All categories

3 years ago

Two pounds of grapes cost 6$ what number belongs in the most right empty cell?

Mathematics

1 answer:

sdas [7]3 years ago

6 0

Answer:

Table shown below

Step-by-step explanation:

Table shown below

To solve this problem let's use proportions.

If 2 pounds of grapes cost $6, half the amount will cost half the dollars, so the last row will have $3 in the price

For the second row, we know the price is $1, that is, one-sixth of the original given price. It should correspond to one-sixth of the amount of grapes or 2/6 pounds.

Simplifying the fraction, we get 1/3 or 0.33 pounds

You might be interested in

Simplify the following equation: 5^2 -4/5+2

I am Lyosha [343]

Answer:

131/5

Step-by-step explanation:

5^2=25

25-4/5 +2

=131/5 or 26.2

7 0

2 years ago

Suppose the initial height of a Pumpkin is 12 feet and the pumpkin is being launched with a velocity of 61 feet per second. Use

iogann1982 [59]

<h2>Hello!</h2>

The answer is:

The maximum height before landing will be 69.7804 feet.

Since there is no information about the angle of the launch, we can safely assume that it's launched vertically.

So, we can calculate the maximum height of the pumpkin using the following formulas:

$y=y_o+v_{o}*t-\frac{1}{2}gt^{2}$

$v=vo-gt$

Where,

y, is the final height

$y_o$ , is the initial height

g, is the acceleration of gravity , and it's equal to:

$g=32.2\frac{ft}{s^{2} }$

t, is the time.

Now, we are given the following information:

$y_{o}=12ft\\\\v=61\frac{ft}{s}$

Then, to calculate the maximum height, we must remember that at the maximum height, the speed tends to 0, so, calculating we have:

Time calculation,

We need to use the following equation,

$v=vo-gt$

So, substituting we have:

$v=61\frac{ft}{s}-32.2\frac{ft}{s^{2}}*t\\\\-61\frac{ft}{s}=-32.2\frac{ft}{s^{2}}*t\\\\t=\frac{-61{ft}{s}}{-32.2\frac{ft}{s^{2}}}=1.8944s$

We know that it will take 1.8944 seconds to the pumpkin to reach its maximum height.

Maximum height calculation,

Now, calculating the maximum height, we need to use the following equation:

$y=y_o+v_{o}*t-\frac{1}{2}gt^{2}$

Substituting and calculating, we have:

$y=y_o+v_{o}*t-\frac{1}{2}gt^{2}$

$y=12ft+61\frac{ft}{s}*1.8944s-\frac{1}{2}32.2\frac{ft}{s^{2}}*(1.8944s)^{2}$

$y=12ft+61\frac{ft}{s}*1.8944s-\frac{1}{2}32.2\frac{ft}{s^{2}}*(1.8944s)^{2}\\\\y_{max}=12ft+115.5584ft-16.1\frac{ft}{s^{2}}*(3.5887s^{2})\\\\y_{max}=127.5584ft-57.7780ft=69.7804ft$