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Jlenok [28]
3 years ago
11

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

5 0
2 years ago
Rational number between -1 and 1<br>​
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<br><br>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.

6 0
3 years ago
Can Someone help and explain how to get this answer?
Kay [80]
44%
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i calculated everything
8 0
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:

<u>System Of Linear Equations </u>

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
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