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3 years ago

Expand and evaluate the power (1.6) ^2 A. 3.6 B. 3.2 C. 2.56 D. 0.8

Mathematics

2 answers:

Citrus2011 [14]3 years ago

5 0

2.56.-----really all you had to do was 1.6*(its self) 1.6 giving you 2.56 (C)----

bonufazy [111]3 years ago

3 0

1.6^2

It's telling you to multiply 1.6 by itself 2 times.

1.6 * 1.6 = 2.56

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3 years ago

Determine the equation of a line through the point (3,-4) that is perpendicular to the line y=3x+7

Lana71 [14]

Answer:

y = - $\frac{1}{3}$ x - 3

Step-by-step explanation:

The equation of a line in slope- intercept fprm is

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

y = 3x + 7 is in slope- intercept form

with slope m = 3

Given a line with slope m then the slope of a line perpendicular to it is

$m_{perpendicular}$ = - $\frac{1}{m}$ = - $\frac{1}{3}$ , thus

y = - $\frac{1}{3}$ x + c ← is the partial equation of the perpendicular line

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3 years ago

Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com

fredd [130]

$f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}$

$f$ has critical points where the derivative is 0:

$2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1$

The second derivative is

$f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}$

and $f''(-1)=\frac2e>0$ , which indicates a local minimum at $x=-1$ with a value of $f(-1)=\frac1e$ .

At the endpoints of [-2, 2], we have $f(-2)=1$ and $f(2)=e^8$ , so that $f$ has an absolute minimum of $\frac1e$ and an absolute maximum of $e^8$ on [-2, 2].

So we have

$\dfrac1e\le f(x)\le e^8$

$\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx$

$\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}$

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3 years ago