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2 years ago

No files please........

Mathematics

1 answer:

FinnZ [79.3K]2 years ago

3 0

√7(-8√10+√14)

-8√70+√98

-8√70+√49.2

-8√70+7√2

Just comment if you want to ask

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The sum of the angle measured in a triangle is 180°.Two angle of a triangle measures 20° and 50°.What did is the measure of the

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Answer:

110 Degrees

Step-by-step explanation:

The sum is when different numbers are added.

When all angles are added it adds up to 180 Degrees for any triangle.

Triangles have three sides and two are given as 50 Degrees and 20 Degrees.

There is one angle remaining.

50 Degrees + 20 Degrees = 70 Degrees

Therefore, the last angle must be 110 Degrees. (180-70)

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Step-by-step explanation:

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2 years ago

What is 6 x 6 x 6 x 6 x 6 written using an exponent.

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Step-by-step explanation:

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2 years ago

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You want to sell triangular kites. Each kite will have a base of 4 feet and a height of 3 feet. Cloth for the kites costs $0.52

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Step-by-step explanation:

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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,

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

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

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2 years ago