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Anarel [89]
3 years ago
5

I need help anyone can help me

Mathematics
1 answer:
REY [17]3 years ago
5 0
I found the answer at another website plus wanna get exposed little boys yall aint girls
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find the local and/or absolute extrema for the function over the specified domain. (Order your answers from smallest to largest
Arlecino [84]

Answer:

Minimum 8 at x=0, Maximum value: 24 at x=4

Step-by-step explanation:

Retrieving data from the original question:

f(x)=x^{2}+8\:over\:[-1,4]

1) Calculating the first derivative

f'(x)=2x

2) Now, let's work to find the critical points

Set this

2x=0\\x=0    

0, belongs to the interval. Plug it in the original function

f(0)=(0)^2+8\\f(0)=8

3)  Making a table x, f(x) then compare

x|  f(x)

-1 | f(-1)=9  

0 | f(0)=8   Minimum

4 | f(4)=24 Maximum

4) The absolute maximum value is 24 at x=4 and the absolute minimum value is 8 at x=0.    

5 0
3 years ago
Helppppppppp due today in 10 mins construct a square by following these steps
trasher [3.6K]

just do what it said and you'll get the answer

8 0
2 years ago
A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece
Alex_Xolod [135]

Answer:

The number of seat when n is odd S_n=\frac{n^2+2n+1}{4}

The number of seat when n is even S_n=\frac{n^2+2n}{4}

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

S_n=\frac{n}{2}[2a+(n-1)d]

    =\frac{n}{2}[a+l]

a = first term of the series.

d= common difference.

n= number of term

l= last term

n^{th} term of a A.P series is

T_n=a+(n-1)d

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let t^{th} of the series is n.

T_n=a+(n-1)d

Here T_n=n, n=t, a=1 and d=2

n=1+(t-1)2

⇒(t-1)2=n-1

\Rightarrow t-1=\frac{n-1}{2}

\Rightarrow t = \frac{n-1}{2}+1

\Rightarrow t = \frac{n-1+2}{2}

\Rightarrow t = \frac{n+1}{2}

Last term l= n,, the number of term =\frac{ n+1}2, First term = 1

Total number of seat

S_n=\frac{\frac{n+1}{2}}{2}[1+n}]

    =\frac{{n+1}}{4}[1+n}]

     =\frac{(1+n)^2}{4}

    =\frac{n^2+2n+1}{4}

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let t^{th} of the series is n.

T_n=a+(n-1)d

Here T_n=n, n=t, a=2 and d=2

n=2+(t-1)2

⇒(t-1)2=n-2

\Rightarrow t-1=\frac{n-2}{2}

\Rightarrow t = \frac{n-2}{2}+1

\Rightarrow t = \frac{n-2+2}{2}

\Rightarrow t = \frac{n}{2}

Last term l= n, the number of term =\frac n2, First term = 2

Total number of seat

S_n=\frac{\frac{n}{2}}{2}[2+n}]

    =\frac{{n}}{4}[2+n}]

     =\frac{n(2+n)}{4}

    =\frac{n^2+2n}{4}  

4 0
2 years ago
NEED HELP PLZ!!!!!!!!!!!!!
podryga [215]

Answer:

1st Option is correct

Step-by-step explanation:

Just keep adding 150 to see the results

8 0
2 years ago
How do you solve 10+|m-6|=7
kondor19780726 [428]
There are no values of
m
m
that make the equation true.
No solution
7 0
3 years ago
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