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sammy [17]
3 years ago
15

How do I solve this problem. I have to find the missing side lengths and lease my sender as radicals in simplest form

Mathematics
1 answer:
nlexa [21]3 years ago
4 0

Answer:

x = y = 2√2

Step-by-step explanation:

Find the diagram attached

To get the unknown side x and y, we need to use the SOH CAH TOA identity

Opposite side = x

Adjacent = y

Hypotenuse = 4

Sin theta = opposite/hypotenuse

sin 45 = x/4

x = 4 sin 45

x = 4 * 1/√2

x = 4 * 1/√2 * √2/√2

x = 4 * √2/√4

x = 4 * √2/2

x = 2√2

Similarly;

cos theta = adjacent/hypotenuse

cos 45 = y/4

y = 4cos45

y = 4 * 1/√2

y = 4 * 1/√2 * √2/√2

y = 4 * √2/√4

y = 4 * √2/2

y = 2√2

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4 years ago
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Given the function f(x)=(x-2)(x+3)(2x+3), which set includes all of the zeros for the polynomial?
Marrrta [24]

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{2, -3, -3/2}

Step-by-step explanation:

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3 years ago
The first term of a geometric sequence is equal to a and the common ratio of the sequence is r.
ololo11 [35]

Answer: (a)  {a, ar, ar², ar³, ar⁴, ar⁵...}, (b)  arⁿ⁻¹

For part (a), the question gives us the first term a, and then asks us to apply the common ratio r six times.

In order for ar = a, the nth term of r will have to equal 0 (this implies that n is an exponent; thus giving us the first term a, as r = 1).

Since we use this method on the first term, we must use it for the next five, in which r gains an additional exponent for every consecutive value (nth term) thereafter.  

Ultimately getting: {a, ar, ar², ar³, ar⁴, ar⁵...}

For part (b), we first have to understand that the sequence does not start at 0, but at 1 for n. In order for ar = a, with n = 1, there needs to be subtraction of -1 within the exponent. So that arⁿ⁻¹

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3 years ago
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Cerrena [4.2K]

The expression -22x + 3y = 6 in the form ax + by + c = 0 is -2x + 3y - 6 = 0

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The given equation is:

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Rewrite -2x + 3y = 6 in the form ax + by + c = 0

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Tasya [4]

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