Ask question

Ask question

All categories

Angelina_Jolie [31]

3 years ago

9

A car travels 45 miles per hour on one gallon I gas write an algebraic expression to find the distance the car can travel using

x gallons of gas

1 answer:

yanalaym [24]3 years ago

4 0

Answer:

$y=45x$

$y$ represent the distance traveled by the car in $x$ gallons of gas.

Step-by-step explanation:

Given:

A car travels 45 miles on one gallon of gas.

We need to calculate distance covered by the car in $x$ gallons of gas.

Let $y$ represent the distance traveled by the car in $x$ gallons of gas.

We know that distance traveled by the car directly varies with the amount of gas.

So, we have $y$ ∝ $x$

Thus $y=kx$

where $k$ represents the distance traveled by car in 1 gallon of gas.

From the data given we know $k=45$

So, the expression to find the distance the car can travel using $x$ gallons of gas is written as:

$y=45x$

You might be interested in

Help ASAP I’ll mark you as brainlister

liraira [26]

Answer:

about 0.10 cents a unit

Step-by-step explanation:

If I'm wrong, I'm sorry!!

I hope you have a great day! Good luck with your math!

6 0

2 years ago

Please help on this one plzzzz

deff fn [24]

We know that Angle Bisector Divides an Angle into Two Equal Angles.

As UP is the Angle Bisector of Angle U, It Divides Angle U into two Equal Parts they are Angle(1) and Angle(2)

⇒ Angle(1) = Angle(2)

Given Angle(1) = 5x + 10 and Angle(2) = 3x + 14

⇒ 5x + 10 = 3x + 14

⇒ 2x = 4

⇒ x = 2

⇒ Angle(1) = 5x + 10 = 5(2) + 10 = 10 + 10 = 20

So : Measure of Angle(1) is 20

8 0

2 years ago

Can you please help me

Dennis_Churaev [7]

Answers A mark me brainliest

6 0

3 years ago

S is between B and C. True or false

Brut [27]

False because I looked it up online

8 0

3 years ago

Read 2 more answers

A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be

irga5000 [103]

Answer:

The statement is now presented as:

$\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}$

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

Step-by-step explanation:

Let $C = (h,k)$ the coordinates of the center of the circle, which must be transformed into $C'=(h', k')$ by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

$(x-h)^{2}+(y-k)^{2} = r^{2}$

Where $r$ is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) <em>Center of the circle is translated 4 units to the right</em> (+x direction):

$C''(x,y) = C(x, y) + U(x,y)$ (Eq. 1)

Where $U(x,y)$ is the translation vector, dimensionless.

If we know that $C(x, y) = (h,k)$ and $U(x,y) = (4, 0)$ , then:

$C''(x,y) = (h,k)+(4,0)$

$C''(x,y) =(h+4,k)$

2) <em>Reflection over the x-axis</em>:

$C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)]$ (Eq. 2)

Where $O(x,y)$ is the reflection point, dimensionless.

If we know that $O(x,y) = (h+4,0)$ and $C''(x,y) =(h+4,k)$ , the new point is:

$C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]$

$C'(x,y) = (h+4, 0)-(0,k)$

$C'(x,y) = (h+4, -k)$

And thus, $h' = h+4$ and $k' = -k$ . The statement is now presented as:

$\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}$

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

<em />

4 0

3 years ago