Find the fraction that is equivalent to 21%

Mathematics

2 answers:

stealth61 [152]3 years ago

8 0

Since percent is essentially the number over 100, we can find this fraction quickly in the same fashion:

$\frac{21}{100}$

This is in simplest form, so there is no more reducing that can take place.

Ratling [72]3 years ago

3 0

21% is the same as 0.21 (move the decimal point to the left two place values to make it a decimal).

Next, to make it a fraction, move the decimal to the right two place values, and place the number over 100

0.21 = 21/100

21/100 is your answer

hope this helps

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Find x please help thank you

Angelina_Jolie [31]

Answer: x=(7√6)/2

Step-by-step explanation:

To find x, we would have to find the hypotenuse of the 45-45-90 triangle. First, we would have to find the hypotenuse by using the 30-60-90 triangle on top to find it.

For a 30-60-90 triangle, the hypotenuse is 2x in length. the x is the same in all sides. All you would have to do is to plug it in. The leg opposite of 60° is x√3 in length. the leg opposite of 30° is x in length.

Since we know that 7 is opposite of the 30° angle, we know that x is 7. Across fron 60° is the hypotenuse of the 45-45-90 triangle. That leg is x√3. We plug in x=7 and get 7√3.

The hypotenuse of the 45-45-90 triangle is x√2 and the legs are both x. We can set 7√3 equal to x√2 to find x of the missing side.

7√3=x√2 [divide both sides by √2]

x=(7√6)/2

Now, we know x=(7√6)/2.

8 0

3 years ago

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.

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