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Hitman42 [59]
3 years ago
11

Evaluate h(x) = -2+ 9 when x = -2, 0, and 5

Mathematics
1 answer:
sasho [114]3 years ago
5 0

Answer: 7

Step-by-step explanation:

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haha 6 days of history, 3 weeks of math, and 1 week of english all to do tn. It's currently 9:27. oh well who cares aha kinda fu
Mashcka [7]

Answer:

oop... glll

Step-by-step explanation:

have u uploaded any? Maybe I can help

4 0
2 years ago
Read 2 more answers
1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat
brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

x = 1 is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

x = 2 is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) f(x) = 6x + 6/x

f\prime(x) = 6 - 6/x^2 = 0 and x \neq 0

So, the roots are x = -1 and x = 1

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum.

2) f(x)=6-4/x+2/x^2

f\prime(x)=4/x^2-4/x^3=0 and x \neq 0

So the root is x = 1

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

x = 1 is absolute minimum.

3) f(x) = 8x^2/(x-4)

f\prime(x) = (8x^2-64x)/(x-4)^2=0 and x \neq 4

So the roots are x = 0 and x = 8

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum.

4) f(x)=6(x-2)^{2/3} +4=0

f\prime(x) = 4/(x-2)^{1/3} has no solution and x = 2 is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

x = 2 is absolute minimum.

5) f(x)=8\sqrt x - 6x for x>0

f\prime(x) = (4/\sqrt x)-6 = 0

So the root is x = 4/9

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum.

5 0
2 years ago
R is the midpoint of QS, if QR=2x and RS=x+3, what is RS?
lbvjy [14]
Q -------------R------------S

suppose this is your line,

R is the mid point

and it's given that

QR = 2x
and
RS = x+3

as R is the mid point of QS
so,
QR = RS
then,
2x = x + 3
2x - x =3
x = 3


as,
it's given that RS = x + 3
then,
RS = 3 + 3
= 6 units.

5 0
3 years ago
I attached the math problem below and I hope u guys help me. Thxs to who's gonna answer!
vova2212 [387]

Answer:

Numbers | least to greatest | opposites | opposites least to greatest

1/4, -1/2    | -1/2, 1/4                | -1/4, 1/2     | -1/4, 1/2

2, -10        | -10, 2                   | -2, 10         | -2, 10

0, 3 1/2     | 0, 3 1/2                | 0, -3 1/2    | -3 1/2, 0

-5, -5.6     | -5.6, -5                 | 5, 5.6       | 5, 5.6

24 1/2, 24 | 24, 24 1/2            | -24 1/2 -24 | -24 1/2 -24

-99.9, -100 | -100, -99.9         | 99.9, 100 | 99.9, 100

-0.05, -0.5  | -0.5, -0.05         | 0.5, 0.05 | 0.05, 0.5

-0.7, 0       | -0.7, 0                  | 0.7, 0       | 0, 0.7

100.02, 100.04 | 100.02, 100.04 | -100.02, -100.04 | -100.04, -100.02

Hope this helps you!!!!

8 0
2 years ago
Hey guys help my please !!Q 4,6,7
padilas [110]
Q4, 4y²=60
y=√15 ( i am not sure)
q6, (5+y) ²=81
5+y=9
y=4
q7 (x+4) (x) =77
x²+4x-77=0
(x+11) (x-7) =0
x=-11(rej) or x=7
therefore ,length= 7+4=11cm
breadth=7cm
6 0
3 years ago
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