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strojnjashka [21]
3 years ago
11

Write an inequality for the graph. Write your answer with y by itself on the left side of the inequality.​

Mathematics
1 answer:
Lunna [17]3 years ago
6 0

Answer:

Y < -1/4x + 2

Step-by-step explanation:

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hat are the solutions to the equation |x| = 2? Choose all answers that are correct. A. B. C. D.
zalisa [80]
Easy, so absolute value makes anything inside into positive
|x|=2
therefor x=-2 or x=2 since  it would be made posotive

answer is -2 and 2

6 0
3 years ago
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If the volume of the cube was 27 (each side length is 3 units), what would the new volume be if each side length of the cube was
Nat2105 [25]

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Step-by-step explanation:

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3 years ago
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Find the mean and median of the data set 3,5,6,2,10,9,7,5,11,6,4,2,5,4
Tanya [424]

Answer:

mean: 5.6 ; median: 5

Step-by-step explanation:

first order them:

2, 2, 3, 4, 4, 5, 5, 5, 6, 6, 7, 9, 10, 11

now we can see that the median is 5 since it's in the middle

and the mean is adding all and dividing by how many numbers we have (AKA 14) so 2+2+3+4+4+5+5+5+6+6+7+9+10+11 = 79 and now we divide 79 by 14 which is:

5.64285714

8 0
4 years ago
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Andrew scored an 89, 92, 97, 81, and 96 on five tests. What is the mean of Andrew's test scores?
natulia [17]
To find the mean, add all the numbers together and divide by the amount of numbers there is

89 + 92 + 97 + 81 + 96 = 455

455/5 = 91

91 is Andrew's mean score, and is your answer

hope this helps
8 0
3 years ago
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Find the value of x of this question ​
mel-nik [20]
<h3><u>Let's</u><u> </u><u>understand</u><u> </u><u>the concept</u><u>:</u><u>-</u></h3>
  • Here angle B is 90°
  • So \triangle ABC and \triangle ABD Are right angled triangle
  • So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>
  • First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

  • We need to find base=b

According to Pythagoras thereon

{\boxed{\sf b^2=h^2-p^2}}

  • Substitutethe values

\longrightarrow\sf b^2=10^2-p^2

\longrightarrow\sf b={\sqrt {10^2-8^2}}

\longrightarrow\sf b={\sqrt{100-64}}

\longrightarrow\bf b={\sqrt {36}}

\longrightarrow\sf b=6

\therefore\overline{BC}=6cm

BD=BC+CD

\longrightarrowBD=9+6

\longrightarrowBD=15cm

  • Now in \triangle ABD

Perpendicular=p=8cm

Base =b=15cm

  • We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

{\boxed {\sf h^2=p^2+b^2}}

  • Substitute the values

\longrightarrow\sf h^2=8^2+15^2

\longrightarrow\sf h={\sqrt {8^2+15^2}}

\longrightarrow\sf h={\sqrt {64+225}}

\longrightarrow\sf h={\sqrt {289}}

\longrightarrow\sf h=17cm

\therefore{\underline{\boxed{\bf x=17cm}}}

8 0
3 years ago
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