All categories

2 years ago

Find 4+ (-1 2/3). Write your answer in simplest form.

Mathematics

2 answers:

kramer2 years ago

7 0

Answer:

The answer is -5 1/3

Step-by-step explanation:

joja [24]2 years ago

3 0

Answer:

Step-by-step explanation:

$-1\frac{2}{3}= -\frac{1*3+2}{3}=-\frac{3+2}{3}=\frac{-5}{3}\\\\\\4+(-1\frac{2}{3})=4 - \frac{5}{3}$

$=\frac{4*3}{1*3} - \frac{5}{3}\\\\= \frac{12}{3} - \frac{5}{3}\\\\=\frac{12-5}{3}\\\\=\frac{7}{3}\\\\=2\frac{1}{3}$

You might be interested in

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

5(x+4)>2x-7 please help

Alik [6]

Answer:

x > -9

Step-by-step explanation:

3 0

2 years ago

33 points!! The line contains the point (4,-2) and is perpendicular to 4x+3y=-6 The equation of the line is:

Deffense [45]

Answer:

y = 3/4 x - 5

Step-by-step explanation:

The slopes of perpendicular lines are negative reciprocals. We find the slope of the given line. Then we can find the slope of the reciprocal.

4x + 3y = -6

Solve for y.

3y = -4x - 6

y = -4/3 x - 2

The slope of the original line is -4/3.

The slope of the perpendicular is 3/4.

Now we need to find the equation of line with slope 3/4 that contains point (4, -2).

y = mx + b

-2 = (3/4)(4) + b

-2 = 3 + b

-5 = b

b = -5

The equation is

y = 3/4 x - 5

5 0

3 years ago

Raymond is making stir-fry rice for dinner. The recipe says to add 3/4 tablespoon of ground ginger, but he only has a 1/8 tables

densk [106]

Answer:

Step-by-step explanation:

3/4 = 6/8.

6/8 divided by 1/8 = 6

brianliest pls

4 0

2 years ago

Calculate the perimeter and area for the parallelogram if a = 2.8 ,b = 16.5 and h = 11.5

Nana76 [90]

Perimeter = 2(a+b)

= 2(2.8+16.5)

=2(19.3)

Perimeter = 38.6

Area=base x height

=2.8 x 11.5

=32.2

3 0

3 years ago

Read 2 more answers