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OLEGan [10]
3 years ago
8

Eduardo counted the square tiles on the floor of his rectangular room the area of each tile is one square foot the floor has 6!r

ows of tiles and 7 tiles in each row what is the area of the floor of Eduardo's room?
Mathematics
1 answer:
Deffense [45]3 years ago
5 0

Answer:

6 X 7= 42

Step-by-step explanation:

to find the area of anything its length times width times height (LXWXH)

since they arent giving a width here its only LXH=?  

so 6X7=42

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Given that (x,y) = (5,15) find tan Ø. <br><br> A. 3 <br> B. 1/3 <br> C. 5 <br> D. 1/5
lesantik [10]
This value is in the positive quadrant meaning that all values remain positive. To answer this problem you must have an understanding of trig ratios being that tangent is opposite/adjacent. When you plot the points and draw a triangle using the origin of the point, you find that the adjacent value is 5 and opposite value is 15
7 0
3 years ago
Jim's backyard is a rectangle that is 15 5/6 yards long and 10 2/5 yards wide. Jim buys sods in pieces that are 1 1/3 yards long
Ne4ueva [31]

We are given

Jim's backyard:

Length is

L=15\frac{5}{6} =\frac{95}{6} yards

width is

W=10\frac{2}{5} =\frac{52}{5} yards

Since, this is rectangle

so, we can find area of rectangle

A=L*W

A=(\frac{95}{6})*(\frac{52}{5})

A=\frac{494}{3} yard^2

Area of one sod:

length is

l=1\frac{1}{3} =\frac{4}{3} yards

width is

w=1\frac{1}{3} =\frac{4}{3} yards

Since, it is rectangle in shape

so,

Area=l*w

A=\frac{4}{3}*\frac{4}{3}

A=\frac{16}{9}yard^2

Number of pieces of sod:

we can use formula

Number of pieces of sod = (area of Jim's backyard)/(area of one sod)

N=\frac{\frac{494}{3} }{\frac{16}{9} }

now, we can simplify it

N=\frac{741}{8} pieces need ..............Answer

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3 years ago
The number doing the dividing in a division problem
Pavlova-9 [17]

Answer:

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Step-by-step explanation:

I just looked it up bro lol

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Natasha2012 [34]

let's bear in mind that B is the midpoint and thus it cuts a segment into two equal halves.

\bf \underset{\leftarrow \qquad \textit{\large 10x-6}\qquad \to }{\boxed{A}\stackrel{4x+2}{\rule[0.35em]{10em}{0.25pt}} B\stackrel{\underline{4x+2}}{\rule[0.35em]{10em}{0.25pt}\boxed{C}}} \\\\\\ AC=AB+BC\implies 10x-6=(4x+2)+(4x+2)\implies 10x-6=8x+4 \\\\\\ 2x-6=4\implies 2x=10\implies x=\cfrac{10}{2}\implies x= 5 \\\\[-0.35em] ~\dotfill\\\\ AC=(4x+2)+(4x+2)\implies AC=[4(5)+2]+[4(5)+2] \\\\\\ AC=22+22\implies AC=44

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Liam has 3/4 gallon of milk in one container and 1/5 gallon in
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Altogether he has 1 and 1/8 gallon of milk
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