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11 days ago

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In July, the average temperature in one US city was 29°C. By December, the average temperature had fallen by 29°C. Explain why t

he average temperature in December was 0°C.

2 answers:

tester [8.8K]11 days ago

10 0

Since the temperature was 29 degrees in July and by December it was 29 degrees lower than the initial 29, this results in zero because 29-29=0.

tester [8.8K]11 days ago

7 0

The temperature increased by 29°C, which indicates a positive change of 29. Then, it decreased by 29°C, reflecting a negative change. The average temperature experienced equal increases and decreases, leading to identical absolute values and resulting in a difference of 0.

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The formula h -15 = 3.2t gives the height h in inches of a plant t weeks after planting.

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Set t = 0 to find the initial height

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Match each pair of points to the equation of the line that is parallel to the line passing through the points.

Svet_ta [9500]

It's known that

When two lines are parallel, their slopes are identical.

The slope between any two points can be calculated using the following formula:

$m=\frac{y2-y1}{x2-x1}$

We will calculate the slope for each case to find the solution to the problem.

Case A) Point $B(5,2)\ C(7,-5)$

Determine the slope of BC

Insert the values into the formula:

$m=\frac{-5-2}{7-5}$

$m=\frac{-7}{2}$

$m=-3.5$

Thus,

The equation $y=-3.5x-15$ is parallel to the line that goes through the points $B(5,2)\ C(7,-5)$

Therefore,

the result for Part A) is

$B(5,2)\ C(7,-5)$ ------> $y=-3.5x-15$

Case B) Point $D(11,6)\ E(5,9)$

Calculate the slope of DE

Plug the values into the formula:

$m=\frac{9-6}{5-11}$

$m=\frac{3}{-6}$

$m=-0.5$

Thus,

The equation $y=-0.5x-3$ is parallel to the line that goes through the points $D(11,6)\ E(5,9)$

Therefore,

the result for Part B) is

$D(11,6)\ E(5,9)$ ------> $y=-0.5x-3$

Case C) Point $F(-7,12)\ G(3,-8)$

Determine the slope of FG

Insert the values into the formula:

$m=\frac{-8-12}{3+7}$

$m=\frac{-20}{10}$

$m=-2$

Thus,

Any linear equation with slope $m=-2$ will be parallel to the line through the points $F(-7,12)\ G(3,-8)$

Case D) Point $H(4,4)\ I(8,9)$

Calculate the slope of HI

Substitute the values in the formula:

$m=\frac{9-4}{8-4}$

$m=\frac{5}{4}$

$m=1.25$

Thus,

The equation $y=1.25x+4$ is parallel to the line through the points $H(4,4)\ I(8,9)$

Therefore,

the result for Part D) is

$H(4,4)\ I(8,9)$ ------> $y=1.25x+4$

Case E) Point $J(7,2)\ K(-9,8)$

Determine the slope of JK

Insert the values into the formula:

$m=\frac{8-2}{-9-7}$

$m=\frac{6}{-16}$

$m=-0.375$

Thus,

Any linear equation characterized by slope $m=-0.375$ will be parallel to the line that runs through the points $J(7,2)\ K(-9,8)$

Case F) Point $L(5,-7)\ M(4,-12)$

Find the slope of LM

Substitute the values in the formula:

$m=\frac{-12+7}{4-5}$

$m=\frac{-5}{-1}$

$m=5$

Thus,

The equation $y=5x+19$ is parallel to the line connecting the points $L(5,-7)\ M(4,-12)$

Therefore,

the result for Part F) is

$L(5,-7)\ M(4,-12)$ ------> $y=5x+19$

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THIS IS THE ACTUAL WORK AND RESULT

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