2 years ago

Solve the equation 3+2/5x=11-2/5x using at least two different sequences of steps. Show your work for each step.

1 answer:

4 0

First combine like terms ,
5/5x=9/5x
move the variable which will cancel out because it is the same
5=9 is no solution sorry if i’m wrong lol

You might be interested in

dalvyx [7]

Multiply straight across and then just simply your answer. I divided by 2.

5 0

2 years ago

slavikrds [6]

Answer:

5n-t/7

Step-by-step explanation:

The product of 5 and n puts them in the same category and product means times and your going to minus t divided by 7

8 0

3 years ago

Dima020 [189]

Split $C$ into two component segments, $C_1$ and $C_2$ , parameterized by

$\mathbf r_1(t)=(1-t)(0,0)+t(6,1)=(6t,t)$

$\mathbf r_2(t)=(1-t)(6,1)+t(7,0)=(6+t,1-t)$

respectively, with $0\le t\le1$ , where $\mathbf r_i(t)=(x(t),y(t))$ .

We have

$\mathrm d\mathbf r_1=(6,1)\,\mathrm dt$

$\mathrm d\mathbf r_2=(1,-1)\,\mathrm dt$

where $\mathrm d\mathbf r_i=\left(\dfrac{\mathrm dx}{\mathrm dt},\dfrac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt$

so the line integral becomes

$\displaystyle\int_C(x+6y)\,\mathrm dx+x^2\,\mathrm dy=\left\{\int_{C_1}+\int_{C_2}\right\}(x+6y,x^2)\cdot(\mathrm dx,\mathrm dy)$

$=\displaystyle\int_0^1(6t+6t,(6t)^2)\cdot(6,1)\,\mathrm dt+\int_0^1((6+t)+6(1-t),(6+t)^2)\cdot(1,-1)\,\mathrm dt$

$=\displaystyle\int_0^1(35t^2+55t-24)\,\mathrm dt=\frac{91}6$

6 0

2 years ago

julsineya [31]

The answer to this problem is 12. If you want me to write it out I can

8 0

3 years ago

Serjik [45]

The inverse x exists

7 0

3 years ago

Solve the equation 3+2/5x=11-2/5x using at least two different sequences of steps. Show your work for each step.​