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natita [175]
3 years ago
9

A water tank holds 486 gallons but leaking at a rate 4 gallons per week. A second water tank holds 648 gallons but is leaking at

a rate of 7 gallons per week. After how many weeks will the amount of water in the 2 tanks? The amount of water that holds the two tanks will be the same in _ weeks.
Mathematics
1 answer:
notka56 [123]3 years ago
6 0

The first tank contains

486 gal - (4 gal/week) <em>t</em>

gal of water after <em>t</em> weeks.

Similarly, the second tank contains

648 gal - (7 gal/week) <em>t</em>

gal of water.

The tanks hold the same amount of water when these two amounts are equal:

486 gal - (4 gal/week) <em>t</em> = 648 gal - (7 gal/week) <em>t</em>

(7 gal/week) <em>t</em> - (4 gal/week) <em>t</em> = 648 gal - 486 gal

(3 gal/week) <em>t</em> = 162 gal

<em>t</em> = (162 gal) / (3 gal/week)

<em>t</em> = 54 weeks

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Leon needs to save more than $350 to buy a new bike. He has$130 saved so far, and he plans to save $20 each week until he has en
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Read 2 more answers
(1 point) (a) Find the point Q that is a distance 0.1 from the point P=(6,6) in the direction of v=⟨−1,1⟩. Give five decimal pla
natima [27]

Answer:

following are the solution to the given points:

Step-by-step explanation:

In point a:

\vec{v} = -\vec{1 i} +\vec{1j}\\\\|\vec{v}| = \sqrt{-1^2+1^2}

    =\sqrt{1+1}\\\\=\sqrt{2}

calculating unit vector:

\frac{\vec{v}}{|\vec{v}|} = \frac{-1i+1j}{\sqrt{2}}

the point Q is at a distance h from P(6,6) Here, h=0.1  

a=-6+O.1 \times \frac{-1}{\sqrt{2}}\\\\= 5.92928 \\\\b= 6+O.1 \times \frac{-1}{\sqrt{2}} \\\\= 6.07071

the value of Q= (5.92928 ,6.07071  )

In point b:

Calculating the directional derivative of f (x, y) = \sqrt{x+3y} at P in the direction of \vec{v}

f_{PQ} (P) =\fracx{f(Q)-f(P)}{h}\\\\

            =\frac{f(5.92928 ,6.07071)-f(6,6)}{0.1}\\\\=\frac{\sqrt{(5.92928+ 3 \times 6.07071)}-\sqrt{(6+ 3\times 6)}}{0.1}\\\\= \frac{0.197651557}{0.1}\\\\= 1.97651557

\vec{v} = 1.97651557

In point C:

Computing the directional derivative using the partial derivatives of f.

f_x(x,y)= \frac{1}{2 \sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{2 \sqrt{22}}\\\\f_x(x,y)= \frac{1}{\sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{\sqrt{22}}\\\\f_{(PQ)}(P)= (f_x \vec{i} + f_y \vec{j}) \cdot \frac{\vec{v}}{|\vec{v}|}\\\\= (\frac{1}{2 \sqrt{22}}\vec{i} + \frac{1}{\sqrt{22}} \vec{j}) \cdot   \frac{-1}{\sqrt{2}}\vec{i} + \frac{1}{\sqrt{2}} \vec{j}

4 0
3 years ago
The points (6, -3) and (7, -10) fall on a particular line. What is its equation in point-slope form? Use one of the specified po
kumpel [21]

Answer:

Step-by-step explanation:

Given the coordinate points (6, -3) and (7, -10), we are to find the equation of a line passing through this two points;

The standard equation of a line is y = mx+c

m is the slope

c is the intercept

Get the slope;

m = Δy/Δx = y2-y1/x2-x1

m = -10-(-3)/7-6

m = -10+3/1

m = -7

Get the intercept;

Substitute the point (6, -3) and m = -7 into the expression y = mx+c

-3 = -7(6)+c

-3 = -42 + c

c = -3 + 42

c = 39

Get the required equation by substituting m = -7 and c= 39 into the equation y = mx+c

y = -7x + 39

Hence the required equation is y = -7x + 39

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Help please! find the value of y​
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Answer:

Step-by-step explanation:

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