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zhannawk [14.2K]
2 years ago
14

It is 76°f at the 6000-foot level of a mountain, and 49°f at the 12,000-foot level of the mountain. write a linear equation to f

ind the temperature tt at an elevation xx on the mountain, where xx is the height in feet. enter answers as decimals rounded to the 4 digits.
Mathematics
1 answer:
valina [46]2 years ago
5 0

Answer:

t(x) = 103 -0.0045x

Step-by-step explanation:

We are given the following in the question:

Let the equation:

t(x) = a + bx

be the linear equation that represent temperature at an elevation x.

Temperature at 6000-foot level = 76 f

76 = a + 6000b

Temperature at 12000-foot level = 49 f

49 = a + 12000b

Solving the two equation, we get,

76-49 = -6000b\\27=-6000b\\\\b = \dfrac{-27}{6000}\\\\b = -0.0045\\76 = a + (-0.0045)6000\\76 = a -27\\a = 76 + 27\\a  =103

Thus, we can write the linear equation as:

t(x) = 103 -0.0045x

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Select two numbers in which the value of the 8 is 10 times the value 8 in 563840
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Answer:

768,543, and 8,451

Step-by-step explanation:

The 8 is in the hundreds place, so when multiplied by 10 it would be in the thousands place.

7 0
3 years ago
I need the answers to a b c and d PLZ DUE AT 11:59PM ITS 9:23PM
Nookie1986 [14]

Answer:

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Step-by-step explanation:

3 0
2 years ago
Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the
Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees wasx+\frac{1}{4} x=\frac{5}{4} x, and for the next three years we have that

                             Start                                          End

Second year     \frac{5}{4}x --------------   \frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x

Third year    (\frac{5}{4} )^{2}x-------------(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x

Fourth year (\frac{5}{4})^{3}x--------------(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.

So  the formula to calculate the number of trees in the fourth year  is  

(\frac{5}{4} )^{4} x, we know that all of the trees thrived and there were 6250 at the end of 4 year period, then  

6250=(\frac{5}{4} )^{4}x⇒x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.

Therefore the number of trees at the begging of the 4-year period was 2560.  

7 0
3 years ago
An object moves in simple harmonic motion with period 8 seconds and amplitude 6 cm. At time t = 0 seconds, its displacement d fr
Vladimir79 [104]

In the equation model, the simple harmonic motion is d=6sin(πt/4)-6 if the object moves in simple harmonic motion with a period of 8 seconds and amplitude of 6 cm.

<h3>What is simple harmonic motion?</h3>

Simple Harmonic Motion is described as a motion in which the restoring force is proportionate to the body's displacement from its mean position.

We have:

Amplitude A = 6 cm

Time period = 8 seconds

T = 2π/w

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w = π/4

Its displacement d from rest is -6 cm, and initially, it moves in a positive direction.

d(0) = -6 cm at t = 0

The equation becomes:

\rm d = 6sin(\dfrac{\pi}{4}t)-6

Thus, in the equation model, the simple harmonic motion is d=6sin(πt/4)-6 if the object moves in simple harmonic motion with a period of 8 seconds and amplitude of 6 cm.

Learn more about the simple harmonic motion here:

brainly.com/question/17315536

#SPJ1

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2 years ago
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stich3 [128]
8/100, 3/5, 7/10

Explanation:
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6 0
3 years ago
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