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3 years ago

Lisa paid $47.82 for 16.9 gallons of gasoline. What was the cost per gallon, rounded to the nearest hundredth?

Mathematics

1 answer:

agasfer [191]3 years ago

3 0

Divide total amount spent by total gallons bought:

47.82 / 16.9 = $2.83 per gallon

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What is −2x3−6x+4 divided by x+1?

stepladder [879]

Answer:

-( 2x^2+6-4/x)

that what i got when you divide

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2 years ago

Dave loves underwater diving. His house boat is 250 feet above sea level. He jumped from that level. He went 600 feet below sea

nikklg [1K]

Answer:

Step-by-step explanation:

<u>I would just like to explain how this works with a visual (please enjoy my drawing <3 haha)</u>

With my drawing we know that the bird is 150 feet above sea level, and the fish is 100 feet below sea level.

From the bird to the water it is 150 feet, from water to fish is 100 feet

<h3>This is equal to 250 feet, </h3>

so we know from the bird (150 feet above sea level) to the fish(100 feet below sea level).

<h2>Knowing this information we can easily answer this question</h2><h3>Dave is 250 feet above sea level, and he went 600 feet nelow sea level, meaning<u> he only went 600 feet under the water.</u></h3>

Download pdf

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2 years ago

PLSSSSSSSSSSSSSSSSSSSSSSSSSS ANSWER THIS

postnew [5]

Step-by-step explanation:

$(a.y^{3} +3y^{2} -3 ) = (ay^{2} +(4a+3)y +4(4a+3))(y-4) +64a+45$

so A = 64a +45

( $2y^{3} -5y+a =(2y^{2} +8y +27)(y-4) +108 +a$

so B = 108 +a

2A-B=0--> 2(64a+45)-108-a=0--> a =18/127

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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.

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3 years ago

What is the equation for the line in slope intercept form PLEASE HELP

aliina [53]

Answer:

y=-4x+5

Step-by-step explanation:

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2 years ago