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Let the cost of a hamburger = x   and the cost of a carton of milk = y

And we have that

2x  +  1y  = 5

1x +   2y  =  4.60

Using the second equation

x  = 4.60 - 2y

Sub this into the first equation

2 (4.60 - 2y) + 1y  = 5    simplify

9.20 - 4y + 1y = 5

-3y  =  -4.2      divide both sides by -3

y = $ 1.40   =   cost of a carton of milk

But it could be worse. I got a 62% on my last math test, so yeah....

If Profit=Revenue-Costs, then \(p(x)=(5x+15,000)-(2x-2,000)\).

First distribute the negative into (2x-2000), then drop the parenthesis because they are no longer needed.

\(p(x)=5x+15,000-2x+2,000\)

Combine like terms.

So, your profit function is \(p(x)=3x+17,000\).

Yes, you are correct.

I really need help on these questions. Any help is appreciated!

P=Pens, p=pencils

6P + 1p =10-1=9

4P + 2p =10-2=8, solve for P, p

P =$1.25 - cost of 1 Pen

p =$1.50 - cost of 1 pencil.

The annual percentage rate is __10.46, but round it to 10.5_______%

The annual percentage rate is _11.75____ %.

1)   FV = PV x e^(0.09t)

$30,000 =$15,000 x e^(.09t), solve for t

t =100ln(2)/9 =~7.7 years.

2)  FV =20,000 x e^(0.0875*20)

      FV =20,000 x 5.754602676....

      FV =$115,092.05 - what James will have when he is 20 years old.

FV = PV x {1 + R}^N

5,800 =115,000 x {1 - 0.22}^N  divide both sides by 115,000

0.0504347826 = 0.78^N             Take the log of both sides

N =log(0.0504347826) / log(0.78)

N = ~ 12 Years to depreciate to about $5,800.

x+y=0

5x-9y=42

+

-5x-5y=0

-14y=42

y=-3

x=3

Is it really 10,000% per month? If it is, then we have:

[1 + 10,000/100]^12 =[1 + 100]^12 =101^12 - 1 x 100 =1.13 x 10^26% per year !!!!!

[1 + 10,000/100]^(1/30) =101^(1/30) =1.1663 - 1 x 100 =16.63% per day  !!!!!.

So that: [1 + 0.1663]^360 =1.13 x 10^24  x  100 =1.13 x 10^26% per year !!!!!.

Note: I used 360 days because we took a month to be 30 days x 12 =360 days.

sorry jeff :(

replace x and f(x)

x=1/2(f(x) +5

solved is 2x-10=f(x)

mmmm i just a ton of tasty chocolate donuts!!! DONUTS ARE LOVE!!!!!!!​​​​​​​​​🍩🍩🍩🍩🍩🍩🍩🍩

well, i did pretty bad on mine too, so dont feel that bad Jeff!!

Just divide:

(3/8) /2 =

(3/8) x 1/2  =

3/16 of the pizza that each will get.

s=sweaters, p=pants

a)  35s + 20p =$150

b)  35*3 + 20p =$150

       105  + 20p =$150 subtract 105 from both sides

        20p =150 - 105

        20p =45                 divide both sides by 20

         p =45/20

         p=2.25 =2 pants that she can buy.

c)   35s + 0p = $150

       35s = $150       divide both sides by 35

        s =150/35

        s=4 2/7 = 4 sweaters that she can buy.

A = 40 x 2^-(12/36)

A =40 x 0.7937

A =31.748 Mg - remaining in the patient's blood at noon on the first day.

20 = 40 x 2^-(t/36), solve for t

t=36 hours(obviously) - because that is the half-life of Valium, which would be the noon hour of the second day.

-2.6=-6.4

1=-2.8

4.2=0.4

A certain element has a half-life of 1.5 billion years. a. You find a rock containing a mixture of the element and lead. You determine that 65 % of the original element remains; the other 35 % decayed into lead. How old is the rock?

Let the original age of the rock =A. Then we have:

65% =100% x 2^-(A/1.5), solve for A

0.65 = 2^(-0.666667 A)

0.65 = 13/20 and 2^(-0.666667 A) = 2^(-2 A/3):
13/20 = 2^(-(2 A)/3)

13/20 = 2^(-(2 A)/3) is equivalent to 2^(-(2 A)/3) = 13/20:
2^(-(2 A)/3) = 13/20

Take reciprocals of both sides:
2^((2 A)/3) = 20/13

Take the logarithm base 2 of both sides:
(2 A)/3 = (log(20/13))/(log(2))

Multiply both sides by 3/2: 

A = (3 ln(20/13))/(2 ln(2)= 0.932 x 10^9 =9.32 x 10^8 - Years - the original age of the rock.

b. Analysis of another rock shows that it contains15​% of its original​ element; the other 85​% decayed into lead. How old is the​ rock?

Let the original age of the rock =A

15% =100% x 2^-(A/1.5), solve for A

0.15 = 2^(-0.666667 A)

0.15 = 3/20 and 2^(-0.666667 A) = 2^(-2 A/3):
3/20 = 2^(-(2 A)/3)

3/20 = 2^(-(2 A)/3) is equivalent to 2^(-(2 A)/3) = 3/20:
2^(-(2 A)/3) = 3/20

Take reciprocals of both sides:
2^((2 A)/3) = 20/3

Take the logarithm base 2 of both sides:
(2 A)/3 = (log(20/3))/(log(2))

Multiply both sides by 3/2:
A = (3 ln(20/3))/(2 ln(2))=4.1 x 10^9 Years - original age of the rock

Here's an explanation...

Let's find  sin θ  using the Pythagorean identity.

sin²θ  +  cos²θ   =   1

sin²θ  +  (3/5)²   =   1

sin²θ   =   1 - (3/5)²      sin is positive in Quad I , so take the positive sqrt of both sides.

sin θ   =   √[ 1 - (3/5)² ]

sin θ   =   4/5

And...

tan θ   =   sin θ / cos θ   =   (4/5) / (3/5)   =   4/3

\(\frac{12}{x}+\frac5y=8\\ \frac8x-\frac3y=-1 \)

                                           This is the same as....

\(12(\frac1x)+5(\frac1y)=8 \\ 8(\frac1x)-3(\frac1y)=-1\)

                                           Multiply the top equation through by  2 , and

                                           multiply the bottom equation through by  -3 .

\(24(\frac1x)+10(\frac1y)=16 \\ -24(\frac1x)+9(\frac1y)=3 \)

                                           Add the two equations together and solve for  y .

\(19(\frac1y)=19 \\ 19=19y\\ \color{Emerald}1=y\)

                            Plug in  1  for  y  in one of the original equations to find  x .

\(\frac{12}{x}+\frac51=8 \\ \frac{12}x=3 \\ 12=3x\\ \color{Emerald}4=x\)

you're a lifesaver, thank you so much, thumbs up to you :^)