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Sliva [168]
3 years ago
7

15 POINTS!!!!! screenshot of question is attached Is this correct?

Mathematics
2 answers:
SVEN [57.7K]3 years ago
3 0
Nah I don't think that's always true. 
VladimirAG [237]3 years ago
3 0
I think sometimes is the answer
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Compute 41, 42, 43, 44, 45, 46, 47, and 48. Make a conjecture about the units digit of 4n where n is a positive integer. Use str
padilas [110]

Answer:

The conjecture is that the unit digit of 4^n = 4 when n = odd also 4^n = 6 when n = even

Step-by-step explanation:

4^1 = 4\\4^2 = 16\\4^3 = 64\\4^4 = 256 \\4^5 = 1024\\4^6 = 4096\\4^7 = 16384\\4^8 = 65536

The conjecture is that the unit digit of 4^n = 4 when n = odd also 4^n = 6 when n = even

To prove this conjecture

4^1 = 4\\4^2 = 16 unit digit = 6

hence the property is true for ; n = 1 and n = 2 and also for every odd and even number ( i.e. from 1  to 8 )

8 0
3 years ago
Helpppp I don’t get the question!!!
avanturin [10]

Answer:

43.75    hope i helped you

4 0
4 years ago
Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into
Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = (\frac{60\pi }{\pi+4}) in

b)the length of the wire for the square = (\frac{240}{\pi+4}) in

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=\frac{y}{4}

Area of the square which is L² can now be said to be;

A_S=(\frac{y}{4})^2 = \frac{y^2}{16}

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

r = \frac{60-y}{2\pi }

Area of the circle which is πr² can now be;

A_C= \pi (\frac{60-y}{2\pi } )^2

     =( \frac{60-y}{4\pi } )^2

Total Area (A);

A = A_S+A_C

   = \frac{y^2}{16} +(\frac{60-y}{4\pi } )^2

For the smallest possible area; \frac{dA}{dy}=0

∴ \frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0

If we divide through with (2) and each entity move to the opposite side; we have:

\frac{y}{18}=\frac{(60-y)}{2\pi}

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

y= \frac{240}{\pi+4}

∴ since y= \frac{240}{\pi+4}, we can determine for the length of the circle ;

60-y can now be;

= 60-\frac{240}{\pi+4}

= \frac{(\pi+4)*60-240}{\pi+40}

= \frac{60\pi+240-240}{\pi+4}

= (\frac{60\pi}{\pi+4})in

also, the length of wire for the square  (y) ; y= (\frac{240}{\pi+4})in

The smallest possible area (A) = \frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0
4 years ago
What is the volume of this square-based pyramid with a base length of 6 feet and a height of 10 feet?
bulgar [2K]

Answer:

60 or 90

Step-by-step explanation:

5 0
3 years ago
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The average length of a ladybug can range from 0.08 to 0.4 inches. Find two lengths that are within the given span.Write them as
Alenkinab [10]
The average length of a ladybug can range from 0.08 to 0.4 inches.

So two lengths within the given span are 0.1 inch and 0.3 inches.

In  \: fraction: \\  \\ 0.1 \: inch =  \frac{1}{10}  \: inch \\  \\ 0.3 \: inch =  \frac{3}{10}  \: inch
7 0
3 years ago
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