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Stells [14]
2 years ago
11

Katie jogged a total distance of 8 and 1 over 2 miles during the months of January and February. If Katie only jogged 1 over 4 m

ile every day, which expression shows the number of days she went jogging? (1 point)
Group of answer choices

8 and 1 over 2 ÷ 1 over 4

8 and 1 over 2⋅ 1 over 4

8 and 1 over 2 + 1 over 4

8 and 1 over 2 − 1 over 4

Flag question: Question 4
Mathematics
2 answers:
sattari [20]2 years ago
8 0

Answer:

the correct answer is....

C) 8 and 1 over 2 + 1 over 4

Hope this helps!!!! :)

**(Bojana)**

Step-by-step explanation:

lesya [120]2 years ago
5 0

Answer:

  (a)  8 1/2 ÷ 1/4

Step-by-step explanation:

Katie's total mileage is the product of the miles per day (1/4) and the number of days (n).

  miles = (1/4)n

If Katie jogged 8 1/2 miles, then we have ...

  8 1/2 = (1/4)n

We find the value of n (the number of days) by dividing both sides of this equation by the coefficient of n:

  (8 1/2) ÷ (1/4) = n

_____

<em>Additional comment</em>

If you consider the units, you know that the only possible operation is division. The number 8 1/2 has units of miles. The number 1/4 has units of miles per day.

  \dfrac{\text{miles}}{\left(\dfrac{\text{miles}}{\text{day}}\right)}=\dfrac{\text{miles}\cdot\text{day}}{\text{miles}}=\text{days}

If you multiply, the product units are miles²/day. You cannot add or subtract numbers with different units. (They are not "like terms".)

As here, "units analysis" often tells you what math operations are needed, and which ones make no sense whatever.

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Solution. Notice that

 

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Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| < 1, so −2 < x < 0. In particular |x| < 2. So

 

|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|

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Step-by-step explanation:

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stepan [7]

Answer:

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Step-by-step explanation:

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Nataly_w [17]

{ \qquad\qquad\huge\underline{{\sf Answer}}}

Here we go ~

In the above question, it is given that :

\qquad \sf  \boxed{ \sf f(x) =  \frac{x + 3}{2} }

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\qquad \sf  \dashrightarrow \: f(2) =  \dfrac{2 + 3}{2}

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or

\qquad \sf  \dashrightarrow \: f(2) = 0.5

B.) Find { \sf {f}^{-1}(x) } :

\qquad \sf  \dashrightarrow \: let \: y = f (x)

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