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Tresset [83]
2 years ago
13

The table represents the function f(x) = 2x+1.

Mathematics
2 answers:
Irina18 [472]2 years ago
6 0

Step-by-step explanation:

f(x) = 2x+1

f(2) = 2(2)+1.

f(2)=4+1

f(2)=5

Serhud [2]2 years ago
3 0

Answer:

5

Step-by-step explanation:

Plug in x = 2 into the equation 2x + 1

2x + 1

2 * 2 + 1

4 + 1

5

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Select all polynomials that have (x-3) as a factor.
Alchen [17]

Answer:

Below.

Step-by-step explanation:

Substitute x = 3 into each polynomial and see if the value of the polynomial is zero. If it is zero then x-3 is a factor.

Try number 1:

a(x) = x^3 - 2x^2 - 4x + 3

a(3) = 3^3 - 2(3)^2 - 4(3) + 3

= 27 - 2(9) - 12 + 3

= 27 - 18 - 12 + 3

= 9 - 18 - 9

= 27 - 27

= 0

So a(x) has (x-3) is a factor of a(x)

The other 3 polynomials are solved in the same way.

8 0
2 years ago
Given that f(x) = x2 − 3x + 3 and g(x) = the quantity of x minus one, over four , solve for f(g(x)) when x = 5. (1 point)
Anna007 [38]

Rewrite g(x) as    x-1

                           ------

                              4

and then substitute this result for x in f(x) = x^2 - 3x + 3:

f(g(x)) = (x-1)^2 / 4^2 - 3(x-1)/4 + 3.

At this point we can substitute the value 5 for x:

f(g(5)) = (5-1)^2 / 4^2 - 3(5-1)/4 + 3

          = 16/16 - 3(4/4) + 3 = 1 - 3 + 3 = 1

Therefore, f(g(5)) = 1.

6 0
3 years ago
If f(x) = x^2- 6 and g(x) = 2x – 1, determine the value of (gºf)(-3).
k0ka [10]
I’m not too sure if I did it correct, but the answer I got was -6x^2+36. I just assumed you’d solve it as usual and then multiply (-3) after distributing-2 to x^2-6. Hope this helps.

5 0
2 years ago
NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 11z​
Roman55 [17]

Answer:

(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}

(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}

Step-by-step explanation:

Given system of equations:

\begin{cases}y=x^2-9\\y=4x-4\end{cases}

To solve by the method of substitution, substitute the first equation into the second equation and rearrange so that the equation equals zero:

\begin{aligned}x^2-9&=4x-4\\x^2-4x-9&=-4\\x^2-4x-5&=0\end{aligned}

Factor the quadratic:

\begin{aligned}x^2-4x-5&=0\\x^2-5x+x-5&=0\\x(x-5)+1(x-5)&=0\\(x+1)(x-5)&=0\end{aligned}

Apply the <u>zero-product property</u> and solve for x:

\implies x+1=0 \implies x=-1

\implies x-5=0 \implies x=5

Substitute the found values of x into the <u>second equation</u> and solve for y:

\begin{aligned}x=-1 \implies y&=4(-1)-4\\y&=-4-4\\y&=-8\end{aligned}

\begin{aligned}x=5 \implies y&=4(5)-4\\y&=20-4\\y&=16\end{aligned}

Therefore, the solutions are:

(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}

(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}

6 0
1 year ago
Read 2 more answers
I’m so confused please help there’s a photo above btw
Strike441 [17]
Wow I don't get it either but my guess would be 9,6...sorry....
8 0
3 years ago
Read 2 more answers
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