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xeze [42]
3 years ago
14

Jack's dog weighs 25 pounds. the vet says that his dog must loose 30% of his weight to be healthier. how much weight is the vet

asking the dog to loose?
Mathematics
1 answer:
zepelin [54]3 years ago
7 0
The weight to be lost = 30 / 100   * 25  = 7.5 pounds
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3 years ago
39 is what percent of 96
ra1l [238]

Answer:

40.625%

Step-by-step explanation:

This is a very simple question.

We have to understand converting word equations to algebraic ones.

"is" means "="

"of" means "*"

So, we can write:

39 = what percent * 96

Now, we let "what percent" be "p" and solve the equation for p:

39 = P * 96

So,

39 = P * 96\\39=96P\\P=\frac{39}{96}\\P=0.40625

Converting this decimal to percentage means multiplying by 100, so we have:

0.40625 * 100

= 40.625%

7 0
3 years ago
A team of 15 workers picked 1,050 apples. Each person picked the same number of apples. How many did each worker pick?
GarryVolchara [31]

Answer:

70 apples

Step-by-step explanation:

Please let me know if you want me to add an explanation as to why this is the answer. I can definitely do that, I just wouldn’t want to write it if you don’t want me to :)

4 0
2 years ago
Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g
AysviL [449]

Answer:

The quantity of salt at time t is m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} }), where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}

Where:

c - The salt concentration in the tank, as well at the exit of the tank, measured in \frac{pd}{gal}.

\frac{dc}{dt} - Concentration rate of change in the tank, measured in \frac{pd}{min}.

V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

V \cdot \frac{dc}{dt} + 6\cdot c = 3

60\cdot \frac{dc}{dt}  + 6\cdot c = 3

\frac{dc}{dt} + \frac{1}{10}\cdot c = 3

This equation is solved as follows:

e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }

\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }

e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt

e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C

c = 30 + C\cdot e^{-\frac{t}{10} }

The initial concentration in the tank is:

c_{o} = \frac{10\,pd}{60\,gal}

c_{o} = 0.167\,\frac{pd}{gal}

Now, the integration constant is:

0.167 = 30 + C

C = -29.833

The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

m_{salt} = V_{tank}\cdot c(t)

m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })

Where t is measured in minutes.

7 0
3 years ago
Answer all questions CORRECTLY BIG POINTS!
hammer [34]
  • y^2+9y+20
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  • y^2+2y+3
  • Step-by-step explanation:
  1. (y+5(y+4) =y(y+4)+5(y+4) =y^2+4y+5y+20 =y^2+9y+20
  2. (y+3(y+7) =y(y+7)+3(y+7) =y^2+7y+3y+21 =y^2+10y+21
  3. (y+3)(y-1)

=y(y-1)+3(y-1)

=y^2-y+3y-3

= y^2+2y+3

5 0
3 years ago
Read 2 more answers
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