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2 years ago

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Jose earned 50 points in a video game. He lost 40 points, earned 87 points, then lost 30 more pints. Write and evaluate an expre

ssion to find his final score in the video game.

1 answer:

Aneli [31]2 years ago

4 0

50-40+87-30=67

hope i helped

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John runs to the market and comes back in 15 minutes. His speed on the way to the market is 5m/s and his speed on the way back i

katen-ka-za [31]

Answer:

The distance to the market is 2000 m

Step-by-step explanation:

∵ John runs to the market and comes back in 15 minutes

→ Change the min. to the sec. because the unit of his speed is m/s

∵ 1 minute = 60 seconds

∴ 15 minutes = 15 × 60 = 900 seconds

→ Assume that t1 is his time to the market and t2 is his time from

the market

∵ t1 + t2 = 15 minutes

∴ t1 + t2 = 900 ⇒ (1)

→ Assume that the distance to the market is d

∵ His speed on the way to the market is 5m/s

∵ Time = Distance ÷ Speed

∴ t1 = d ÷ 5 ⇒ (1 ÷ 5 = 0.2)

∴ t1 = 0.2d ⇒ (2)

∵ His speed on the way back is 4m/s

∴ t2 = d ÷ 4 ⇒ (1 ÷ 4 = 0.25)

∴ t2 = 0.25d ⇒ (3)

→ Substitute (2) and (3) in (1)

∵ 0.2d + 0.25d = 900

∴ 0.45d = 900

→ Divide both sides by 0.45

∴ d = 2000 m

∴ The distance to the market = 2000 m

8 0

3 years ago

A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in the solution. Water containing1 lb

devlian [24]

Answer:

(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is $\left(1-\frac{4000000}{\left(200+t\right)^3}\right)\left(200+t\right)$ .

(b) The concentration (in lbs per gallon) when it is at the point of overflowing is $\frac{121}{125}\:\frac{lb}{gal}$ .

(c) The theoretical limiting concentration if the tank has infinite capacity is $1\:\frac{lb}{gal}$ .

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If <em>Q(t)</em> gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for <em>Q(t)</em>.

The main equation that we’ll be using to model this situation is:

Rate of change of <em>Q(t)</em> = Rate at which <em>Q(t)</em> enters the tank – Rate at which <em>Q(t)</em> exits the tank

where,

Rate at which <em>Q(t)</em> enters the tank = (flow rate of liquid entering) x (concentration of substance in liquid entering)

Rate at which <em>Q(t)</em> exits the tank = (flow rate of liquid exiting) x (concentration of substance in liquid exiting)

Let $C$ be the concentration of salt water solution in the tank (in $\frac{lb}{gal}$ ) and $t$ the time (in minutes).

Since the solution being pumped in has concentration $1 \:\frac{lb}{gal}$ and it is being pumped in at a rate of $3 \:\frac{gal}{min}$ , this tells us that the rate of the salt entering the tank is

$1 \:\frac{lb}{gal} \cdot 3 \:\frac{gal}{min}=3\:\frac{lb}{min}$

But this describes the amount of salt entering the system. We need the concentration. To get this, we need to divide the amount of salt entering the tank by the volume of water already in the tank.

$V(t)$ is the volume of brine in the tank at time t. To find it we know that at t = 0 there were 200 gallons, 3 gallons are added and 2 are drained, and the net increase is 1 gallons per second. So,

$V(t)=200+t$

Therefore,

The rate at which $C(t)$ enters the tank is

$\frac{3}{200+t}$

The rate of the amount of salt leaving the tank is

$C\:\frac{lb}{gal} \cdot 2 \:\frac{gal}{min}+C\:\frac{lb}{gal} \cdot 1\:\frac{gal}{min}=3C\:\frac{lb}{min}$

and the rate at which $C(t)$ exits the tank is

$\frac{3C}{200+t}$

Plugging this information in the main equation, our differential equation model is:

$\frac{dC}{dt} =\frac{3}{200+t}-\frac{3C}{200+t}$

Since we are told that the tank starts out with 200 gal of solution, containing 100 lb of salt, the initial concentration is

$\frac{100 \:lb}{200 \:gal} =0.5\frac{\:lb}{\:gal}$

Next, we solve the initial value problem

$\frac{dC}{dt} =\frac{3-3C}{200+t}, \quad C(0)=\frac{1}{2}$

$\frac{dC}{dt} =\frac{3-3C}{200+t}\\\\\frac{dC}{3-3C} =\frac{dt}{200+t} \\\\\int \frac{dC}{3-3C} =\int\frac{dt}{200+t} \\\\-\frac{1}{3}\ln \left|3-3C\right|=\ln \left|200+t\right|+D\\\\$

We solve for C(t)

$C(t)=1+D(200+t)^{-3}$

D is the constant of integration, to find it we use the initial condition $C(0)=\frac{1}{2}$

$C(0)=1+D(200+0)^{-3}\\\frac{1}{2} =1+D(200+0)^{-3}\\D=-4000000$

So the concentration of the solution in the tank at any time t (before the tank overflows) is

$C(t)=1-4000000(200+t)^{-3}$

(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is just the concentration of the solution times its volume

$(1-4000000(200+t)^{-3})(200+t)\\\left(1-\frac{4000000}{\left(200+t\right)^3}\right)\left(200+t\right)$

(b) Since the tank can hold 500 gallons, it will begin to overflow when the volume is exactly 500 gal. We noticed before that the volume of the solution at time t is $V(t)=200+t$ . Solving the equation

$200+t=500\\t=300$

tells us that the tank will begin to overflow at 300 minutes. Thus the concentration at that time is

$C(300)=1-4000000(200+300)^{-3}\\\\C(300)= \frac{121}{125}\:\frac{lb}{gal}$

(c) If the tank had infinite capacity the concentration would then converge to,

$\lim_{t \to \infty} C(t)= \lim_{t \to \infty} 1-4000000\left(200+t\right)^{-3}\\\\\lim _{t\to \infty \:}\left(1\right)-\lim _{t\to \infty \:}\left(4000000\left(200+t\right)^{-3}\right)\\\\1-0\\\\1$

The theoretical limiting concentration if the tank has infinite capacity is $1\:\frac{lb}{gal}$

4 0

3 years ago

Find the area of each figure s<br> (10 Points)<br> 12 mi<br> 1<br> 8 mi<br> 18 mi<br> 5 mi

Ksenya-84 [330]

I think it is 135 mi to the 2 power

6 0

3 years ago

Select the correct answer. A simulated game of chess is programmed between two computers. The game is supposed to be biased in f

pishuonlain [190]

Answer:

C is the most suspicious set.

Step-by-step explanation:

We would expect that B wins 4/5 * 30 = 24 games.

In set A, B wins 23 games.

In B, B wins 24 games.

In C, B wins 15 games.

In D, B wins 25 games.

So result C is very suspicious.

7 0

2 years ago

Read 2 more answers

How many 4' x 8' sheets of plywood are required to cover a floor that is 9 m by 16 m?

Xelga [282]

Assuming no border or waste, 9m by 16 m translates to 29.52' x 52.49'.
use 32/8=4 sheets of plywood to cover the width on the 8' side.
For the length, use 52/4=13 pieces on the 4' side.
This gives a total of 4*13=52 sheets.
The extra 0.49' x 29.52 can be covered using the extra from the last four sheets for the width.

Check:
Area to be covered: 29.52*52.49=1550 sq. ft.
Area of plywood supplied = 52*4'x8'= 1664 sq. ft, > 1550 sq.ft.

8 0

2 years ago