What is the product of the binomials (4a – 1) and (2b + 3)?

Mathematics

2 answers:

galben [10]3 years ago

8 0

Answer:

D. 8ab+12a-2b-3

Step-by-step explanation:

You are basically combining like terms

Lesechka [4]3 years ago

3 0

Answer:

D.) is the correct answer 8ab + 12a – 2b – 3

Step-by-step explanation:

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Help! How would I solve this trig identity?

NeTakaya

Using simpler trigonometric identities, the given identity was proven below.

<h3>How to solve the trigonometric identity?</h3>

Remember that:

$sec(x) = \frac{1}{cos(x)} \\\\tan(x) = \frac{sin(x)}{cos(x)}$

Then the identity can be rewritten as:

$sec^4(x) - sen^2(x) = tan^4(x) + tan^2(x)\\\\\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)} = \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)} \\\\$

Now we can multiply both sides by cos⁴(x) to get:

$\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)} = \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)} \\\\\\\\cos^4(x)*(\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}) = cos^4(x)*( \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)})\\\\1 - cos^2(x) = sin^4(x) + cos^2(x)*sin^2(x)\\\\1 - cos^2(x) = sin^2(x)*sin^2(x) + cos^2(x)*sin^2(x)$