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Soloha48 [4]
2 years ago
13

Sara lost 12 pounds over the summer by jogging each week. By winter time she had gained 4 3/8 pounds. Represent this situation w

ith an expression involving signed numbers. What is the overall change in Sara’s weight?.
Mathematics
1 answer:
Vika [28.1K]2 years ago
3 0

The situation with an expression involving signed numbers is w - 12 + 4 3/8 and the overall change in Sara’s weight is -7 5/8 pounds

<h3>Represent this situation with an expression involving signed numbers. </h3>

Let Sara's initial weight be w.

The given parameters are:

Lost = 12 pounds

Gain = 4 3/8 pounds

The situation with an expression involving signed numbers is represented as:

Weight = Initial - Lost + Gain

So, we have:

Weight = w - 12 + 4 3/8

<h3>What is the overall change in Sara’s weight?.</h3>

In (a), we have:

Weight = w - 12 + 4 3/8

The overall change is:

Change = - 12 + 4 3/8

Evaluate

Change = -7 5/8

Hence, the overall change in Sara’s weight is -7 5/8 pounds

Read more about sum and difference at:

brainly.com/question/17695139

#SPJ1

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Answer:

Step-by-step explanation:

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3 years ago
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sweet-ann [11.9K]

Answer:

  • Question 1a. i) x=1.8m

  • Question 1a. ii)   Volume=27.6m^3

  • Question 1b) Volume=65.9m^3

Explanation:

<u><em>Question 1 a. i) Find the value of x.</em></u>

         tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}

For the smalll triangle you can write:

        tan(\theta )=\dfrac{x}{1m}

For tthe big triangle:

      tan(\theta )=\dfrac{x+2.7m}{2.5m}

Substitute:

        \dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}

Solve for x:

        2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m

<u><em>Question 1a ii) Find the volume of the frustrum</em></u>

  • Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

         Volume=(1/3)\pi \times radius^2\times height

Substitute:

         Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3

  • Find the volume of a cone with heigth = 1.8m and radius = 1m

        Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3

  • Subtract the volume of the small cone from the volume of the big cone

        Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3

<u><em>Question 1b. Calculate the volume of the bin</em></u>

<u>i) Upper frustrum</u>

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

<u>ii) Lower frustrum</u>

            \dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}

           2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m

        Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)

       Volume=38.3m^3

<u>iii) Add the volume of the two frustrums</u>

  • Volume=27.6m^3+38.3m^3=65.9m^3

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Answer:

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Answer:

Percentage: 115%

Decimal: 1.15

Fraction: \frac{23}{20}\\

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Now we express 115% as a decimal, simply dividing by 100:

115% = 115/100 = 1.15

To express it as a fraction, we simplify the division instead of calculating it:

115% = \frac{115}{100}

Simplifying by 5:

115% = \frac{23}{20}

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