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3 years ago

Find the average rate of change for the given function from x = −1 to x = 2.

Mathematics

1 answer:

GREYUIT [131]3 years ago

8 0

Well, if you look at the graph, when x = -1, y or f(x) = 4.

and when x = 2, y = 0.

$\bf slope = m = \cfrac{rise}{run} \implies 
\cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby 
\begin{array}{llll}
average~rate\\
of~change
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
x_1=-1\\
x_2=2
\end{cases}\implies \cfrac{f(2)-f(-1)}{2-(-1)}\implies \cfrac{0-4}{2+1}\implies \cfrac{-4}{3}$

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kherson [118]

4: A school has 314 boys & 310 girls. If they are to be grouped into equal classes of 26 each , how many classes will there be?

314 boys + 310 girls = 624

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Therefore

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260 x 2 = 520 = 20 classes

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Number of classes = 24

Question 5).

Jim worked 48 weeks last year. Each week he worked 38 hours. If he worked an additional 240 hours of overtime, how many hours did he work in all?

Solution:

48 weeks

24hrs × 7 = 168hrs

168 x 48 = 8,064

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= 168hrs = 7 days (a week)

240hrs - 168 = 72 (3days)

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Step-by-step explanation:

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3 years ago

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valkas [14]

Answer: The mat is 4.33 ft high off the ground.

Step-by-step explanation:

Since we have given that

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Angle of elevation with the second triangle = 60°

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so, As shown in the figure:

We need to find the height of the mat off the ground.

If CD = 5 ft,

Let, AB = y, DC = x.

In Δ ABC,

$\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}$

Similarly, in Δ ACD,

$\tan 30^\circ=\frac{AB}{BD}\\\\\frac{1}{\sqrt{3}}=\frac{y}{x+5}\\\\\frac{1}{\sqrt{3}}=\frac{y}{\frac{y}{\sqrt{3}}+5}\\\\\frac{1}{\sqrt{3}}=\frac{y\sqrt{3}}{y+5\sqrt{3}}\\\\3y=y+5\sqrt{3}\\\\2y=5\sqrt{3}\\\\y=\frac{5\sqrt{3}}{2}\\\\y=4.33\ ft$

Hence, the mat is 4.33 ft high off the ground.

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3 years ago

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