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4 years ago

Complete the division problem by determining the number that should be placed in the box.

Mathematics

1 answer:

S_A_V [24]4 years ago

5 0

560 is the miss one.
do 658-560 = 98

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How many once’s are equal to 7 pounds

JulsSmile [24]

112 ounce’s are in 7 pounds

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3 years ago

The number of that grow sugar beet =12

erastova [34]

Answer:

I think question is not proper

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3 years ago

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

hram777 [196]

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.

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8 0

2 years ago

HELP! VERY DESPERATE! MIGHT FAIL!

dem82 [27]

Hope this helps you note only if it is the correct question then the answer will be correct

5 0

3 years ago

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Given the system of equations, what is the solution?<br><br> x + 3y = 5<br> x - 3y = -1

mr_godi [17]

$\begin{cases}x+3y=5\\x-3y=-1 \end{cases}\\---------(+)\\\\x+x+3y-3y=5+(-1)\\\\2x=4\qquad|:2\\\\\boxed{x=2}\\\\\\\\x+3y=5\\\\2+3y=5\qquad|-2\\\\2+3y-2=5-2\\\\3y=3\qquad|:3\\\\\boxed{y=1}$

The solution is (2; 1)

3 0

3 years ago

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