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Ahat [919]
3 years ago
12

Three classes at a junior high school raised money to buy new computers. Ms. Moore’s class raised $249.00. Ms. Aguilar raised $3

96.62 more than Ms. Moore’s class. Mr. Barry raised $430.43 less than Ms. Aguilar’s class. What is the total amount of money raised by all three classes?
Mathematics
1 answer:
Olin [163]3 years ago
3 0

Answer:

The final amount is $1109.81

Step-by-step explanation:

In order to find the total amount, start with the know amount, which is Ms. Moore's class. Her class raised $249. Now we can use that to find the amount from Ms. Aguilar's class.

$249 + $396.62 = $645.62

Now we can use the amount from Ms. Aguilar's class to find the amount from Ms. Barry's class

$645.62 - $430.43 = $215.19

Now we can add the three amounts together to find the total amount.

$249 + $645.62 + $215.19 = $1109.81

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Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.
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The center of mass is mathematically given as

\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \  \ \  x \in(0,1)$ \\\end

We calculate the total mass of the system.

\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}

Step 03: Calculate the moment of the system.

\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}

we calculate the center of mass.

\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}

Read more about the center of mass.

brainly.com/question/27549055

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8 0
2 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cunderline%7B%20%5Cunderline%7B%20%5Ctext%7Bquestion%7D%7D%7D%20%3A%20" id="TexFormula1"
steposvetlana [31]

Answer:

See Below.

Step-by-step explanation:

In the given figure, AP = BP = PC.

And we want to prove that ∠ABC is a right angle.

Since AP = BP and BP = PC, we can create two isosceles triangles: ΔAPB and ΔCPB.

By the definition of isosceles triangles, in ΔAPB, ∠PAB and ∠PBA are equivalent. Let the measure of each of them be <em>x°</em>.

Likewise, in ΔCPB, ∠PCB and ∠PBC are equivalent.

And since AP = BP = PC, each of the angles∠PCB and ∠PBC will also be equivalent to <em>x°.</em>

<em />

And since the sum of the interior angles of a triangle total 180°, we acquire:

\angle PAB+\angle PBA+\angle PCB+\angle PBC=180

Since they are all equivalent:

4x=180

Hence:

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∠ABC is the sum of ∠PBA and ∠PBC, each of which measures 45°. Hence:

\angle ABC=\angle PBA+\angle PBC=45+45=90^\circ

8 0
2 years ago
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yKpoI14uk [10]
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Answer:

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