3 years ago

Choose the friendly number that is closest to 25,689

Mathematics

1 answer:

Flauer [41]3 years ago

8 0

25,688 sonygfkdjfjdjfjgj needed 20 lol

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Find the measure of the arc TDB in the circle P shown below:

andrey2020 [161]

$\huge{ \mathfrak{ \underline{ Answer \: \: ✓ }}}$

The Angle formed by an arc on the centre of the circle is double the angle made by the same arc on the circumference of the circle.

So,

$\boxed{ \mathrm{ \angle TDB = \dfrac{1}{2} \angle TPB }}$

$\mathrm{\angle TDB = \dfrac{1}{2} \times 90 \degree}$

$\mathrm{\angle TDB =45 \degree }$

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White raven [17]

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Kitty [74]

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(Squreroot x+2)(squreoot x-2)=0

yuradex [85]

Answer:

$\boxed{ \ x = 2 \ or \ x = -2 \ }$

Step-by-step explanation:

I understand that you want to solve this equation

$\sqrt{x+2}\sqrt{x-2}=0$

<=>

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<=>

$(x+2)(x-2)=0$

<=>

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if your question is $(\sqrt{x}+2)(\sqrt{x}-2) = 0$

then this is $(\sqrt{x})^2-4 = 0\\ x = 4$

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