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ratelena [41]
3 years ago
10

Who is taking in more calories if the small bag of snickers are the exact same size: A person in a group where 5 people are shar

ing 3 small bags of snickers or a person in a group where 7 people are sharing 2 small bags of snickers
Mathematics
1 answer:
MrRa [10]3 years ago
3 0

Answer:

A person in a group where 5 people are sharing 3 small bags of snickers

Step-by-step explanation:

Let the number of calories in one bag of snickers =x

Number of calories in 3 small bags of snickers =3x

  • If 5 people share 3 small bags of snickers, 1 person's share  =\dfrac{3x}{5}

Number of calories in 2 small bags of snickers =2x

  • If 7 people share 2 small bags of snickers, 1 person's share  =\dfrac{2x}{7}

We then compare the two.

Let x=1

\dfrac{3}{5}=0.6$ calories per person\\\dfrac{2}{7}=0.29$ calories per person

Therefore, a person in a group where 5 people are sharing 3 small bags of snickers gets more calories.

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Is this right idk but i need work fir it
bija089 [108]

Answer:

The answer is C 5/4

Step-by-step explanation:

When lines are parallel and you know there is a scalar factor, make the ratio by putting the original over the dilation 15:12 or 15/12 this is the ratio and after you need to reduce by dividing by the greatest common factor. 15 and 12's GCF is 3 so (15/3)/(12/3) = 5/4.

7 0
3 years ago
Find the product -a2 b2 c2 (a+b-c)
mina [271]
Use the distributive property:
a(c+b)=ab+ac
And
a^n\cdot a^m=a^{n+m}\\\\a=a^1

-a^2b^2c^2(a+b-c)=-a^2ab^2c^2-a^2b^2bc^2+a^2b^2c^2c=-a^3b^2c^2-a^2b^3c^2+a^2b^2c^3

7 0
3 years ago
What’s the product of 0.54(8)
maw [93]

The answer to your problem is 4.32.

6 0
3 years ago
Plz help me Quick l need helpppp​
Angelina_Jolie [31]
I’m not really sure but I would have to guess 4.5 because all of the other numbers don’t have decimals
5 0
3 years ago
A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force F⃗ =( 35 N )i^−( 37 N )j^ to
Trava [24]

Answer:

-189.8 J

Step-by-step explanation:

We are given that f

Force applied=F=35 Ni-37 Nj

Displacement, s=(-8.7m)i-(3.1m)j

We have to find the work done .

We know that

Work done=F\cdot s

Using the formula

Work done=(35i-37j)\cdot (-8.7-3.1)

Work done=-304.5+114.7

Using i\cdot i=j\cdot j=k\cdot k=1,i\cdot j=j\cdot i=j\cdot k=k\cdot j=i\cdot k=k\cdot i=0

Work done=-189.8J

Hence, the work done =-189.8 J

5 0
3 years ago
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