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Papessa [141]
2 years ago
14

HELP QUICK PLEASE Determine the following values of the function. f(-3) = f(2) =

Mathematics
2 answers:
Tatiana [17]2 years ago
6 0
It says if x is less than or equal to 0 use 3x + 1
so -3 being less than 0, 3*-3 + 1 = -9 + 1 = -8
and if x is bigger or equal to 2, use 4
2 being equal to 2, you're value is 4
hence the answer for the funcion is -8 and 4
astra-53 [7]2 years ago
4 0
F(-3) would be -8, because you would use the first equation. f(2) would be 4 apparently cause theres no x after the 4, or in the equation of the 4. Hope this helps. Please rate, leave a thanks, and mark a brainliest answer(not necessarily mine). Thanks, it really helps! :D
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Which ordered pairs represent points on the graph of f(x)= Negative RootIndex 3 StartRoot x EndRoot? Check all that apply.
DedPeter [7]

Answer:

  • (0, 0)
  • (1, –1)
  • (8, –2)

Step-by-step explanation:

We normally think of the ordered pairs as ...

  (x, y) = (x, -∛x)

Another way to think about them is ...

  (x, y) = (-y³, y)

I find it a little easier to think about the cube of a number, rather than its cube root.

For the y-values given, -2, -1, 0, 2, the points that will be on the graph are ...

  (8, -2), (1, -1), (0, 0), (-8, 2)

The first three of these points are listed in your answer choices.

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What is negative 9 over 5 divided by 2?
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Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st
Naily [24]

Answer:

\boxed{2^{\frac{802}{27}} \cdot 3^9}

Step-by-step explanation:

<u>I will try to give as many details as possible. </u>

First of all, I just would like to say:

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

\left(\dfrac{1}{3^2}  \cdot \dfrac{1}{4^5}  \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3

Also,

\boxed{a^{0} = 1, a\neq 0 }

\left(\dfrac{1}{9}  \cdot \dfrac{1}{1024}  \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27

Note

\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq  0}

\left(\dfrac{1}{9216}   \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27

\left(\dfrac{1}{9216}   \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)

\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)

\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)

Note

\boxed{\dfrac{1}{\dfrac{1}{a} }  = a}

9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)

\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)

Once

9216=2^{10}\cdot 3^2 \implies  9216^3=2^{30}\cdot 3^6

\boxed{(a \cdot b)^n=a^n \cdot b^n}

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27} $

We have

\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)

Also, once

\boxed{\dfrac{c^a}{c^b}=c^{a-b}}

2^{30-\frac{8}{27}} \cdot 3^6\cdot 27

As

30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27}  =\dfrac{802}{27}

2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3

2^{\frac{802}{27}} \cdot 3^9

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