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Serjik [45]
2 years ago
11

Stuck on this question, can someone please help

Mathematics
2 answers:
Rufina [12.5K]2 years ago
6 0

Answer:

16/36 = 12/27

Step-by-step explanation:

16/36 = 12/27 ( TRUE bc its gonna leave both side of 0.44444.........)

12/ 15 = 47/50 ( FALSE bc The left side  0.8  does not equal to the right side  

0.94)

4/16 = 2/14 ( FALSE bc The left side  0.25  does not equal to the right side  0. 142857 142857142857......)

15/20 = 24/36 ( FALSE bc The left side  0.75  does not equal to the right side  

0. 6666666...........)

Therefore, 16/36 = 12/27 is the answer

xxTIMURxx [149]2 years ago
4 0

Answer:

no

Step-by-step explanation:

asdsadfdddddddddddddddddddddddddddjkdf

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Please HelP with both​
Tomtit [17]

Answer:

Step-by-step explanation:

Number 5 is third option

and number 6 is second option

3 0
3 years ago
F(x)= -2 |x-2|+ 4 vertically stretched by a factor of 2.
Veronika [31]

Answer:

f(x) = 4x^2 +2,

 is the function

Step-by-step explanation:

if i get this wrong imma tie a brick to my feet and jump in the ocean

6 0
2 years ago
What is the answer please.... need help
mina [271]

Answer:

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

f\left(x\right)=x^3-6x^2+3x+10

As the highest power of the x-variable is 3 with the leading coefficients of 1.

  • So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

f\left(x\right)=x^3-6x^2+3x+10

0=x^3-6x^2+3x+10              ∵  f(x)=0

as

Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0

so

\left(x+1\right)\left(x-2\right)\left(x-5\right)=0    

Using the zero factor principle

if  ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x=-1,\:x=2,\:x=5

Therefore, the zeros of the function are:

x=-1,\:x=2,\:x=5

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.    

8 0
2 years ago
Suppose p" must approximate p with relative error at most 10-3 . Find the largest interval in which p* must lie for each value o
goblinko [34]

Answer:

[p-|p|*10^{-3} \, , \, p+|p|* 10^-3]

Step-by-step explanation

The relative error is the absolute error divided by the absolute value of p. for an approximation p*, the relative error is

r = |p*-p|/|p|

we want r to be at most 10⁻³, thus

|p*-p|/|p| ≤ 10⁻³

|p*-p| ≤ |p|* 10⁻³

therefore, p*-p should lie in the interval [ - |p| * 10⁻³ , |p| * 10⁻³ ], and as a consecuence, p* should be in the interval  [p - |p| * 10⁻³ , p + |p| * 10⁻³ ]

8 0
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A) f(x)<br> B) g(x)<br> C) h(x)
jeka57 [31]

Answer:

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3 0
2 years ago
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