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tigry1 [53]
3 years ago
11

Six equilateral triangles are connected to create a regular hexagon. The area of the hexagon is 24a2 – 18 square units. Which is

an equivalent expression for the area of the hexagon based on the area of a triangle?
6(4a2 – 3)
6(8a2 – 9)
6a(12a – 9)
6a(18a – 12)
Mathematics
1 answer:
shusha [124]3 years ago
3 0
It would be 6(4a2 - 3)
you would factor out just the 6 because the 18 doesn’t a a term with it like 34a2 does. then just divide normally to get ur answer 6(4a2 - 3)
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In the city of Carmen, there is a drawbridge that is opened twice per hour over the summer. The graph below shows the number of
n200080 [17]

Answer:

1) Between 0 and 1 minute the rate of change is 25 ft./min

2) Between 1 and 2 minute the rate of change is 0 ft./min

3) Between 2 and 4 minute the rate of change is -12.5 ft./min

Step-by-step explanation:

1) Between 0, and 1 feet, we have;

The rate of change = (Final height - Initial height)/(Final time - Initial time)

The rate of change = (40 - 15)/(1 - 0) = 25 ft/min

Between 0 and 1 minute the rate of change = 25 ft./min

2) Between 1, and 2 feet, we have;

The rate of change = (40 - 40)/(2 - 1) = 0 ft/min

Between 1 and 2 minute the rate of change = 0 ft./min

3) Between 2, and 4 feet, we have;

The rate of change = (15 - 40)/(4 - 2) = -12.5 ft/min

Between 2 and 4 minute the rate of change = -12.5 ft./min

7 0
3 years ago
Please help me!!! Thanks!
Vlada [557]

Answer:

I believe it's "D" 13.1

Step-by-step explanation:

3 0
3 years ago
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vampirchik [111]
\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h

Employ a standard trick used in proving the chain rule:

\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h

The limit of a product is the product of limits, i.e. we can write

\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)

The rightmost limit is an exercise in differentiating \sqrt x using the definition, which you probably already know is \dfrac1{2\sqrt x}.

For the leftmost limit, we make a substitution y=\sqrt x. Now, if we make a slight change to x by adding a small number h, this propagates a similar small change in y that we'll call h', so that we can set y+h'=\sqrt{x+h}. Then as h\to0, we see that it's also the case that h'\to0 (since we fix y=\sqrt x). So we can write the remaining limit as

\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}

which in turn is the derivative of \tan y, another limit you probably already know how to compute. We'd end up with \sec^2y, or \sec^2\sqrt x.

So we find that

\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}
7 0
3 years ago
Which value of x makes the equation 0.5(x + 16) = 3 + 0.25(x – 4) true?
Fynjy0 [20]

Answer:it is confusing

Step-by-step explanation:

5 0
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Solve the following function for y when x=5 round to the nearest tenth
Wittaler [7]

In the given question , we have a function given , which is

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And we need to find the value of y when x=5.

So we substitute 5 for x and solve for y, that is

y=13(0.5^5) = 0.40625

And on rounding to the nearest tenths, we will get

y=0.4

So when x=5, y=0.4

7 0
3 years ago
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