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12345 [234]
3 years ago
7

Which point is on the interior of DEF? point F point H. point D point G​

Mathematics
2 answers:
algol133 years ago
8 0

Answer:

Point G

Step-by-step explanation:

Interior is the inside of the geometric shape, which is Point G.

Point H is the exterior of DEF.

Alenkasestr [34]3 years ago
3 0
F will be your interior
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PilotLPTM [1.2K]

Answer:

Bc equals 17.

you must solve for BA and CA, because this is a isosceles triangle and isosceles have 2 equal sides.

BA = CA

then plug in what you find in for x

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In triangle DEF, find m
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You would have to share a picture or the details of this triangle in order for me to help
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Phil has 7 quarts of hot chocolate to give his class mates. how many of phils friend's can have one cup of hot chocolate
Rudik [331]
4 cups per quarter

4 x 7 = 28

28 kids can have a cup of hot chocolate.
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3 years ago
Find the slope of the like y = 1/8x -1<br><br> anser choices <br> -8/1 <br> 1/8<br> 1<br> -1
otez555 [7]

Answer:

The equation is not linear. So the slope does not exist. Not linear.


Maybe you wrote the equation wrong? If so, let me know and I'll tell you what the answer is.


6 0
3 years ago
The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t
shtirl [24]

Answer:

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

Step-by-step explanation:

As the given Augmented matrix is

\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 1 :

r_{1}↔r_{1} - r_{2}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 2 :

r_{3}↔r_{3} - 8r_{1}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]

Step 3 :

r_{2}↔\frac{r_{2}}{7}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]

Step 4 :

r_{1}↔r_{1} + 14r_{2} , r_{3}↔r_{3} - 124r_{2}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]

Step 5 :

r_{3}↔\frac{r_{3}. 7}{254}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]

Step 6 :

r_{1}↔r_{1} - 4r_{3} , r_{2}↔r_{2} + \frac{1}{7} r_{3}

\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]

∴ we get

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

6 0
3 years ago
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