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Ksenya-84 [330]

2 years ago

9

Can you pls help me with this question?

1 answer:

8090 [49]2 years ago

4 0

Answer:

heyyyyy so basically the answers would be 8 and 5 on the first and 2nd one Im thinking ...

Step-by-step explanation:

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I had fell asleep but this for the nikkas det wanted to see what i look like <3

exis [7]

Answer:

yes

Step-by-step explanation:

5 0

3 years ago

Given f(x)=-3x+3solve for x when f(x)=6

natali 33 [55]

Answer:

x = -1

Step-by-step explanation:

Given: $f(x) = - 3x + 3$

Solve for x, When: $f(x) = 6$

Step by step:

$- 3x + 3 = 6$

$- 3x = 6 - 3$

$- 3x = 3$

$x = 3 \div - 3$

$\boxed{\green{x = - 1}}$

3 0

2 years ago

How are the real solutions of a quadratic equation related to the graph of the quadratic function?

Anika [276]

The solutions you get when you solve the formula are the corresponding y coordinates to your x value. So say a point on your graph is (2,3). The first number is x and the second is y. (x,y). The number you plug into your function is x,or in this case: 2. The solution to the equation when the x value is plugged in is y, or 3. Therefore, giving you a point on your graph.

7 0

3 years ago

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

3 years ago

There are 10 chocolates in a box.

marusya05 [52]

What’s the question?

6 0

3 years ago