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dusya [7]
3 years ago
8

What are all the possible rectangles with whole-number side

Mathematics
1 answer:
11111nata11111 [884]3 years ago
5 0

Answer:

So (L,W) possibilities are:

(1,4),(4,1),(2,3),(3,2)

That makes 4 possibilities.

Step-by-step explanation:

The perimeter of a rectangle is P=2L+2W where L is the length and W is the width.

We have that P=10, so 10=2L+2W.

10=2L+2W

10=2(L+W)  By factoring using the distributive property.

2(5)=2(L+W)  I factored 10 as 2(5).

If 2(5)=2(L+W), then 5=L+W.

Whole numbers are {0,1,2,3,4,5,6,7,8,9,10,...}. They are your counting numbers and 0.

I think they want natural numbers {1,2,3,4,...}.  This is also just called the counting numbers. The reason I think they want this because if one of the dimensions is 0, we won't actually have a rectangle.

So now looking for numbers from this set that satisfy: L+W=5.

L+W=5

1+4=5

4+1=5

2+3=5

3+2=5

So (L,W) possibilities are:

(1,4),(4,1),(2,3),(3,2)

That makes 4 possibilities.

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James uses 1/3 of his land for growing durains, 1/4 for bananas,3/8 for Guavas and the remaining 9 hectares for mangoes. What is
dusya [7]
Greetings and Happy Holidays! 

Let x represent the total area of land

We can create an equation to solve for the unknown, this being the total area: 
x= \frac{1}{3}x+ \frac{1}{4}x+ \frac{3}{8}x+9

Combine like terms (terms with the same variable and degree):
x= \frac{1}{3}x+ \frac{2}{8}x+ \frac{3}{8}x+9

x= \frac{1}{3}x+\frac{5}{8}x+9

x= \frac{8}{24}x+\frac{15}{24}x+9

x=\frac{23}{24}x+9

Add 
-\frac{23}{24}x to both sides.
(x)+(-\frac{23}{24}x)=(\frac{23}{24}x+9)+(-\frac{23}{24}x)

x-\frac{23}{24}x=9

\frac{24}{24}x-\frac{23}{24}x=9

\frac{1}{24}x=9

Divide 
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\frac{\frac{1}{24}x}{\frac{1}{24}}=\frac{9}{\frac{1}{24}}

x=(9)(\frac{24}{1})

x=(\frac{9}{1})(\frac{24}{1})

x=\frac{216}{1}

x=216

The Answer Is:
\left[\begin{array}{ccc}x=216\end{array}\right]

James has 216 square units of land.

Hope this helps!
-Benjamin


3 0
3 years ago
If x and y are both negative when is x-y positive
jarptica [38.1K]
Hmm

let's try some numbers

we have 3 scenarios
x>y
x=y
x<y

x>y
x=-1 and y=-2
x-y=-1-(-2)=-1+2=1
it's positive when x>y

x=y
x=-1, y=-1
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nope

x<y
x=-2 and y=-1
-2-(-1)=-2+1=-1
nope


x-y is positive when 0>x>y
4 0
3 years ago
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6 0
3 years ago
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