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3 years ago

Which expression is equivalent to log29x ?

Mathematics

2 answers:

yarga [219]3 years ago

8 0

Answer:

$log_29 +3log_2x$

Step-by-step explanation:

$log_29x^3 \\ = log_29 +log_2x^3 \\ = log_29 +3log_2x \\ \huge \red{= log_29+3log_2x}\\$

atroni [7]3 years ago

7 0

The answer would be the first option.

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Plz help I need it or I’m gonna fail this test

quester [9]

Answer:

-2

Step-by-step explanation:

I think this is the answer

4 0

2 years ago

Please help i’m confused

Pavel [41]

Answer:

The answer is A. t = 0.4n

Step-by-step explanation:

See how on the graph, when n is 2, t is around 1? We just plug in 2 into n into the equation, t=0.4n. 0.4 times 2 is 0.8, so t=0.8. 0.8 is around 1, so a is right. If we tried the same thing with one of the other choices, like B. t=2.5n, then plugging in 2 into n would give us 2 times 2.5, which is 5. 5 is nowhere around 1, making it wrong.

6 0

2 years ago

I’ve been stuck on this for a min now

dedylja [7]

Answer:

3/14

Step-by-step explanation:

1st subtract the y's

the subtract the x's in the same order

-9 - 5 = -14

put the first result over the second

-3/-14 = 3/14

4 0

2 years ago

Fitness Classes cost £6 per person. The table shows the number of people who go to the classes on Thursday, Friday and Saturday.

Pachacha [2.7K]

Answer:

£216

Step-by-step explanation:

Cost = £6 per person
Teacher earns = 90% of total

Total number of people:

13 + 11 + 16 = £40

Total payment:

40*6= £240

Teacher earns in total per given week:

240*90/100 = £216

7 0

3 years ago

Anyone know the answer to this algebra problem?

ikadub [295]

Answer: $\bold{a=1\qquad b=\dfrac{1}{16}\qquad c=\dfrac{1}{64}\qquad d=1\qquad e=\dfrac{4}{9}\qquad f=\dfrac{16}{81}}$

Step-by-step explanation:

$\begin{array}{c|l}\underline{\quad x\quad}&\underline{\quad 4^{-x}\qquad \qquad}\\-1&4^{-(-1)}=4^1=4\\\\0&4^{-(0)}=4^0=1\\\\2&4^{-(2)}=\dfrac{1}{4^2}=\dfrac{1}{16}\\\\4&4^{-(4)}=\dfrac{1}{4^4}=\dfrac{1}{64}\end{array}$

$\begin{array}{c|l}\underline{\quad \bigg x \quad}&\underline{\quad \bigg(\dfrac{2}{3}\bigg)^x\qquad \qquad}\\\\-1&\bigg(\dfrac{2}{3}\bigg)^{-1}=\dfrac{3}{2}\\\\0&\bigg(\dfrac{2}{3}\bigg)^{0}=1\\\\2&\bigg(\dfrac{2}{3}\bigg)^{2}=\dfrac{2^2}{3^2}=\dfrac{4}{9}\\\\4&\bigg(\dfrac{2}{3}\bigg)^{4}=\dfrac{2^4}{3^4}=\dfrac{16}{81}\end{array}$

3 0

3 years ago