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11111nata11111 [884]
3 years ago
10

40 POINTS!!!!Which of the following is not a polynomial?

Mathematics
1 answer:
ElenaW [278]3 years ago
8 0

Answer:

5x²

Step-by-step explanation:

because polynomials are the expressions where many terms are involved

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What do I do here cause I'm confused :/ ​
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Matt simply divided the 8 students in the class by the amount of vinegar the teacher has (15). this is incorrect because in order to determine how many ounces each student should get you would divide the amount of vinegar by the amount of students, thus ending up with 15/8.
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Can someone solve this and explain?
Anarel [89]
Answer: (y-c)/m
Reasoning:
Subtract c from both sides [ y-c=mx+c-c ]
Divide m from both sides [ (y-c)/m=mx/m ]
7 0
3 years ago
A rectangle has the length of
dangina [55]

Answer:

\large\boxed{C.\ 9\ inches\ squared}

Step-by-step explanation:

\text{Use}\\\\a^\frac{m}{n}=\sqrt[n]{a^m}\\\\a^n\cdot a^m=a^{n+m}\\-----------------\\\\length=\sqrt[3]{81}=\sqrt[3]{3^4}=3^\frac{4}{3}\\\\width=3^\frac{2}{3}\\\\\text{The area of a rectangle:}\ A=width\cdot length.\\\text{Substitute:}\\\\A=3^\frac{4}{3}\cdot3^\frac{2}{3}=3^{\frac{4}{3}+\frac{2}{3}}=3^{\frac{4+2}{3}}=3^\frac{6}{3}=3^2=9

8 0
3 years ago
Find (f −1)'(a).<br><br> f(x) = 4 + x2 + tan(πx/2), −1 &lt; x &lt; 1, a = 4
Vinvika [58]

Answer:

The  solution  is   (f^{-1})'(a) = \frac{2}{\pi }

Step-by-step explanation:

From the question we are told that

      The  function is  f(x) =  4 +  x^2  + tan [\frac{ \pi x}{2} ] ,     -1 <  x  <  1  a =  4

Here we are told find  (f^{-1}) (a)

Let equate

     f(x) =  a

So  

      4 +  x^2  + tan[\frac{\pi x }{2} ] =  4

       x^2  + tan[\frac{\pi x }{2} ]  =  0

For the equation above to be valid x must be equal to 0

  Now when x = 0

       f(0) = 4+0^2 + tan [\frac{ \pi * 0}{2} ]

=>      f(0) =  4

=>  0 =  f^{-1} (4)

Differentiating  f(x)

     f(x)'  =  0 + 2x  + sec^2 (\frac{\pi x}{2} )\cdot \frac{\pi}{2}

Now

    since 0 =  f^{-1} (4)

We have

      f(0)'  =  0 +  \frac{\pi }{2}  sec^2 (0)

       f(0)'  = \frac{\pi }{2}

Now  

        (f^{-1})'(a) =  \frac{1}{(\frac{\pi}{2} )}

        (f^{-1})'(a) = \frac{2}{\pi }

       

3 0
3 years ago
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