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4 years ago

What is -4R-2r+5 as a simpler exspression

2 answers:

8 0

$\sf-4R-2r+5$ is already in simplest form.

But if you meant to say $\sf-4r-2r+5$ , we would combine the first two terms.

Adding/subtracting like terms is the same as adding/subtracting whole numbers.

$\sf-4-2=-6$

Therefore:

$\sf-4r-2r=-6r$

Which gives us:

$\boxed{\sf-6r+5}$

Leni [432]4 years ago

6 0

-4-2=-6 so its -6r+5

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Which point-slope equation represents a line that passes through (3, –2) with a slope of negative StartFraction 4 Over 5 EndFrac

pshichka [43]

Answer:

The equation is y = (-4/5)x + 2/5.

Step-by-step explanation:

In a slope form equation, y = mx + b where m represents gradient/slope and b is the y-intercept. Given that the slope is -4/5 so you have to find the value of b by substituing the coordinates into the equation :

$y = - \frac{4}{5} x + b$

$let \: x = 3,y = - 2$

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

$- 2 = - \frac{12}{5} + b$

$b = - 2 + \frac{12}{5}$

$b = \frac{2}{5}$

$y = - \frac{4}{5} x + \frac{2}{5}$

7 0

4 years ago

Let V be the volume of the solid obtained by rotating about the y-axis the region bounded y = 4x and y = x2/4 . Find V by slicin

aliya0001 [1]

Answer:

Step-by-step explanation:

Consider the graphs of the $y = 4x$ and $y = \frac{x^{2} }{4}$ .

By equating the expressions, the intersection points of the graphs can be found and in this way delimit the area that will rotate around the Y axis.

$4x = \frac{x^{2} }{4} \\ x^{2} = 16x \\ x^{2} - 16x = 0 \\ x(x-16) = 0$ then $x=0$ o $x=16$ . Therefore the integration limits are:

$y = 4(0) = 0$ and $y = 4(16) = 64$

The inverse functions are given by:

$x = 2 \sqrt{y}$ and $x = \frac{y}{4}$ . Then

The volume of the solid of revolution is given by:

$\int\limits^{64}_ {0} \, [2\sqrt{y} - \frac{y}{4}]^{2} dy = \int\limits^{64}_ {0} \, [4y - y^{3/2} + \frac{y^{2}}{16} ]\ dy = [2y^{2} - \frac{2}{5}y^{5/2} + \frac{y^{3}}{48} ]\limits^{64}_ {0} = 546.133 u^{2}$