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3 years ago

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A tree is growing on a hill. In an attempt to make the tree grow straight up, a 12-foot-long wire is attached to a tree at a poi

nt 6 feet off the ground. The wire is anchored to the ground 10 feet from the base of the tree. At what angle is the tree growing in relation to the hill?

2 answers:

nasty-shy [4]3 years ago

5 0

Answer:

Its 93.8 on ed

melisa1 [442]3 years ago

3 0

For this case what we must do is use the law of cosines.
We have then:
12 ^ 2 = 6 ^ 2 + 10 ^ 2 - 2 * (6) * (10) cos (x)
Clearing we have:
cos (x) = (12 ^ 2- (6 ^ 2 + 10 ^ 2)) / (- 2 * (6) * (10))
x = acos ((12 ^ 2- (6 ^ 2 + 10 ^ 2)) / (- 2 * (6) * (10)))
x = 93.82 degrees
Answer:
The tree is growing in relation to the hill at:
x = 93.82 degrees

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Ivan

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</span></span>Let's solve your equation step-by-step.<span><span><span>5−x</span>−<span>(<span><span>2x</span>+3</span>)</span></span>=<span>3+<span>8x</span></span></span>Step 1: Simplify both sides of the equation.<span><span><span>5−x</span>−<span>(<span><span>2x</span>+3</span>)</span></span>=<span>3+<span>8x</span></span></span>Simplify:<span><span><span>−<span>3x</span></span>+2</span>=<span><span>8x</span>+3</span></span>Step 2: Subtract 8x from both sides.<span><span><span><span>−<span>3x</span></span>+2</span>−<span>8x</span></span>=<span><span><span>8x</span>+3</span>−<span>8x</span></span></span><span><span><span>−<span>11x</span></span>+2</span>=3</span>Step 3: Subtract 2 from both sides.<span><span><span><span>−<span>11x</span></span>+2</span>−2</span>=<span>3−2</span></span><span><span>−<span>11x</span></span>=1</span>Step 4: Divide both sides by -11.<span><span><span>−<span>11x</span></span><span>−11</span></span>=<span>1<span>−11</span></span></span><span>x=<span><span>−1</span>11</span></span>Answer:<span>x=<span><span>−1</span>11</span></span>
<span>Espero que esto ayude</span>

4 0

2 years ago

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The city is replacing all the streets, curbs, and gutters in Middleville. The homeowners on both sides of the street are charged

bija089 [108]

Mr. Kyle will be expected to pay $8,100.

The total cost for his land would be 120 x 90 = 10,800.

Since the city is paying 25%, he must pay the other 75%.

0.75 x 10800 = 8100

8 0

3 years ago

you measure the period of a mass oscillating on a vertical spring ten times as follows: period (s): 1.06, 1.31, 1.28, 0.99, 1.48

lidiya [134]

The mean and (sample) standard deviation σ = 0.2098.

<h3>What exactly would the standard deviation indicate?</h3>

The term "standard deviation" (or "") refers to the degree of dispersion of the data from the mean. Data are grouped around the mean when the standard deviation is low, and are more dispersed when the standard deviation is high.

<h3>According to given information:</h3>

The mean is the product of the dataset's total and the sample size. Mathematically.

$\bar{x}=\frac{\sum X_i}{N}$

The individual periods are Xi.

The sample size is N.

$\sum X i$ = 1.06 + 1.31 + 1.28 + 0.99,+ 1.48 + 1.37+ 0.98 + 1.31 + 1.59 + 1.55

$\sum X i$ = 12.92

N = 10

While substituting the value we get:

x = 12.96/10

x = 1.292

The samples' average is 1.292.

The standard deviation:

$\sigma=\sqrt{\frac{\sum(x-\bar{x})^2}{N}}$

$\sum(x-\bar{x})^2$ = (1.48-1.292)^2+(1.37-1.292)^2+(0.98-1.292)^2+(1.31-1.292)^2+(1.59-1.292)^2+(1.55-1.292)^2.

$\sum(x-\bar{x})^2$ = 0.43996

Putting into the formula we get:

$\sigma=\sqrt{\frac{0.43996}{10}}$

σ = √(0.043996)

σ = 0.2098

The mean and (sample) standard deviation σ = 0.2098.

To know more about standard deviation visit:

brainly.com/question/18521100

#SPJ4

I understand that the question you are looking for is:

You measure the period of a mass oscillating on a vertical spring ten times as follows:

Period (s): 1.06, 1.31, 1.28, 0.99, 1.48, 1.37, 0.98, 1.31, 1.59, 1.55

Required:

What are the mean and (sample) standard deviation?

a. Mean: 1.228, Standard Deviation: 0.2135

b. Mean: 1.325, Standard Deviation: 0.1674

c. Mean: 1.292. Standard Deviation: 0.2211

d. Mean: 1.228, Standard Deviation: 0.2098

e. Mean: 1.292, Standard Deviation: 0.2135

7 0

1 year ago

A sample of a radioactive substance decayed to 97% of its original amount after a year. (Round your answers to two decimal place

Andrej [43]

Answer:

a) The half life of the substance is 22.76 years.

b) 5.34 years for the sample to decay to 85% of its original amount

Step-by-step explanation:

The amount of the radioactive substance after t years is modeled by the following equation:

$P(t) = P(0)(1-r)^{t}$

In which P(0) is the initial amount and r is the decay rate.

A sample of a radioactive substance decayed to 97% of its original amount after a year.

This means that:

$P(1) = 0.97P(0)$

Then

$P(t) = P(0)(1-r)^{t}$

$0.97P(0) = P(0)(1-r)^{0}$

$1 - r = 0.97$

So

$P(t) = P(0)(0.97t)^{t}$

(a) What is the half-life of the substance?

This is t for which P(t) = 0.5P(0). So

$P(t) = P(0)(0.97t)^{t}$

$0.5P(0) = P(0)(0.97t)^{t}$

$(0.97)^{t} = 0.5$

$\log{(0.97)^{t}} = \log{0.5}$

$t\log{0.97} = \log{0.5}$

$t = \frac{\log{0.5}}{\log{0.97}}$

$t = 22.76$

The half life of the substance is 22.76 years.

(b) How long would it take the sample to decay to 85% of its original amount?

This is t for which P(t) = 0.85P(0). So

$P(t) = P(0)(0.97t)^{t}$

$0.85P(0) = P(0)(0.97t)^{t}$

$(0.97)^{t} = 0.85$

$\log{(0.97)^{t}} = \log{0.85}$

$t\log{0.97} = \log{0.85}$

$t = \frac{\log{0.85}}{\log{0.97}}$

$t = 5.34$

5.34 years for the sample to decay to 85% of its original amount

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3 years ago

What is the value of y

zvonat [6]

The answer is 40. Hope this helps

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3 years ago

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