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luda_lava [24]
3 years ago
15

In a group of 80 children there are three times as many boys as girls how many girls are there ?

Mathematics
2 answers:
Neporo4naja [7]3 years ago
8 0
If I assume the number of girls is x, then the number of boys will be 3x
We know there are 80 children is 80, therefore
x+3x=80
4x=80
x=20
There are 20 girls.

Brut [27]3 years ago
7 0
Hi there! ~

Okay, so lets say that 
x = The total number of girls
and
y = The total number of boys. 

In this equation it says "there are three times as many boys as there are girls."

So we know we would have 

3 * x = y

So now we need to solve;

We first have x + y = 80 (80 being the total number of children, as stated in the question) 

Now we want to add x to both sides.

3 * x + x = 80 + x

Now we can do

4 * x = 80/4

x = 80/4

and then we would know that 80/4 = 20

So

x = 20

20 TOTAL GIRLS.

Now we have 

y= 3 * x = 3 * 20 - = 60

60 TOTAL BOYS.

Now we can check by doing

60+20=80

So in conclusion we know that we have a total of 20 girls and 60 boys. 
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Alona [7]
300 is the answer to the question


4 0
3 years ago
Read 2 more answers
AB bisects CD at point M, CD bisects AB at point M, and AB = 4 x CM. Describe the relationship between AM and CD
padilas [110]

\text{Given that AB bisects the line segment CD at the point M,}\\
\text{And CD bisects the line segment AB at the point M. so M is the }\\
\text{midpoint of both the line segements AB and CD.}\\
\\
\text{also we have given a relation that, }AB=4\times CM\\
\text{and we need to find a relation between AM and CD.}\\
\\
\text{observe that since M is the midpoint of CD, so we have }CD=2\times CM\\
\\
\Rightarrow CM=\frac{CD}{2}

\text{and M is midpoint of AB, so }AB=2\times AM\\
\\
\text{so with these substitutions, the given realtion gives}\\
\\
AB=4\times CM\\
\\
\Rightarrow 2\times AM=4\times \frac{CD}{2}\\
\\
\Rightarrow 2\times AM=2\times CD\\
\\
\Rightarrow AM=CD

Hence the relationship between AM and CD is : AM=CD

4 0
3 years ago
How many of the numbers from 10 through 97 have the sum of their digits equal to a perfect​ square?
Anarel [89]
Perfect squares are:
1,4,9,16,25,36,49,64,81,100,....

the sum of the digits of our biggest number is 16 so any perfect square bigger than 16 doesn't work for us
1-
1+0=1 so any number containing the digits will work(keep in mind we only will look into whole numbers because digits can't be negative or have fractions or be irrational)
thereful 10 works for our category
2-
0+4=4
1+3=4
2+2=4
22 13 31 and 40 will work two
3-
0+9
1+8
2+7
3+6
4+5
90 18 81 27 72 36 63 45 54
4-
0+16
1+15
2+14
3+13
4+12
5+11
6+10
7+9
8+8
79 97 88

so our set of numbers contain:
10 22 13 31 40 90 18 81 27 72 36 63 45 54 79 97 88
3 0
4 years ago
Please help on this one
Margarita [4]

For the first inequality, subtract 5 from both sides to get

x \geq -2

For the second inequality, subtract 1 from both sides to get

6x < -30

Divide both sides by 6:

x < -5

So, x has to be greater than or equal to -2, or less than -5, which is option D

6 0
3 years ago
The sum of two consecutive cube numbers is 341. Work out the two numbers. (2 marks)
Kobotan [32]

Answer:

63 and 53

Step-by-step explanation:

you could use trail and error. So:

13+23=9

33+43=91

63+73=559

53+63=341

3 0
3 years ago
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