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2 years ago

The in center is the center of the circle of a triangle

Mathematics

1 answer:

Lana71 [14]2 years ago

4 0

Answer:

inscribed

Step-by-step explanation:

For any given triangle, the circle inside of it is called the Inscribed circle

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Save me the headache

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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Oksanka [162]

Answer:

11 tshirts would cost £106.15

Step-by-step explanation:

67.55 divided by 7 = 9.65

9.65 x 11 = 106.15

Therefore the total would be £106.15.

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3 years ago

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4(1+ 2y) = 12y. Solve for y.

Alja [10]

Answer:

y=1

Step-by-step explanation:

4(1+2y)=12y

4*1 + 4*2y=12y

4 + 8y=12y

4=12y - 8y

4=4y

4/4=y

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2 years ago

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julsineya [31]

Answer:

Power is the zero at the end of the number like this 56 small 0 at the top right

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bezimeni [28]

Divide maple street’s length by 3

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