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2 years ago

Given the diameter of a circle is 17 1/2cm, find its area. (use π=22/7)

Mathematics

1 answer:

Leokris [45]2 years ago

6 0

Step-by-step explanation:

$area \: of \: circle = \pi{r}^{2} \\ = 22 \div 7{ \times (35 \div 2)}^{2} \\ = 3.14 \times 306.25 \\ = 961.625 {cm}^{2}$

<h2>hope it helps.</h2><h2>stay safe healthy and happy...</h2>

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Mrs Smith has 12 times as many markers as color pencils the total number of markers and colored pencils is 78 how many markers d

Ainat [17]

Let the colored pencil amount be p.
Then this is the equation:
12p+p=78
Simplifying, we can see that
13p=78
Divide both sides by 13 to get
p=6

Since there are 12 times as many markers as colored pencils, there are 72 markers.

6 0

3 years ago

Prove it please <br>answer only if you know

deff fn [24]

Part (c)

We'll use this identity

$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\\\\$

to say

$\sin(A+45) = \sin(A)\cos(45) + \cos(A)\sin(45)\\\\\sin(A+45) = \sin(A)\frac{\sqrt{2}}{2} + \cos(A)\frac{\sqrt{2}}{2}\\\\\sin(A+45) = \frac{\sqrt{2}}{2}(\sin(A)+\cos(A))\\\\$

Similarly,

$\sin(A-45) = \sin(A + (-45))\\\\\sin(A-45) = \sin(A)\cos(-45) + \cos(A)\sin(-45)\\\\\sin(A-45) = \sin(A)\cos(45) - \cos(A)\sin(45)\\\\\sin(A-45) = \sin(A)\frac{\sqrt{2}}{2} - \cos(A)\frac{\sqrt{2}}{2}\\\\\sin(A-45) = \frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\$

-------------------------

The key takeaways here are that

$\sin(A+45) = \frac{\sqrt{2}}{2}(\sin(A)+\cos(A))\\\\\sin(A-45) = \frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\$

Therefore,

$2\sin(A+45)*\sin(A-45) = 2*\frac{\sqrt{2}}{2}(\sin(A)+\cos(A))*\frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\2\sin(A+45)*\sin(A-45) = 2*\left(\frac{\sqrt{2}}{2}\right)^2\left(\sin^2(A)-\cos^2(A)\right)\\\\2\sin(A+45)*\sin(A-45) = 2*\frac{2}{4}\left(\sin^2(A)-\cos^2(A)\right)\\\\2\sin(A+45)*\sin(A-45) = \sin^2(A)-\cos^2(A)\\\\$

The identity is confirmed.

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Part (d)

$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\\\\\sin(45+A) = \sin(45)\cos(A) + \cos(45)\sin(A)\\\\\sin(45+A) = \frac{\sqrt{2}}{2}\cos(A) + \frac{\sqrt{2}}{2}\sin(A)\\\\\sin(45+A) = \frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\\\\$

Similarly,

$\sin(45-A) = \sin(45 + (-A))\\\\\sin(45-A) = \sin(45)\cos(-A) + \cos(45)\sin(-A)\\\\\sin(45-A) = \sin(45)\cos(A) - \cos(45)\sin(A)\\\\\sin(45-A) = \frac{\sqrt{2}}{2}\cos(A) - \frac{\sqrt{2}}{2}\sin(A)\\\\\sin(45-A) = \frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\\\\$

-----------------

We'll square each equation

$\sin(45+A) = \frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\\\\\sin^2(45+A) = \left(\frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\right)^2\\\\\sin^2(45+A) = \frac{1}{2}\left(\cos^2(A)+2\sin(A)\cos(A)+\sin^2(A)\right)\\\\\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\frac{1}{2}*2\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

and

$\sin(45-A) = \frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\\\\\sin^2(45-A) = \left(\frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\right)^2\\\\\sin^2(45-A) = \frac{1}{2}\left(\cos^2(A)-2\sin(A)\cos(A)+\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\frac{1}{2}*2\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

--------------------

Let's compare the results we got.

$\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

Now if we add the terms straight down, we end up with $\sin^2(45+A)+\sin^2(45-A)$ on the left side

As for the right side, the sin(A)cos(A) terms cancel out since they add to 0.

Also note how $\frac{1}{2}\cos^2(A)+\frac{1}{2}\cos^2(A) = \cos^2(A)$ and similarly for the sin^2 terms as well.

The right hand side becomes $\cos^2(A)+\sin^2(A)$ but that's always equal to 1 (pythagorean trig identity)

This confirms that $\sin^2(45+A)+\sin^2(45-A) = 1$ is an identity

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Given the diameter of a circle is 17 1/2cm, find its area. (use π=22/7)​

Given the diameter of a circle is 17 1/2cm, find its area. (use π=22/7)