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jasenka [17]
2 years ago
6

URGENT!! YOU WILL GET BRAINLIEST!

Mathematics
1 answer:
KiRa [710]2 years ago
4 0
Let Xo = original length of the deck
Yo = original width of deck
Therefore:
Yo = (1/6)Xo
The original deck area is:
Ao = XoYo
Ao = Xo(1/6)Xo = (1/6)Xo2
The length of the new deck is 4ft longer than the original deck, or XN = Xo + 4, and the width is 2ft longer than the original, or YN = Yo + 2, and the new area of the deck is 56ft2 than the original area, or AN = Ao + 56.
Therefore: AN = XNYN
Ao + 56 = (Xo + 4)(Yo + 2)
(1/6)Xo2 + 56 = (Xo + 4)(1/6Xo + 2)
(1/6)Xo2 + 56 = (1/6)Xo2 + 2Xo + (4/6)Xo +8
56 = 2Xo + (2/3)Xo +8
56 - 8 = (2 + 2/3)Xo
48 = (6/3 + 2/3)Xo
48 = (8/3)Xo
48(3/8) = Xo
Xo = 18ft
Yo = (1/6)(18)
Yo = 3ft
Check: Ao = XoYo = (18)(3) = 54ft2
XN = Xo + 4 = 18 + 4 = 22ft
YN = Yo + 2 = 3 + 2 = 5ft
AN = (22)(5) = 110ft2
Ao + 56 = AN
54 + 56 = 110
110 = 110 check------->ANSWER: original length, Xo = 18ft
original width, Yo = 3ft
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3 years ago
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vekshin1
Simple you get the area of the inner rectangle, then the area of the 2 semi circles then add them.
First, you need to notice that the diameter of each circle is same as the side of the rectangle (46m), also not that the 2 semi circles will make a complete one circle with radius (46/2)
so now we are left with the area of the rectangle+the area of a circle with radius of 32m
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3 0
3 years ago
Refer to the accompanying TI-83/84 Plus calculator display of a 95% confidence interval. The sample display results from using a
nikklg [1K]

Answer:

The confidence interval is 22.805

Step-by-step explanation:

We have given,

The Z interval (23.305,25.075)  

Mean \bar{x}=23.69

Sample n=30

To find : The confidence interval?

Solution :

We know, The confidence interval is in the format of

\bar{x}-E

E denotes the margin of error,

E=\frac{U-L}{2}

Where, U is the upper limit U=25.075

L is the lower limit L=23.305

E=\frac{25.075-23.305}{2}

E=\frac{1.77}{2}

E=0.885

Substitute the value of E and \bar{x} in the formula,

23.69-0.885

22.805

Therefore, The confidence interval is 22.805

8 0
3 years ago
Please help <br><br> Consider the matrices.
Andrei [34K]

_ _

| - 17 |

PQ = | - 3 |

| - 37 |

|_ 49 _|

7 0
2 years ago
A cube block has sides 8cm long. A cylindrical hole of radius 3.6 is drilled through the block. Find the volume of the block tha
djverab [1.8K]

Answer:

186.45 cm^3

Step-by-step explanation:

Given data

Side lenght of cube block= 8cm

Radius of cylindrical hole= 3.6cm

Volume of entire Shape= l^3

Volume of entire Shape= 8*8*8

Volume of entire block= 512cm^3

Volume of cylindrical hole= πr^2h

Volume of cylindrical hole= 3.14*3.6^2*8

Volume of cylindrical hole= 3.14*12.96*8

Volume of cylindrical hole= 325.55cm^3

Hence the volume of the block

=volume of entire shape- volume of cylinder

=512-325.55

=186.45 cm^3

7 0
2 years ago
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