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3 years ago

Which relation is a function? A. B. C. D.

Mathematics

1 answer:

SIZIF [17.4K]3 years ago

6 0

A function is relationship where a member of the Domain maps to one and only one member of the Range.

To prove a relationship is a function, a VERTICAL LINE TEST may be performed. If a vertical line cuts the graph of a relationship only once then it is a function if not, then it is not.

View the image attached for the test.

The answer is option A

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Problem 9

lapo4ka [179]

Answer:

3:36PM

Step-by-step explanation:

Leon starts at 12PM with 12 gallons of gas, and after 2 hours he has used 5 gallons of gas. This means that every 2 hours he uses 5 gallons of gas.

Next we will find at what point Leon will stop to get gas. Since he will stop when the tank is at $\frac{1}{4}$ capacity, we can use the equation:

$\frac{1}{4} * 12 = \boxed{3}$

This shows $\frac{1}{4}$ of his tank's capacity ( $12$ ) is equal to $3$ gallons. This means he will stop for gas when $3$ gallons are remaining.

Now we need to find how many gallons of gas he uses, but as a unit rate. (This will allow us to find what time Leon will stop to get gas.) To find the unit rate, we will need to find how many gallons of gas he uses per hour.

$\frac{\mbox{5 gallons}}{\mbox{2 hours}} = \frac{\mbox{2.5 gallons}}{\mbox{1 hours}}$

This is a simple proportion, and now we know he uses $2.5$ gallons of gas per hour.

Now we can how many hours of gas Leon has left.

He has $7$ gallons of gas left at 2PM, so we can divide to find how many hours left of gas he has.

$\frac{7-3}{2.5} = 1.6$

The $7-3$ is because Leon doesn't stop when his tank is empty, he stops $3$ gallons earlier. We are dividing by $2.5$ because that is how much gas he uses per hour, meaning the result of this division ( $1.6$ ) is how many hours he has left.

Now we can solve for what time Leon will stop to get gas.

12PM + $2$ hours of driving + the remaining $1.6$ hours = 3:36PM

( $1.6$ hours is equal to 1 hour and 36 minutes)

Therefore, Leon will stop for gas at 3:36PM

8 0

2 years ago

A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the

Basile [38]

Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= $\pi r^2h$

According to the problem,

$\pi r^2 h=25$

$\Rightarrow h=\frac{25}{\pi r^2}$

The surface area of the base of the can is = $\pi r^2$

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× $\pi r^2$ )

The lateral surface area of the can is = $2\pi rh$

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× $2\pi rh$ )

$=\$2.5 \pi r h$

Total cost of metal is C= 2.00 $\pi r^2$ + $2.5 \pi r h$

Putting $h=\frac{25}{\pi r^2}$

$\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}$

$\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}$

Differentiating with respect to r

$C'=4\pi r- \frac{62.5}{ r^2}$

Again differentiating with respect to r

$C''=4\pi + \frac{125}{ r^3}$

To find the minimize cost, we set C'=0

$4\pi r- \frac{62.5}{ r^2}=0$

$\Rightarrow 4\pi r=\frac{62.5}{ r^2}$

$\Rightarrow r^3=\frac{62.5}{ 4\pi}$

⇒r=1.71

Now,

$\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0$

When r=1.71 cm, the metal cost will be minimum.

Therefore,

$h=\frac{25}{\pi\times 1.71^2}$

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

6 0

3 years ago

Solve: 6 – x = 4(5 + x) – 4

frez [133]

Answer:

x= -2

Step-by-step explanation:

3 0

3 years ago

Kite E F G H is inscribed in a rectangle. Points F and H are midpoints of sides of the rectangle, and creates a side length of x

Hoochie [10]

Answer:

5 units

Step-by-step explanation:

Let point O be the point of intersection of the kite diagonals.

|OF| = 2, |OH| = 5

|FH| = |OF| + |OH| = 2 + 5 = 7

FH and EG are the diagonals of the kite. Hence the area of thee kite is:

Area of kite EFGH = (FH * EG) / 2

Substituting:

35 = (7 * |EG|) / 2

|EG| * 7 = 70

|EG| = 10 units

The longer diagonal of a kite bisects the shorter one, therefore |GO| = |EO| = 10 / 2 = 5 units

x = |GO| = |EO| = 5 units

7 0

3 years ago

Read 2 more answers

Can some one explain please

inna [77]

What kind of question is this lol ? i don’t even understand it

7 0

3 years ago