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kherson [118]
3 years ago
5

June gavyn and Alex share some sweets in the ratio 3:5:4 June gets 39 sweets how many sweets are there altogether?

Mathematics
2 answers:
NeX [460]3 years ago
6 0

Answer:

146

Step-by-step explanation:

j=3 13 = 39

g=5 13 = 65

a=4 13 = 42

katovenus [111]3 years ago
3 0

Answer:

156 sweets

Step-by-step explanation:

3+4+5=12

12/12 x 39 x 12/3=156 sweets

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Jace purchases a new motorcycle for $8000. The bike depreciates 12% every year.
Alex777 [14]

Answer:

We know that the original price of the bike was $8,000.

After one year, the price of the bike decays a 12%, then the new price of the bike is:

P(1) = $8,000*( 1 - 12%/100%) = $8,000*(1 - 0.12)

After another year, the price again decays a 12%, the new price will be:

P(2) = $8,000*(1 - 0.12)*(1 - 12%/100%) = $8,000*(1 - 0.12)*(1 - 0.12) = $8,000*(1 - 0.12)^2

We already can see the pattern here, after t years, the price of the bike will be:

P(t) = $8,000*(1 - 0.12)^t

With this equation, we can find the price of the bike after 5 years, where we only need to replace t by 5

We will get:

P(5) = $8,000*(1 - 0.12)^5 = $4,221.9

The price of the bike after 5 years is  $4,221.9.

Now we want two equations to represent this, so we need another.

This is kinda trivial, because there is only one equation that represents this, so we only can rewrite our equation in another form.

We could say that the price of the bike after t years is:

P(t) = $8,000*(0.88)^t

or something like:

P(t) = $7040*(0.88)^(t - 1)

Where i multiplied the initial value by 0.88, and then reduced the exponent by one.

3 0
3 years ago
This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex
noname [10]

L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)

L_x=10+10\lambda x=0\implies1+\lambda x=0

L_y=10+10\lambda y=0\implies1+\lambda y=0

L_z=2+4\lambda z=0\implies1+2\lambda z=0

L_\lambda=5x^2+5y^2+2z^2-42=0

\begin{cases}L_x=0\\L_y=0\\L_z=0\end{cases}\implies1+\lambda x=1+\lambda y=1+2\lambda z=0\implies x=y=2z

5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1

z^2=1\implies z=\pm1\implies x=y=\pm2

There are two critical points, at which we have

f\left(2,2,1\right)=42\text{ (a maximum value)}

f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}

3 0
3 years ago
2.67 repeating as a simplified fraction
MAXImum [283]

Answer:

2 67/100

don't forget to say thanks and rate!

5 0
3 years ago
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Stells [14]
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X=2/5y
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