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3 years ago

F(x) = -4x – 6 g(x) = x - 3 find f(g(3))

Mathematics

1 answer:

qaws [65]3 years ago

6 0

Answer:

F (X-3) =4X-18

Step-by-step explanation:

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What is the solution to the system shown?

Mice21 [21]

Answer:d

Step-by-step explanation:

16-4=12 and 16+4=20

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2 years ago

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iris [78.8K]

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2 years ago

Please help me !!!!!!

cestrela7 [59]

Given:

The set of pair and graphs.

To find:

The domain and range.

Solution:

We know that,

Domain is the set of x-values or input values.

Range is the set of y-values or output values.

(a)

The given set of ordered pairs is

{(-3,3),(5,5),(-3,2),(5,3)}

Here, the x-coordinates are -3, 5, -3, 5.

A set contains distinct values.

Therefore, the domain is {-3,5}.

(b)

The graph is given.

From the given graph the set of ordered pairs is

{(-2,1),(-1,0.5),(-1,3),(0,0),(0,2),(1,0.5),(1,3)(2,1)}

Here, the y-values are 1, 0.5, 3, 0, 2, 0.5, 3, 1.

Therefore, the range is {0, 0.5, 1, 2, 3}.

(c)

The graph is given.

From the given graph the set of ordered pairs is

{(-2,3),(-1,3),(0,1),(2,4)}

Here, the x-values are -2, -1,0, 2.

Therefore, the domain is {-2,-1,0,2}.

4 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

What is the length of AC?

BartSMP [9]

Your answer would be 18. AC must be longer than 9

5 0

3 years ago