Can someone explain to me how to do this? (inverse of function)

On a coordinate plane, the x-axis is labeled bags of trail mix and the y-axis is labeled ounces of almonds. Line a is labeled y

Musya8 [376]

Answer:

Line B

Step-by-step explanation:

Only one line on the graph shows 2 ounces of almonds for 1 bag of trail mix: line B.

7 0

3 years ago

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Sandeep used 8 litres of fuel to travel 76 miles.

OleMash [197]

Answer:

I got, when calculated, 209 miles. (It may be wrong)

Step-by-step explanation:

First, I calculated how many miles can Sandeep travel with 1L, therefore I divided 76 miles by 8L and got 9.5 miles on 1L.

Then I multiplied 9.2 miles by 22L and got 209 miles.

This may be wrong but if anyone else is able to work it out much appreciated if it's correct.. Your most welcome!

4 0

3 years ago

jessica weighs x+34 pounds and Ronda weighs 12 pounds less. If Jessica gains 5 pounds and Ronda loses 2 pounds, what is the sum

Vladimir [108]

Answer:

2x+59.

Step-by-step explanation:

Let J represent Jessica's weight and R represent Ronda's weight.

Jessica weighs x+34 pounds. Thus:

$J=x+34$

Ronda weighs 12 pounds less than Jessica. In other words:

$R=J-12=(x+34)-12\\R=x+22$

The sum of their weights, therefore, is:

$J+R\\=(x+34)+(x+22)=2x+56$

Now, if Jessica gains 5 pounds and Ronda loses 2 pounds, the net gain of the total weight would be 3 pounds. Thus, we only need to add 3 to the original total to find the sum of their new weights:

$2x+56+3=2x+59$

The sum of the new [weights] is represented by 2x+59.

8 0

3 years ago

What is the reciprocal of 3 1/5?

aivan3 [116]

if u like it,u can start following me...

5 0

3 years ago

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A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe

adoni [48]

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

$\frac{dC}{dt}=Ci*Qi-Co*Qo$

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

$\frac{dC}{dt}=-Co*Qo$

Rearranging the equation, it becomes

$\frac{dC}{C}=-Qo*dt$

Integrating both sides

$\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}$

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

$C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\$

The final equation for the concentration of salt at any given time is

$C=exp^{-3*t-0.693}$

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

$C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}= -\frac{-0.266}{3}=0.088$

5 0

3 years ago