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Consider the equation Ax+By=−36. If the x-intercept is (−3,0) and the y-intercept is (0,9), what are the values of A and B?

Grace [21]

Answer:

A= 12

B = -4

Step-by-step explanation:

( -3,0) & (0,9) should satisfy the equation Ax + By = -36

when ( -3,0)

-3A+ 0 = -36

A = 12

when (0,9)

0 + 9B = -36

B = -4

5 0

3 years ago

Can someone help me with this math homework please!

slega [8]

Answer:

3. The number of students that rode on each bus

Step-by-step explanation:

The equation they give us is:

2b + 6 = 70

The question already tells us that in this equation:

2 is the 2 buses that the school hired

6 is the 6 leftover kids who couldn't fit on the 2 buses

The question also tells us that the 2 buses only have a total of 64 seats, so 32 seats on 1 bus (64/2 = 32).

So that means b would equal the 32 seats on 1 bus, or the number of students that rode on each bus.

Hope it helps (●'◡'●)

3 0

3 years ago

Help me please help me

Anon25 [30]

It’s represents the greatest number of balloon bunches and that Alice can make

3 0

3 years ago

Read 2 more answers

Bobby catches 8 passes in 3 football games. At this rate, how many passes does he

olasank [31]

Answer:

40 passes

Step-by-step explanation:

we know that

Bobby catches 8 passes in 3 football games

so

using proportion

Find out how many passes does he catch in 15 games

$\frac{8}{3}=\frac{x}{15}\\\\x=8(15)/3\\\\x=40\ passes$

7 0

3 years ago

Read 2 more answers

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

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