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3 years ago

What is this can you guys help

Mathematics

1 answer:

dangina [55]3 years ago

3 0

Answer:

see below

Step-by-step explanation:

The graph of y=10 is the graph of all the points (x, 10) for any value of x. Each one of those points is 10 units above the x-axis. Together, those points make a horizontal line where y = 10 (!).

You might be interested in

What is the probability that a student who plays an instrument is in band

m_a_m_a [10]

Answer:

20/25

Step-by-step explanation:

Here, we want to calculate the probability that a student who plays an instrument is in band

The number of students who play instruments are;

Amongst this,

20 are in a band

So the probability of having a student who plays instrument and is in a band will be;

20/25 = 4/5

3 0

3 years ago

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

3 years ago

Five computer program modules are ranked as M1, M2, M3, M4, and M5 according to the ascending order of effort required to debug

gulaghasi [49]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Technician selects three out of 5 systems

In C(5,3)=10ways, this can be achieved

In part a:

Space sample chooses 3 of a 5 systems

$(M_1, \ M_2,\ M_3),$ $(M_1,M_2,M_4) \ (M_1,M_2,M_5) \ (M_1,M_3,M_4) \ (M_1,M_3,M_5),$ $(M_1,M_4,M_5) \ (M_2,M_3,M_4)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5)}$

In point b:

A =MODULE WHICH INCLUDE M1 minimal amount of effort

Outcomes probable =

$(M_1,M_2,M_3),\ (M_1,M_2,M_4) \ (M_1,M_2,M_5)\ (M_1,M_3,M_4)\\\\(M_1,M_3,M_5),\ (M_1,M_4,M)5)\ =\ 6$

$\to p(A)=\frac{6}{10}\\\\$

$=0.6$

In point c:

B = highest effort that is $M_5$

Potential result=

$(M_1,M_2,M_5) \ (M_1,M_3,M_5) \ (M_2,M_3,M_5)\(M_2,M_4,M_5) \\ (M_2,M_4,M_5), \ (M_3,M_4,M_5) \ =\ 6 \\\\$

$\to B= \frac{6}{10} \\\\$

$=0.6$

$\to P(B)=10$

In point d:

$\to \ A \ intersection \ B=(M_1,M_2,M_5), \ (M_1,M_3,M_5) \ ,(M_1,M_4,M_5)$

$\to A (A \ intersection \ B) = \frac{3}{10} \\\\\ \ \ \ \ \$

$=0.3$

In point e:

$\to (A \cup B) = (M_1,M_2,M_3),\ (M_1,M_2,M_4)\ (M_1,M_2,M_5)(M_1,M_3,M_4)\ (M_1,M_3,M_5), \\ (M_1,M_4,M_5)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5) \ = \ 9$ $\to P(A \cap B)=\frac{9}{10}$