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3 years ago

PLS ANSWER QUICK! Thanks!

Mathematics

1 answer:

ziro4ka [17]3 years ago

5 0

Answer:

The correct answer would be $a=\sqrt{c^2-b^2}$

Step-by-step explanation:

given $a^2+b^2=c^2$ we want to solve for a

How?

We can do this by using basic algebra ( isolating the variable (a ))

Step 1 subtract $b^2$ from each side

$a^2+b^2-b^2=a^2\\c^2-b^2=c^2-b^2$

now we have $a^2=c^2-b^2$

step 2 take the square root of each side

$\sqrt{a^2} =a\\\sqrt{c^2-b^2} =\sqrt{c^2-b^2}$

we're left with $a=\sqrt{c^2-b^2}$

Hence your answer is A

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3 years ago

If you have the following equation, 2/3x+5/6-x=2-3/4x what would you multiply each term by to get rid of all the fractions

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Step-by-step explanation:

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3 years ago

(12x + y + z = 26

mel-nik [20]

Option D. D has the matrix of constants [[12], [11], [4]].

Step-by-step explanation:

Step 1:

With the given equations, we can form matrices to represent them.

The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.

Step 2:

The linear system A is represented as

$\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}26\\17\\23\end{array}\right]$ .

Step 3:

The linear system B is represented as

$\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}23\\17\\26\end{array}\right]$ .

Step 4:

The linear system C is represented as

$\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}4\\11\\12\end{array}\right]$ .

Step 5:

The linear system D is represented as

$\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}12\\11\\4\end{array}\right]$ .