WILL GIVE BRAINLIEST!

The graph below shows the value of Edna's profits f(t), in dollars, after t months:

umka21 [38]

Since we know that the graph of our quadratic function has x-intercepts at (6,0) and (18,0), $x=6$ and $x=18$ are the zeroes of our quadratic. To find our quadratic we are going to factor each zero backwards and multiply them:
$x=6$
$x-6=0$
$x=18$
$x-18=0$

$(x-6)(x-18)=x^{2}-6x-18x+108$
$=x^{2}-24x+108$

Now that we have our quadratic function, we are going to use the average formula: $m= \frac{f(9)-f(3)}{9-3}$
$where$
$f(9)$ is the function evaluated at the 9th month
 $f(3)$ is the function evaluated at the 3rd month

$m= \frac{f(9)-f(3)}{9-3}$
$m= \frac{[9^{2}-24(9)+108]-[3^{2}-24(3)+108]}{9-3}$
$m= \frac{-27-45}{6}$
$m= \frac{-72}{6}$
$m=-12$

We can conclude that the closest approximate average rate of change for Edna's profits from the 3rd month to the 9th month is: B. −11.63 dollars per month.

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3 years ago

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A landscaper needs 348 pounds of plant food. He has 114 pounds in his truck, and another 46 pound at his shop.

Black_prince [1.1K]

Answer:

$\frac{19}{12}\ pounds$

or

$1\frac{7}{12}\ pounds$

Step-by-step explanation:

The correct question is

A landscaper needs 3 4/8 pounds of plant food. He has 1 1/4 pounds in his truck, and another 4/6 pound at his shop. How many more pounds of plant food does the landscaper need?

Let

x ----> the additional pounds of plant food needed for the landscaper

we know that

The additional pounds of plant food needed for the landscaper plus the pounds in his truck plus the pounds in his shop must be equal to the total pounds of plant food needed

so

The linear equation that represent this problem is

$x+1\frac{1}{4}+\frac{4}{6}=3\frac{4}{8}$