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lawyer [7]
3 years ago
7

The depth of a lake is 15.8m

Mathematics
1 answer:
Leya [2.2K]3 years ago
4 0

a. Jada got the measurement 16m.

b. The measured depth differ 0.2m from the actual depth.

c. There is a 12.66% error in calculation.

Step-by-step explanation:

Given,

The depth of lake = 15.8 m

a. Jada accurately measured the depth of the lake to the nearest meter. What measurement did Jada get

When a digit after the decimal point is 5 or more than that, the number before decimal is rounded to the next number.

Therefore,

15.8 rounded to nearest number is 16.

The depth of lake measured by Jada = 16 m

Jada got the measurement 16m.

b. By how many meters does the measured depth differ from the actual depth?

Difference = Approx - Exact

Difference = 16 - 15.8 = 0.2m

The measured depth differ 0.2m from the actual depth.

c. Express the measurement error ad a percentage of the actual depth.

Percent error = \frac{|approx-exact|}{exact}*100

Percent error = \frac{|2|}{15.8}*100 = \frac{200}{15.8}

Percent error = 12.66%

There is a 12.66% error in calculation.

Keywords: percent, error

Learn more about percent at:

  • brainly.com/question/9323337
  • brainly.com/question/9103248

#LearnwithBrainly

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Mumz [18]
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2) 4/6 = 1/6 + 1/6 + 1/6 + 1/6
3) 3/5 = 1/5 + 1/5 + 1/5
4) 3/3 = 1/3 + 1/3 + 1/3
5) 7/8 = 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8
6) 6/2 = 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2
7) 5/6 = 1/6 + 1/6 + 1/6 + 1/6 + 1/6
8) 9/5 = 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5
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3 years ago
i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.
Natali [406]

Answer:

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = 7

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = -5\sqrt 2

Step-by-step explanation:

Given

N = (-7,-1) --- terminal side of \varnothing

Required

Determine the values of trigonometric functions of \varnothing.

For \varnothing, the trigonometry ratios are:

sin\ \varnothing = \frac{y}{r}       cos\ \varnothing = \frac{x}{r}       tan\ \varnothing = \frac{y}{x}

cot\ \varnothing = \frac{x}{y}       sec\ \varnothing = \frac{r}{x}       csc\ \varnothing = \frac{r}{y}

Where:

r^2 = x^2 + y^2

r = \sqrt{x^2 + y^2

In N = (-7,-1)

x = -7 and y = -1

So:

r = \sqrt{(-7)^2 + (-1)^2

r = \sqrt{50

r = \sqrt{25 * 2

r = \sqrt{25} * \sqrt 2

r = 5 * \sqrt 2

r = 5 \sqrt 2

<u>Solving the trigonometry functions</u>

sin\ \varnothing = \frac{y}{r}

sin\ \varnothing = \frac{-1}{5\sqrt 2}

Rationalize:

sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

sin\ \varnothing = \frac{-\sqrt 2}{5*2}

sin\ \varnothing = \frac{-\sqrt 2}{10}

sin\ \varnothing = \frac{-1}{10}\sqrt 2

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cos\ \varnothing = \frac{-7}{5\sqrt 2}

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cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}

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cos\ \varnothing = \frac{-7}{10}\sqrt 2

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tan\ \varnothing = \frac{-1}{-7}

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cot\ \varnothing = \frac{x}{y}

cot\ \varnothing = \frac{-7}{-1}

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sec\ \varnothing = \frac{r}{x}

sec\ \varnothing = \frac{5\sqrt 2}{-7}

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = \frac{r}{y}

csc\ \varnothing = \frac{5\sqrt 2}{-1}

csc\ \varnothing = -5\sqrt 2

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