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drek231 [11]
3 years ago
11

7. What is the perimeter of a square with sidelengths of 4 inches?​

Mathematics
1 answer:
irinina [24]3 years ago
7 0

Answer:

perimeter is 16 inches

Step-by-step explanation:

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Arrange the expressions in increasing order of their values.
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Answer:

The 4th one is the least so you would put that as the first one, then the 2nd one, then the 1st one, then the 3rd one.

Step-by-step explanation:

The fourth one is first because 10 to the power of 0 is 0 times 10 to the power of 1 is still 0 minus 1 to the power of 10 is 1 so 0 minus 1 is -1. The second one is second because 10 to the power of 0 times 10 to the power of 1 is 0, times 1 to the power of 10 is still 0. The first one is third in line because 10 to the power of 0 plus 10 to the power of 1 is 10 times 1 to the power of 10 is 10. The third one is last because its the greatist, 10 to the power of 0 plus 10 to the power of 1 is 10 plus 1 to the power of 10 is 11.

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The standard diameter of a golf ball is 42.67 mm.
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B should be the only correct answer.

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4 years ago
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How many times will 1/14 go into 6 7/10?<br> a. 6 1/14<br> b. 6 1/20<br> c. 93 4/5<br> d. 93 7/10
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Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck
Nimfa-mama [501]
\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}
\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}

Now, isolate the variables, so you can integrate.
(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}
\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}
2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C


4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C
y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C
(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C
(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C


y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}
y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}

At x = 1, y = 0.
0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}
-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}

4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0
\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}


4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C
0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C
C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}
C = -\frac{\pi}{2}

\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}
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