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2 years ago

Fifty percent of the surface area of the bread is crust. What is the height h?

Mathematics

1 answer:

Vikentia [17]2 years ago

5 0

Answer:

The height h is $5\ cm$

Step-by-step explanation:

we know that

The surface area of the bread is equal to

$SA=2B+Ph\\$

where

B is the area of the base

P is the perimeter of the base

h is the height

In this problem

If fifty percent of the surface area of the bread is crust

then

$2B=Ph\\$

substitute the values

$2(10)(10)=4(10)h\\$

$200=40h\\$

$h=200/40=5\ cm$

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Hassan's soccer team is trying to raise $688 to travel to a tournament in Florida, so they decided to host a pancake breakfast.

Maslowich

Answer:

5 people

Step-by-step explanation:

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2 years ago

Read 2 more answers

If 8 identical blackboards are to be divided among 4 schools,how many divisions are possible? How many, if each school mustrecei

MAXImum [283]

Answer:

There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.

Step-by-step explanation:

Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.

The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is ${11 \choose 3} = 165 .$ As a result, we have 165 ways to distribute the blackboards.

If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is ${7 \choose 3} = 35$ . Thus, there are only 35 ways to distribute the blackboards in this case.

4 0

2 years ago

A sports company wants to package a ball with a 1.5-inch radius in sets of two. They have two options: a cylinder or a square pr

guajiro [1.7K]

Step 1

Find the Volume of the two balls

we know that

the Volume of a sphere is equal to

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

In this problem we have

$r=1.5\ in$

Substitute

$V=\frac{4}{3} \pi 1.5^{3}=4.5\pi\ in^{3}$

the volume of the two balls is equal

$V=2*4.5\pi=9 \pi\ in^{3}$

Step 2

Find the volume of a cylinder

we know that

The Volume of a cylinder is equal to

$V=\pi r^{2}h$

In this problem we have

$r=1.5\ in$

$h=4*r=4*1.5=6\ in$

Substitute

$V=\pi (1.5^{2})(6)=13.5 \pi\ in^{3}$

Step 3

Find the amount of wasted space in the cylinder

Subtract the volume of the two balls from the volume of the cylinder

$13.5\pi\ in^{3}-9\pi\ in^{3}=4.5\pi\ in^{3}=14.14\ in^{3}$

Step 4

Find the volume of the square prism

The volume of the prism is equal to

$V=Bh$

where

B is the area of the base of the prism

h is the height of the prism

In this problem we have

$B=(2*1.5)^{2}=9\ in^{2}$ -----> the base is equal to the diameter

$h=4*r=4*1.5=6\ in$

Substitute

$V=9*6=54\ in^{3}$

Step 5

Find the amount of wasted space in the prism

Subtract the volume of the two balls from the volume of the prism

$54\ in^{3}-9\pi\ in^{3}=25.73\ in^{3}$

Step 6

Compare the amount of wasted space

the amount of wasted space in the cylinder is $14.14\ in^{3}$

the amount of wasted space in the prism is $25.73\ in^{3}$

Find the difference

$25.73\ in^{3}-14.14\ in^{3}=11.59\ in^{3}=11.6\ in^{3}$

The package that has the least amount of wasted space is the cylinder

therefore

<u>the answer is the option</u>

the cylinder because it has approximately 11.6 in.3 less wasted space than the prism

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2 years ago

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What is the value of n in the equation –negative StartFraction one-half EndFraction left-parenthesis 2 n plus 4 right-parenthesi

fgiga [73]

Answer:

Solution to equation: $n = 1$

Step-by-step explanation:

We are given the following equation:

$-\dfrac{1}{2}(2n + 4) + 6 = -9 + 4(2n + 1)$

We have to solve the equation to find the value of n.

Solving the equation, we get,

$-\dfrac{1}{2}(2n + 4) + 6 = -9 + 4(2n + 1)\\\\-\dfrac{1}{2}(2n + 4)-4(2n+1) = -9-6\\\\-n-2-8n-4 = -15\\\Rightarrow -9n = -15 + 6\\\Rightarrow -9n = -9\\\Rightarrow n = 1$

Thus, value of n is 1.

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2 years ago

A student wants to know how far above the ground the top of a leaning flagpole is. At high noon, when the sun is almost directly

Sonbull [250]

Answer:

16.30 ft ≈ 16 ft 4 in

Step-by-step explanation:

Please see the image attached. According to the figure, AB is the leaning flag. When the sun is directly overhead at noon, the shadow of the flag is the line AC in the figure. The length of AC is 5ft. The length of the string of plumb bob is 3ft and this is represented by the line XY. Y is the point of the plumb bob and this is at a distance of 11 in ( = $\frac{11}{12} = 0.92$ ft from the base of the flag, i.e, AY = 0.92ft.

Now, the angle between the AB and AC is taken as ∅. In traingle AXY, tan(∅) = $\frac{XY}{AY}$ . In triangle ABC, for the same ∅, tan(∅) = $\frac{BC}{AC}$ . Therefore, tan(∅) = $\frac{XY}{AY} = \frac{BC}{AC}$ . Here, XY = 3ft, AY = 0.92ft, AC = 5ft.

$\frac{XY}{AY} = \frac{BC}{AC} == \frac{3}{0.92} = \frac{BC}{5}$

Therefore, $BC = \frac{3}{0.92}$ ×5 = 16.30 ft ≈ 16 ft 4 in.

7 0

2 years ago