PLS HELP Find an equation of the line with a y-intercept of -3 and an x-intercept of -4.5

Mathematics

1 answer:

kari74 [83]3 years ago

3 0

Answer:

y = - $\frac{2}{3}$ x - 3

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (0, - 3) and (x₂, y₂ ) = (- 4.5, 0 ) ← coordinates of intercepts

m = $\frac{0-(-3)}{-4.5-0}$ = $\frac{0+3}{-4.5-0}$ = $\frac{3}{-4.5}$ = - $\frac{2}{3}$

The line crosses the y- axis at (0, - 3 ) ⇒ c = - 3

y = - $\frac{2}{3}$ x - 3 ← equation of line

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To find, we must differentiate this same given curve r ( t ) with respect to t to estimate |r'(t)|.

The given integration limits in this case are a = 0, b = 2π.

Now, as per the question;

The equation of the curve is given as;

r(t) = (4cost)i + (4sint)j + 4tk

Now, differentiate this same given curve r ( t ) with respect to t.

$\begin{aligned}\left|r^{\prime}(t)\right| &=\sqrt{(-4 \sin t)^2+(4 \cos t)^2+4^2} \\&=\sqrt{16 \sin ^2 t+16 \cos ^2 t+16} \\&=\sqrt{16\left(\sin t^2+\cos ^2 t\right)+16}\end{aligned}$

Further simplifying;

$\begin{aligned}&=\sqrt{16(1)+16} \\&=\sqrt{16+16} \\&=\sqrt{32} \\\left|r^{\prime}(t)\right| &=4 \sqrt{2}\end{aligned}$

Now, use integration to find the mass of the wire;

$\begin{aligned}&=\int_a^b \delta\left|r^{\prime}(t)\right| d t \\&=\int_0^{2 \pi} 54 \sqrt{2} d t \\&=20 \sqrt{2} \int_0^{2 \pi} d t \\&=20 \sqrt{2}[t]_0^{2 \pi} \\&=20 \sqrt{2}[2 \pi-0] \\&=40 \pi \sqrt{2}\end{aligned}$