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Nutka1998 [239]
2 years ago
6

A cake has a mass of 2.5kg. It contains 275g of fruit. What percentage of fruit is there in the cake

Mathematics
2 answers:
Travka [436]2 years ago
8 0
The answer is:  "11 %" . 
__________________________________
There is 11% of fruit in the cake.
______________________________
Explanation:
____________________________________

275g is what percent of 2.5 kg?  

First, convert "275 g" into "kg".

Note the exact conversion:  1000 g = 1 kg .  

So 275 g = (275/1000) kg  = 0.275 kg .


0.275 kg = (n/100) * 2.5 kg ; 

→ (n/100) * 2.5 = 0.275 ;
____________________________
Divide each side by "(2.5)" 
______________________
→ (n/100) = (0.275) / (2.5) ;

→ (n/100) = 0.11 ;

Multiply each side by "100" ; 

n = 11 . 
________________________________________
The answer is:  11 % .
________________________________________
postnew [5]2 years ago
8 0
275g=0.275kg
___________
2.5kg———100%
0.275kg—x.
___________
2.5•x=0.275•100
2.5x=27.5
x=11%
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Using the technique in the model above, find the missing segments in this 30°-60°-90° right triangle. AB = 8 BC = 4 CD =
zimovet [89]
For a 30-60-90 triangle the sides always have the same relationship
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6 0
2 years ago
Read 2 more answers
A cube-shaped water tank having 6 ft side lengths is being filled with water. The bottom is solid metal but the sides of the tan
Mekhanik [1.2K]

Answer:

0.033 ft

Step-by-step explanation:

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Pressure is distributed from 0 to maximum at the bottom like the following equation:

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4 0
2 years ago
The equation 1.5r + 15 = 2.25r represents the number r of movies you must rent to spend the same amount at each movie store. how
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1.5r+15=2.25r
First of all, you have to arrange the numbers according to those similar to them. So that will be;
15=2.25r-1.5r
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2.25r-1.5r is the same as 2.25-1.5. And that is 0.75r
  Therefore;
15=0.75r
To find out what r is, you have to divide the both sides by 0.75. This is done to remove the 0.75 close to r to finally reveal what r is. Anyway;
15/0.75=0.75r/0.75
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4 0
2 years ago
Square $ABCD$ has area $200$. Point $E$ lies on side $\overline{BC}$. Points $F$ and $G$ are the midpoints of $\overline{AE}$ an
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A_{BFGC}=\dfrac{FG+BC}{2}\cdot \dfrac{AB}{2}=\dfrac{5\sqrt{2}+10\sqrt{2}}{2}\cdot \dfrac{10\sqrt{2}}{2}=75.

4.

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