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3 years ago

Given brainliest answer

Mathematics

2 answers:

Stels [109]3 years ago

3 0

Answer:

Domain all real numbers Range (3,6,12,24,48)

Step-by-step explanation:

The domain is all real numbers so its either the first one or the second one. The range is multiplyed by 2 so the range is 3,6,12,24,48

lidiya [134]3 years ago

3 0

i guess it will be (C) hopefully i helped

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3 years ago

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What’s the quotient of 5 divided into 68

marishachu [46]

Answer:

13.6

Step-by-step explanation:

5 divided into 68 is basically the same thing as 68 divided by 5.

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3 years ago

Given the function ƒ(x ) = 3x + 1, evaluate ƒ(a + 1).

Setler79 [48]

Answer:

$\Huge \boxed{3a+4}$

Step-by-step explanation:

The function is given.

$f(x)=3x+1$

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3 years ago

Which expression uses the greatest common factor and distributive property to find 10+50?

Rudiy27

Ab+ac=a(b+c) where a is the greatest common factor
find greatest common factor of 10 and 50
10=1,2,5,10
50=1,2,5,10,25,50
greatest common is 10
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3 0

3 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

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

7 0

2 years ago