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kobusy [5.1K]
3 years ago
15

Learning Task 1: Find the value of th

Mathematics
1 answer:
shepuryov [24]3 years ago
8 0

Answer:

a=5your answer right answer right answer

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In 4 games, Larry averaged 23 PPG (points per game). How many points does he need to score in the last game to have an average o
ivann1987 [24]
Larry needs to score 33 points in the last game to have an average of 25 PPG.

To find the answer, we can solve the equation:

\frac{(4)23+p}{5}=25

<em />(where <em>p</em> is how many points he needs to score in the last game)

To find the mean of a group of numbers, you add them all up and divide the total by the number of numbers there are. Since Larry averaged 23 PPG in 4 games, we can multiply 23 by 4 to get the total of the first 4 games from the data. Then, we find <em>p</em>, which we will add to get our final total. Then, you divide by the 5 games.

\frac{(4)23+p}{5}=25
\frac{92+p}{5}=25
{92+p}=125
p=33

First, I simplified 4 x 23 to get 92. Then, I multiplied each side by 5 to get rid of the denominator. Finally, I subtracted 92 from each side to isolate <em>p</em>, and found that <em>p</em> = 33.


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3 years ago
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777dan777 [17]
Am not sure but Y=-x+2
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I will give out brainliest answer this
Archy [21]

Answer:

Step-by-step explanation:

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A number plus 18 is less than 28
andrezito [222]
Written as an inequality- x+18<28
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Lim t^4 - 6 / 2t^2 - 3t + 7
Harman [31]

I think you meant to say

\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}

(as opposed to <em>x</em> approaching 2)

Since both the numerator and denominator are continuous at <em>t</em> = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}

Because these expressions are continuous at <em>t</em> = 2, we can compute the limits by evaluating the limands directly at 2:

\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}

6 0
3 years ago
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