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Whitepunk [10]
3 years ago
5

What is the area of the figure in square millimeters and in square centimeters?

Mathematics
1 answer:
arlik [135]3 years ago
7 0
The answer to this equation is C, 2925 mm^2 and 29.25 cm^2.  
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Mekhanik [1.2K]

I used Desmos Graphing Calculator. Helps so much!

Hope this helps!

5 0
2 years ago
How do you round 28.26 to the nearest square foot
scZoUnD [109]
28.26 is closest to 28 so you round 28.26 to 28, not 29.
8 0
3 years ago
Water is added or drained from a tank each day. The first day, 910 of a gallon is added to the empty tank. The second day, 710 o
dimulka [17.4K]

Answer:

The quantity of water in the tank after 15 days is 1610.0 gallons OR 1.61 × 10³ gallons.

Step-by-step explanation:

The amount of water in the tank after 15 days is given by the series

910+(−710)+810+(−610)+⋯+310+(−110)+210

From the series, we can observe that, if water is added for a particular day then water will be drained the following day.

Also, for a day when water is to be added, the quantity to be added will be 100 gallon lesser than the quantity that was last added. Likewise, for a day when water is to be drained, the quantity to be drained will be 100 gallons lesser than the quantity that was last drained.

Hence, we can complete the series thus:

910+(−710)+810+(−610)+710(-510)+610(-410)+510(-310)+410(-210)+310+(−110)+210

To evaluate this, we  get

910-710+810-610+710-510+610-410+510-310+410-210+310-110+210

= 1610.0 gallons

Hence, the quantity of water in the tank after 15 days is 1610 gallons OR 1.61 × 10³ gallons.

8 0
3 years ago
The perimeter of an isosceles triangle is p = 2a + b. Find the numerical value of a if p = 10 and b = 3.
bagirrra123 [75]
P = 2a + b
10= 2a + 3
2a = 10-3
a = 7/2 = 3.5


Therefore, numerical value of a is 3.5
7 0
3 years ago
39-50 find the limit.<br> 41. <img src="https://tex.z-dn.net/?f=%5Clim%20_%7Bt%20%5Crightarrow%200%7D%20%5Cfrac%7B%5Ctan%206%20t
Katyanochek1 [597]

Write tan in terms of sin and cos.

\displaystyle \lim_{t\to0}\frac{\tan(6t)}{\sin(2t)} = \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)}

Recall that

\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1

Rewrite and expand the given limand as the product

\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)

Then using the known limit above, it follows that

\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}

4 0
1 year ago
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