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seropon [69]
3 years ago
13

The function h ( x ) = 1 x − 1 h ( x ) = 1 x - 1 can be expressed in the form f ( g ( x ) ) f ( g ( x ) ) , where g ( x ) = ( x

− 1 ) g ( x ) = ( x - 1 ) , and f ( x ) f ( x ) is defined as:
Mathematics
1 answer:
Radda [10]3 years ago
4 0

Answer: f(x) = 1^(x + 1)

Step-by-step explanation:

we have that h(x) = 1^x

and h(x) = f(g(x))

This mean that we are evaluating the function f(y) in the point y = g(x)

where g(x) = x - 1

then:

f(g(x) = f(x - 1) = h(x) = 1^x

then we should have that:

f(x) = 1^(x + 1)

then:

f(x - 1) = 1^(x - 1 + 1) = 1^x

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3 years ago
Is (-3,4) a solution of the inequality y> - 2x – 3?
jeyben [28]

Answer:

  (-3, 4) is a solution

Step-by-step explanation:

The point (-3, 4) is inside the shaded area of the graph, so is a solution.

You can check in the inequality

  y > -2x -3

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8 0
3 years ago
Use the given transformation to evaluate the integral.
GuDViN [60]

For the transformation

\begin{cases}x=7u+v\\y=u+7v\end{cases}

the Jacobian is

\dfrac{\partial(x,y)}{\partial(u,v)}=\begin{bmatrix}7&1\\1&7\end{bmatrix}

with determinant

\det\left(\begin{bmatrix}7&1\\1&7\end{bmatrix}\right)=48

The vertices of the triangle in the u,v-plane are

(x,y)=(0,0)\implies(u,v)=(0,0)

(x,y)=(7,1)\implies(u,v)=(1,0)

(x,y)=(1,7)\implies(u,v)=(0,1)

Then the integral is

\displaystyle\iint_R(x-8y)\,\mathrm dA=-48\int_0^1\int_0^{1-v}(u+55v)\,\mathrm du\,\mathrm dv=\boxed{-448}

3 0
3 years ago
Given point P (3, 4). What is the distance of point P from (a) x axis (b) y axis?
vova2212 [387]

Answer:

a) 3

b) 4

Step-by-step explanation:

The point P is(3,4)

which means that 3 is the point P's x- coordinate and 4 is it's y-coordinate

Because we use the axes as our reference point when plotting points,

the distance of P from x-axis and y-axis would be the point itself.

6 0
3 years ago
Please help me please
Sergeu [11.5K]

Answer:

0.09,70%,5/8, 0.8

7 0
2 years ago
Read 2 more answers
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