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2 years ago

Which expression is equivalent to 2(9 + 7)? (1 point) Select one: a. 18 + 14 b. 18 + 7 c. 11 + 9 d. 11 + 7

Mathematics

2 answers:

Rom4ik [11]2 years ago

7 0

2(9 + 7) = 2*9 + 2*7 = 18 + 14

So it is a.

Leni [432]2 years ago

7 0

Answer:

Option a. 18 + 14

Step-by-step explanation:

Thinking process:

The expression : 2 (9+7) gives 2(9) + 2(7)

= 18 + 14

= 32

Looking at the answers and comparing, a is the best option

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Answer:

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Step-by-step explanation:

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3 years ago

Two friends, Autumn and Angel, had just bought their first cars. The equation

Arlecino [84]

Answer:

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Step-by-step explanation:

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2 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

2 years ago

plz this is easy just answer Use < or > to compare the integers. –14 and –3 A. –14 > –3 B. –14 < –3

oee [108]

It would be B. because if you put the numbers on a number line, you notice that -14 is further down than -3 since -3 is closer to 0, so -14 is less than -3

3 0

2 years ago

Identify the vertex of g(x) = (x + 14)? - 8.

Aleks04 [339]

Do you think you can send a picture of the problem I’m not really understanding

6 0

3 years ago