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Natasha2012 [34]
3 years ago
12

5/6h - 7/12 = -3/4h - 13/6

Mathematics
2 answers:
Makovka662 [10]3 years ago
7 0

Answer:

h = -1

Step-by-step explanation:

vovangra [49]3 years ago
3 0
\frac{5}{6} h - \frac{7}{12} = -\frac{3}{4} h - \frac{13}{6}

First, simplify \frac{5}{6} h to \frac{5h}{6} / Your problem should look like: \frac{5h}{6} - \frac{7}{12} = -\frac{3}{4} h - \frac{13}{6}
Second, simplify \frac{3}{4} h to \frac{3h}{4} / Your problem should look like: \frac{5h}{6} - \frac{7}{12} = - \frac{3h}{4} - \frac{13}{6}
Third, multiply both sides by 12, which is the LCM of 6 and 4. / Your problem should look like: 10h - 7 = -9h - 26
Fourth, add 9h to both sides. / Your problem should look like: 10h - 7 + 9h = -26
Fifth, simplify 10h - 7 + 9h to 19h - 7 / Your problem should look like: 19h - 7 = -26
Sixth, add 7 to both sides. / Your problem should look like: 19h = -26 + 7
Seventh, simplify -26 + 7 to -19. / Your problem should look like: 19h = -19
Eighth, divide both sides by 19. / Your problem should look like: h = - \frac{19}{19}
Ninth, cancel out 19. / Your problem should look like: h = -1

Answer: h = -1

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Read 2 more answers
Find the inverse of the given​ matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3
BabaBlast [244]

Answer:

A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]

Step-by-step explanation:

We want to find the inverse of A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Subtract row 1 multiplied by 4 from row 2

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Add row 1 multiplied by 4 to row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]

  • Subtract row 2 multiplied by 2 from row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]

  • Divide row 3 by −27

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Add row 3 multiplied by 2 to row 1

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Subtract row 3 multiplied by 11 from row 2

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

As can be seen, we have obtained the identity matrix to the left. So, we are done.

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3 years ago
State whether the given measurements determine zero, one, or two triangles. C = 31°, a = 13, c = 11
mina [271]

9514 1404 393

Answer:

  2

Step-by-step explanation:

The given angle is opposite the shorter of the given sides, so there are two possible triangles.

__

<em>Additional comment</em>

The law of sines tells you that sin(A) = a/c·sin(C). There will be 0 triangles if that is greater than 1. There will be 1 triangle if that is exactly 1 (unlikely with rational values of sides and angles). There will be 2 triangles if sin(A) < 1, as it is here.

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Subtract. [5 0 − 2]− [− 1 3 8 ] Enter your answer in the boxes. [ ]
Fiesta28 [93]

Answer:

i don't understand the question

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4 years ago
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