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Rzqust [24]
3 years ago
15

I need to find the surface area on questions 1-8 , please help!​

Mathematics
1 answer:
ser-zykov [4K]3 years ago
4 0

Answer:

2.486

3.1734

4.726

5.1350

6.canot see number in picture

7.864

8.54

Step-by-step explanation:

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Z roms
andre [41]

Answer:

answer is $28.87

Step-by-step explanation:

8 0
3 years ago
-6x ≤ 12<br> solve and show work
kipiarov [429]

Answer:

x = -2

Step-by-step explanation:

- x - = positive

-6 x -2 = 12

3 0
3 years ago
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Va rog e urgent am nevoie de raspuns
Lady_Fox [76]

Answer: 4

<u>Explanation:</u>

f(x) = 2x - 1

f(√2) = 2√2 - 1

f(1) = 2(1) - 1

     =  2 - 1

     =     1

f(√3)  = 2√3 - 1

*******************************************************

\frac{f(\sqrt{2})-f(1)}{\sqrt{2}-1} +\frac{f(\sqrt{3})-f(\sqrt{2})}{\sqrt{3}-\sqrt{2}}

= \frac{(2\sqrt{2}-1)-1}{\sqrt{2}-1} +\frac{(2\sqrt{3}-1)-(2\sqrt{2}-1)}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1} +\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1}(\frac{\sqrt{2}+1}{\sqrt{2}+1})+\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}})

= \frac{4+2\sqrt{2}-2\sqrt{2}-2}{2 - 1} + \frac{6 +2\sqrt{6}-2\sqrt{6}-4}{3-2}

= \frac{2}{1} +\frac{2}{1}

= 4

3 0
3 years ago
Help me plz !!!?!?!?!
olganol [36]
I believe the answer is the second one I am in the same class.
5 0
3 years ago
A method for finding the volume of any solid for which cross sections by parallel planes have equal area is called
lapo4ka [179]
The appropriate response is Caliveire's <span>Principle</span> It is a technique, with the recipe given beneath, of finding the volume of any strong for which cross-segments by parallel planes have measured up to ranges. This incorporates, yet is not constrained to, chambers and crystals.
7 0
3 years ago
Read 2 more answers
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