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2 years ago

What is the solution to the system of linear equations represented by the matrix equations below. Please help!!!!!

Mathematics

1 answer:

Evgen [1.6K]2 years ago

5 0

Answer: c no solution
Explanation I did the test and got it right

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Calculus, question 5 to 5a

Llana [10]

5. Let $x = \sin(\theta)$ . Note that we want this variable change to be reversible, so we tacitly assume 0 ≤ θ ≤ π/2. Then

$\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}$

and $dx = \cos(\theta) \, d\theta$ . So the integral transforms to

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \int \frac{\sin^3(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta = \int \sin^3(\theta) \, d\theta$

Reduce the power by writing

$\sin^3(\theta) = \sin(\theta) \sin^2(\theta) = \sin(\theta) (1 - \cos^2(\theta))$

Now let $y = \cos(\theta)$ , so that $dy = -\sin(\theta) \, d\theta$ . Then

$\displaystyle \int \sin(\theta) (1-\cos^2(\theta)) \, d\theta = - \int (1-y^2) \, dy = -y + \frac13 y^3 + C$

Replace the variable to get the antiderivative back in terms of x and we have

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\cos(\theta) + \frac13 \cos^3(\theta) + C$

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\sqrt{1-x^2} + \frac13 \left(\sqrt{1-x^2}\right)^3 + C$

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\frac13 \sqrt{1-x^2} \left(3 - \left(\sqrt{1-x^2}\right)^2\right) + C$

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \boxed{-\frac13 \sqrt{1-x^2} (2+x^2) + C}$

6. Let $x = 3\tan(\theta)$ and $dx=3\sec^2(\theta)\,d\theta$ . It follows that

$\cos(\theta) = \dfrac1{\sec(\theta)} = \dfrac1{\sqrt{1+\tan^2(\theta)}} = \dfrac3{\sqrt{9+x^2}}$

since, like in the previous integral, under this reversible variable change we assume -π/2 < θ < π/2. Over this interval, sec(θ) is positive.

Now,

$\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \int \frac{27\tan^3(\theta)}{\sqrt{9+9\tan^2(\theta)}} 3\sec^2(\theta) \, d\theta = 27 \int \frac{\tan^3(\theta) \sec^2(\theta)}{\sqrt{1+\tan^2(\theta)}} \, d\theta$

The denominator reduces to

$\sqrt{1+\tan^2(\theta)} = \sqrt{\sec^2(\theta)} = |\sec(\theta)| = \sec(\theta)$

and so

$\displaystyle 27 \int \tan^3(\theta) \sec(\theta) \, d\theta = 27 \int \frac{\sin^3(\theta)}{\cos^4(\theta)} \, d\theta$

Rewrite sin³(θ) just like before,

$\displaystyle 27 \int \frac{\sin(\theta) (1-\cos^2(\theta))}{\cos^4(\theta)} \, d\theta$

and substitute $y=\cos(\theta)$ again to get

$\displaystyle -27 \int \frac{1-y^2}{y^4} \, dy = 27 \int \left(\frac1{y^2} - \frac1{y^4}\right) \, dy = 27 \left(\frac1{3y^3} - \frac1y\right) + C$

Put everything back in terms of x :

$\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac1{\cos^3(\theta)} - \frac3{\cos(\theta)}\right) + C$

$\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac{\left(\sqrt{9+x^2}\right)^3}{27} - \sqrt{9+x^2}\right) + C$

$\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \boxed{\frac13 \sqrt{9+x^2} (x^2 - 18) + C}$

2(b). For some constants a, b, c, and d, we have

$\dfrac1{x^2+x^4} = \dfrac1{x^2(1+x^2)} = \boxed{\dfrac ax + \dfrac b{x^2} + \dfrac{cx+d}{x^2+1}}$

3(a). For some constants a, b, and c,

$\dfrac{x^2+4}{x^3-3x^2+2x} = \dfrac{x^2+4}{x(x-1)(x-2)} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac c{x-2}}$

5(a). For some constants a-f,

$\dfrac{x^5+1}{(x^2-x)(x^4+2x^2+1)} = \dfrac{x^5+1}{x(x-1)(x+1)(x^2+1)^2} \\\\ = \dfrac{x^4 - x^3 + x^2 - x + 1}{x(x-1)(x^2+1)^2} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac{cx+d}{x^2+1} + \dfrac{ex+f}{(x^2+1)^2}}$

where we use the sum-of-5th-powers identity,

$a^5 + b^5 = (a+b) (a^4-a^3b+a^2b^2-ab^3+b^4)$

4 0

2 years ago

Could some one please help me with this

Elden [556K]

Answer:

x=40°

Step-by-step explanation:

Firstly, lets look at some things that we know based on this image:

We have a equilateral triangle(The triangle on the left has 3 tick marked on the sides, so they are equal. It also has 3 of the same angle, so it must be equilateral) and a isosceles triangle (There are two tick marks showing that two of the sides are equal length), the measure of each of the equilateral triangle's angles must be 60° each, the measure of these two triangles together must be 360°, and angle x and the unmarked angle must be the same size as this triangle is isosceles.

To solve this, we can set up an equation to solve for x. To do this, we can add up all of the known angles and set it equal to 360.

$(100+60)+60+(60+x)+x=360\\\\160+60+60+2x=360\\\\280+2x=360\\\\2x=80\\\\x=40$

4 0

3 years ago

A king company requires 20 hours of labor to produce a standard table, and a chair requires 12 hours of labor. The labor availab

Dimas [21]

Since x represents tables, and tables require 20 hours, 20x is the number of hours spent on tables. The total hours must not exceed 565, so must be less than or equal to that value. The first selection is appropriate.

4 0

2 years ago

Plz help ill give brainleist

sergij07 [2.7K]

Answer:

Step-by-step explanation:

there isn't much of a harsh trend, but the graph shows that generally when the unemployment is lower, the life expectancy is higher, and when the unemployment is higher, the life expectancy is lower.

3 0

2 years ago

Using the Completing the Square method, what are the zeros of the quadratic function f(x) = 2x2 + 8x - 3?

Ipatiy [6.2K]

Answer:

Step-by-step explanation:

1) to rewrite the given equation:

f(x)=2(x²+4x)-3; ⇔ f(x)=2(x²+4x+4)-3-8; ⇔ f(x)=2(x²+4x+4)-11; ⇔ f(x)=2(x+2)²-11;

2) the roots are:

$\left[\begin{array}{ccc}x=-\sqrt{\frac{11}{2}}-2 \\x=\sqrt{\frac{11}{2}}-2 \end{array}$

3) finally, the correct answer is A.

4 0

1 year ago