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2 years ago

A Mercedes can go 78 miles on 3 gallons of gas. How far can it go with a full tank of 16 gallons?

Mathematics

1 answer:

Alex787 [66]2 years ago

5 0

Answer:

416 miles

Step-by-step explanation:

78 divided by 3 = 26

26 x 16 = 416

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Whats 8 5/7 plus 5 13/14 ASAP

lozanna [386]

14.6428 in decimal , in fraction 14 16/25

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2 years ago

Rob notices that 55 percent of the people leaving the supermarket chhose plastic bags instead of paper bags. out of 600 people,

Bumek [7]

To solve this answer, you are trying to find 55% of 600, because that is the number of people who chose plastic bag.

55% can be re-written as 55/100 OR 0.55 since percentages are always out of 100.

So, we do 0.55*600, which gives us 330. 330 people carry plastic bags.

5 0

3 years ago

(factor out) 3x(to the power of 2) -39x+36

motikmotik

3x² - 39x + 36
3(x²) - 3(13x) + 3(12)
3(x² - 13x + 12)
3(x² - 12x - x + 12)
3(x(x) - x(12) - 1(x) + 1(12))
3(x(x - 12) - 1(x - 12))
3(x - 1)(x - 12)

7 0

2 years ago

You have 100 feet of fencing and decide to make a rectangular garden. What is the largest area enclosed

k0ka [10]

Answer:

625 ft^2

Step-by-step explanation:

Given

$P = 100$ --- perimeter

Required

The largest area

The perimeter is calculated as:

$P = 2 * ( L + W)$

So, we have:

$2 * ( L + W) = 100$

Divide both sides by 2

$L + W = 50$

Make L the subject

$L = 50 - W$

The area is calculated as:

$A= L * W$

Substitute $L = 50 - W$

$A= (50 - W) * W$

Open bracket

$A = 50W - W^2$

Differentiate with respect to W

$A' = 50 -2W$

Set to 0; to get the maximum value of W

$50 - 2W = 0$

Collect like terms

$-2W = -50$

Divide by -2

$W = 25$

So, the maximum area is:

$A = 50W - W^2$

$A = 50 * 25 - 25^2$

$A = 1250 - 625$

$A = 625$

3 0

2 years ago

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago