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I am Lyosha [343]
3 years ago
5

In a class of 30 students there are 18 girls. what is the ratio of boys to girls in the class.

Mathematics
2 answers:
tekilochka [14]3 years ago
6 0
Ratio:girls:boys

18:12 

divide both by 6

3:2


Bess [88]3 years ago
3 0
18 girls ⇒ 30-18=12 boys

boys/girls = \frac{12}{18}=\frac{2}{3}
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Apply the rules for order of operstions to simplify 2 + 3 - 4 (5 × 4)
Lerok [7]
Just use PEMDAS 
2+3-4 (5x4) 

Parenthesis
Exponents 
Multiplication
Division
Addition 
Subtraction    So first is Parenthesis 5x4=20 next will be Addition because there are no exponents 2+3= 5-4= 1 so now we go back to multiplication 20 x 1= 20 so 20 is your answer. 

Hope I helped out ^^


7 0
3 years ago
Solve -3 = x + x + x.
kozerog [31]
-3 = x + x + x
-3 = 3x
-3/3 = x
x = -1
6 0
3 years ago
Read 2 more answers
.…: need this answers.....
Leto [7]
Sorry, cannot just give you answers without your input.
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3 0
2 years ago
Harvey the wonder hamster can run 3 1/6 km in 1/4 hour harvey runs at a constant rate what is his average speed in kilometers pe
vampirchik [111]


1/4 times 4 is 1 hour so
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3 0
3 years ago
Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP
erik [133]

Answer:

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

Step-by-step explanation:

we have

y=-x^{2} +2x+10 ----> equation A

y=x+2 ----> equation B

Solve the system by substitution

substitute equation B in equation A

x+2=-x^{2} +2x+10

solve for x

-x^{2} +2x+10-x-2=0

-x^{2} +x+8=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-x^{2} +x+8=0

so

a=-1\\b=1\\c=8

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}

x=\frac{-1\pm\sqrt{33}} {-2}

x=\frac{-1+\sqrt{33}} {-2}  -----> x=\frac{1-\sqrt{33}} {2}  

x=\frac{-1-\sqrt{33}} {-2}  -----> x=\frac{1+\sqrt{33}} {2}  

<em>Find the values of y</em>

For x=\frac{1-\sqrt{33}} {2}  

y=x+2

y=\frac{1-\sqrt{33}} {2}+2  ---->y=\frac{5-\sqrt{33}} {2}  

For x=\frac{1+\sqrt{33}} {2}  

y=x+2

y=\frac{1+\sqrt{33}} {2}+2  ---->y=\frac{5+\sqrt{33}} {2}  

therefore

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

5 0
2 years ago
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