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3 years ago

There are 3,785 milliliters in 1 gallon, and there are 4 quarts in 1 gallon. For each question, explain or show reasoning.

Mathematics

2 answers:

irakobra [83]3 years ago

5 0

11356.235 946.353 is your answer

Novay_Z [31]3 years ago

4 0

Answer:

There are 11,355 milliliters in 3 gallons, and 946.25 milliliters in 1 quart.

Step-by-step explanation:

If there are 3,785 milliliters in 1 gallon, multiply each side by 3 because you want the product to equal 3 gallons.3,785 times 3 is equal to 11,355, and 1 times 3 is 3. So, 3 gallons are equal to 11,355 milliliters. If there are 3,785 milliliters in 1 gallon, and there are 4 quarts in 1 gallon, that means that there are 3,785 milliliters in 4 quarts. Now, you want to find the amount of milliliters in 1 quart so you should divide both the 3,785 and the 4 by 4. So, there are 946.25 milliliters in 1 quart.

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Joseph received a $20 gift card for downloading music. Each downloaded song costs $1.29. Explain how to write and solve an inequ

beks73 [17]

Answer:

Step-by-step explanation:

Lets take the number of songs he download = x

So $1.29x = $20

So "x" = 20/4.29 = 15.5

So songs he can purchase = 15 songs

4 0

3 years ago

Read 2 more answers

Which of the following numbers are greater than or equal to 4/7? Is it 2/5, 0.57, or 2/3. Use explanation why.

Sergeeva-Olga [200]

Answer:

2/3

Step-by-step explanation:

2/5=0.4 and 4/7=0.571428

0.57<0.571428

---------------------------

2/3=0.6667

which is greater than or equal to 4/7

5 0

3 years ago

You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.