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Mathematics

1 answer:

zimovet [89]3 years ago

4 0

Answer:

The SF is 3

Step-by-step explanation:

The sides are 3x the length of the sides of ∆ABC

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What is the equation of the line that passes through (0, 3) and (7, 0)?

r-ruslan [8.4K]

The answer should be (C):y=(-3/7)x+3 because:

if you use the slope formula: (Y2-Y1)/(X2-X1) it would look like this:

(0-3)/(7-0) and the slope of the line would be -3/7

then using the slope you just found, plug it into the y=mx+b equation along with one set of coordinates to find b (the y-intercept):

y=mx+b
0=(-3/7)(7)+b
b=3

SO, overall if you put the slope (m) and the y-intercept (b) that you found together, you get the line formula of y=(-3/7)+3

5 0

3 years ago

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify

irina [24]

Answer:

x-coordinates of relative extrema = $\frac{-1}{2}$

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

$f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}$

Differentiate with respect to x

$f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}$

$f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}$

Differentiate f'(x) with respect to x

$f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1$