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allochka39001 [22]
3 years ago
7

Which of the following is the value of a when the function f(x)=3|xl is written in the standard form of an absolute value

Mathematics
1 answer:
Monica [59]3 years ago
7 0
F(x)=|x| is the simplified version of an absolute function such as the parent function
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I'm stuck on this one:( help me please ​
ZanzabumX [31]

Answer:

Step-by-step explanation:

f-1(x) is the inverse of f(x)

to find the inverse let’s set the equation up like this

y = 4x -2

now we’ll switch x and y

x = 4y -2

and solve for y

x + 2 = 4y

x/4 + 2/4 = y

the inverse is

y = x/4 + 1/2

this equation is written in y = mx + b form

where b = y intercept

so the y intercept is 1/2

6 0
3 years ago
Graph the functions and approximate an x-value in which the exponential function exceeds the polynomial function. y = 4x y = 7x2
nordsb [41]

Answer:

The Exponential function is

y=4^x

And the polynomial function is

y=7 x^2 +4 x -2

And, we have to find the value of x for which, exponential function exceeds the polynomial function which can be written as

4^x> 7 x^2 +4 x -2

1. When , x= -1

LHS

4^{-1}=\frac{1}{4}=0.25

RHS

=7*(-1)^2 +4 *(-1)-2\\\\=7-4-2=1

2. When , x=0

L HS

 4^0=1

RHS

7*0+4*0-2= -2

3. When ,x= 0.5

L HS

 4^{0.5}=2

RHS

=7*0.25 +4*0.5 -2\\\\= 1.75+2-2\\\\=1.75

4. When , x=2

LHS

 4^{2}=16

RHS

=7*4^2+4*4-2\\\\=112+16-2\\\\=126

The Minimum value for which  exponential function exceeds the polynomial function is , x= 0.5

But,there is other value for which  exponential function exceeds the polynomial function is , x=2.

5 0
3 years ago
Find the inverse function of h(x)=-2/3x+6
storchak [24]

Answer:

-3/2 (x - 6)

Step-by-step explanation:

7 0
3 years ago
The fuel efficiency (mpg rating) for cars has been increasing steadily since 1980. The formula for a car's fuel efficiency for a
Lisa [10]

Answer:

a) In the interval of [1980,1983].

b) In the interval of [1991, 1996].

Step-by-step explanation:

Fuel efficiency:

The fuel efficiency, for the cars, in x years after 1980, is given by:

E(x) = 0.36x + 15.9

15.9 is the fuel efficiency in 1980.

a. In what years was the average fuel efficiency for cars less than 17 mpg (in interval form)?

From 1980 to:

E(x) < 17

0.36x + 15.9 < 17

0.36x < 1.1

x < \frac{1.1}{0.36}

x < 3.06

3.06 = 1980 + 3 = 1983. So

In the interval of [1980,1983].

b. In what years was the average fuel efficiency for cars more than 20 mpg (in interval form)?

From x until 1996.

E(x) > 20

0.36x + 15.9 > 20

0.36x > 4.1

x > \frac{4.1}{0.36}

x > 11.38

11.38 = 1980 + 11 = 1991. So

In the interval of [1991, 1996].

4 0
3 years ago
what is the simplified form of the following expression? Assume X 0 and y 0. 2(^4 sqrt 16 x) -2(^4 sqrt 2y )+3 (^4 sqrt 81x ) -4
eduard

For this case we have the following expression:

2(\sqrt[4]{16x})-2(\sqrt[4]{2y})+3(\sqrt[4]{81x})-4(\sqrt[4]{32y})

Rewriting the numbers within the roots we have:

2(\sqrt[4]{2*2*2*2x})-2(\sqrt[4]{2y})+3(\sqrt[4]{3*3*3*3x})-4(\sqrt[4]{2*2*2*2*2y})

Then, by properties of powers we have:

2(\sqrt[4]{2^4x})-2(\sqrt[4]{2y})+3(\sqrt[4]{3^4x})-4(\sqrt[4]{2^42y})

Then, by radical properties we have:

2(2\sqrt[4]{x})-2(\sqrt[4]{2y})+3(3\sqrt[4]{x})-4(2\sqrt[4]{2y})

Rewriting the expression we have:

4\sqrt[4]{x}-2\sqrt[4]{2y}+9\sqrt[4]{x}-8\sqrt[4]{2y}

Finally, adding similar terms we have:

(4+9)\sqrt[4]{x}-(2+8)\sqrt[4]{2y}

13\sqrt[4]{x}-10\sqrt[4]{2y}

Answer:

The simplified form of the expression is:

13\sqrt[4]{x}-10\sqrt[4]{2y}

7 0
3 years ago
Read 2 more answers
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