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2 years ago

Which is the better unit to use to measure the capacity of a bathtub. milliliters or liters?

Mathematics

1 answer:

Lana71 [14]2 years ago

3 0

Liters is the best. A bathtub can hold a lot of water.

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What is the quotient? <br> X+5/3x^2 + 4x+5

nasty-shy [4]

Answer:

3x - 11

Step-by-step explanation:

x + 5/ 3x-11

3x - 11 is the quotient

5 0

1 year ago

Women's heights are normally distributed with a mean given by p = 63.6 in. and a standard deviation given by o = 2.5 in. (a) If

Ganezh [65]

Step-by-step explanation:

I am sorry question samajh Nahin a Raha question dijiye

4 0

2 years ago

Find the value of the function f (x) = -3x + 17 for x = 5.

ss7ja [257]

Answer: 17/3

Step-by-step explanation:

7 0

2 years ago

Use the graph below to fill in the blank with the correct number:

Alik [6]

Let
$A(-3,0) \\ B(-2,2) \\ C(0,1) \\ D(1,-2)$

we know that

Point A
$A(-3,0) \\ for x=-3 \\ f(-3)=0$

Point B
$B(-2,2) \\ for x=-2 \\ f(-2)=2$

Point C
$C(0,1) \\ for x=0 \\ f(0)=1$

Point D
$D(1,-2) \\ for x=1 \\ f(1)=-2$

therefore

the answer is
$f(0)=1$

4 0

2 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