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3 years ago

6

I need the answer now. Which of the following values in the set below will make the equation 4x + 2 = 18 true? (Only input the n

umber.) {0, 1, 2, 3, 4} (5 points)

1 answer:

Whitepunk [10]3 years ago

7 0

Answer:

x=4

Step-by-step explanation:

at x=4

$4(4)+2=18\\16+2=18$

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4 years ago

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3 years ago

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If p(x)=x2-1 and q(x)=5(x-1 which expression is equivalent to (p-q)(x)

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3 years ago

Consider the graph of f(x) below. The graph contains the points (0, 1) and (5, 4).

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3 years ago

ITS TIMED PLEASE HELP

Lubov Fominskaja [6]

Answer:

The graph of the function $f(x)=\frac{1}{2}x^{2}-4x+5$ has a minimum located at (4,-3)

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

$f(x)=a(x-h)^{2}+k$

where

a is a coefficient

(h,k) is the vertex of the parabola

If a > 0 the parabola open upward and the vertex is a minimum

If a < 0 the parabola open downward and the vertex is a maximum

In this problem

The coefficient a must be positive, because we need to find a minimum

therefore

Check the option C and the option D

Option C

we have

$f(x)=\frac{1}{2}x^{2}-4x+5$

Convert to vertex form

$f(x)-5=\frac{1}{2}x^{2}-4x$

Factor the leading coefficient

$f(x)-5=\frac{1}{2}(x^{2}-8x)$

$f(x)-5+8=\frac{1}{2}(x^{2}-8x+16)$

$f(x)+3=\frac{1}{2}(x^{2}-8x+16)$

$f(x)+3=\frac{1}{2}(x-4)^{2}$

$f(x)=\frac{1}{2}(x-4)^{2}-3$

The vertex is the point (4,-3) ( is a minimum)

therefore

The graph of the function $f(x)=\frac{1}{2}x^{2}-4x+5$ has a minimum located at (4,-3)

5 0

4 years ago