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Eva8 [605]
3 years ago
5

A neighborhood in Shiloh is building a treehouse. It has a rectangle floor of 6 ft by 4 ft. What is the area of the tree house f

loor?
Mathematics
1 answer:
Helen [10]3 years ago
3 0

Answer:

24

Step-by-step explanation:

Shiloh is building a tree house

It's measurement is 6ft by 4ft

Therefore the area can be calculated as follows

= 6×4

= 24

Hence the area of the rectangle is 24

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(f+g)(x)=5x-6+x^2-4x-8
A.(f+g)(x)=x^2+x-14 is correct:)
5 0
3 years ago
Please help explain the steps
Gennadij [26K]

Step-by-step explanation:

3a^2+24a+45=0

×3/3( first ×3)

9a^2+24(3a)+135=0

(3a+15)(3a+9)=0

now /3

3(a+5)(a+3)

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3 0
3 years ago
The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The d
zaharov [31]

Answer:

P(61≤ X≤94) = 49.85%

Step-by-step explanation:

From the given information:

The mean of the bell shaped fluorescent light bulb μ = 61

The standard deviation σ = 11

The objective of this question is to determine the approximate percentage of light bulb replacement requests numbering between 61 and 94 i.e P(61≤ X≤94)

Using the empirical (68-95-99.7)rule ;

At 68% , the data lies between  μ - σ and μ + σ

i.e

61 - 11 and 61 + 11

50 and 72

At  95%, the data lies between  μ - 2σ and μ + 2σ

i.e

61 - 2(11) and 61 + 2(11)

61 - 22 and 61 +22

39   and   83

At 99.7%, the data lies between  μ - 3σ and μ + 3σ

i.e

61 - 3(11) and 61 + 3(11)

61 - 33 and 61 + 33

28   and   94

the probability equivalent to 94 is when  P(28≤ X≤94) =99.7%

This implies that ,

P(28≤ X≤94) + P(61≤ X≤94) = 99.7%

P(28≤ X≤94) = P(61≤ X≤94) = 99.7 %    

This is so because the distribution is symmetric about the mean

P(61≤ X≤94) = 99.7 %/2

P(61≤ X≤94) = 49.85%

5 0
3 years ago
The radius of a circle is changing at the rate of 1/π inches per second. At what rate, in square inches per second, is the circl
MAXImum [283]

Answer:

10 square inches per second.

Step-by-step explanation:

The radius of the circle is given by the equation:

r(t) = (1/π  in/s)*t

Where time in seconds.

Remember that the area of a circle of radius R is written as:

A = π*R^2

Then the area of our circle will be:

A(t) = π*( (1/π  in/s)*t)^2 = π*(1/π  in/s)^2*(t)^2

Now we want to find the rate of change (the first derivation of the area) when the radius is equal to 5 inches.

Then the first thing we need to do is find the value of t such that the radius is equal to 5 inches.

r(t) = 5 in =  (1/  in/s)*t

       5in*(π s/in) = t

        5*π s = t

So the radius will be equal to 5 inches after 5*π seconds, let's remember that.

Now let's find the first derivate of A(t)

dA(t)/dt = A'(t) = 2*(π*(1/π  in/s)^2*t = (2*π*t)*(1/π  in/s)^2

Now we need to evaluate this in the time such that the radius is equal to 5 inches, we will get:

A'(5*π s) = (2*π*5*π s)*((1/π  in/s)^2

              = (10*π^2  s)*(1/π^2  in^2/s^2) = 10 in^2/s

The rate of change is 10 square inches per second.

4 0
3 years ago
What is 89% of 260 sec?<br> pls help
ryzh [129]

Answer:

3.85666667 minutes of 260sec

8 0
2 years ago
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