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Pani-rosa [81]
2 years ago
15

Consider two circular swimming pools. Pool A has a radius of 22 feet, and Pool B has a diameter of 13.6 meters. Complete the des

cription for which pool has a greater circumference. Round to the nearest hundredth for each circumference. 1 foot ≈ 0.305 meters
The diameter of Pool A is ...
meters. So, the diameter of Pool... is greater, and the circumference is... by...
meters.
Mathematics
1 answer:
maw [93]2 years ago
8 0

Answers:

13 42 m; B; 0.57 m

Step-by-step explanation:

Data:

Pool A:  r = 22 ft

Pool B: D = 13.6 m  

Calculations:

1. Radius of Pool A

r = 22 ft × (0.305 m/1 ft) = 6.71 m

2. Diameter of Pool A

D =2r = 2 × 6.71 = 13.42 m

The diameter of Pool A is 13.42 m.

3. Compare pool diameters

The diameter of Pool B is 13.6 m.

So, the diameter of Pool <u>B</u> is greater.

4. Compare circumferences

The formula for the circumference of a circle is

C = 2πr or C = πD

Pool A: C = 2π × 6.71 = 42.16 m

Pool B: r =    π × 13.6 = 42.73 m

Pool B - Pool A = 42.73 - 42.16 = <u>0.57 m </u>

The circumference is greater by <u>0.57 m.</u>

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Chris is behind in 9 classes of school he has a total time of 25 days to make up 75 hours and 4 minutes worth of late work how m
Vlad1618 [11]

Answer:

Step-by-step explanation:

First, convert the 4 mins into hours by dividing by 60;

4/60 =0.067

Add 0.067 to 75 hrs to get the total = 75+0.067 = 75.067

Therefore the total time Chris needs to make up for is 75.067 hrs

Total number of days Chris has = 25

Next,

IF in 25 days = he can cover 75.067hrs

then in 1 day he will cover = (1 * 75.067)/ 25

        = 3.002

Therefore, Chris needs to do 3hrs each day to be able to complete the deadline

3 0
3 years ago
Find the midpoint of side EF. (4 points)
umka2103 [35]

Answer:

Midpoint of side EF would be (-.5,4.5)

Step-by-step explanation:

We know that the coordinates of a mid-point C(e,f) of a line segment AB with vertices A(a,b) and B(c,d) is given by:

e=a+c/2,f=b+d/2

Here we have to find the mid-point of side EF.

E(-2,3) i.e. (a,b)=(2,3)

and F(1,6) i.e. (c,d)=(1,6)

Hence, the coordinate of midpoint of EF is:

e=-2+1/2, f=3+6/2

e=-1/2, f=9/2

e=.5, f=4.5

SO, the mid-point would be (-0.5,4.5)

 

3 0
1 year ago
Please help!
laila [671]
Because
\frac{f(x)}{x-3} = 2x^{2} + 10x - 1
therefore
f(x) = (x-3)(2x² + 10x - 1) + k, where k =  constant.

Because f(3) = 4, therefore k =4.
The polynomial is
f(x) = 2x³ + 10x² - x - 6x² - 30x + 3 + 4
      = 2x³ + 4x² - 31x + 7

Answer:  f(x) = 2x³ + 4x² - 31x + 7

6 0
2 years ago
Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2
Verdich [7]

Answer:

(a) f'(x)=-\frac{2}{x^3}

(b) y=-0.25x+0.75

Step-by-step explanation:

The given function is

f(x)=\frac{1}{x^2}                  .... (1)

According to the first principle of the derivative,

f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}

f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}

f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}

f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}

f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}

Cancel out common factors.

f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}

By applying limit, we get

f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}

f'(x)=\frac{-2x)}{x^4}

f'(x)=\frac{-2)}{x^3}                         .... (2)

Therefore f'(x)=-\frac{2}{x^3}.

(b)

Put x=2, to find the y-coordinate of point of tangency.

f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}

Substitute x=2 in equation 2.

f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

y-y_1=m(x-x_1)

y-0.25=-0.25(x-2)

y-0.25=-0.25x+0.5

y=-0.25x+0.5+0.25

y=-0.25x+0.75

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0
3 years ago
Which is the graph for y=(x)-2? ​
Alex
The shape of this graph is

8 0
2 years ago
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