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3 years ago

3 а. A = 6 cm Calculate the diameter of the circle.

Mathematics

1 answer:

jeyben [28]3 years ago

6 0

Answer:

2.739... cm

Step-by-step explanation:

The formula for finding diameter from area is

d = 2√(a/pi)

Replace the a with 6, and the answer becomes

2.739

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maxonik [38]

$(9\sin2x+9\cos2x)^2=81$

Taking the square root of both sides gives two possible cases,

$9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1$

$9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1$

Recall that

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

If $\alpha=2x$ and $\beta=\dfrac\pi4$ , we have

$\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}$

so in the equations above, we can write

$\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1$

Then in the first case,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi$

(where $n$ is any integer)

$\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi$

$\implies x=n\pi\text{ or }\dfrac\pi4+n\pi$

and in the second,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi$

$\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi$

$\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi$

Then the solutions that fall in the interval $[0,2\pi)$ are

$x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4$

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3 years ago

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3 years ago

Factor both quadratic expressions. (x 4 + 5x 2 - 36)(2x 2 + 9x - 5) = 0

Sonbull [250]

So focusing on x^4 + 5x^2 - 36, we will be completing the square. Firstly, what two terms have a product of -36x^4 and a sum of 5x^2? That would be 9x^2 and -4x^2. Replace 5x^2 with 9x^2 - 4x^2: $x^4+9x^2-4x^2-36$

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One tennis ball can holds 3 balls. Jenna filled 5 cans and had 1 ball left over.

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3 years ago

3 а. A = 6 cm Calculate the diameter of the circle.​

3 а. A = 6 cm Calculate the diameter of the circle.