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3 years ago

5

help!!!! I'm going to be doing quick questions. you answer them, you get a brainly. Starting with this one!!!!!!!! I think it's

B.

1 answer:

stepladder [879]3 years ago

7 0

Answer:

the answer is C

Step-by-step explanation:

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How do you solve this

AlladinOne [14]

Answer:

domain: all terms are defined for <em>all real numbers</em>
solution: x = 6

Step-by-step explanation:

Rewrite the equation as a single exponential. After taking the log, the solution becomes obvious.

$5^{x-2}-7^{x-3}=7^{x-5}+11\cdot 5^{x-4}\\\\5^{x-2}-11\cdot 5^{x-4}=7^{x-3}+7^{x-5}\\\\5^{x-2}(1-11\cdot 5^{-2})=7^{x-3}(1+7^{-2})\\\\5^{x-2}\dfrac{14}{25}=7^{x-3}\dfrac{50}{49}\\\\1=\dfrac{2\cdot 7^{x-3}5^4}{2\cdot 5^{x-2}7^3}=\left(\dfrac{7}{5}\right)^{x-6}\\\\0=(x-6)\log(1.4)\\\\x=6$

6 0

3 years ago

Solve the system by using a matrix equation.<br> --4x - 5y = -5<br> -6x - 8y = -2

evablogger [386]

Answer:

Solution : (15, - 11)

Step-by-step explanation:

We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )

$\begin{bmatrix}-4&-5&|&-5\\ -6&-8&|&-2\end{bmatrix}$

Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )

Row Echelon Form :

$\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}$

Step # 1 : Swap the first and second matrix rows,

$\begin{pmatrix}-6&-8&-2\\ -4&-5&-5\end{pmatrix}$

Step # 2 : Cancel leading coefficient in row 2 through $R_2\:\leftarrow \:R_2-\frac{2}{3}\cdot \:R_1$ ,

$\begin{pmatrix}-6&-8&-2\\ 0&\frac{1}{3}&-\frac{11}{3}\end{pmatrix}$

Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.

$\begin{bmatrix}1&0&|&15\\ 0&1&|&-11\end{bmatrix}$

As you can see our solution is x = 15, y = - 11 or (15, - 11).

4 0

3 years ago

Combining like term 6k+7k

Nadya [2.5K]

13k is the answer :)

8 0

3 years ago

Hurry hurry ! I really need this one

kondor19780726 [428]

Answer:

The third option.

4 0

4 years ago

Read 2 more answers

Which expressions are equivalent to 9+h+ (j+k)? Check all that apply.

GaryK [48]

Answer:3

the 3rd one

5 0

3 years ago