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3 years ago

Given that sinx=3/5 where x is an acute angle for cosx

1 answer:

4 0

The correct question is
Given that sinx=3/5 where x is an acute angle, what is cosx and tanx<span>?

we know that

</span>since x is acute 0<x<<span>π/2
</span><span>then the trigonometric rations will be positive</span>

sin ²x+cos²x=1------------> cos²x=1-sin²x-------> cos²x=1-(3/5)²----> 1-(9/25)
cos²x=16/25----------> cos x=4/5

the answer part A) is
cos x=4/5

tan x=sin x/cos x
so
tan x=(3/5)/(4/5)---------> tan x=3/4

the answer part B) is
tan x=3/4

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Which value is closest to (2.4×103)+(5.7×102)?

astraxan [27]

Answer:

Step-by-step explanation:

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3 years ago

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mr_godi [17]

Answer:

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Step-by-step explanation:

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3 years ago

Solve for t <br><br> 2t^2-14t+3=3

Dvinal [7]

Answer:

t=7,t=0

Step-by-step explanation:

2t^2-14t+3=3

Subtract 3 from both sides

2t^2 - 14t +3 - 3 = 3-3

Simpilfy

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7 0

3 years ago

Simplify 425xy⁴/25xy²

beks73 [17]

We have the expression

$\frac{425xy^4}{25xy^2}$

We can already simplify x because it's both on the numerator and denominator

$\frac{425y^4}{25y^2}$

Now we can simplify 425/25 = 17

$\frac{17y^4}{y^2}$

Remember that

$\frac{a^n}{a^m}=a^{n-m}$

Then

$17y^{4-2}=17y^2$

The final result is

$17y^2$

3 0

2 years ago

For each sequence, write an explicit and a recursive formula. NO LINKS!!!!

expeople1 [14]

Answer:

a. explicit: an = 16(1/4)^(n-1); recursive: a[1] = 16, a[n] = 1/4·a[n-1]

b. explicit: an = -2 +7(n -1); recursive: a[1] = -2, a[n] = 7+a[n-1]

Step-by-step explanation:

Finding sequence formulas is all about finding and matching patterns. It is generally pretty easy to subtract one term from the next to see if there is a common difference. If the differences are not constant, but terms have the same ratio. then the sequence is geometric.

The forms for explicit and recursive formulas for arithmetic and geometric sequences are shown here. All you need to do is fill in the values of the first term (a1) and common difference (d) or common ratio (r).

<h3>Arithmetic sequence formulas:</h3>

<u>explicit formula</u>:

$a_n=a_1+d(n-1)$

<u>recursive formula</u>:

$a_1 = \text{[first term]};\ a_n=d+a_{n-1}$

<h3>Geometric sequence formulas:</h3>

<u>explicit formula</u>:

$a_n=a_1\cdot r^{n-1}$

<u>recursive formula</u>:

$a_1 = \text{[first term]};\ a_n=r\cdot a_{n-1}$

The first term is 16. There is no common difference, but the common ratio is 4/16 = 1/4.

explicit formula: an = 16(1/4)^(n-1)

recursive formula: a[1] = 16; a[n] = 1/4·a[n-1]

The first term is -2. The common difference is 5-(-2) = 7.

explicit formula: an = -2 +7(n -1)

recursive formula: a[1] = -2; a[n] = 7 +a[n-1]

_____

<em>Additional comment</em>

Term numbers are generally identified by subscripts on the variable. When writing in plain text, subscripts are difficult to create, so we simply write the subscript next to the variable name: an, a1. This is less satisfactory when there is arithmetic in the subscript, as in the recursive formulas. For those, we have written the plain text formulas using square brackets to show the term numbers: a[1], a[n], a[n-1].

The recursive formula for a given term is defined in terms of the previous term in the sequence. In order for that to work, there needs to be a definition of the first term. We have referred to that generically as a1, but the meaning of a1 = a1 in the recursive formula is maybe no so clear. That is why we used a1 = [first term]. The intention is that [first term] be understood to mean the numerical value of the first term of the sequence.

In short, these problems are worked by finding the common ratio or difference, identifying the first term of the sequence, and filling those two values into the generic formulas.

5 0

2 years ago