1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
NNADVOKAT [17]
3 years ago
7

Solve for x: –|2x+3|=1

Mathematics
1 answer:
Anna [14]3 years ago
8 0

Step-by-step explanation:

-|2x+3|=1

-|2x+3|/-1=1/-1

|2x+3|=-1

2x=-1-3

2x=-4

2x/2=-4/2

x= -2

You might be interested in
Please hurry!!!! I’m on a strict time limit!!!!
shusha [124]
Number one is B Number two is also B
3 0
3 years ago
Read 2 more answers
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
Which relation defines y as a function of x?
UkoKoshka [18]

Answer:

A function is a relation in which each input has only one output. In the relation, y is a function of x, because for each input x (1, 2, 3, or 0), there is only one output y.

Step-by-step explanation:

For every x there is exactly one y value. We can forgive a function if some values of x do not have a y, but if there is more than one y for even one value of x, then the relation is not a function.

8 0
2 years ago
A room has a maximum capacity of 396 people. How many people are allowed in the room at anytime?
Volgvan
396 very simple math
4 0
2 years ago
Image down below plz help
Mariana [72]

Answer:

44?

Step-by-step explanation:

I realy just have a feeling

8 0
2 years ago
Other questions:
  • Find the equation of the line that contains the point P(−4, −3) and is perpendicular to the graph of y = − 4 5 x − 4.
    8·1 answer
  • 11/x^2+3x+2 and 1/x+2 find the lcd
    6·1 answer
  • I’m confused on this one
    7·1 answer
  • Laura  earns $ 255 by working 20 hours each week .
    9·1 answer
  • What is 24/60 simplified
    11·2 answers
  • Suppose you are a salaried employee paid monthly. You currently receive $5,200 net income each month. The 20-50-30 budget model
    14·1 answer
  • Algebra, please help. 2 phone companies prices.
    5·1 answer
  • Two stores sell the same shirt at the same price. Both stores decide to sell the shirt at a discounted price.
    15·1 answer
  • Means and SDs. For each part, compare distributions (1) and (2) based on their means and standard deviations. You do not need to
    13·1 answer
  • A farm allows customers to cut their own Christmas trees. The farm uses the equation y = 15x + 10 to determine the total charge,
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!