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noname [10]
2 years ago
12

Write the names of 5 object which are longer than your hand- span in your surrounding

Mathematics
1 answer:
Semenov [28]2 years ago
4 0

5 object which are longer than your hand- span in my surrounding is Note book, Laptop, Window, Curtain, Table.

Hand span is a measure of distance from the tip of the thumb to the tip of the little finger with the hand fully extended.

Hand span is a measure that has been used for many years. By placing the hand on the edge of a piece of paper and marking the tips of the thumb and little finger, the student can measure a straight line. This is a better method than placing the hand directly on the ruler.

Hand span and cubit are not used as standard units of length because their sizes vary from person to person. So, two different persons may give different measurements for the same length, which is not desirable for a standard unit.

Measure using yours hands and write as per your hand-span measurement.

Therefore,

5 object which are longer than your hand- span in my surrounding is Note book, Laptop, Window, Curtain, Table.

Find out more information about hand span here

brainly.com/question/993795

#SPJ4

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F(x) = x2 − x − ln(x) (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the inte
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Answer:

(a) Decreasing on (0, 1) and increasing on (1, ∞)

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Step-by-step explanation:

ƒ(x) = x² - x – lnx

(a) Intervals in which ƒ(x) is increasing and decreasing.

Step 1. Find the zeros of the first derivative of the function

ƒ'(x) = 2x – 1 - 1/x = 0

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We reject the negative root, because the argument of lnx cannot be negative.

There is one zero at (1, 0). This is your critical point.

Step 2. Apply the first derivative test.

Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.

(1) x = ½

ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1

ƒ'(x) < 0 so the function is decreasing on (0, 1).

(2) x = 2

ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½  = ⁵/₂

ƒ'(x) > 0 so the function is increasing on (1, ∞).

(b) Local extremum

ƒ(x) is decreasing when x < 1 and increasing when x >1.

Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.

(c) Inflection point

(1) Set the second derivative equal to zero

ƒ''(x) = 2 + 2/x² = 0

             x² + 2 = 0

                   x² = -2

There is no inflection point.

(2). Concavity

Apply the second derivative test on either side of the extremum.

\begin{array}{lccc}\text{Test} & x < 1 & x = 1 & x > 1\\\text{Sign of f''} & + & 0 & +\\\text{Concavity} & \text{up} & &\text{up}\\\end{array}

The function is concave up on (0, ∞).

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