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3 years ago

How do I solve (6^-2)^2 ?

1 answer:

8 0

(6^-2)^2 = 1/1296 = 0.0007716049382716 hope this is what you were looking for. the math is right tho for (6^-2)^2 most calculators can do this easy just do 6^-2 then after just do ^2

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Multiplicando √3 por √27 podemos obter a resposta A-81 B-9 C-3 D-12

ki77a [65]

Answer:

Step-by-step explanation:

ROOT 3 * 3 ROOT 3

7 0

2 years ago

Im confused on how to use distance formula on this

zheka24 [161]

Answer:

Not sure what you need ----see below

Step-by-step explanation:

Leg 1 of left triangle = 8 units

Leg 2 = 6

Hypotenuse = 10 ( via Pythag theorem....or distance formula)

R triangle Leg1 = 8 units

leg 2 = 6 units

hypotenuse = 10 units so the triangles are congruent by SSS

distance from right angle of triangle 1 ( at 0,0) to

right angle of triangle 2 ( at 6,-3)

using distance formula

from 0,0 to 6, -3

d^2 = ( 6-0)^2 + (-3 -0)^2

d^2 = 45

d = sqrt 45

4 0

2 years ago

Consider the equation below. (If an answer does not exist, enter DNE.)

storchak [24]

Answer:

a. f is increasing in the interval $(-4,0)$ and decreasing in the intervals $\left ( -\infty ,-4 \right )\,,\,\left ( 0,4 \right )$

b. local maximum value of the function is 6, -250 and Local minimum value of the function is -250

c. inflexion points are $\left ( \frac{-4\sqrt{3}}{3},\frac{-1226}{9} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\frac{-1226}{9} \right )$

f is concave up in intervals $\left ( -\infty ,\frac{-4\sqrt{3}}{3} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\infty \right )$ and concave down in interval $\left (\frac{-4\sqrt{3}}{3},\frac{4\sqrt{3}}{3} \right )$

Step-by-step explanation:

Given: $f(x)=x^4-32x^2+6$

To find: interval on which f is increasing and decreasing, local minimum and maximum values of f, inflection points and interval on which f is concave up and concave down

Solution:

A function f is increasing in the interval in which $f'>0$ and decreasing in the interval in which $f''$

If a function is increasing before a point and decreasing after that point, the point is said to be a point of local maxima.

If a function is decreasing before a point and increasing after that point, the point is said to be a point of local minima.

An inflection point is a point on the graph of a function at which the concavity changes. Put second derivative equal to zero as check if concavity changes at the points obtained. Such points are said to be points of inflexion.

A function f is concave up in the interval in which $f''>0$ and concave down in the interval in which $f''$

$f(x)=x^4-32x^2+6\\f'(x)=4x^3-64x\\=4x(x^2-16)\\=4x(x+4)(x-4)$

$f'(x)=0\\4x(x+4)(x-4)=0$

Observe the attached table.

So, f is increasing in the interval $(-4,0)$ and decreasing in the intervals $\left ( -\infty ,-4 \right )\,,\,\left ( 0,4 \right )$

From the table,

a function f has a local maxima at $x=0,4$ and local minima at $x=-4$

$f(-4)=(-4)^4-32(-4)^2+6=256-512+6=-250\\f(0)=0^4-32(0)^2+6=6\\f(4)=4^4-32(4)^2+6=256-512+6=-250$

So, local maximum value of the function is 6, -250

Local minimum value of the function is -250

$f'(x)=4x^3-64x\\f''(x)=12x^2-64=4(3x^2-16)\\f''(x)=0\Rightarrow 4(3x^2-16)=0\\3x^2-16=0\\x^2=\frac{16}{3}\\x=\pm \frac{4}{\sqrt{3}}=\pm \frac{4\sqrt{3}}{3}$

See the attached table

So, f is concave up in intervals $\left ( -\infty ,\frac{-4\sqrt{3}}{3} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\infty \right )$

and concave down in interval $\left (\frac{-4\sqrt{3}}{3},\frac{4\sqrt{3}}{3} \right )$

Also,

$f\left ( \frac{-4\sqrt{3}}{3} \right )=\left ( \frac{-4\sqrt{3}}{3} \right )^4-32\left ( \frac{-4\sqrt{3}}{3} \right )^2+6=\frac{-1226}{9}\\f\left ( \frac{4\sqrt{3}}{3} \right )=\left ( \frac{4\sqrt{3}}{3} \right )^4-32\left ( \frac{4\sqrt{3}}{3} \right )^2+6=\frac{-1226}{9}$