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3 years ago

What is the length of the diagonal of a TV that measures 2.25 feet high and 4 feet wide?

Mathematics

1 answer:

NikAS [45]3 years ago

6 0

Answer: Rounded, the diagonal is 4.6 feet

Step-by-step explanation: Use the Pythagorean Theorem: c² = a² + b²

Then take the square root of c² to find the length

2.25² =5.0625 This is a²

4² = 16 This is b² Add a² + b² to get c²

5.o625 + 16 = 21.0625 This is c²

√21.0625 = 4.589389938 This is the square root.

Round this to common measurements or "significant digits" 4.6 feet.

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3^2 - 2x + 4 and g(x) = 5x^2 + 6x - 8 find f-g(x)

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Answer:

For the given functions f(x) and g(x) $(f-g)(x) = -2x^{2} - 8x +12$

Step-by-step explanation:

Here, the given function are:

$f(x) = 3x^{2} - 2x+ 4\\g(x) = 5x^{2} + 6x -8$

Now, simplifying for (f-g)(x) = f(x) - g(x):

$f(x) - g(x) = (3x^{2} - 2x+ 4) - (5x^{2} + 6x -8)$

$= 3x^{2} - 2x+ 4- 5x^{2} - 6x +8$

Doing operation on LIKE TERMS, we get

f(x) - g(x) = $( 3x^{2}- 5x^{2}) + (-2x - 6x) +( 4 +8) = -2x^{2} - 8x + 12$

Hence, for the given functions f(x) and g(x) $(f-g)(x) = -2x^{2} - 8x +12$

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