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Bingel [31]
3 years ago
12

In a school there are 1000 boys and 800 girls. Calculate the percentage of boys and girls

Mathematics
1 answer:
Arisa [49]3 years ago
8 0

Answer:

55.55%= boys & 44.44% = girl

Step-by-step explanation:

1000/(1000+800)=1000/1800=.5555 or 55.55% are boys

800/(1000+800)=800/1800=.4444 or 44.44% are girls.

Therefore, 55,55% are boys and 44.44% are girls.

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How much greater than m2+3mn−n2 is 4m2+5mn+6n2?<br> PLEASE HELP
tiny-mole [99]

Answer:

3m² + 2mn + 7n²

Step-by-step explanation:

Subtract m² + 3mn - n² from 4m² + 5mn + 6n², that is

4m² + 5mn + 6n² - (m² + 3mn - n²)

= 4m² + 5mn + 6n² - m² - 3mn + n² ← collect like terms

= 3m² + 2mn + 7n²


5 0
3 years ago
Find the shaded region of the figure below​
lilavasa [31]

Answer:

-x³ + 3x² - 14x + 12

Step-by-step explanation:

Area of outer rectangle = (x² + 3x - 4) * (2x - 3)

      = (x² + 3x - 4) * 2x  + (x² + 3x - 4) * (-3)

     =x²*2x + 3x *2x - 4*2x  + x² *(-3) + 3x *(-3)  - 4*(-3)

     =2x³ + 6x² - 8x - 3x² - 9x + 12

    = 2x³ + <u>6x² - 3x²</u>   <u>- 8x - 9x</u> + 12     {Combine like terms}

    = 2x³ + 3x² - 17x + 12

Area of inner rectangle = (x² - 1)* 3x

                                       = x² *3x - 1*3x

                                       = 3x³ - 3x

Area of shaded region = area of outer rectangle - area of inner rectangle

         = 2x³ + 3x² - 17x + 12 - (3x³ - 3x)

         = 2x³ + 3x² - 17x + 12 -3x³ + 3x

        = 2x³ - 3x³ + 3x² - 17x + 3x + 12

        = -x³ + 3x² - 14x + 12

4 0
3 years ago
It costs Charlie's Car Repair Shop $27 to change the oil. The shop charges their costumers $40 for an oil change. What is the pe
frosja888 [35]
Cost Price for oil change =$27 
Charge for the customers =$40 
Markup cost = Difference between selling price and cost price =$40-$27=$13  
To find the percent of markup, divide the markup cost by the cost price and multiply it by 100. 
Markup percent = (13/27)*100 
 =48.15%
3 0
3 years ago
I need help with percent and tax
sasho [114]

Hello there

56•25/100=

1400÷100=

14$

56-14=42$

He will pay 42$

:D

6 0
4 years ago
Find all the solutions for the equation:
Contact [7]

2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0

Divide both sides by x^2\,\mathrm dx to get

2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}

Substitute v(x)=\dfrac{y(x)}x, so that \dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}. Then

x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}

x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}

x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}

The remaining ODE is separable. Separating the variables gives

\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x

Integrate both sides. On the left, split up the integrand into partial fractions.

\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}

\implies v^2+2v+1=a(v^2+1)+(bv+c)v

\implies v^2+2v+1=(a+b)v^2+cv+a

\implies a=1,b=0,c=2

Then

\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v

On the right, we have

\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C

Solving for v(x) explicitly is unlikely to succeed, so we leave the solution in implicit form,

\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C

and finally solve in terms of y(x) by replacing v(x)=\dfrac{y(x)}x:

\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C

\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C

\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}

7 0
3 years ago
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