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podryga [215]
2 years ago
15

30% of 50 what’s the easiest and fastest way to solve it?

Mathematics
2 answers:
madam [21]2 years ago
4 0
Step 1: Our output value is 50.

Step 2: We represent the unknown value with $x$.

Step 3: From step 1 above,$50=100\%$.

Step 4: Similarly, $x=30\%$.

Step 5: This results in a pair of simple equations:

$50=100\%(1)$.

$x=30\%(2)$.

Step 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both
equations have the same unit (%); we have

$\frac{50}{x}=\frac{100\%}{30\%}$

Step 7: Again, the reciprocal of both sides gives

$\frac{x}{50}=\frac{30}{100}$

$\Rightarrow x=15$

Therefore, $30\%$ of $50$ is $15$
Olegator [25]2 years ago
4 0

Answer:

15

Step-by-step explanation:

change the percent to a decimal

30% to .30

.30 x 50= 15

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N-2=10n+4 over 2 what is the value of n
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Consider the given equation:

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3 years ago
Find the measure of the missing angle.
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Answer:

the answer is 138

Step-by-step explanation:

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2 years ago
Orthogonally diagonalize the​ matrix, giving an orthogonal matrix P and a diagonal matrix D. To save​ time, the eigenvalues are
alexgriva [62]

Answer:

P=\left(\begin{array}{ccc}-\frac{2}{3}&-\frac{2}{3}&\frac{1}{3}\\\frac{1}{\sqrt{5}}&0&\frac{2}{\sqrt{5}}\\-\frac{4}{3\sqrt{5}}&\frac{\sqrt{5}}{3}&\frac{2}{3\sqrt{5}}\end{array}\right)

Step-by-step explanation:

It is a result that a matrix A is orthogonally diagonalizable if and only if A is a symmetric matrix.  According with the data you provided the matrix should be

A=\left(\begin{array}{ccc}-9&-4&2\\ -4&-9&2\\2&2&-6\\\end{array}\right)

We know that its eigenvalues are \lambda_{1}=-14, \lambda_{2}=-5, where \lambda_{2}=-5 has multiplicity two.

So if we calculate the corresponding eigenspaces for each eigenvalue we have

E_{\lambda_{1}=-14}=\langle(-2,-2,1)\rangle,E_{\lambda_{2}=-5}=\langle(1,0,2),(-1,1,0)\rangle..

With this in mind we can form the matrices P, D that diagonalizes the matrix A so.

P=\left(\begin{array}{ccc}-2&-2&1\\1&0&2\\-1&1&0\\\end{array}\right)

and

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Observe that the rows of P are the eigenvectors corresponding to the eigen values.

Now you only need to normalize each row of P dividing by its norm, as a row vector.

The matrix you have to obtain is the matrix shown below

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3 years ago
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Answer:

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\boxed{\green{y = 0}}

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