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3 years ago

Im giving free p0ints. thanks for helping everyone

Mathematics

1 answer:

kondaur [170]3 years ago

7 0

Answer:

really?! thank you so much! enjoy your day :)

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Javier invests money in two accounts. He invests twice as much in an account paying 11% as he does in an account paying 3%. If h

77julia77 [94]

PRT+PRT=Total Interest

0.03x+0.11(2x)=104.5

0.25x=104.5
x=418

x+2x, 418+2(418)=1254

Total Investment=1254 dollars

7 0

3 years ago

max said that 36594 is less than 5980 because 3 is less then 5 describe max's error and give the correct answer

Verizon [17]

He was wrong because, although three is greater than five, the value of the three is in the ten thousands place, while the five is in the thousands place. Therefore, the correct answer would be that 36594 is greater than 5980.

4 0

3 years ago

Simplify The Following: -4h + 22 + 6h -7

Vikentia [17]

Answer:

2h + 15

Step-by-step explanation:

-4h + 22 + 6h -7

2h + 22 - 7

2h + 15

3 0

3 years ago

The percent of area associated with z = 1.4 is ____%

Solnce55 [7]

The answer is A. 38.49

7 0

3 years ago

Read 2 more answers

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

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

7 0

2 years ago