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Musya8 [376]
3 years ago
8

Mr. Conners put a fence around the outside of his rectangular yard shown at the right. He put a fence post every 6 feet. How man

y fence posts did he use?

Mathematics
2 answers:
seraphim [82]3 years ago
4 0
A rectangular has four sides (6+6=12 + 6 = 18+6= 24) So, the correct answer is “Mr.Conners use 4 fence posts outside the yard”
[Please correct me if I am wrong]
Jet001 [13]3 years ago
3 0

Answer:

Mr. Conners will use 144 fence posts.

Step-by-step explanation:

Given question is incomplete; please find the question attached herewith.

As we can see from the attachment, Mr Corner's yard is in the rectangular shape.

Total distance around the yard = Perimeter of a rectangle = 2(Length + width)

Length of the yard = 330 ft

And width of the yard = 102 ft

So perimeter of the yard = 2(330 + 102)

                                         = 864 ft

Mr. Conners wants to put a fence post at every 6 feet.

So number of posts used = \frac{\text{Perimeter of the yard}}{\text{Distance between two fence pots}}

                                          = \frac{864}{6}

                                          = 144

Therefore, Mr. Conners will use 144 fence posts.

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Answer:

20.3cm^2

Step-by-step explanation:

Bullseye has a circular shape. So area of a bulleyse = Pi x (radius) ^2

Díameter is 2 inches = 5. 08cm

But radius = diameter /2

Radius = 5.08cm /2

Radius = 2.54cm

Hence Area of bulleyse = 3.14 x (2.54) ^2

= 3.14 x 6. 451 cm^2

= 20.256 cm^2

=20.3cm^2 (to nearest tenth)

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Use the following function rule to find f(2).<br> f(x) = 10x<br> f(2)=
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a mixture of 40 liters of paint is 25% red tint, 30% yellow tint and 45% water. 6 liters of yellow tint are added to the origina
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which of the following is the algebraic expression that best describes the sequence 3, 9, 27, 81, 243,...?
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Approximate area under the curve f(x) =-x^2+2x+4 from x=0 to x=3 by using summation notation with six rectangles and use the the
bekas [8.4K]

Answer:

Summation notation:

\frac{1}{2}\sum_{k=1}^6f((.5k))

or after using your function part:

\frac{1}{2}\sum_{k=1}^6(-(.5k)^2+2(.5k)+4)

After evaluating you get 11.125 square units.

Step-by-step explanation:

The width of each rectangle is the same so we want to take the distance from x=0 to x=3 and divide by 6 since we want 6 equal base lengths for our rectangles.

The distance between x=0 and x=3 is (3-0)=3.

We want to divide that length of 3 units by 6 which gives a length of a half per each base length.

We are doing right endpoint value so I'm going to stat at x=3. The first rectangle will be drawn to the height of f(3).

The next right endpoint is x=3-1/2=5/2=2.5, and the second rectangle will have a height of f(2.5).

The next will be at x=2.5-.5=2, and the third rectangle will have  a height of f(2).

The fourth rectangle will have a height of f(2-.5)=f(1.5).

The fifth one will have a height of f(1.5-.5)=f(1).

The last one because it is the sixth one will have a height of f(1-.5)=f(.5).

So to find the area of a rectangle you do base*time.

So we just need to evaluate:

\frac{1}{2}f(3)+\frac{1}{2}f(2.5)+\frac{1}{2}f(2)+\frac{1}{2}f(1.5)+\frac{1}{2}f(1)+\frac{1}{2}f(.5)

or by factoring out the 1/2 part:

\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

To find f(3) replace x in -x^2+2x+4 with 3:

-3^2+2(3)+4

-9+6+4

1

To find f(2.5) replace x in -x^2+2x+4 with 2.5:

-2.5^2+2(2.5)+4

-6.25+5+4

2.75

To find f(2) replace x in -x^2+2x+4 with 2:

-2^2+2(2)+4

-4+4+4

4

To find (1.5) replace x in -x^2+2x+4 with 1.5:

-1.5^2+2(1.5)+4

-2.25+3+4

4.75

To find f(1) replace x in -x^2+2x+4 with 1:

-1^2+2(1)+4

-1+2+4

5

To find f(.5) replace x in -x^2+2x+4 with .5:

-.5^2+2(.5)+4

-.25+1+4

4.75

Now let's add those heights.  After we obtain this sum we multiply by 1/2 and we have our approximate area:

\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

\frac{1}{2}(1+2.75+4+4.75+5+4.75)

\frac{1}{2}(22.25)

11.125

Okay now if you wanted the summation notation for:

\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

is it

\frac{1}{2}\sum_{k=1}^{6}(f(.5+.5(k-1)))

or after simplifying a bit:

\frac{1}{2}\sum_{k=1}^6 f((.5+.5k-.5))

\frac{1}{2}\sum_{k=1}^6f((.5k))

If you are wondering how I obtain the .5+.5(k-1):

I realize that 3,2.5,2,1.5,1,.5 is an arithmetic sequence with first term .5 if you the sequence from right to left (instead of left to right) and it is going up by .5 (reading from right to left.)

6 0
3 years ago
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