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2 years ago

Please help me ASAP im stressing a lot right now

Mathematics

1 answer:

Alexeev081 [22]2 years ago

4 0

Answer: 1/3 or 0.333333 etc.

Step-by-step explanation:

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What is the answer = 8w- 8 - 6w = 4w - 7<br> w= ??

ololo11 [35]

Answer:

Step-by-step explanation:

so first, simplify by combining like terms.

so 8w-6w = 2w

so now we have

2w - 8 = 4w - 7

now we have to get the w's to one side of the equation so we can solve for it. so that mean we have to subtract 2w from both sides. 2w - 2w is 0. and then 4w - 2w = 2w. so now we have

-8 = 2w - 7

then to slowly start isolating w, add 7 to both sides of th equation. so now its

-1 = 2w

finally to get w by itself divide 2 from both sides.

and the answer is

- $\frac{1}{2}$ = w

5 0

2 years ago

The difference between 45 and y is 9

faltersainse [42]

Answer:

y = 36

Step-by-step explanation:

8 0

2 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$