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3 years ago

Which equation is equivalent to f(x) = 16x4 – 81 = 0?

Mathematics

2 answers:

Vesna [10]3 years ago

5 0

Answer:

The equivalent expression is:

$16x^4-81=(4x^2-9)(4x^2+9)$

Step-by-step explanation:

We have been given the equation:

$f(x)=16x^4-81=0$

Equivalent expression can be computed by solving the expression ai its maximum.

We can factorize the given equation:

By using $a^2-b^2=(a+b)(a-b)$

Here, $a=4x^2,b=9$

Hence, we get the equation below:

$16x^4-81=(4x^2-9)(4x^2+9)$

Therefore, the equivalent expression is:

$16x^4-81=(4x^2-9)(4x^2+9)$

Alenkasestr [34]3 years ago

5 0

Answer:

The answer is (4x2+9)(4x2-9)=0

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2 years ago

Write a linear equation that contains the points shown in the table.

Afina-wow [57]

Answer:

y=7x+1

Step-by-step explanation:

From the table, the difference in y is 7 and the difference in x is 1.

Let the linear equation be

$y = mx + b$

Where m is the constant difference in y divided by the constant difference in x.

This means m=7/1=7

Our equation now becomes:

y=7x+b

To find b, we substitute any ordered pair from the table.

From the to table, when x=1, y=8.

This implies that,

8=7(1)+b

b=8-7=1

Therefore the equation is y=7x+1

4 0

2 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$