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3 years ago

SOLVE: 6^3x+2 = 4^x-5

Mathematics

1 answer:

Lilit [14]3 years ago

6 0

There's an app called photomath that would be useful for these kinds of problems!

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A walker has travelled 9 km along a trail. If he has completed 80% of the trail, how much further does he still have to go?

Anit [1.1K]

Answer:

2.25 km to go

Step-by-step explanation:

In order to get this answer, you have to figure out how many km 10% is, 0.1125. Then multiply that by the remaining 20% because he already finished 80% of the trail. So, 0.1125 x 20 = 2.25.

Hope this helps! :)

3 0

2 years ago

Select the true statement. A. Neither function is even. B. Only function f is even. C. Only function g is even. D. Both function

astraxan [27]

Answer:

The answer is "Option C".

Step-by-step explanation:

The graph of the even function will have both ends pointed in the same direction, while the graph of the odd function will have both ends pointed inside the reverse way. Furthermore, as the y-axis approaches, and also regular expression graph becomes symmetric. The graph of an unusual function, on either hand, is symmetric around the origin and also has rotational symmetry around the origin. They could see that the graph of function g meets a function requirement by looking at it. That graph of function f, but on the other hand, is not symmetric about its y-axis, implying that function f is not even. As both a result, g is a function that is even.

5 0

3 years ago

Read 2 more answers

Vito has a fair 20-sided die with the faces

Hitman42 [59]

Answer:

15, C

Step-by-step explanation:

Any number is equally likely to appear. Therefore, we have 3/20 * 100 = 15. This is C.

Hope this helped!

~clouddragon

5 0

2 years ago

A clock was reading the time accurately on Friday at noon. On Monday at 6pm the clock was running late by 468 seconds. On averag

Setler [38]

The clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

The clock was still accurate by Friday noon. The clock was late by 468 seconds by Monday, 6 pm.

To solve the problem, we must:

Know how many 30-minutes have passed during the time period.

1 day = 24 hours

1 hour = 60 minutes = 2 × (30 minutes)

1 day = 24 hours × 2 × (30 minutes)

1 day = 48 × (30 minutes)

Thus, there are 48, 30-minutes in a day. On Friday, however, we start counting at noon, which is half of the day. Moreover, on Monday, the mark is only up to 6 pm, which is three-fourths of the day.

Friday = 48 × $\frac{1}{2}$ = 24

Saturday = 48

Sunday = 48

Monday = 48 × $\frac{3}{4}$ = 36

TOTAL = 24 + 48 + 48 + 36 = 156

Therefore, the total number of 30-minutes that have passed is 156. There were 156, 30-minutes that passed during the time period.

Divide the number of total seconds late by the number of 30-minutes passed.

That is, the number of total seconds late= 468 seconds ÷ 156

= 3 seconds

Therefore, the clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

To learn more about clock problems visit:

brainly.com/question/27122093.

#SPJ1

3 0

2 years ago

Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. r = 2 cos 5θ

murzikaleks [220]

Looking at the graph, we'll notice that there is a local maximum at x=0 and it looks similar on both sides of the y axis, therefore making it symmetric around the y axis given the options

3 0

3 years ago

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