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loris [4]
3 years ago
14

The front view of a podium is shown. Bryan wants to paint the front of the podium blue. He decomposed the shape into two triangl

es and a rectangle to find the area.

Mathematics
2 answers:
Shalnov [3]3 years ago
8 0

Answer:

Step-by-step explanation:

Given:

Base length:

A = 8.5 m

B = 4 m

Height, h = 2 ft

1 gallon = 350 ft^2

Area of trapezoid = 1/2 × (A + B) × h

= 1/2 × (8.5 + 4) × 2

= 12.5 ft^2

1 gallon = 350 ft^2

2 gallons = 700 ft^2

1 podium = 12.5 ft^2

= 700/12.5 podium

= 56 podiums will be painted with 2 gallons of paint.

alexira [117]3 years ago
7 0

Complete Question

The front of a podium is in the shape of a trapezoid with base lengths 4 and 8.5 feet. the height is 2 feet. a gallon of paint covers about 350 square feet. Bryan wants to paint the front of the podium blue. He decomposed the shape into two triangles and a rectangle to find the area.

How many front frames of a podium can Bryant paint with 2 gallons of paint?

Answer:

56 front podiums

Step-by-step explanation:

A proper look at the image attached will provide the explanations

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Read 2 more answers
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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}
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