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Scilla [17]
2 years ago
7

I need help with this question I got 1 as an answer but it’s saying it’s wrong

Mathematics
1 answer:
Debora [2.8K]2 years ago
5 0

Answer:

see image

Step-by-step explanation:

I did two different problems depending on whether the 4 beside the log is a base or not.

If the log IS the base, I got x = 1 as you did. But if the 4 is not the base then there is no base written and that means the base is 10.

For log problems that have log on one side of the equation only, then a good strategy is to change the equation from logarithmic form to exponential form. See image for work. Hope this helps.

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A go-cart racetrack has 100-meter straightaways and semicircular ends with diameters of 40 meters. Calculate the average speed i
emmainna [20.7K]

Answer:

Average speed of go cart is 217.07 meters per minute.

Step-by-step explanation:

We are given the following in the question:

Length of straightway racetrack = 100 m

Diameter of  semicircular end = 40 m

Radius =

r = \dfrac{d}{2} = \dfrac{40}{2} = 20\text{ m}

Perimeter of racetrack =

Perimeter of semicircle + 2(Length of straightway) + Perimeter of semicircle

=\pi r + 2(100) + \pi r\\=2\pi r + 2(200)\\=(2\times 3.14\times 20) + 200\\=325.6\text{ m}

Speed =

s = \dfrac{\text{Distance}}{\text{Time}}\\\\s = \dfrac{4\times \text{Perimeter of track}}{6}\\\\s = \dfrac{4\times 325.6}{6}\\\\s = 217.06\text{ meter per minute}

Thus, average speed of go cart is 217.07 meters per minute.

4 0
3 years ago
What is this answer.Estimated pls help now!i need it in 1-3 minutes
Shtirlitz [24]

Answer:

11.271

Step-by-step explanation:

just multiply

3 0
2 years ago
Read 2 more answers
Need help please. Thank u
a_sh-v [17]
The answer is the third option choice
3 0
3 years ago
GIVING BRAINLIEST!! HELP PLEASE!! you don't have to make it perfect. (no links/spam/or stealing points. i will report your accou
choli [55]
<h2>Answer:</h2>

C = 2πr

r = C/2π ...(1)

A = πr²

A = π(C/2π)² = πC²/4π²

A = C²/4π.

<u>Correct choice</u> - [A] A = C²/4π.

6 0
3 years ago
In a random sample of 25 ?people, the mean commute time to work was 30.4 minutes and the standard deviation was 7.1 minutes. ass
seropon [69]

In this problem, the given values are:

x = average = 30.4

s = standard deviation = 7.1

n = number of samples = 25

Degrees of freedom = n -1 =24

Using the t-distribution table for a normal curve at 90% CI, we get t-crit:

t-crit = 1.711

Now the Margin of Error (E) is calculated as:

E = t-crit*(s/ \sqrt{n})

By substituting the known values into the equation:

E = 1.711 * (7.1/ \sqrt{8})

E = +- 4.30

Therefore, the margin of error is 4.36. Hence, the commute time is between 26.1 minutes and 34.7 minutes.

<span> </span>

7 0
3 years ago
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