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BlackZzzverrR [31]
3 years ago
6

Select all of the situations which can be represented by the expression 25.75x + 10.

Mathematics
1 answer:
Volgvan3 years ago
4 0

Answer:

1,2,4

Explanation to the answer:

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What is the solution to the system of equations 2x+y=2 and y= -2x-2?
finlep [7]

Answer:

NO SOLUTION

Step-by-step explanation:

Solve this system by substitution:

Replace y in the first equation with -2x - 2:

2x + (-2x - 2) = 2, or

2x - 2x - 2 = 2

0 = 4      This is not possible, so this system has NO SOLUTION.

5 0
3 years ago
If f(x) = log2 (x + 1), what is f-1(2)? (2 points)
Soloha48 [4]

as you already know, to get the inverse of any expression, we start off by doing a quick switcheroo on the variables, and then solve for "y".

\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{f(x)}{y}=\log_2(x+1)\implies \stackrel{\textit{quick switcheroo}}{\underline{x}=\log_2(\underline{y}+1)}\implies 2^x=2^{\log_2({y}+1)} \\\\\\ 2^x=y+1\implies 2^x-1=\stackrel{f^{-1}(x)}{y} \\\\[-0.35em] ~\dotfill\\\\ 2^2-1=f^{-1}(2)\implies 3=f^{-1}(2)

3 0
3 years ago
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5 0
2 years ago
From 1st principle find the derivative of log (ax+b)
OLga [1]

Answer:

Step-by-step explanation:

The parent function here is y = log x, where 10 is the base.

The derivative of y = log x is dy/dx = (ln x) / ln 10.

The derivative of y = log (ax+b) is found in that manner, but additional steps are necessary:  differentiate the argument ax + b:

The derivative with respect to 10 of log (ax + b) is:

dy/dx = [ 1 / (ax + b) ] / [ ln 10 ] *a, where a is the derivative of (ax + b).

Alternatively, we could express the answer as

dy/dx = [ a / (ax + b) ] / [ ln 10 ]

5 0
3 years ago
Franklin rode his bike 6 5/6 miles on Tuesday. He also rode his bike 5 5/6 miles on Thursday. What is the difference between the
11Alexandr11 [23.1K]
The difference = miles rode on Tuesday - miles rode on Thursday.
6 5/6 - 5 5/6 = 1 mile


3 0
3 years ago
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