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tiny-mole [99]
3 years ago
9

(+5)+(-2)+(+20) solve

Mathematics
2 answers:
Veseljchak [2.6K]3 years ago
5 0

Answer:

5(x-2)^2=20

5(x^2-4)=20

5x^2-20=20

5x^2-20-20=0

5x^2-40=0

5x^2+0x-40=0

Step-by-step explanation:

hope this is helpfull

Darina [25.2K]3 years ago
4 0

Answer:

23

Step-by-step explanation:

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Please help me with this please.
natali 33 [55]

Answer:

f(-2) = -1

f(0) = -5

f(4) = -1

Step-by-step explanation:

The number inside of f( ) is telling you which equation to use to get the answer.

Example:

f(-2) since -2 is less than 0 you would use (x^2) - 5. So, ((-2)^2) - 5 = 4-5 = -1

f(0) since 0 is less than or equal to 0 you would use (x^2) - 5 again. So, ((0)^2) - 5 = 0-5 = -5

f(4) since 4 is grater than 3 you would use (2^(x-1)) - 9. So, (2^(4-1)) - 9 =  -1

<u><em>The Graph, try to make it as straight as possible</em></u>

<u><em></em></u>

6 0
3 years ago
Rewrite 0.169 in expanded form using powers of ten. A) (1×103) + (6×102) + (9×101) B) (1×101) + (6×102) + (9×103) C) (1×10-1) +
lubasha [3.4K]
That would be C....(1 x 10^-1) + (6 x 10^-2) + (9 x 10^-3)
5 0
3 years ago
The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic m
amm1812

Answer:

%82.5

Step-by-step explanation:

  1. The final exam of a particular class makes up 40% of the final grade
  2. Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam.

From point 1 we know that Moe´s grade just before taking the final exam represents 60% of the final grade. Then, using the information in the point 2 we can compute Moe´s final grade as follows:

FG=0.40*FE+0.60*0.45,

where FG is Moe´s Final Grade and FE is Moe´s final exam grade. Then,

\frac{ FG-0.60*0.45}{0.40}=FE.

So, in order to receive the passing grade average of 60% for the class Moe needs to obtain in his exam:

FE=\frac{ 0.60-0.60*0.45}{0.40}=0.825

That is, he need al least %82.5 to obtain a passing grade.

6 0
3 years ago
Pamela and Debbie are shelving books at a public library. Pamela shelves 26 books at a time, whereas Debbie shelves 7 at a time.
Ira Lisetskai [31]

If both Pamela and Debbie can shelf the same number of books at the end, the smallest number of books each could have shelved is 182

The number of books that Pamela can shelf at a time = 26

The number of books that Debbie can shelf at a time = 7

If both Pamela and Debbie can shelf the same number of books at the end, the smallest number of books each could have shelved is the least common multiple of 26 and 7

The Least common multiple of 26 and 7  = 182

Therefore, if both Pamela and Debbie can shelf the same number of books at the end, the smallest number of books each could have shelved is 182

Learn more on Least Common Multiple here: brainly.com/question/363238

8 0
2 years ago
What are the solutions to the equation? x^3 - 8x^2 - x + 8 =0
kogti [31]

Answer:

Step-by-step explanation:

x^3 - 8x² - x + 8 =0

x^3 - 8x² - (x - 8 )=0

x²(x-8) - (x-8) = 0

(x-8)(x² -1) =0

x-8 = 0  or x² - 1 = 0

x-8=0 or (x-1)(x+1)=0

x=8 x=1 or x=-1

8 0
3 years ago
Read 2 more answers
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