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10 months ago

If f(x)3(=- Vx-3, complete the following statement:x + 2f(19) ==Answer here

Mathematics

1 answer:

Digiron [165]10 months ago

8 0

Explanation

This exercise is about evaluating a function at a particular argument. To do that, we replace the variable with the argument in the formula of the function, and simplify.

Let's do that:

$\begin{gathered} f(19)=\frac{3}{19+2}-\sqrt[]{19-3}, \\ \\ f(19)=\frac{3}{21}-\sqrt[]{16}, \\ \\ f(19)=\frac{1}{7}-4, \\ \\ f(19)=\frac{1-28}{7}, \\ \\ f(19)=-\frac{27}{7}\text{.} \end{gathered}$ Answer $f(19)=-\frac{27}{7}\text{.}$

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Read 2 more answers

P(x)=3x^2+12x+7 find coordinates of vertex and intercepts

kotegsom [21]

Te x intercepts are where the graph crosses the x axis or wher y=0
the y intercept is where the graph crosses the y axis or where x=0

to find vertex, here is a hack
the x coordinate of the vertex for an equation in form ax^b+bx+c=y is -b/2a

so
y=3x^2+12x+7
-b/2a=-12/2(3)=-12/6=-2
sub that back
y=3(-2)^2+12(-2)+7
y=3(3)-24+7
y=9-17
y=-8

vertex is (-2,-8)

intercepts
x intercept is where y=0

0=3x^2+12x+7
using quadratic formula
x=(-6-√15)/3 and (-6+√15)/3
xints at ((-6-√15)/3,0) and ((-6+√15)/3,0)

yint is where x=0
set x=0
y=7
yint is at y=7 or (0,7)

vertex at (-2,-8)
xints at x= $\frac{-6+ \sqrt{15} }{3}$ and $\frac{-6- \sqrt{15} }{3}$ or at the points $( \frac{-6+ \sqrt{15} }{3} , 0)$ and $(\frac{-6- \sqrt{15} }{3},0)$
yint at y=7 or at (7,0)

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2 years ago