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2 years ago

Find the distance between (-4,-3) and (-2,0). Round your answer to the nearest tenth.

Mathematics

2 answers:

Kazeer [188]2 years ago

8 0

Answer:

just count the spaces between (-4, -3) (-2,0)

Step-by-step explanation:

use rise and run or sum else

dlinn [17]2 years ago

4 0

Answer:

$\sqrt{13}$

Step-by-step explanation:

$\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$\mathrm{The\:distance\:between\:}\left(-4,\:-3\right)\mathrm{\:and\:}\left(-2,\:0\right)\mathrm{\:is\:}$

$=\sqrt{\left(-2-\left(-4\right)\right)^2+\left(0-\left(-3\right)\right)^2}$

$Simplify$

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The newspaper in Haventown had a circulation of 80,000 papers in the year 2000. In 2010, the circulation was 50,000. With x = 0

mixas84 [53]

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2 years ago

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Part A: During what interval(s) of the domain is the water balloon's height increasing?

Advocard [28]

Answer:

The answer is below

Step-by-step explanation:

The linear model represents the height, f(x), of a water balloon thrown off the roof of a building over time, x, measured in seconds: A linear model with ordered pairs at 0, 60 and 2, 75 and 4, 75 and 6, 40 and 8, 20 and 10, 0 and 12, 0 and 14, 0. The x axis is labeled Time in seconds, and the y axis is labeled Height in feet. Part A: During what interval(s) of the domain is the water balloon's height increasing? (2 points) Part B: During what interval(s) of the domain is the water balloon's height staying the same? (2 points) Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest? Use complete sentences to support your answer. (3 points) Part D: Use the constraints of the real-world situation to predict the height of the water balloon at 16 seconds.

Answer:

Part A: During what interval(s) of the domain is the water balloon's height increasing?

Between 0 and 2 seconds, the height of the balloon increases from 60 feet to 75 feet

Part B: During what interval(s) of the domain is the water balloon's height staying the same?

Between 2 and 4 seconds, the height remains the same at 75 feet. Also from 10 seconds the height of the balloon is at 0 feet

Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest?

Between 4 and 6 seconds, the height of the balloon decreases from 75 feet to 40 feet (i.e. -17.5 ft/s)

Between 6 and 8 seconds, the height of the balloon decreases from 40 feet to 20 feet (i.e. -10 ft/s)

Between 8 and 10 seconds, the height of the balloon decreases from 20 feet to 0 feet (i.e. -10 ft/s)

Hence it decreases fastest from 4 to 6 seconds

Part D: Use the constraints of the real-world situation to predict the height of the water balloon at 16 seconds

From 10 seconds, the balloon is at the ground, so it remains at the ground (0 feet) even at 16 seconds

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2 years ago

Four equivalent forms of a quadratic function are given. Which form displays the zeros of function h?

galina1969 [7]

A). because the roots are shown. The roots of this quadratic function would be x=2, and x=-2

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2 years ago

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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.

Paraphin [41]

we know that

If two lines are parallel, then, their slopes are equal.

The formula to calculate the slope between two points is equal to

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

we will proceed to calculate the slope in each case, to determine the solution of the problem

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

Find the slope BC

Substitute the values in the formula

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

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

$m=-3.5$

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

therefore

the answer Part A) is

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

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

Find the slope DE

Substitute the values in the formula

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

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

$m=-0.5$

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

therefore

the answer Part B) is

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

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

Find the slope FG

Substitute the values in the formula

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

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

$m=-2$

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

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

Find the slope HI

Substitute the values in the formula

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

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

$m=1.25$

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

therefore

the answer Part D) is

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

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

Find the slope JK

Substitute the values in the formula

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

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

$m=-0.375$

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

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

Find the slope LM

Substitute the values in the formula

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

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

$m=5$

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

therefore

the answer Part F) is

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

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2 years ago

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What is another way to group the factors (3x2)x5

tia_tia [17]

Answer

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Explanation

Multiplication of numbers is associative. For example,

(a×b)×c = a×(b×c)

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For the equation given above, (3x2)x5, it can also be grouped as 3×(2×5).

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2 years ago

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