What is the least denominator do -97 and -156 have in common

Mathematics

1 answer:

masha68 [24]2 years ago

6 0

Answer:

the fraction can't be further simplified, but you can simplify the negatives, so it would be left as 97 / 156

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I’m confused on this one

dedylja [7]

It's a parallelogram so if one angle is 90 degrees they all will be because of transverse angles and all that good stuff.

So we're given the diagonal of a rectangle and one side and we're asked to find the other. The diagonal of a rectangle is the hypotenuse of the right triangle whose legs are the sides of the rectangle. So this is a Pythagorean Theorem question in disguise:

$AB^2 + AD^2 = BD^2$

$AD^2 = BD^2 - AB^2$

$AD = \sqrt{ BD^2 - AB^2 }$

$AD = \sqrt{149^2 - 51^2} = \sqrt{19400} = 140$

Answer: 140 cm

I don't recall seeing this Pythagorean Triple before.

3 0

2 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

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