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3 years ago

Which is f(6) for the quadratic function graphed? 0 -2 * -0.5 1.5 4

Mathematics

1 answer:

rusak2 [61]3 years ago

4 0

Answer:

Step-by-step explanation:

I'm pretty sure theres a graph to this so....

The given parabolic graph of quadratic function

To find:What is F(6) for the quadratic function

Solution:We have given graph of quadratic function

there is a different value for Y and a different value for X

For f(6)

here X=6, need to find graph of quadratic function

X=6

f(6)=4

Y=4

Your welcome :)

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A particle is moving on a straight line in such a way that its velocity v is given by v ( t ) = 2 t + 1 for 0 ≤ t ≤ 5 where t is

Lady_Fox [76]

Answer:

The total distance traveled by the particle is S = 30.

Step-by-step explanation:

Given that velocity,

v(t) = 2t + 1

To find the total distance travel, we integrate the velocity function, v(t), to obtain the distance function s(t), and evaluate the resulting distance at the interval given. That is at t = 0 to t = 5.

Integrating v(t) with respect to t, we have

s(t) = t² + t + C.

At t = 5

s(5) = 5² + 5 + C

= 25 + 5 + C

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At t = 0

S(0) = 0 + 0 + C

= C

The required distance is now

S(5) - S(0)

= 30 + C - C

= 30.

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3 years ago

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Bezzdna [24]

A. (-20)4
Hope this helped! With love, Eilyssa! :)

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3 years ago

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Hey how do you get from standard form to vertex form?

vlabodo [156]

Explanation:

Conversion of a quadratic equation from standard form to vertex form is done by completing the square method.

Assume the quadratic equation to be $\mathbf{ax^{2}+bx+c=0}$ where x is the variable.

Completing the square method is as follows:

send the constant term to other side of equal $\mathbf{ax^{2}+bx=-c}$
divide the whole equation be coefficient of $\mathbf{x^{2}}$ , this will give $\mathbf{x^{2}+\frac{b}{a}x=- \frac{c}{a}}$
add $\mathbf{(\frac{b}{2a})^{2}}$ to both side of equality $\mathbf{x^{2}+2\times\frac{b}{2a}x+\frac{b}{2a}^{2}=-\frac{c}{a}+\frac{b}{2a}^{2}}$
Make one fraction on the right side and compress the expression on the left side $\mathbf{(x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}}$
rearrange the terms will give the vertex form of standard quadratic equation $\mathbf{a(x+\frac{b}{2a})^{2}-\frac{b^{2}-4ac}{4a}=0}$