HELP!!!

1 answer:

7 0

Trigonometry as taught is mostly about memorizing two lousy triangles, namely half a square and half an equilateral triangle. This one is half an equilateral triangle.

The respective opposite sides of the 30/60/90 triangle are in ratio $1:\sqrt{3}:2$

A sine of √3/2 is an opposite of √3 and a hypotenuse of 2, so opposite and hypotenuse of the 60 degree angle. (It's the mama bear, in the middle, just right.)

$\sin 60^\circ = \dfrac{\sqrt{3}}{2}$

In the end you just have memorize the two tired triangles of trig.

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the axis of symmetry of a quadratic equation is x = –3. if one of the zeroes of the equation is 4, what is the other zero?

FinnZ [79.3K]

ANSWER

The other zero is

$x = - 10$

EXPLANATION

The axis of symmetry serves as the midpoint of the two zeroes.

We were given that the axis of symmetry of the quadratic equation is
$x = - 3$

We were also given that, one of the zeroes of the quadratic equation is
$x = 4$

Let the other zero of the quadratic equation be
$x = p$

,then, we apply the midpoint formula to find the value of
$p$

Since it is the x-values of the intercepts that gives the solution, we use only the x-value part of the midpoint formula which is given by,

$\frac{x_1 + x_2}{2} = axis \: of \: symmetry$

We substitute to obtain,

$\frac{p + 4}{2} = - 3$

We multiply through by 2 to get,

$p + 4 = - 6$

This implies that,

$p = - 6 - 4$

$p = - 10$

3 0

3 years ago

Read 2 more answers

John bought a tv for 1280 and had to pay7% tax. he also recieved a 5% rebate what did he pay

Alex777 [14]

Let's assume the rebate is applied only to the 1280 and not the tax, so John pays a net 2% additional, tax less rebate.

1.02 × 1280 = 1305.06

Answer: 1305.06

8 0

4 years ago

What the recursive formula for the sequence

klemol [59]

$\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})$

so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.

$\bf \begin{cases} a_1=2\\ a_n=a_{n-1}-5 \end{cases}$

just a quick note on notation:

$\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_{n-1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_{5-1}\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_{12}}\qquad \quad \stackrel{\textit{term before it}}{a_{12-1}\implies a_{11}}$