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2 years ago

13

Help please!! Which are factor pairs of 3580? (Hint: Look at the first factor in each pair and use divisibility rules.) Select a

ll that apply. O 2 x 1790 O 6 x 597 3 x 1193 8 x 448 5 x 716 4 x 895 O 9 x 398 O 10 x 358

1 answer:

lawyer [7]2 years ago

3 0

Answer:

2×1790

5×716

4×895

10×358

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Answer:

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Prove that tan 10+tan 70'+taniou = tanio tanfo'tam 1oo.

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Question:

Prove that:

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Answer:

Proved

Step-by-step explanation:

Given

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Required

Prove

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Subtract tan(10) from both sides

$- tan(10)+tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)$

$tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)$

Factorize the right hand size

$tan(70) + tan(100) = -tan(10)(-tan(70). tan(100) + 1)$

Rewrite as:

$tan(70) + tan(100) = -tan(10)(1-tan(70). tan(100))$

Divide both sides by $1-tan(70). tan(100)$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = \frac{-tan(10)(1-tan(70). tan(100))}{1-tan(70). tan(100))}$

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In trigonometry:

$tan(A + B) = \frac{tan(A) + tan(B)}{1 - tan(A)tan(B)}$

So:

$\frac{tan(70) + tan(100)}{1 - tan(70)tan(100)}$ can be expressed as: $tan(70 + 100)$

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$tan(70 + 100) = -tan(10)$

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In trigonometry:

$tan(180 - \theta) = -tan(\theta)$

So:

$tan(180 - 10) = -tan(10)$

Because RHS = LHS

Then:

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$ has been proven

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Answer:

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