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Alik [6]
1 year ago
8

Which point corresponds to the real zero of the graph of y = log3 (x+2) - 1?

Mathematics
1 answer:
brilliants [131]1 year ago
8 0

Answer:

(1, 0 )

Step-by-step explanation:

using the rule of logarithms

log_{b} x = n ⇒ x = b^{n} , then

y = log_{3} (x + 2) - 1

to find the zero let y = 0 , that is

log_{3} (x + 2) - 1 = 0 ( add 1 to both sides )

log_{3} (x + 2) = 1 , then

x + 2 = 3^{1} = 3 ( subtract 2 from both sides )

x = 1

the zero is (1, 0 )

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2 years ago
Find the range of the function for the given domain. ​
Mice21 [21]

Answer: {5, -7, -19, -27, -35}

Step-by-step explanation:

In order solve this, we need to plug in the values of x into the table.

For spaces on the left of the equals sign, you need to write each x from the domain. You can then match that x-value with its function value by putting that on the right side.

For each equation, we are simply plugging a number from the domain into the function and replacing the x-value:

f(-1) = -4(-1) + 1 = 5\\f(2) = -4(2) + 1 = -7\\f(5) = -4(5) + 1 = -19\\f(7) = -4(7) + 1 = -27\\f(9) = -4(9) + 1 = -35

I hope this helps. If you need any extra explanation on how the functions are set up, please let me know.

5 0
1 year ago
and 2008 Classic Car Wash estimated its business would increase by 20% each year. if they washed 19,300 in 2008, how many cars c
professor190 [17]
Well, from 2008 to 2010 is just 2 years, so let's say the initial amount is the 19300 from 2008, how many will it be in 2010?

\bf \qquad \textit{Amount for Exponential Growth}\\\\
A=I(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &19300\\
r=rate\to 20\%\to \frac{20}{100}\to &0.20\\
t=\textit{elapsed time}\to &2\\
\end{cases}
\\\\\\
A=19300(1+0.2)^2
6 0
2 years ago
Which table shows y as a function of x
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3 0
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A school P is 16km due west of a school Q. what is the bearing of Q from P​
Irina-Kira [14]

Answer:

16Km due east of school P

Step-by-step explanation:

Given

A school P is 16km due west of a school Q

Thus, we can say that distance PQ = 16 km.

________________________________

now we have to find  the bearing of Q from P​

As distance is same

distance PQ = distance QP

Thus,

Distance will remain same of 16 km.

For direction,

If Q is west of P, then P will be east of QP------------------>Q

as shown in figure P is west of Q,

now from point P , Q is west P.

Thus,

Bearing of  School Q from P is 16Km due east of school P

7 0
3 years ago
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