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Pachacha [2.7K]
3 years ago
15

Will mark as Brainiest answer if you help ASAP!

Mathematics
1 answer:
spayn [35]3 years ago
6 0
Vertex is (x+5)^2+12 and minimum value of f(x) is 12 
Hope this helps 
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PLEASE HELP ASAP!!!!
Setler [38]

Piecewise Function is like multiple functions with a speific/given domain in one set, or three in one for easier understanding, perhaps.

To evaluate the function, we have to check which value to evalue and which domain is fit or perfect for the three functions.

Since we want to evaluate x = -8 and x = 4. That means x^2 cannot be used because the given domain is less than -8 and 4. For the cube root of x, the domain is given from -8 to 1. That meand we can substitute x = -8 in the cube root function because the cube root contains -8 in domain but can't substitute x = 4 in since it doesn't contain 4 in domain.

Last is the constant function where x ≥ 1. We can substitute x = 4 because it is contained in domain.

Therefore:

\large{  \begin{cases} f( - 8 ) =   \sqrt[3]{ - 8}  \\ f(4) = 3 \end{cases}}

The nth root of a can contain negative number only if n is an odd number.

\large{  \begin{cases} f( - 8 ) =   \sqrt[3]{ - 2 \times -  2 \times   - 2}  \\ f(4) = 3 \end{cases}} \\  \large{  \begin{cases} f( - 8 ) =  - 2\\ f(4) = 3 \end{cases}}

Answer

  • f(-8) = -2
  • f(4) = 3
6 0
2 years ago
Translate to algebra; the quotient of a number less than ten and the number less two<br>​
Alja [10]

Answer:

please can u give full question

5 0
2 years ago
How do you work out px+py=a?
natka813 [3]
\bf px+py=a\implies \stackrel{\textit{common factor}}{p}(x+y)=a\implies p=\cfrac{a}{x+y}
5 0
3 years ago
Help Please?<br> Solve for x. <br><br> x − 2.7 ≥ 10.3<br><br> Solve for x. <br><br> −0.84x ≤ 168
stiks02 [169]
For the first one, we add 2.7 to both sides to get x≥13. Next, we divide both sides by -0.84 to get x≥ 168/-0.84. SInce we multiplied or divided by a negative number, we switch the sign around and end up with x≥200
4 0
3 years ago
Help need help with this
vaieri [72.5K]
Maybe use a math app
6 0
2 years ago
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