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kirza4 [7]
3 years ago
15

The area of a rectangle is 36 cm2. The length of the rectangle is 8 cm.

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
4 0
<span>What is the width of the rectangle? My answer is 32</span>
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A function T multiplies the input z by 2/3 and divides that by the quantity (z-1/2) .
ratelena [41]
Thank you for posting you question here at brainly. I hope the answer will help you. 

T (z) = 2/3z over z - 1/2
T(3) = 2/2.5 = 4/5 or 0.8
T(1/2) is indeterminate, he reason being that the denominator becomes zero. Division by zero gives an indeterminate result.
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3 years ago
PLEASE HELP!! I can't figure this out
9966 [12]

Answer:

how do i use this site im new

Step-by-step explanation:

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3 years ago
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Use the graph of the function. Determine Over what interval(s) the function is positive or negative.
san4es73 [151]

Answer:

I can't see the numbers well, but the graph is decreasing from (-inf, minimum) and increasing from (minimum, inf).  The minimum  is the vertex.  Hope this helps!

Step-by-step explanation:

4 0
3 years ago
What is the difference?
worty [1.4K]

Answer:

4th option

Step-by-step explanation:

Given

\frac{x}{x^2+3x+2} - \frac{1}{(x+2)(x+1)} ← expand denominator using FOIL

= \frac{x}{x^2+3x+2} - \frac{1}{x^2+3x+2}

Since the denominators are common, then subtract the numerators

= \frac{x-1}{x^2+3x+2}

5 0
2 years ago
Consider the function f ( x ) = − 2 x 3 + 27 x 2 − 108 x + 9 . For this function there are three important open intervals: ( − [
Darya [45]

Answer:

<em>A=3 and B=6</em>

Step-by-step explanation:

<u>Increasing and Decreasing Intervals of Functions</u>

Given f(x) as a real function and f'(x) its first derivative.

If f'(a)>0 the function is increasing in x=a

If f'(a)<0 the function is decreasing in x=a

If f'(a)=0 the function has a critical point in x=a

As we can see, the critical points may define open intervals where the function has different behaviors.

We have

f ( x ) = - 2 x^3 + 27 x^2 - 108 x + 9

Computing the first derivative:

f' ( x ) = - 6 x^3 + 54 x - 108

We find the critical points equating f'(x) to zero

- 6 x^3 + 54 x - 108=0

Simplifying by -6

x^2 -9 x +18=0

We get the critical points

x=3,\ x=6

They define the following intervals

(-\infty,3),\ (3,6),\ (6,+\infty)

Thus A=3 and B=6

3 0
3 years ago
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