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3 years ago

Amber is solving the inequality |x+6|-12<13 by graphing. Which equations should Amber graph?

Mathematics

2 answers:

Virty [35]3 years ago

8 0

Answer:

A. y₁ = |x + 6| and y₂ = 25

Step-by-step explanation:

Got it on Edg 2020

DENIUS [597]3 years ago

3 0

Answer:

y₁ = |x + 6| and y₂ = 25

Step-by-step explanation:

<u>Given inequality</u>

|x+6| - 12 < 13
|x+6| < 13 + 12
|x+6| < 25

To solve this inequality graphically you need two graphs according left and right sides of inequality

<u>So the graphs should be:</u>

y₁ = |x + 6| and y₂ = 25

See attached for the graph

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A bug completes one lap along the edge of a circular planter of radius 11.7 cm in 13.91 s. How fast was it traveling? 1. 10.8518

IrinaK [193]

Answer:

8.5.28492 cm/s

Step-by-step explanation:

We are given that

Radius of circular planter=11.7 cm

Time taken by bug =13.91 s

We have to find the speed of bug.

We know that

Speed = $\frac{distance}{time}$

Distance traveled by bug = $2\pi r$

Distance traveled by bug = $2\times 3.14\times 11.7$

Because $\pi=3.14$

Speed of bug= $\frac{2\times 3.14\times 11.7}{13.91}=5.28224 cm/s$

Hence, speed of bug=5.2822 4 cm/s

Answer:8.5.28492 cm/s

8 0

3 years ago

The table lists how poverty level income cutoffs (in dollars) for a family of four have changed over time. Use the midpoint for

frez [133]

The poverty level cutoff in 1987 to the nearest dollar was $10787.

<h3>How to find a midpoint?</h3>

The midpoint as the point that divides the line segment exactly in half having two equal segments. Therefore, the midpoint presents the same distance between the endpoints for the line segment. The midpoint formula is: $\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)$ .

For solving this exercise, first you need plot the points in a chart. See the image.

Your question asks to approximate the poverty level cutoff in 1987 to the nearest dollar using the midpoint formula. Note that the year 1987 is between 1980 and 1990, thus you should apply the midpoint formula from data for this year (1987).

$\mathrm{Midpoint\:of\:}\left(1980,8429),\:\left(1990,13145):\quad \left(\frac{1990+1980}{2},\:\:\frac{13145+8429}{2}\right)\\\\ \\$

$\mathrm{Midpoint\:of\:}\left(1980,8429)\:=\left(\frac{1990+1987}{2},\:\frac{13145+3843}{2}\right)\\ \\ =\left(\frac{3977}{2},\:\frac{21574}{2}\right)$

$\mathrm{Midpoint\:of\:}\left(1980,8429)\:=\=(\frac{3977}{2},10787)$

The answer for your question will be the value that you calculated for the y-coordinate. Then, the poverty level cutoff in 1987 to the nearest dollar was $10787.

Read more about the midpoint segment here:

brainly.com/question/11408596