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Damm [24]
2 years ago
7

What is the surface area of this design?

Mathematics
1 answer:
Archy [21]2 years ago
6 0

Answer:

You probably already know that the tall rectangles take up 4×(5×8)=160 of surface area, and that the short rectangles take up 4×(2×10)=80 of surface area. This looks like an office building so let's call it one.

You probably also saw the big, 10×10=100 inch area underneath the design.

BUT, did you know that the small, 5×5 inch square on top can be COMBINED with the weird, L-shaped flat-roof part in the middle? That's right! the 5×5 inch square can be seen as part of a full 10×10 inch square sitting on top of the short, wide rectangles.

If you look at the L-shaped flat roof, you can see that it's actually 3 more 5×5 squares. You need to connect the 10-inch dimension on the bottom with the 5-inch one on the top.

SO, the answer is gonna be:

160 in² in tall rectangles, +

80 in² in short rectangles, +

100 in² on the bottom, +

100² on the combined top small square and corner roof

Add those all up using some quick maths, and you get 440 in².

Some more details if you're still confused:

The sides of the tall rectangle? It's 5 inches by 8 inches, so each side 8×5=40 in². They go all around the tall box, so there are 4 of them. 40+40+40+40=160, or in other words, 40×4=160.

Do you see the short, wide rectangles on the bottom? they're 10 inches by 2 inches each, so each one is 2×10=20 in². They also go all around the short box, so they add up to 20+20+20+20=80, or in other words, 20×4=80.

Now I want you to think about the bottom of the design, underneath. It is 10 inches by 10 inches, so it is 10×10=100.

Finally, the small square on top of the tall rectangle, and that L-shaped part. Together, they make another 100, or 5×5 on top plus 5×10 + 5×5 again for the parts of the L-shaped roof, or 25+50+25, which equals 100.

Step-by-step explanation:

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Answer:

\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\  \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right

Step-by-step explanation:

a) \lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}

b) \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}

c) \lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0

d) \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}

e) \lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2

f) \lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}

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