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1 year ago

How many solutions does the equation have? |h| − 8 = 4 no solution, one solution, two solutions

Mathematics

1 answer:

Gennadij [26K]1 year ago

5 0

Since the solution to the system of equation is 12 and -12, hence the equation has 2 solutions

<h3>Modulus functions</h3>

Modulus functions are functions that can either be positive or negative. Given the modulus function below;

|h| − 8 = 4

The modulus of h can both be negative and positive

If the modulus of h is positive then;

h - 8 = 4

Add 8 to both sides

h-8 + 8 = 4 + 8

h = 4 + 8

h = 12

If the value of h is negative

-h - 8 = 4

-h = 4 + 8

h =-12

Since the solution to the system of equation is 12 and -12, hence the equation has 2 solutions

Learn more on modulus function here: brainly.com/question/23450491

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Read 2 more answers

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:

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

Now we can use the identity:

sin²(x) + cos²(x) = 1

$1 - cos^2(x) = sin^2(x)*(sin^2(x) + cos^2(x)) = sin^2(x)\\\\1 = sin^2(x) + cos^2(x) = 1$

Thus, the identity was proven.

If you want to learn more about trigonometric identities:

brainly.com/question/7331447

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1 year ago