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irinina [24]
2 years ago
8

I need help with number 15

Mathematics
2 answers:
goblinko [34]2 years ago
6 0
X= 75 bc the two angles are congruent
solmaris [256]2 years ago
5 0

Answer:

X=12 have a good day!

Step-by-step explanation:

hope this helped

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Which property is illustrated by the following statement if ABC=DEF and DEF=XYZ then ABC=XYZ
Harlamova29_29 [7]

Answer:

Transitive property of equality

Step-by-step explanation:

Let A be any non empty set and R is any subset of the Cartesian product A × A. Then, R is a relation on A.

The relation R is said to be a transitive relation if (a, b) ∈ R, (b, c) ∈ R, then (a, c) ∈ R.

It is given that ABC = DEF and DEF = XYZ, then ABC = XYZ.

This shows the transitive property of equality.

3 0
3 years ago
Read 2 more answers
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
2 years ago
Can someone help not just whatever
bogdanovich [222]
4.) (the next prime function)
2x2x3x2
2x2x3x2x2
2x2x3x2x2x2
7 0
3 years ago
Hey can you please help me posted picture of question
SpyIntel [72]
We are to find the Probability the someone buys a book that is paperback and fiction.

Let P(F) represents the event that the book is fiction and P(P) represents the event that the book is paperback. We are to find P(F∩P)

P(F∩P) = P(F) x P(P)

From the tree diagram we can see that:

P(F) = 0.45
P(P) = 0.65

Using the values, we get:

P(F∩P) = 0.45 x 0.65 = 0.2925

So, the Probability the someone buys a book that is paperback and fiction is 0.2925.

So, option B gives the correct answer
3 0
3 years ago
I’ll give brainliest:)
Natasha2012 [34]

Answer:

ok ok ok ok ok ok okok ok ok

6 0
3 years ago
Read 2 more answers
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