-meter circumference</em></u>

$9.73*(92)=\$895.16$

therefore

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

<u><em>Find the cost of sphere with a 7.85-meter circumference</em></u>

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

therefore

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

6 0

2 years ago

The graph shows a proportional relationship.

koban [17]

The unit rate is 4.5

8 0

3 years ago

A square has a side of 5 ft. What is<br> the perimeter of the triangle?

luda_lava [24]

I'll draw it for you cause im nice.

Ok so if you have 5ft on a square, all the sides are 5ft.

That mean that if its in half, one of the sides will be 2.5ft.

Now you can add 2.5 + 2.5 + 5+5 = 15.

So its 15 ft.

8 0

2 years ago

The sum of two numbers is 50 and the difference is 4

juin [17]

Step-by-step explanation:

let the 2 numbers be x and x + 4 as their deference is 4.

their sum = 50

now,

→ x + x + 4 = 50

→ 2x + 4 = 50

→ 2x = 50 - 4 = 46

→ x = 46/2 = 23

therefore two numbers are,

x = 23

x + 4 = 23 + 4 = 27

hope this answer helps you dear...take care and may u have a great day ahead!

3 0

2 years ago

Read 2 more answers

Select the correct answer from each drop-down menu. the endpoints of the longest chord on a circle are (4, 5.5) and (4, 10.5). t

torisob [31]

The equation of this circle in standard form is equal to (x - 4)² + (y - 8)² = 5².

Based on the calculations, the equation of this circle in standard form is equal to (x - 4)² + (y - 8)² = 5²

The equation of a circle.

Mathematically, the standard form of the equation of a circle is given by;

(x - h)² + (y - k)² = r²

Where:

h and k represent the coordinates at the center.

r is the radius of a circle.

The midpoint of the given points represents the center of this circle:

h = (4 + 4)/2 = 4

k = (5.5 + 10.5)/2 = 8

Next, we would determine the radius by using the distance formula for coordinates:

r = √[(x₂ - x₁)² + (y₂ - y₁)²]

r = √[(4 - 4)² + (10.5 - 5.5)²]

r = √[0² + 5²]

r = √25

r = 5 units.

Therefore, the equation of this circle in standard form is equal to (x - 4)² + (y - 8)² = 5².

Learn more about circle here: brainly.com/question/12823137

#SPJ4

6 0

1 year ago