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3 years ago

A turkey costs $12.75. You have 10$. How many much more do you need ?

Mathematics

2 answers:

Illusion [34]3 years ago

8 0

Answer:

$2.75

Step-by-step explanation:

I hope this helps you

Leona [35]3 years ago

6 0

Answer:

$2.75

Step-by-step explanation:

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cody spent two thirds of his weekly allowance on candy. to earn more money his parents let him wash wash the car for $5. what is

anastassius [24]

Explanation is in the file

tinyurl.com/wtjfavyw

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3 years ago

Hey i need help with system of equations.<br> i uploaded a doxs file you can onpen it in google docs

Art [367]

Answer:

Step-by-step explanation:

PART A:

The first equation is $x + y = 7\\y = 7 - x .............(a)$

The second equation is $3x - 2y = 1\\2y = 3x - 1$

Putting the value of y from (a) in the above equation, we get $2y = 3x - 1\\2(7 - x) = 3x - 1\\14 - 2x = 3x - 1\\x = 15$

From (a) we get y = 7 - x = 7 - 15 = -8

PART B:

Here the system of equations are $-x + 4y = 17........(a) and 3x - 5y = -30.........(b)$

From (a) we get, $x = 4y - 17$

Putting the above value of x in (b) we get, $3x - 5y = -30\\3(4y - 17) - 5y = -30\\12y - 51 - 5y = -30\\7y = 21\\y = 3$

Hence, $x = 4y - 17 = 12 - 17 = -5$

3 0

3 years ago

Can someone help me on this please

Anika [276]

Answer:

sin ∅ = 16 /34

cos ∅ = 30 /34

tan ∅ = 16 /30

Step-by-step explanation:

SOH CAH TOA

sin = opp/hyp

cos = adj/hyp

tan = opp/adj

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3 0

3 years ago

A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan

Artyom0805 [142]

Answer:

1) $\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) $y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let $y$ represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: $\frac{dy}{dt}=rate\:in-rate\:out$

The amount coming in is $0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}$

The rate going out depends on the concentration of salt in the tank at time t.

If there is $y(t)$ pounds of salt and there are $100+2t$ gallons at time t, then the concentration is: $\frac{y(t)}{2t+100}$

The rate of liquid leaving is is 3gal\min, so rate out is $=\frac{3y(t)}{2t+100}$

The required differential equation becomes:

$\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) We rewrite to obtain:

$\frac{dy}{dt}+\frac{3}{2t+100}y=2.5$

We multiply through by the integrating factor: $e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }$

to get:

$(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }$

This gives us:

$((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }$

We integrate both sides with respect to t to get:

$(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C$

Multiply through by: $(50+t)^{-\frac{3}{2}}$ to get:

$y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }$

$y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}$

We apply the initial condition: $y(0)=0$

$0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}$

$C=-12500\sqrt{2}$

The amount of salt in the tank at time t is:

$y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

$y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}$

$y(50)=98.23$

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

$\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})$

As $t\to \infty$ , $50+t\to \infty$ and $\frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0$

This implies that:

$\lim_{t \to \infty}y(t)=\infty- 0=\infty$

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

6 0

3 years ago

Show me a graph of 6x + 5y < 30

Fittoniya [83]

Here is a graph of <span>6x + 5y < 30</span>

7 0

3 years ago