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3 years ago

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Elliot spent $16.15 on ground beef at a grocery store. He purchased 3 and 2 over 5 pounds of ground beef. What is the price per

pound of the ground beef at that store?

1 answer:

Andrej [43]3 years ago

4 0

We want to know how much is the price per pound (or unit rate) of ground beef, given that we know the price for Elliot's purchase. We will see that each pound costs $4.75

So we know that (3 + 2/5) lb of ground beef costs $16.15, the cost of a single pound will be given by the quotient between the total cost and the amount purchased, then we need to solve:

$P =\frac{ \$ 16.15}{(3 + 2/5)lb}$

We can rewrite the denominator as:

(3 + 2/5)lb = (15/5 + 2/5)lb = (17/5) lb

Replacing that we get:

$P =\frac{ \$ 16.15}{(17/5)lb} = \frac{ \$ 16.15*5}{17lb} = \$4.75 \ per \ lb$

So each pound of ground beef costs $4.75

If you want to learn more about unit rates, you can read:

brainly.com/question/2375289

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Barbara drives between Miami, Florida and West Palm Beach, Florida. she drives 45 miles in clear weather and then encounters a t

WINSTONCH [101]

Answer:

Barbara's speed in clear weather is $60mph$ and in the thunderstorm is $38mph$ .

Step-by-step explanation:

Let $v_1$ be the speed and $t_1$ be the time Barbara drives in clear weather, and let $v_2$ be the speed and $t_2$ be the time she drives in the thunderstorm.

Barbara drives 22 mph lower in the thunderstorm than in the clear weather; therefore,

(1). $v_2 = v_1 -22$

Also,

(2). $v_1t_1 = 45miles$

(3). $v_2 t_2 = 57$ ,

and

(4). $t_1+t_2 = 2.25hr$

From equations (2) and (3) we get:

$t_1 = \dfrac{45}{v_1}$

$t_2 =\dfrac{57}{v_2}$

putting these in equation (4) we get:

$\dfrac{45}{v_1}+\dfrac{57}{v_2}=2.25$

and substituting for $v_2$ from equation (1) we get:

$\dfrac{45}{v_1}+\dfrac{57}{v_1-22}=2.25$

This equation can be rewritten as

$2.25v_1^2-151.5v_1+990=0$

which has solutions

$v_1 = 60$

$v_1 = 7.33$

We take the first solution $v_1 =60$ because it gives a positive value for $v_2:$

$v_2 = v_1 -22$

$v_2 = 60 -22\\$

$v_2 = 38$ .

Thus, Barbara's speed in clear weather is $60mph$ and in the thunderstorm is $38mph$ .

4 0

4 years ago

What is 3/5 estimated to

Ber [7]

0.6 am i correct?????

4 0

3 years ago

Read 2 more answers

The recommended heart rate for weight management exercise and improving cardio fitness, in beats per minute, depend on a persons

Korolek [52]

The answer to this question is true. If i'am right this from k12 and edg-enuity. have a good day bro. ;)

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3 years ago

Any chance anyone can help me if the last question I had post

german

Yes i can the answer would be question

6 0

3 years ago

An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object

Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

$1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2 + \frac{1}{2} at^2\\3) v = v_0 + at$

What do we already know?

$v_0 = 29.4 ms^-1$
The launch is straight up
$a = -9.81 ms^-2$ this is the gravitational acceleration $g$

$s_0 = 0 m$ , since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, $v = 0 ms^-2$

$v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m$

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

$s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2$

this is the equation of the whole trajectory that object makes.

and if you solve this by making $s = 0$ , you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

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3 years ago