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2 years ago

11

How can you determine if a piece-wise relation is a function?

1 answer:

uysha [10]2 years ago

7 0

Use the vertical line rule to test.
Imagine you are drawing lines parallel to the y axis, if any of your lines touches the graph twice, the graph does not represent a function. otherwise, it does.

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Help me quickly!!!!!!!!!

Assoli18 [71]

Answer:273

Step-by-step explanation:

52 x 5.25 = 273 miles

15 minutes is 1/4 of an hour so it is .25 so you multiply 52 times 5.25 to find your answer.

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2 years ago

Read 2 more answers

Today everything at the store is on sale. The store is offering a 20% discount on all items.

katovenus [111]

Answer:

Number 1: $14.40

Number 2: $21.60

Step-by-step explanation:

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3 years ago

You originally draw a design for an art contest on a 2 inch by 5 inch card. The second phrase of the contest requires the drawin

Bingel [31]

4.2.
If the 2 inch side was scaled to 10.5 inches the 5 inch side would be scaled by 10.5/2 = 5.25, so would be 25.125 inches which is too big.

So the 10.5 inches is the 5 inch side on the original, and so it 10.5/5 = 2.1 times bigger.

Therefore the smaller side would be 2*2.1/= 4.2 inches.

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3 years ago

A tank has the shape of a surface generated by revolving the parabolic segment y = x2 for 0 ≤ x ≤ 3 about the y-axis (measuremen

Darina [25.2K]

Answer:

$100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy$ ft-lbs.

Step-by-step explanation:

Given:

The shape of the tank is obtained by revolving $y=x^2$ about y axis in the interval $0\leq x\leq 3$ .

Density of the fluid in the tank, $D=100\ lbs/ft^3$

Let the initial height of the fluid be 'y' feet from the bottom.

The bottom of the tank is, $y(0)=0^2=0$

Now, the height has to be raised to a height 5 feet above the top of the tank.

The height of top of the tank is obtained by plugging in $x=3$ in the parabolic equation . This gives,

$H=3^2=9\ ft$

So, the height of top of tank is, $y(3)=H=9\ ft$

Now, 5 ft above 'H' means $H+5=9+5=14$

Therefore, the increase in height of the top surface of the fluid in the tank is given as:

$\Delta y=(14-y)$ ft

Now, area of cross section of the tank is given as:

$A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section$

Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.

We have, $y=x^2$

So, $x=\sqrt y$

Therefore, radius, $r=\sqrt y$

Now, area of cross section is, $A(y)=\pi (\sqrt y)^2$

Work done in pumping the contents to 5 feet above is given as:

$W=D\int\limits^{y(3)}_{y(0)} {A(y)(\Delta y)} \, dy$

Plug in all the values. This gives,

$W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}$

7 0

2 years ago

Cesar, Carlos, David, and Edwin decided to have a push-up contest. The following table describes the push-up speeds of the four

vitfil [10]

Answer:

Edwin (55 push-ups per minute)

Step-by-step explanation:

We need to find the actual number of pushups per one minute in order to compare them on the same base (per ONE minute). So we do"

Cesar: 200 / 4 = 50 pu/min

Carlos: 240 / 5 = 48 pu/min

David: 150 / 3 = 50 pu/min

Edwin: 110 / 2 = 55 pu/min

Therefore, Edwin is the fastest since he does more pushups per minute.

4 0

3 years ago