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3 years ago

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I need help anyone can help me

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