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3 years ago

Two functions are graphed on the coordinate plane.

Mathematics

1 answer:

Keith_Richards [23]3 years ago

6 0

Answer:

f(4) = g(4) and f(0) = g(0)

Step-by-step explanation:

In order for f(x) = g(x), the value of x must be the same in both functions:

f(4) = g(4) . . . corresponds to x=4

f(0) = g(0) . . . corresponds to x=0

The graph is not shown here, so we cannot say if these are the appropriate solutions. We can only say that the other choices are not.

f(x) = g(x) if ...

f(4) = g(4) and f(0) = g(0)

Something like f(0) = g(4) is useless for finding solutions to f(x) = g(x).

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Una tienda vende un artículo por $35 después de un margen de beneficio del 25%. ¿Cuál es el precio que pagó la tienda?

nadya68 [22]

Answer:

43.75

Step-by-step explanation:

25 dividido por 100=.25

.25*35=8.75

8,75+35=43,75

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2 years ago

Rational number between -1 and 1

Nonamiya [84]

Answer:

$\dfrac{-1}{2}$

Step-by-step explanation:

We need to find ration numbers between -1 and 1.

Multiplying and dividing both -1 and 1 by 2.

$\dfrac{-1\times 2}{2}=\dfrac{-2}{2}$

and $\dfrac{1\times 2}{2}=\dfrac{2}{2}$

Now, the rational number between -2/2 and 2/2 is $\dfrac{-1}{2}$ .

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2 years ago

X^2-6x+9 quadratic equations

kipiarov [429]

think the answer is
x=6+sqrt0/2 x=6-sqrt0/2 both equaling into 3.

But I might be wrong.

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3 years ago

Can Someone help and explain how to get this answer?

Kay [80]

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3 years ago

Read 2 more answers

Write the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10).

Serjik [45]

Answer:

$\displaystyle f(x)=x^2+2x+2$

Step-by-step explanation:

System Of Linear Equations

In this problem, we'll need to solve a 3x3 system of linear equations because we have three unknowns and three conditions.

We are required to find the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10)

The general quadratic function can be written as

$\displaystyle f(x)=ax^2+bx+c$

We need to find the values of a,b, and c. Let's use the first condition, i.e. f(-1)=1

$\displaystyle f(-1)=a(-1)^2+b(-1)+c$

$\displaystyle f(-1)=a-b+c$

$\displaystyle a-b+c=1.....[eq\ 1]$

Now we use the second condition f(1)=5

$\displaystyle f(1)=a(1)^2+b(1)+c$

$\displaystyle f(1)=a+b+c$

$\displaystyle a+b+c=5.......[eq\ 2]$

Finally, we use the third condition f(2)=10

$\displaystyle f(2)=a(2)^2+b(2)+c$

$\displaystyle f(2)=4a+2b+c$

$\displaystyle 4a+2b+c=10....[eq\ 3]$

We put together eq 1, eq 2, and eq 3 to form the system

$\displaystyle \left\{\begin{matrix}a-b+c=1\\ a+b+c=5\\ 4a+2b+c=10\end{matrix}\right.$

Adding the first two equations we have

$\displaystyle 2a+2c=6$

$\displaystyle a+c=3$

And also

$\displaystyle b=2$

Using the above equation and the value of b in the third equation, we have

$\displaystyle \left\{\begin{matrix}a+c=3\\ 4a+c=6\end{matrix}\right.$

Subtracting the first equation from the second

$\displaystyle 3a=3$

$\displaystyle a=1$

And therefore

$\displaystyle c=2$

Now we have all the values, the quadratic function is

$\displaystyle \boxed{f(x)=x^2+2x+2}$

6 0

3 years ago