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harina [27]
3 years ago
11

1. what is the setting of home sick by jean fritz

Mathematics
1 answer:
galben [10]3 years ago
8 0

Answer:

2. Jean  a ten-year-old girl in the town of Hankow. Her mother and father work for the YMCA doing relief work. She attends a British school there and hates her teacher, Miss Williams, along with a couple of the other students. In general, Jean feels culturally conflicted, the product of two cultures but not belonging to either one, and she longs for a long-promised trip to America in a couple of years.

Step-by-step explanation:

Homesick is author Jean Fritz's story of her childhood growing up in China in a town called Hankow. While Jean adopts a first-person perspective and includes mostly real-life details from her life, some parts of Homesick are fictionalized in order to add dramatic interest.

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A = L * W
A / L = W <==

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A / L = W
42/16.8 = W
2.5 = W <=== width = 2.5 inches
4 0
3 years ago
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Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

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There are 7 bags. 2 bags are big. The rest are small.How many bags are small?
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• take the left endpoints of these intervals and the values of <em>f(x)</em> these endpoints, so <em>x</em> ∈ {0, 2, 4} and <em>f(x)</em> ∈ {0, 0.48, 0.84}

• approximate the area under <em>f(x)</em> on [0, 6] with the sum of the areas of rectangles with dimensions 2 × <em>f(x)</em> (that is, width = 2 and height = <em>f(x)</em> for each rectangle)

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∫₀⁶ <em>f(x)</em> d<em>x </em>≈ 2 ∑ <em>f(x)</em>

for <em>x</em> ∈ {0, 2, 4}, which is about

∫₀⁶ <em>f(x)</em> d<em>x </em>≈ 2 (0 + 0.48 + 0.84) = 2.64

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We have an equation: 2/1= ?/2.25

Cross multiply:
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The final answer is 4.5~
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3 years ago
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