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2 years ago

What is the tan of 30 degrees

Mathematics

2 answers:

Verdich [7]2 years ago

7 0

The answer to your problem is
0.57735026919

Ira Lisetskai [31]2 years ago

4 0

Answer:

The value of tan 30 degrees is 0.57736.

Step-by-step explanation:

Tan 30 degrees in radical form is = $\frac{1}{\sqrt{3} }$

We know $\sqrt{3}$ = 1.732

So, tan 30 degrees = $\frac{1}{1.732}$

= 0.57736

The answer is 0.57736.

Please find attached the exact value reference triangle .

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HELP for (ii)…..how do we find a?

Mariana [72]

By using the information given on the graph, we can see that a = 3.

<h3>How to find the value of a?</h3>

Here we have an exponential of the form:

$y = p*a^x$

And by looking at the graph, it passes through (0, 4) and (2, 36), so we can write:

$4 = p*a^0\\\\36 = p*a^2$

From the first equation we get that p = 4, replacing that on the second one we get:

$36 = 4*a^2\\\\36/4 = 9 = a^2\\\\\sqrt{9} = 3 = a$

So we conclude that a = 3.

If you want to learn more about exponential equations:

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7 0

2 years ago

CORRECT GETS BRAINLIEST

Dmitrij [34]

C. 2/3 is the correct answer. I worked this question out and that is what I got! I hope you have a nice day! (mark as brainliest!)

6 0

2 years ago

Report Error Suppose $P(x)$ is a polynomial of smallest possible degree such that: $\bullet$ $P(x)$ has rational coefficients $\

motikmotik

Answer:

We want a polynomial of smallest degree with rational coefficients with zeros in $\sqrt{7}$ , $1 - \sqrt{6}$ and -3. The last root gives us the factor (x+3). Hence, our polynomial is

$P(x) =(x+3)q(x)$

where $q$ is a polynomial with rational coefficients and roots $\sqrt{7}$ and $1 - \sqrt{6}$ . The root $\sqrt{7}$ gives us a factor $x-\sqrt{7}$ , but in order to obtain rational coefficients we must consider the factor $x^2-7$ .

An analogue idea works with $1 - \sqrt{6}$ . For convenience write $x - 1 + \sqrt{6} = ( x - 1) + \sqrt{6}$ . This gives the factor $(x-1)^2-6$ . Hence,

$P(x) = (x+3)(x^2-7)((x-1)^2-6)=x^5+x^4-18x^3-22x^2+77x+105$

Notice that $P(-1)=24$ . So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is

$P(x) =(1/3)x^5+(1/3)x^4-6x^3-(22/3)x^2+(77/3)x+35$

Step-by-step explanation:

We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type $x-\sqrt7$ will introduce in the expression, we need to multiply by its conjugate $x+\sqrt7$ . Hence, we will obtain $x^2-7$ that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.