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2 years ago

7

Three cans of Additive are regularly priced at 5 for $2.95. If you buy them at a special sale, you get a factory rebate of $1. W

hat is the net cost per can the sale price?

1 answer:

slamgirl [31]2 years ago

3 0

The regular price is 5 for $2.95. One can is $2.95/5 = $0.59. Three cans are $1.77. With the $1 rebate, the final price would be $0.77 for 3 cans, $0.25 per can.

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Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

so

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

8 0

2 years ago

2. Solve the system of equations. Write your answer in parametric vector notation. X, +5x2 + 4x3 + 3x4 +9xs = 18 2x, +6x2 + 4x3

ratelena [41]

Answer:

The system is inconsistent.

Step-by-step explanation:

The system is

$x_1+5x_2+4x_3+3x_4+9x_5=18\\2x_1+6x_2+4x_3+6x_4+6x_5=28\\3x_1+7x_2+4x_3+10x_4+x_5=41\\4x_1+6x_2+2x_3+7x_4+4x_5=29$

The associated matrix to the system is

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\2&6&4&6&6&28\\3&7&4&10&1&41\\4&6&2&7&4&29\end{array}\right]$

Now we use row operations to find the echelon form of the matrix:

1. We substract to row 2, two times the row 1.

We substract to row 3, three times the row 1.

We substract to row 4, four times the row 1 and obtain the matrix

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&-4&-4&0&-12&-8\\0&-8&-8&1&-26&-13\\0&-14&-14&-5&-32&-43\end{array}\right]$

2. We multiply the second row of the preview step by -1/4. We obtain the matrix

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&-8&-8&1&-26&-13\\0&-14&-14&-5&-32&-43\end{array}\right]$

3.

We add to row 3, eight times the row 2.

We add to row 4, fourtheen times the row 2 and obtain the matrix

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&0&0&1&-2&3\\0&0&0&-5&10&-155\end{array}\right]$

4. We add to row 4 of the preview step, five times the row 3 and obtain the matrix

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&0&0&1&-2&3\\0&0&0&0&0&-140\end{array}\right]$

Using backward substitution we have that

$0x_5=-140$ , then $0=-140$ and this is absurd. Then The system is inconsistent.

4 0

3 years ago

X>0 and 2x^2+3x-2=0 what is the value of x

vladimir2022 [97]

Step-by-step explanation:

2x²+3x-2=0

(2x-1)(x+2)

x+2=0

x=-2 (no, because x must be >0)

2x-1=0

2x=1

x=½ (yes, because x>0)

3 0

2 years ago

At the circus, a clown is shot from a cannon. This situation can be modeled by the function: h = – 16t2 + 45t + 15, where 'h' is

weeeeeb [17]

Answer:

The maximum height is 46.64 feet.

Step-by-step explanation:

If we take the derivative of h whit respect to t and equal this to zero we would find the value of t which corresponds to the maximum h.

So, we have the function h(t):

$h(t)=-16t^{2} + 45t + 15$

Taking the derivative, we have:

$\frac{dh(t)}{dt}=-32t + 45=0$

Now, we solve it for t:

$t=\frac{45}{32}=1.4\: s$

Finally, we put this value of t into the original equation.

$h(t)=-16(1.4)^{2} + 45(1.4) + 15$

$h_{max}=46.64\: ft$

Therefore, the maximum height is 46.64 feet. All the given options are wrong, the one that comes closest is option A.

I hope it helps you!

4 0

2 years ago

Can you help me? Please

Anni [7]

9514 1404 393

Answer:

a.

Step-by-step explanation:

For an even-index radical, ...

$\sqrt[n]{x^n}=|x|\qquad\text{n even}$

This lets us simplify the given expression as follows:

$\sqrt[4]{81x^6y^4}-|y|\sqrt[4]{x^6}-\sqrt[4]{16x^2}=\sqrt[4]{(3xy)^4x^2}-|y|\sqrt[4]{x^4x^2}-\sqrt[4]{2^4x^2}\\\\=3|xy|\sqrt[4]{x^2}-|xy|\sqrt[4]{x^2}-2\sqrt[4]{x^2}=\boxed{2|xy|\sqrt[4]{x^2}-2\sqrt[4]{x^2}}$

4 0

2 years ago