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ololo11 [35]
3 years ago
7

A car moving at a constant speed passed a timing device at t=0. After 9 seconds, the car has

Mathematics
1 answer:
Montano1993 [528]3 years ago
3 0
Lichraly have no idea mate maybe 3?
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Two-fifths of one less than a number is less than three-fifths of one more than that number. What numbers are in the solution se
Soloha48 [4]
Let us say, the number is "a"

one less than "a" is ... well a - 1
2/5 of that will be... well, 2/5 * ( a - 1)

one more than "a" is... well a + 1
3/5 of that is.. hmmm 3/5 * (a + 1)

now, we're told that, the first expression is "less than" the second one, thus

\bf \cfrac{2}{5}(a-1)\ \textless \ \cfrac{3}{5}(a+1)

solve for "a", or make it "x" if you wish, same thing anyway
4 0
3 years ago
HELP ASAP!!!!!
8_murik_8 [283]

Answer:

Well, the answer to 17 divided by 2635 is 155, however, the number in the box must be 850, I believe.

935 subtracted by 850 equals 85 divided by 17 equals 0.

I hope this was helpful!

7 0
4 years ago
What is 0.45 as a fraction how do I show my work
g100num [7]
45/100 = 9/20

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7 0
4 years ago
Read 2 more answers
How do you know if a coordinate on a graph is positive or negative?
Annette [7]
Negative Slope<span>. </span>If<span> a line has a </span>positive slope<span> (i.e. m > 0 ), then y always increases when x increases and y always decreases when x decreases. Thus, the </span>graph<span> of the line starts at the bottom left and goes towards the top right. Thus, as x increases by 3 , y decreases by 4 , and as x decreases by 3 , y increases by 4 .

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5 0
4 years ago
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Answer the following questions about the function whose derivative is f primef′​(x)equals=x Superscript negative three fifths Ba
Elden [556K]

Answer:

(a) The critical points of f are x=0 and x=3.

(b)f is decreasing on (0,3) and f is decreasing on (3,\infty).

(c) Therefore the local minimum of f is at x=3

Step-by-step explanation:

Given function is

f'(x)= x^{-\frac35}(x-3)

(a)

To find the critical point set f'(x)=0

\therefore  x^{-\frac35}(x-3)=0

\Rightarrow x=0,3

The critical points of f are 0,3.

(b)

The interval are (0,3) and (3,\infty).

To find the increasing or decreasing, taking two points one point from the interval (0,3) and another point (3,\infty).

Assume 1 and 4.

Now f'(1)=(1)^{-\frac35}(1-3)

and f'(4)=(4)^{-\frac35}(4-3)>0

Since 1∈(0,3) , f'(x)<0  and 4∈(3,\infty) , f'(x)>0

∴f is decreasing on (0,3) and f is decreasing on (3,\infty).

(c)

f'(x)= x^{-\frac35}(x-3)

Differentiating with respect to x

f''(x)=-\frac35x^{-\frac 85}(x-3)+x^{-\frac35}

Now

f''(0)=-\frac35(0)^{-\frac 85}(0-3)+(0)^{-\frac35}=0

and

f''(3)=-\frac35(3)^{-\frac 85}(3-3)+3^{-\frac35}

        =0.517>0

Since f''(x)>0 at x=3

Therefore the local minimum of f is at x=3

8 0
3 years ago
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