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rodikova [14]
3 years ago
8

The Inverse of G(x) is a function. True or False

Mathematics
1 answer:
Paha777 [63]3 years ago
5 0

Answer:

Inverse of G(x) is not a function

Step-by-step explanation:

We have been given with graph and given graph is of the function y=x^2-4

We need to tell whether the inverse of given function is function or not.

We need to interchange the variables to find inverse of a function we will get

x=y^2-4

Now, after simplifying we will get  y^2=x+4

Hence, y=\pm\sqrt{x+4}

Here for each value of x there will not be different value of y

Therefore inverse of G(x) is not a function.

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Please help me with vivid explanation​
Ivanshal [37]

Answer:

864 m²

Step-by-step explanation:

  • First calculate the total area of the rectangular field

The area of a rectangle is given by the product of the length and the width

let A be the total area

A = 100*120

A = 12000 m²

Calculate the area of the small rectangles

  • Let A' be the total area of the four small rectangles and A" the area of one small rectangle
  • A' = 4 A"
  • A' = 4 [(\frac{120-4}{2})*(\frac{100-4}{2})]
  • A' = 4*58*48
  • A' = 11136 m²

  • Substract the A' from A to get the area of the road

Let A"' be the area of the road

A"' =A-A'

A"' = 12000-11136

A"' = 864 m²

3 0
3 years ago
How would I solve 16x^4-41x^2+25=0 ???
sveticcg [70]
16{ x }^{ 4 }-41{ x }^{ 2 }+25=0

{ x }^{ 4 }={ ({ x }^{ 2 }) }^{ 2 }\\ \\ 16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0



First of all to make our equation simpler, we'll equal x^{2} to a variable like 'a'.

So,

{ x }^{ 2 }=a

Now let's plug x^{2} 's value (a) into the equation.

16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0\\ \\ { x }^{ 2 }=a\\ \\ 16{ (a) }^{ 2 }-41{ a }+25=0

Now we turned our equation into a quadratic equation.

(The variable 'a' will have a solution set of two solutions, but 'x' , which is the variable of our first equation will have a solution set of four solutions since it is a quartic equation (<span>fourth-degree <span>equation) )

Let's solve for a.

The formula used to solve quadratic equations ;

\frac { -b\pm \sqrt { { b }^{ 2 }-4\cdot t\cdot c }  }{ 2\cdot t }

The formula is used in an equation formed like this :
</span></span>
t{ x }^{ 2 }+bx+c=0

In our equation,

t=16 , b=-41 and c=25

Let's plug the values in the formula to solve.

t=16\quad b=-41\quad c=25\\ \\ \frac { -(-41)\pm \sqrt { -(41)^{ 2 }-4\cdot 16\cdot 25 }  }{ 2\cdot 16 } \\ \\ \frac { 41\pm \sqrt { 1681-1600 }  }{ 32 } \\ \\ \frac { 41\pm \sqrt { 81 }  }{ 32 } \\ \\ \frac { 41\pm 9 }{ 32 }

So the solution set :

\frac { 41+9 }{ 32 } =\frac { 50 }{ 32 } \\ \\ \frac { 41-9 }{ 32 } =\frac { 32 }{ 32 } =1\\ \\ a\quad =\quad \left\{ \frac { 50 }{ 32 } ,\quad 1 \right\}

We found a's value.

Remember,

{ x }^{ 2 }=a

So after we found a's solution set, that means.

{ x }^{ 2 }=\frac { 50 }{ 32 }

and

{ x }^{ 2 }=1

We'll also solve this equations to find x's solution set :)

{ x }^{ 2 }=\frac { 50 }{ 32 } \\ \\ \frac { 50 }{ 32 } =\frac { 25 }{ 16 } \\ \\ { x }^{ 2 }=\frac { 25 }{ 16 } \\ \\ \sqrt { { x }^{ 2 } } =\sqrt { \frac { 25 }{ 16 }  } \\ \\ x=\quad \pm \frac { 5 }{ 4 }

{ x }^{ 2 }=1\\ \\ \sqrt { { x }^{ 2 } } =\sqrt { 1 } \\ \\ x=\quad \pm 1

So the values x has are :

\frac { 5 }{ 4 } , -\frac { 5 }{ 4 } , 1 and -1

Solution set :

x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\}

I hope this was clear enough. If not please ask :)



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3 years ago
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Sean has worked 66 hours over the last 10 days without a day off. He worked either a 6 hour day or an 8 hour day. During the 10
marysya [2.9K]
He worked an 8 hour day 3 days.

Let x be the number of 8 hour days he works.  Then 10-x is the number of 6 hour days he works.

8x + 6(10-x) = 66

Using the distributive property,
8x + 6*10 - 6*x = 66
8x + 60 - 6x = 66

Combining like terms,
2x + 60 = 66

Subtract 60 from both sides:
2x + 60 - 60 = 66 - 60
2x = 6

Divide both sides by 2:
2x/2 = 6/2
x = 3

There were 3 days he worked 8 hours.
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Answer:

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3 years ago
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