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3 years ago

Select the descriptions that are true of the equation y - 6= - 3(x+2). Select all that apply.

Mathematics

1 answer:

Lapatulllka [165]3 years ago

3 0

The point-slope form:

$y-y_1=m(x-x_1)$

We have

$y-6=-3(x+2)\to y-6=-3(x-(-2))$

Therefore

<h3>slope = -3 and the point (-2, 6)</h3>

------------

2x - y = 4 <em>it's standard form Ax + By = C</em>

y = 2x - 4 <em>it's slope-intercept form y = mx + b</em>

y - x = 4 →<em> </em>x - y = -4 <em>t's standard form Ax + By = C</em>

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Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

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Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
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Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
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Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

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Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

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When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$