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Troyanec [42]
2 years ago
7

• There are 20 pounds of the mixture.

Mathematics
1 answer:
katrin [286]2 years ago
7 0

Answer:

6.63 pounds

Step-by-step explanation:

Total pounds of mixture = 20

The contents of the mixture are; peanuts and almonds. Of which;

Peanuts cost = $2.95 per pound

Almonds cost = $5.95 per pound

The mixture cost = $4.00 per pound

Total cost of contents of the mixture = $8.90

The pounds of peanuts = \frac{2.95}{8.90} x 20

                     = 6.63

Therefore, 6.63 pounds of peanuts was used in producing 20 pounds of the mixture.

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Can someone please help me solve 10y-2y=64
sashaice [31]

Answer:

y=8

Step-by-step explanation:

first you want to subtract 2y from 10y which will give you 8y and then divide both sides by 8 to get y alone and you will get y=8

6 0
3 years ago
Read 2 more answers
Oliver treated his sister to lunch while visiting her in Cedarburg. Their lunch cost $275, and the sales tax in Cedarburg is 12%
Rudiy27

Answer:

Oliver paid a total of $363.

Step-by-step explanation:

The total amount paid by Oliver on this dinner is given by the cost of their lunch (275) plus the tax (12%) plus the tip (20%). So in order to calculate this value we first need to know how much was paid in tax and the tip, these calculations are shown bellow:

tax = 275*12% = 275*(12/100) = 33

tip = 275*20% = 275*(20/100) = 55

So the total amount paid was:

total = 275 + 33 + 55 = $363

Oliver paid a total of $363.

8 0
3 years ago
How many times can 80 go in to 100
stich3 [128]
The answer will be 1 because 80+80=160 not 100 so it cant be added twice so it has to be one.
4 0
3 years ago
A mass weighing 16 pounds stretches a spring (8/3) feet. The mass is initially released from rest from a point 2 feet below the
mezya [45]

Answer with Step-by-step explanation:

Let a mass weighing 16 pounds stretches a spring \frac{8}{3} feet.

Mass=m=\frac{W}{g}

Mass=m=\frac{16}{32}

g=32 ft/s^2

Mass,m=\frac{1}{2} Slug

By hook's law

w=kx

16=\frac{8}{3} k

k=\frac{16\times 3}{8}=6 lb/ft

f(t)=10cos(3t)

A damping force is numerically equal to 1/2 the instantaneous velocity

\beta=\frac{1}{2}

Equation of motion :

m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}+f(t)

Using this equation

\frac{1}{2}\frac{d^2x}{dt^2}=-6x-\frac{1}{2}\frac{dx}{dt}+10cos(3t)

\frac{1}{2}\frac{d^2x}{dt^2}+\frac{1}{2}\frac{dx}{dt}+6x=10cos(3t)

\frac{d^2x}{dt^2}+\frac{dx}{dt}+12x=20cos(3t)

Auxillary equation

m^2+m+12=0

m=\frac{-1\pm\sqrt{1-4(1)(12)}}{2}

m=\frac{-1\pmi\sqrt{47}}{2}

m_1=\frac{-1+i\sqrt{47}}{2}

m_2=\frac{-1-i\sqrt{47}}{2}

Complementary function

e^{\frac{-t}{2}}(c_1cos\frac{\sqrt{47}}{2}+c_2sin\frac{\sqrt{47}}{2})

To find the particular solution using undetermined coefficient method

x_p(t)=Acos(3t)+Bsin(3t)

x'_p(t)=-3Asin(3t)+3Bcos(3t)

x''_p(t)=-9Acos(3t)-9sin(3t)

This solution satisfied the equation therefore, substitute the values in the differential equation

-9Acos(3t)-9Bsin(3t)-3Asin(3t)+3Bcos(3t)+12(Acos(3t)+Bsin(3t))=20cos(3t)

(3B+3A)cos(3t)+(3B-3A)sin(3t)=20cso(3t)

Comparing on both sides

3B+3A=20

3B-3A=0

Adding both equation then, we get

6B=20

B=\frac{20}{6}=\frac{10}{3}

Substitute the value of B in any equation

3A+10=20

3A=20-10=10

A=\frac{10}{3}

Particular solution, x_p(t)=\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

Now, the general solution

x(t)=e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

From initial condition

x(0)=2 ft

x'(0)=0

Substitute the values t=0 and x(0)=2

2=c_1+\frac{10}{3}

2-\frac{10}{3}=c_1

c_1=\frac{-4}{3}

x'(t)=-\frac{1}{2}e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+e^{-\frac{t}{2}}(-c_1\frac{\sqrt{47}}{2}sin(\frac{\sqrt{47}t}{2})+\frac{\sqrt{47}}{2}c_2cos(\frac{\sqrt{47}t}{2})-10sin(3t)+10cos(3t)

Substitute x'(0)=0

0=-\frac{1}{2}\times c_1+10+\frac{\sqrt{47}}{2}c_2

\frac{\sqrt{47}}{2}c_2-\frac{1}{2}\times \frac{-4}{3}+10=0

\frac{\sqrt{47}}{2}c_2=-\frac{2}{3}-10=-\frac{32}{3}

c_2==-\frac{64}{3\sqrt{47}}

Substitute the values then we get

x(t)=e^{-\frac{t}{2}}(-\frac{4}{3}cos(\frac{\sqrt{47}t}{2})-\frac{64}{3\sqrt{47}}sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

8 0
3 years ago
A firefighter places a 29 feet ladder, so that the base of the ladder is 21 feet from the wall. What is the height at which the
Tju [1.3M]
Pythagorean theorem:
a^2+b^2=c^2 (C^2 always is the hypotenuse)

21^2+b^2=29^2
441+b^2=841
841-441=400
square root of 400 is 20
Your answer is 20
6 0
2 years ago
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