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3 years ago

(x³-3x²+2x+1/x²-4x+4)dx(dx/Dy)+(x²+4y-y²)/(²√4y+y²) solve for dx/dy

Mathematics

1 answer:

Furkat [3]3 years ago

4 0

It’s 22/3 I think so that’s the answer

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A fence must be built to enclose a rectangular area of 45 comma 000 ftsquared. Fencing material costs $ 3 per foot for the two s

Tatiana [17]

Answer:

150 feet by 300 feet.

Step-by-step explanation:

The fence is to enclose a rectangular area of 45,000 ft squared.

If the dimensions of the rectangle are x and y

Area of a rectangle = xy

xy=45000
$x=\frac{45000}{y}$

Perimeter of the Rectangle =2x+2y

Fencing material costs $ 3 per foot for the two sides facing north and south and $6 per foot for the other two sides.

Cost of Fencing, C=$(6*2x+3*2y)=$(12x+6y)

Substitute $x=\frac{45000}{y}$ into the Cost to get C(y)

C=12x+6y

$C(y)=12(\frac{45000}{y})+6y\\C(y)=\frac{540000+6y^2}{y}$

The value at which the cost is least expensive is at the minimum point of C(y), when the derivative is zero.

$C^{'}(y)=\dfrac{6y^2-540000}{y^2}$

$\dfrac{6y^2-540000}{y^2}=0\\6y^2-540000=0\\6y^2=540000\\y^2=\frac{540000}{6} =90000\\y=\sqrt{90000}=300$

Recall,

$x=\frac{45000}{y}=\frac{45000}{300}=150$

Since x=150, y=300

The dimensions that will be least expensive to build is 150 feet by 300 feet.

6 0

2 years ago

What is 7/p=8/9? Solve for p

irina [24]

Answer:

Exact form, p = 63/8

decimal form, p = 7.875

Mixed number form, p = 7 7/8

Step-by-step explanation:

Solve for p by cross multiplying

7 0

2 years ago

Read 2 more answers

How do you calculate the LCM for 128 and 256

Neporo4naja [7]

Answer:

I think its 256

Step-by-step explanation:

The LCM of 128 and 256 is 256. Steps to find LCM. Find the prime factorization of 128 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2; Find the prime factorization of 256.

6 0

1 year ago

Read 2 more answers

The point P(7, −2) lies on the curve y = 2/(6 − x). (a) If Q is the point (x, 2/(6 − x)), use your calculator to find the slope

NARA [144]

Answer:

a) (i) $m = 2.22$ , (ii) $m = 2$ , (iii) $m = 2$ , (iv) $m = 2$ , (v) $m = 1.82$ , (vi) $m = 2$ , (vii) $m = 2$ , (viii) $m = 2$ ; b) $m \approx 2$ ; c) The equation of the tangent line to curve at P (7, -2) is $y = 2\cdot x + 12$ .

Step-by-step explanation:

a) The slope of the secant line PQ is represented by the following definition of slope:

$m = \frac{\Delta y}{\Delta x} = \frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}$

(i) $x_{Q} = 6.9$ :

$y_{Q} =\frac{2}{6-6.9}$

$y_{Q} = -2.222$

$m = \frac{-2.222 + 2}{6.9-7}$

$m = 2.22$

(ii) $x_{Q} = 6.99$

$y_{Q} =\frac{2}{6-6.99}$

$y_{Q} = -2.020$

$m = \frac{-2.020 + 2}{6.99-7}$

$m = 2$

(iii) $x_{Q} = 6.999$

$y_{Q} =\frac{2}{6-6.999}$

$y_{Q} = -2.002$

$m = \frac{-2.002 + 2}{6.999-7}$

$m = 2$

(iv) $x_{Q} = 6.9999$

$y_{Q} =\frac{2}{6-6.9999}$

$y_{Q} = -2.0002$

$m = \frac{-2.0002 + 2}{6.9999-7}$

$m = 2$

(v) $x_{Q} = 7.1$

$y_{Q} =\frac{2}{6-7.1}$

$y_{Q} = -1.818$

$m = \frac{-1.818 + 2}{7.1-7}$

$m = 1.82$

(vi) $x_{Q} = 7.01$

$y_{Q} =\frac{2}{6-7.01}$

$y_{Q} = -1.980$

$m = \frac{-1.980 + 2}{7.01-7}$

$m = 2$

(vii) $x_{Q} = 7.001$

$y_{Q} =\frac{2}{6-7.001}$

$y_{Q} = -1.998$

$m = \frac{-1.998 + 2}{7.001-7}$

$m = 2$

(viii) $x_{Q} = 7.0001$

$y_{Q} =\frac{2}{6-7.0001}$

$y_{Q} = -1.9998$

$m = \frac{-1.9998 + 2}{7.0001-7}$

$m = 2$

b) The slope at P (7,-2) can be estimated by using the following average:

$m \approx \frac{f(6.9999)+f(7.0001)}{2}$

$m \approx \frac{2+2}{2}$

$m \approx 2$

The slope of the tangent line to the curve at P(7, -2) is 2.

c) The equation of the tangent line is a first-order polynomial with the following characteristics:

$y = m\cdot x + b$

Where:

$x$ - Independent variable.

$y$ - Depedent variable.

$m$ - Slope.

$b$ - x-Intercept.

The slope was found in point (b) (m = 2). Besides, the point of tangency (7,-2) is known and value of x-Intercept can be obtained after clearing the respective variable:

$-2 = 2 \cdot 7 + b$

$b = -2 + 14$

$b = 12$

The equation of the tangent line to curve at P (7, -2) is $y = 2\cdot x + 12$ .

7 0

2 years ago

A NASA spacecraft measures the rate R of at which atmospheric pressure on Mars decreases with altitude. The result at a certain

AysviL [449]

Answer:

$R = 0.0000498 \frac{KJ}{ m m^3}=0.0000498 \frac{KJ}{m^4}=4.98x10^{-5}\frac{KJ}{m^4}$

Step-by-step explanation:

For this case we have the following value:

$R = 0.0498 \frac{Kpa}{Km}$

We can convert this first to $\frac{Kpa}{m}$ like this:

$R=0.0498 \frac{Kpa}{Km} *\frac{1km}{1000m}=0.0000498 \frac{Kpa}{m}$

Now we use the fact the the pressure is defined as $P =\frac{F}{A}$ , whre P is the pressure, F the force and A the area, so then $Kpa= \frac{KN}{m^2}$ and then we can replace this:

$R=0.0000498 \frac{KN}{m^3}$

Now from definition of work we know that $W= Fd$ where W is the work, F the force and d the distance, so then is equivalent $KN =\frac{KJ}{m}$

And if we replace this into the equation we got:

$R = 0.0000498 \frac{KJ}{ m m^3}=0.0000498 \frac{KJ}{m^4}=4.98x10^{-5}\frac{KJ}{m^4}$

4 0

2 years ago

(x³-3x²+2x+1/x²-4x+4)dx(dx/Dy)+(x²+4y-y²)/(²√4y+y²) solve for dx/dy​

(x³-3x²+2x+1/x²-4x+4)dx(dx/Dy)+(x²+4y-y²)/(²√4y+y²) solve for dx/dy