Shifting a graph vertically and horizontally is only possible for certain types of functions.

The art teacher has 7 bottles of glue that are 2/5 pull he combines them so he will have fewer bottles how many bottles of glue

vlabodo [156]

Hello!

You have to do 7 * 2/5

Which is the same as 7/1 * 2/5

Which gives us the answer 2 4/5 bottles of glue.

Hope this helps!

3 0

3 years ago

A manufacturing company has a policy that states managers are paid a salary that is proportional to the number of employees they

user100 [1]

From the problem description, we can construct an algebraic equation for the situation. Let S be the salary and x be the number of employees.

S = kx
where k is the proportionality constant

Using variation,

k = S₁/x₁ = S₂/x₂

Substituting,

34000/12 ? 36000/16
2833.33 ? 2250
Thus,
2833.33 ≠ 2250
<em>It does not follow the company's policy.</em>

7 0

3 years ago

Find general solutions of the differential equation. Primes denote derivatives with respect toÂ x.

zepelin [54]

Answer:

$\mathbf{3x^2y^3+2xy^4=C}$

Step-by-step explanation:

From the differential equation given:

$6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0$

The equation above can be re-written as:

$6xy^3 +2y^4 +(9x^2y^2+8xy^3)\dfrac{dy}{dx}=0$

$(6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0$

Let assume that if function M(x,y) and N(x,y) are continuous and have continuous first-order partial derivatives.

Then;

M(x,y) dx + N (x,y)dy = 0; this is exact in R if and only if:

$\dfrac{{\partial M }}{{\partial y }}= \dfrac{\partial N}{\partial x}}} \ \ \text{at each point of R}$

relating with equation M(x,y)dx + N(x,y) dy = 0

Then;

$M(x,y) = 6xy^3 +2y^4\ and \ N(x,y) = 9x^2 y^2 +8xy^3$

So;

$\dfrac{\partial M}{\partial y }= 18xy^3 +8y^3$

$\dfrac{\partial N}{\partial y }$

Let's Integrate $\dfrac{\partial F}{\partial x}= M(x,y)$ with respect to x

Then;

$F(x,y) = \int (6xy^3 +2y^4) \ dx$

$F(x,y) = 3x^2 y^3 +2xy^4 +g(y)$

Now, we will have to differentiate the above equation with respect to y and set $\dfrac{\partial F}{\partial x}= N(x,y)$ ; we have:

$\dfrac{\partial F}{\partial y} = \dfrac{\partial}{\partial y } (3x^2y^3+2xy^4+g(y)) \\ \\ = 9x^2y^2 +8xy^3 +g'(y) \\ \\ 9x^2y^2 +8xy^3 +g'(y) =9x^2y^2 +8xy^3 \\ \\ g'(y) = 0 \\ \\ g(y) = C_1$

Hence, $F(x,y) = 3x^2y^3 +2xy^4 +g(y) \\ \\ F(x,y) = 3x^2y^3 + 2xy^4 +C_1$

Finally; the general solution to the equation is:

$\mathbf{3x^2y^3+2xy^4=C}$

5 0

2 years ago

Help please if it’s wrong is okay ig no links

Mkey [24]

1 2 3 4 & 5 would be the answers!

6 0

3 years ago

Please help me Simplify 5+(z+2) <br><br> Maybe try to simplify

dolphi86 [110]

Answer:

7+z

Step-by-step explanation:

you can open the parentheses right away because it is addition. so you are left with 5+z+2. you can add both 5 and 2 together, so the final result will be 7+z

6 0

3 years ago