All categories

4 years ago

Use the vertical line test to determine if the graph represents a function. Explain.

Mathematics

2 answers:

tatuchka [14]4 years ago

6 0

Answer:

Is it possible to pass a single vertical line through more than one point on the horizontal line?The answer to this question is: No it is not possibleSince we cannot have a vertical line cross through more than one point, this graph passes the vertical line test making it a function. Any input (x) leads to exactly one output (y)

Step-by-step explanation:

nadezda [96]4 years ago

5 0

Ask yourself this: Is it possible to pass a single vertical line through more than one point on the horizontal line?

The answer to this question is: No it is not possible

Since we cannot have a vertical line cross through more than one point, this graph passes the vertical line test making it a function. Any input (x) leads to exactly one output (y)

You might be interested in

Paco bought 3 CDs that cost d dollars each and a pack of gum for C cents. Write an expression for the total cost of his purchase

cupoosta [38]

Answer:

The correct option is A) $3d+C$ .

Step-by-step explanation:

Consider the provided information.

Paco bought 3 CDs that cost d dollars each.

Let d is the cost of each CDs.

The cost of 3 CDs will be 3 times of d.

This can be written as:

$3d$

He bought a pack of gum for C cent. Thus, we can say that the cost of pack of gum is C.

Now add the cost of gum in above expression.

$3d+C$

Hence, the required expression is $3d+C$ .

Thus, the correct option is A) $3d+C$ .

3 0

3 years ago

Pls help will give 5 stars and heart

Tamiku [17]

So the circle is 39.25
And the area of triangle 30 together about 39.25

5 0

3 years ago

Two teams A and B play a series of games until one team wins four games. We assume that the games are played independently and t

crimeas [40]

Answer:

4pq(p³+q³)

Step-by-step explanation:

Exactly 5 game would be when 4 wins and 1 loss of a particular person

loss has to be one of the first 4 games

A wins: qp⁴ + pqp³ + p²qp² + p³qp

= qp⁴ + qp⁴ + qp⁴ + qp⁴ = 4qp⁴

B wins: pq⁴ + qpq³ + q²pq² + q³pq

= pq⁴ + pq⁴ + pq⁴ + pq⁴ = 4pq⁴

A wins or B wins:

4pq⁴ + 4qp⁴ = 4pq(q³+p³)

6 0

3 years ago

Two pizza delivery drivers compared the mean numbers of deliveries they completed in one day.

marishachu [46]

Answer:

Step-by-step explanation:

I just did it Red Riot out

8 0

3 years ago

Read 2 more answers

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$

Recall that when t=0, A(t)=3000 (our initial condition)

$3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}$

3 0

3 years ago