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Eva8 [605]
3 years ago
6

If a room needs 7 rolls of wire, how much will 7/10 of 1 roll do

Mathematics
1 answer:
Ksivusya [100]3 years ago
3 0

1/10th of the room

Step-by-step explanation:

Step 1 :

Given that one room needs 7 rolls of wire.

We need to find how much will 7/10 of 1 roll will do

Step 2:

7 rolls of wire does 1 room, so we have

1 roll of wire will do 1/7 of the room

We can see that as the number of roll increases, the size of the room also increases and when the number of roll decreases the size of the room decreases.  So this is a case of direct proportion.

Step 3 :

Using direct proportion we have,

1 roll of wire does 1/7 of the room , so

7/10 of one roll of wire will do 7/10 *1/7 = 1/10th of the room

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GrogVix [38]
1/20= 1 divided by 20= 0.05

4 0
3 years ago
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(Squreroot x+2)(squreoot x-2)=0
yuradex [85]

Answer:

\boxed{ \ x = 2  \ or \  x = -2 \ }

Step-by-step explanation:

I understand that you want to solve this equation

\sqrt{x+2}\sqrt{x-2}=0

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\sqrt{(x+2)(x-2)}=0

<=>

(x+2)(x-2)=0

<=>

x = 2 or x = -2

hope this helps

if your question is (\sqrt{x}+2)(\sqrt{x}-2) = 0

then this is (\sqrt{x})^2-4 = 0\\ x = 4

7 0
3 years ago
Jake's water bill is $24.40 per month plus $2.20 per ccf (hundred cubic feet) of water. What is the maximum number of ccf Jake c
siniylev [52]

Answer:the maximum number of ccf Jake can use if he wants his bill to be no more than $58.00 is $15.3

Step-by-step explanation:

Let x represent the number of ccf of water that Jake uses in a month.

Jake's water bill is $24.40 per month plus $2.20 per ccf (hundred cubic feet) of water. This means that if she uses x ccf of water in a month, her total water bill would be

24.40 + 2.2x

Therefore, the maximum number of ccf Jake can use if he wants his bill to be no more than $58.00 would be

24.40 + 2.2x ≤ 58

2.2x ≤ 58 - 24.4

2.2x ≤ 33.6

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4 0
4 years ago
A greeting card has a perimeter of 80 cm and an area of 384 square cm what are the dimensions for the greeting card
Ulleksa [173]

Step-by-step explanation:

it is a rectangle (usually).

so,

2×length + 2×width = 80

therefore, length + width = 40

length×width = 384

length = 384/width

384/width + width = 40

384 + width² = 40×width

width² - 40×width + 384 = 0

the general solution for such a square equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = width

a = 1

b = -40

c = 384

width = (40 ± sqrt((-40)² - 4×1×384))/(2×1) =

= (40 ± sqrt(1600 - 1536))/2 =

= (40 ± sqrt(64))/2 = (40 ± 8)/2 =

= 20 ± 4

width1 = 20 + 4 = 24 cm

length1 = 40 - 24 = 16 cm

width2 = 20 - 4 = 16 cm

length2 = 40 - 16 = 24 cm

so, both solutions are the same.

the greeting card is 24×16 cm. or 16×24, depending on how you are holding it.

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