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3 years ago

13

I need help on everything

1 answer:

Arisa [49]3 years ago

6 0

I can't try to help you.

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How to solve an algebraic expression by putting values of unknown

andrey2020 [161]

Answer:

Step-by-step explanation:

Algebraic expressions tend to have atleast one unknown variable in it. In order to solve the expression we need to assign a fixed value to that variable or isolate it if the expression is equal to another value. For example, in the following expression we have the variable x, if we give it a value of 5 we simply solve like a regular expression...

5x + 3

5(5) + 3

25 + 3

28

Therefore, if we swap the variable x for the value of 5 we would get a value of 28 in this algebraic expression.

7 0

3 years ago

Your parents offer you 50%<br> interest every day if you<br> invest your 4 dollars with<br> them.

N76 [4]

Answer:

b

Step-by-step explanation:

6 0

3 years ago

The area of a square is 16 cm2. The square is the base of a cube.

choli [55]

the answer is 96 cm ^ 2

16 × 6 =96

3 0

3 years ago

In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is

zheka24 [161]

Answer:

$f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

Step-by-step explanation:

Let:

$M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y$

This is and exact equation, because:

$\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}$

So, define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)$

The solution will be given by:

$f(x,y)=C_1$

Where C1 is an arbitrary constant

Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y):

$f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)$

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}$

Substitute into $\frac{\partial f(x,y)}{\partial y} =N(x,y)$

$2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y$

Integrate $\frac{dg(y)}{dy}$ with respect to y:

$g(y)=\int\ {8y} \, dy =4y^2$

Substitute g(y) into f(x,y):

$f(x,y)=2x^2+4y^2+2xy$

The solution is f(x,y)=C1

$f(x,y)=2x^2+4y^2+2xy=C_1$

Solving y using quadratic formula:

$y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

4 0

4 years ago

Which piecewise defined function is shown on the graph?

egoroff_w [7]

You gotta show the pic of the graph

6 0

4 years ago