-2+12-2^3 divided by 2^0 times 3

Mathematics

1 answer:

nikklg [1K]2 years ago

6 0

Answer:

The correct answer is -14

Step-by-step explanation:

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If f(x)=4x+3 and g(x)=x^2-3 then f(g(2))

grandymaker [24]

Answer:

f(g(2))=7

Step-by-step explanation:

The Composite Function

Given f(x) and g(x) as real functions, the composite function $(f\circ g)(x)$ is defined as:

$(f\circ g)(x)=f(g(x))$

It can be found by substituting g into f.

The given functions are:

$f(x)=4x+3$

$g(x)=x^2-3$

It's required to find f( g( 2 ) )

Find g(2):

$g(2)=2^2-3$

Operating:

$g(2)=4-3 =1$

g(2)=1

Now find f(1):

$f(1)=4(1)+3$

$f(1)=4+3 =7$

Thus:

f( g( 2 ) )=7

4 0

3 years ago

Evaluate the iterated integral 2 0 2 x sin(y2) dy dx. SOLUTION If we try to evaluate the integral as it stands, we are faced wit

nignag [31]

Answer:

Step-by-step explanation:

Given that:

$\int^2_0 \int^2_x \ sin (y^2) \ dy dx \\ \\ \text{Using backward equation; we have:} \\ \\ \int^2_0\int^2_0 sin(y^2) \ dy \ dx = \int \int_o \ sin(y^2) \ dA \\ \\ where; \\ \\ D= \Big\{ (x,y) | }0 \le x \le 2, x \le y \le 2 \Big\}$

$\text{Sketching this region; the alternative description of D is:} \\ D= \Big\{ (x,y) | }0 \le y \le 2, 0 \le x \le y \Big\}$

$\text{Now, above equation gives room for double integral in reverse order;}$

$\int^2_0 \int^2_0 \ sin (y^2) dy dx = \int \int _o \ sin (y^2) \ dA \\ \\ = \int^2_o \int^y_o \ sin (y^2) \ dx \ dy \\ \\ = \int^2_o \Big [x sin (y^2) \Big] ^{x=y}_{x=o} \ dy \\ \\= \int^2_0 ( y -0) \ sin (y^2) \ dy \\ \\ = \int^2_0 y \ sin (y^2) \ dy \\ \\ y^2 = U \\ \\ 2y \ dy = du \\ \\ = \dfrac{1}{2} \int ^2 _ 0 \ sin (U) \ du \\ \\ = - \dfrac{1}{2} \Big [cos \ U \Big]^2_o \\ \\ = - \dfrac{1}{2} \Big [cos \ (y^2) \Big]^2_o \\ \\ = - \dfrac{1}{2} cos (4) + \dfrac{1}{2} cos (0) \\ \\$