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sergeinik [125]
3 years ago
6

Please solve question 6, doing surds

Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
7 0

See the attached diagram, it has all the information you need.

(a) If the green radii are all 1, then the orange diameters are all 2 + √2, so that the orange radii are (2 + √2)/2 = 1 + √2/2.

This is because we can join the radii of two adjacent green circles to form the sides of a square with side length equal to twice the radius - i.e. the diameter - of the green circles. The diagonal of any square occurs in a ratio to the side length of √2 to 1. Then we get the diameter of an orange circle by summing this diagonal length and two green radii, and hence the radius by dividing this by 2.

(b) We get the blue diameter in the same way. It has length (2 + √2) (1 + √2/2) = 3 + 2√2, so that the blue radius is (3 + 2√2)/2 = 3/2 + √2.

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5 0
3 years ago
P is a prime number.<br> Show that p²+6 could be either even or odd
storchak [24]

<em>p</em> ² + 6 is even if <em>p</em> ² is even (the sum of two even numbers is even), and this happens only if <em>p</em> itself is even, i.e. if <em>p</em> = 2 since 2 is the only even prime.

All other primes <em>p</em> are odd, so <em>p</em> ² would be odd, and adding 6 would not change that (odd + even = odd).

6 0
3 years ago
Find the general solution to 1/x dy/dx - 2y/x^2 = x cos x, y(pi) = pi^2
Finger [1]

Answer:

\frac{y}{x^2}=\sin x+\pi

Step-by-step explanation:

Consider linear differential equation \frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)

It's solution is of form y\,I.F=\int I.F\,q(x)\,dx where I.F is integrating factor given by I.F=e^{\int p(x)\,dx}.

Given: \frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x

We can write this equation as \frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x

On comparing this equation with \frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x), we get p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x

I.F = e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}      { formula used: \ln a^b=b\ln a }

we get solution as follows:

\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C

{ formula used: \int \cos x\,dx=\sin x }

Applying condition:y(\pi)=\pi^2

\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C

So, we get solution as :

\frac{y}{x^2}=\sin x+\pi

4 0
4 years ago
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scoundrel [369]

Answer:

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Step-by-step explanation:

cause they have the same slope

7 0
3 years ago
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kondor19780726 [428]
80 centimeters by far
6 0
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