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Kryger [21]
3 years ago
14

Find f(m - 1) when f(×) = 5×^2+4× - 5​

Mathematics
1 answer:
Nina [5.8K]3 years ago
5 0

Answer:

5m² - 6m + 1

Step-by-step explanation:

f(x) = 5x² + 4x - 5

substituting value,

f(m-1) = 5(m-1)² + 4(m-1) - 5

so,

5(m-1)² + 4(m-1) - 5

= 5(m²-2m+1) + 4m -1 - 5

= 5m² - 10m + 5 + 4m - 4

= 5m² - 6m + 1

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Answer:

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Step-by-step explanation:

Start (A3)

x is equal to 141 because they are alternate interior angles.

A2. x is equal to 39 because they are corresponding angles.

B2. x would be supplementary to 41 because the angle that x supplements is corresponding to 41.

41 + x = 180 due to the linear pair postulate. Therefore, x = 139.

B1. x would be supplementary to 82 because they are consecutive exterior angles.

82 + x = 180 due to the linear pair postulate. Therefore, x = 98.

C1. x = 102 due to the vertical angles theorem.

C2. x would be supplementary to 130 because the angle that x supplements is equal to 130 (Alternate Exterior Angles).

130 + x = 180, x = 50.

D2. x = 74, corresponding angles.

D3. x = 83, corresponding angles.

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x = 162

END

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4 years ago
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6 0
3 years ago
An entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 5 years, the
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Answer:

We conclude that the daily average revenue was actually $675.

Step-by-step explanation:

We are given that the  current owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75.

A sample of 30 days reveals a daily average revenue of $625.

<u><em>Let </em></u>\mu<u><em> = daily average revenue.</em></u>

So, Null Hypothesis, H_0 : \mu = $675     {means that the daily average revenue was $675}

Alternate Hypothesis, H_A : \mu \neq $675     {means that the daily average revenue was different from $675}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                       T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample daily average revenue = $625

            \sigma = population standard deviation = $75

            n = sample of days = 30

Since, we are given that we have decided not to reject the null hypothesis which leads us to the conclusion that the daily average revenue was actually $675.

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