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3 years ago

Please help me? Thank youuu

Mathematics

1 answer:

cricket20 [7]3 years ago

8 0

Answer: The first choice, the one with a circle.

Step-by-step explanation: The verticial line test proves that the circle isnt a function, it fails the vertical line test. Hope this helps :)

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Solve 6x + 2 = 10 for x using the change of base formula log base b of y equals log y over log b.

kodGreya [7K]

Answer:

The answer is -0.715

Step-by-step explanation:

Firstly, we use the change of base formula. The base must be 6 because the "x" variable is a power of it.

$6^x^+^2=10\\\\log_{6} (6^x^+^2)=log_{6} (10)\\$

Then, we can use the property logarithm power rule:

$log_{b}(x^y)=y* log_{b}(x)$

So,

$(x+2)*log_{6} (6)=log_{6} (10)\\log_{6} (6)=1\\x+2=log_{6} (10)\\\\log_{6} (10)=\frac{log(10)}{log(6)}\\log(10)=1\\\\x+2=\frac{1}{log(6)} \\x=\frac{1}{log(6)}-2$

Finally, the value of $\frac{1}{log(6)} =1.285$ , therefore:

$x=1.285-2=-0.715$

3 0

3 years ago

Read 2 more answers

Jerome got in the elevator on the first floor of his office building. He went up 18 floors to his office. Then, Jerome took the

steposvetlana [31]

He starts on the 1st floor. He then goes up 18 floors:

1 + 18 = 19

He took the stairwell down 3 floors

19 - 3 = 16

He then takes the elevator down 8 floors

16 - 8 = 8

He then takes the stairwell up 5 floors for his afternoon meeting

8 + 5 = 13

C. thirteenth is your answer

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7 0

3 years ago

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The following stem-and-leaf plot represents the scores earned by Mr. Roberts's class on their most recent science test.

NNADVOKAT [17]

Total 19 students in class

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4 0

4 years ago

Read 2 more answers

For science class, a student recorded the high and low temperatures, in Fahrenheit, over a ten day period in September. The data

statuscvo [17]

Answer:

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5 0

3 years ago

Roger can run one mile in 8 minutes. jeff can run one mile in 6 minutes. if jeff gives roger a 1 minute head start, how long wi

Anastasy [175]

Recall your d = rt, distance = rate * time.

now, if Roger can do 1 mile in 8 minutes, so in 1 minute, he has done then 1/8 of a mile, so his rate is 1/8 miles per minute.

if Jeff can do 1 mile in 6 minutes, he's faster, in 1 minute he has done 1/6 of a mile, so his rate is 1/6 miles per minute.

now, when Jeff catches up with Roger, the distance covered by both will be the same, say "d" miles, because, at that millisecond, Jeff will be neck and neck with Roger, and their covered distance will be the same.

now, Jeff is generous and let Roger roll on for 1 minute before him, so, by the time time Roger has covered "d" miles, he has been running for say "t" minutes.

however, since Jeff started later by 1 minute, he hasn't been running for "t" minutes, but for "t - 1" minutes.

$\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mpm}{rate}&\stackrel{minutes}{time}\\
&------&------&------\\
Roger&d&\frac{1}{8}&t\\\\
Jeff&d&\frac{1}{6}&t-1
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=\frac{1}{8}t\\\\
d=\frac{1}{6}(t-1)\\
------\\
\boxed{\frac{1}{8}t}=\cfrac{t-1}{6}
\end{cases}
\\\\\\
\cfrac{t}{8}=\cfrac{t-1}{6}\implies 6t=8t-8\implies 8=2t\implies \cfrac{8}{2}=t\implies \boxed{\stackrel{mins}{4}=t}$

4 0

3 years ago