<img src="https://tex.z-dn.net/?f=%203log_%7B3%7D%285%29%20%20%5Ctimes%20%20log_%7B3%7D%284%20%29%20" id="TexFormula1" title=" 3

All categories

shtirl [24]

3 years ago

log_{3}(5) \times log_{3}(4 ) " alt=" 3log_{3}(5) \times log_{3}(4 ) " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

8 0

Answer:

log3 (500)

Step-by-step explanation:

3 log3 (5) * log3(4)

We know that a log b(c) = log b(c^a)

log3 (5)^3 * log3(4)

We know that log a(b) * log a (c) = loga( b*c)

log3 ((5)^3 * 4)

log3 (125*4)

log3 (500)

You might be interested in

X + 2.3 = 11.7* 2.3 14 9.4 11.7 13.10

SCORPION-xisa [38]

Answer:

9.4

Step-by-step explanation:

X + 2.3 = 11.7

Subtract 2.3 from both sides

X = 9.4

3 0

3 years ago

The area of a trapezium is 34 cm and the length one of its parallel side is

Natasha_Volkova [10]

Answer:

6.6 cm.

Step-by-step explanation:

The area of a trapezium A = (h/2)(a + b) where h = the height, a and b are the length of the parallel sides.

So substituting the given values in this formula:

34 = (4/2)(10.4 + b)

34 = 2(10.4 + b

34 = 20.8 + 2b

2b = 34 - 20.8 = 13.2

b = 6.6 cm.

3 0

3 years ago

Four w plus ten equals negative w

ohaa [14]

Answer:

4w+10=-w

w=-2

Step-by-step explanation:

First, write out the equation in the algebraic form to make it easier to solve. This equation written out is 4w+10=-w. Then, isolate the variable and solve. Do this by subtracting 4w from both sides. Then, divide both sides by -5.

4w+10=-w

10=-5w
-2=w

4 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

If line a is parallel to line c, and line c is parallel to line f, and lines c and f are different lines, which of the following

Citrus2011 [14]

Line a is parallel to line f

5 0

4 years ago

Read 2 more answers