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Helen [10]
2 years ago
15

Are all squares proportional?

Mathematics
1 answer:
Stels [109]2 years ago
7 0

Answer:

Yes each side will always be of the same size unless you are purposely trying to make an uneven shape

Step-by-step explanation:

You might be interested in
Rewrite the system of linear equations as a matrix equation AX = B.
iren2701 [21]

Answer:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

7 0
3 years ago
Which of the following is the quotient of the rational expression (x-2/x+3) divided by (2/x)
zloy xaker [14]

Answer:

(x²-2x)/(2x+6)

Step-by-step explanation:

When you are asked to divide two fractions just do keep-change-flip. Keep the first fraction, change the ÷ to x and flip the second fraction upside down. So the problem becomes ((x-2)/(x+3)) times (x/2). Then just multiply numerators and multiply denominators.

6 0
3 years ago
Read 2 more answers
What is the inverse of the given equation y=7x^2-3?
NARA [144]

Answer:

The inverse of this equation would be y = \sqrt{\frac{x - 3}{7} }

Step-by-step explanation:

To find the inverse of any equation, start by switching the y and x values. Then solve for the new y value. That equation will be your inverse.

y = 7x^2 - 3

x = 7y^2 - 3

x + 3 = 7y^2

\frac{x-3}{7} = y^2

\sqrt{\frac{x - 3}{7} } = y

5 0
3 years ago
Is this all correct?
expeople1 [14]
Yes it looks like it would be right
3 0
3 years ago
Read 2 more answers
Does anybody know what the answer to this question is?
crimeas [40]

Answer:

equation 15+8=

solution= 23

Step-by-step explanation:

its ez

7 0
2 years ago
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