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mario62 [17]
3 years ago
11

Variable g is 5 more than variable w. Variable g is also 2 less than w. Which pair of equations best models the relationship bet

ween g and w?
g = w + 5
g = w − 2

g = w − 5
g = w + 2

w = 5g
w = g − 2

w = 5g
w = g + 2
Mathematics
2 answers:
kirill115 [55]3 years ago
6 0
G=w+5
g=w-2
This confused me at first as well, but just try and make a picture in your mind, anyway, if you add 5 to W, then it is equal to g because "Variable g is 5 more than variable w" 
If you subtract 2 from w it is equal to g because Variable g is also 2 less than w."

jeyben [28]3 years ago
3 0
Your answer is the first choice, g = w + 5    &    g = w − 2, because if variable g is 5 MORE THAN variable w, that would mean you would have to add 5 to w in order to get g, (g = w + 5), but if it said it was 5 TIMES MORE THAN, THHHHEEEENNNNN that would mean w times 5.
Now da second problem is g = w - 2, because they said that g is 2 LESS THAN w, so you would have to subtract 2 from w to get g, also known as g = w - 2.
I hope I helped! =D
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a picture 10 1/4 feet long is to be centered on a wall that is 14 1/2 feet long. How much more space is there from the edge of t
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Answer:

Length from the edge will be =2{\frac{1}{8}} feet

Step-by-step explanation:

To find length we have to conver mixed fraction to improper fraction


The length of the wall = \frac{29}{2}


Length of picture =\frac{41}{4}


First we find the total length of wall remain after picture was hung to the wall.


Length of wall remain vacant =\frac{29}{2}-\frac{41}{4}


Taking LCM and solving we get


=\frac{58-41}{4} =\frac{17}{4}


Total length of wall remain vacant =\frac{17}{4}


As the picture to be centered in wall, therefore space feom one edge of wall will be half that of length remain vacant as the same length to be left vacant feom both sides.


Space from one edge =\frac{17}{4} \times \frac{1}{2}


= \frac{17}{8}

=2{\frac{1}{8}}


Length from the edge will be =2{\frac{1}{8}} feet

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4 years ago
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Step-by-step explanation:

5 0
3 years ago
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To snails are 189 cm apart. They crawl towards each other at rates that differ by 1 cm a minute. After 27 minutes they meet. How
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Speed of snail 1 = 3.795 cm/minute

Speed of snail 2 = 4.795 cm/minute

<em><u>Solution:</u></em>

Given that,

Total distance = 189 cm

They crawl towards each other at rates that differ by 1 cm a minute

Let the speed of snail 1 be "x"

Let the speed of snail 2 be "x + 1"

Total time = 22 minutes

Since the snails are moving towards each other,

Speed = x + x + 1 = 2x + 1

<em><u>How fast is each snail crawling?</u></em>

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Thus,

Speed of snail 1 = 3.795 cm/minute

Speed of snail 2 = 3.795 + 1 = 4.795 cm/minute

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