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3 years ago

-17 + 5x = 5(x - 5) + 8

Mathematics

1 answer:

Troyanec [42]3 years ago

8 0

Answer:

x= -1.83

Step-by-step explanation:

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Given the function g(x) = 6^x, Section A is from x = 1 to x = 2 and Section B is from x = 3 to x = 4.

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Section A and B
$\frac{36 - 6}{2 - 1} = 30 \\ \frac{1296 - 216}{4 - 3} = 1080$

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4 years ago

Determine the factors of x2 − 7x − 10. (5 points) a (x + 2)(x − 5) b Prime c (x − 2)(x + 5) d (x + 10)(x − 1)

Savatey [412]

For this case we must factor the following expression:

$x ^ 2-7x-10$

To factor, we must find two numbers that when added together give -7 and when multiplied by -10.

It is observed that there are not two whole numbers that comply with the aforementioned.

Thus, the following equation applies:

$x = \frac {-b \pm\sqrt{b ^ 2-4 (a) (c)}} {2 (a)}$

Where:

$a = 1\\b = -7\\c = -10$

We replace:

$x = \frac {- (- 7) \pm \sqrt {(- 7) ^ 2-4 (1) (- 10)}} {2 (1)}\\x = \frac {7 \pm \sqrt {49 + 40}} {2}\\x = \frac {7 \sqrt {89}} {2}$

So, the roots are:

$x_ {1} =\frac {7+ \sqrt {89}} {2}\\x_ {2} =\frac {7- \sqrt {89}} {2}$

Answer:

$x_ {1} =\frac {7+ \sqrt {89}} {2}\\x_ {2} =\frac {7- \sqrt {89}} {2}$

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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

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

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3 years ago

Aiden put two-fifths of his Money into his piggy bank. He had $15 left in his pocket to buy a toy. How much money did aiden have

Travka [436]

Answer:25

Step-by-step explanation:

15x 2/5=25

5 0

4 years ago

Read 2 more answers