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Nataly_w [17]
1 year ago
8

How many solutions does this problem have? | x - 7 | + 3 = - 15

Mathematics
1 answer:
mixer [17]1 year ago
4 0

There are no solutions. By definition of absolute value, |x-7|\ge0, so |x-7|+3\ge3. The left side will always be positive, and a positive number cannot also be negative.

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I need help I have searched everywhere and nothing
Marta_Voda [28]

Answer:

See attached

Step-by-step explanation:

<em>All answers in one graph, refer to attached</em>

<u>The function we need is</u>

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Table included as well

The vertex is about x = 19 ft which corresponds the minimum cost

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3 0
3 years ago
The number of bacteria (b) in a batch of refrigerated food is given by the function b(t) = 20t2 − 70t + 300, where t is the temp
nata0808 [166]
We want to combine b(t) and t(h) to create a function that described the number of bacteria, b, as a function of time, h. 
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This describes the number of bacteria, given the temperature.

t(h) described the temperature as a function of time, specifically hours after refrigeration:
t(h) = 2h + 3

Since the time h can tell us the temperature t, and the temperature t can tell us the # of bacteria b, we can create a function that tells us the number of bacteria, b, given hours following refrigeration, h.

To do this, we plug t(h) in for every t in the b(t) function:
b(t(h)) = 20 (2h+3)² - 70(2h+3) + 300
We can also call this function b(h), since we now can express b as a function of h.

Simplify the function:
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The Answer is B

7 0
3 years ago
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marshall27 [118]

Please consider the graph.

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We can see from our graph that mean of the weights is 9.5 and standard deviation in 0.5.

The data point that would be below two standard deviation is: 9.5-2\times 0.5 that is 9.5-1=8.5.

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Now we need to check the data points that lie within 8.5 and 10.5.

Upon looking at our given choices, we can see that 8.9, 9.5 and 10.4 pounds lie within 2 standard deviation of the mean.

Therefore, 8.9 lbs, 9.5 lbs and 10.4 lbs are correct choices.

8 0
3 years ago
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Andre45 [30]
2=6
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dexar [7]
10 feet per 5 seconds
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