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vitfil [10]
3 years ago
10

A grocer wants to make a 10-pound mixture of peanuts and cashews that he can sell for $4.75 per pound. If peanuts cost $4.00 per

pound and cashews cost $6.50 per pound, how many pounds of each should he use? Let p = pounds of peanuts and let c = pounds of cashews. Write a system of equations that could be used to solve the problem.
will mark as brainliest
Mathematics
1 answer:
andrew11 [14]3 years ago
7 0

Answer:

ANSWER 1: He should use 3 pounds of cashews.

ANSWER 2: He should use 7 pounds of peanuts.

Step-by-step explanation:

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Read 2 more answers
Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv
Ira Lisetskai [31]

Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

   = (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = \dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}

Let dD/dT = 0 to find the critical value we will get

\dfrac{70000000\left(291T^2-24298T+91800\right)}  = 0

Using formula of quadratic, we get the roots:

T =  79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8)    a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

3 0
3 years ago
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