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3 years ago

15 + 100 • [12 – (12 – 3 • 2)] + 6

Mathematics

1 answer:

lara31 [8.8K]3 years ago

4 0

Multiple: 3 * 2 = 6

Subtract: 12 - 6 = 6

Multiple: 100 * 6 = 600

Add: 15 + 600 = 615

Add: 615 + 6 = 621

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Find the common ratio of the geometric sequence 6, 48, 384, ...

gregori [183]

Answer:

r = 8

Step-by-step explanation:

The common ratio r of a geometric sequence is

r = $\frac{a_{2} }{a_{1} }$ = $\frac{a_{3} }{a_{2} }$ = ..... = $\frac{a_{n} }{a_{n-1} }$

Then

r = $\frac{48}{6}$ = $\frac{384}{48}$ = 8

8 0

3 years ago

4r-2=4r-2 what's the solution? can have either have many solutions or none

butalik [34]

That "equation" has an infinite number of "solutions", because
no matter what 'r' is, it's always a true statement.

The statement is equivalent to saying "I am I." or "You are you."

7 0

3 years ago

Read 2 more answers

Plz help<br>urgent !!!!<br>will give the brainliest !!

Mkey [24]

Answer:

$X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}$

Step-by-step explanation:

The question we have at hand is, in other words,

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}$ - where we have to solve for the value of $X$

If we have to isolate $X$ here, then we would have to take the inverse of the following matrix ...

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}$ ... so that it should be as follows ... $\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}$

Therefore, we can conclude that the equation as to solve for " $X$ " will be the following,

$X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ - First find the 2 x 2 matrix inverse of the first portion,

$\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}$ = $\frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}$ = $\frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix}$ = $\begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}$

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

$X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ ,

$X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}$ - Simplifying this, we should get ...