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3 years ago

Please help me! And please no scam answers

Mathematics

1 answer:

Yuri [45]3 years ago

6 0

Answer:

Average rate of change: $-3$

Step-by-step explanation:

Remember:

The average rate of change of a function over an interval $[a,b]$ is $\frac{f(b)-f(a)}{b-a}$

Given:

$[a,b]=[1,7]$

$f(b)=f(7)$

$f(a)=f(1)$

Calculation:

$f(b)=f(7)=-(7)^2+5(7)+14=-49+35+14=-49+49=0$

$f(a)=f(1)=-(1)^2+5(1)+14=-1+5+14=4+14=18$

$\frac{f(b)-f(a)}{b-a}=\frac{0-18}{7-1}=\frac{-18}{6}=-3$

Therefore, the average rate of change of the function $g(x)=-x^2+5x+14$ over the interval $1\leq x\leq7$ is $-3$ .

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Which doubles fact helps you solve 8+7=15? Write the number sentence.

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8 + 8 = 16
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7 + 7 = 14
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3 years ago

Read 2 more answers

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.