Simplify the follow15- (8+3)

Mathematics

2 answers:

lara31 [8.8K]2 years ago

6 0

8 plus 3 is 11, 15 minus 11 is 4

aleksandrvk [35]2 years ago

6 0

Answer:

Step-by-step explanation:

First you would do what is inside the parenthesis which in this case is 8+3 = 11. After that you would do 15-11 which will get you to the answer 4. :D

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What is the slope OF THE GRAPH, not what is A slope, what is THE slope

PtichkaEL [24]

You can find the slope either by just looking at the line or using the slope formula.

#1: The slope formula is:

$m=\frac{y_2-y_1}{x_2-x_1}$ Find two points and plug it into the formula

I will use (0, 2) and (1, -1)

(0, 2) = (x₁, y₁)

(1, -1) = (x₂, y₂)

$m = \frac{y_2-y_1}{x_2-x_1}$

$m = \frac{2-(-1)}{0-1}$ [two negatives cancel each other out and become positive]

$m=\frac{2+1}{-1} =\frac{3}{-1}$

m = -3

#2: To find the slope without having to do the work, you use this:

$m=\frac{rise}{run}$

Rise is the number of units you go up(+) or down(-) from each point

Run is the number of units you go to the right from each point

If we start at a defined/obvious point, like (0, 2), find the next point and see how many units it goes up or down and to the right. The next point is (1, -1), so from each point, you go down 3 units and to the right 1 unit. So your slope is -3/1 or -3. You can make sure the slope is right by looking at another point.

5 0

3 years ago

Read 2 more answers

Let C = C1 + C2 where C1 is the quarter circle x^2+y^2=4, z=0,from (0,2,0) to (2,0,0), and where C2 is the line segment from (2,

trapecia [35]

Not much can be done without knowing what $\mathbf F(x,y,z)$ is, but at the least we can set up the integral.

First parameterize the pieces of the contour:

$C_1:\mathbf r_1(t_1)=(2\sin t_1,2\cos t_1,0)$
$C_2:\mathbf r_2(t_2)=(1-t_2)(2,0,0)+t_2(3,3,3)=(2+t_2, 3t_2, 3t_2)$

where $0\le t_1\le\dfrac\pi2$ and $0\le t_2\le1$ . You have

$\mathrm d\mathbf r_1=(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1$
$\mathrm d\mathbf r_2=(1,3,3)\,\mathrm dt_2$

and so the work is given by the integral

$\displaystyle\int_C\mathbf F(x,y,z)\cdot\mathrm d\mathbf r$
$=\displaystyle\int_0^{\pi/2}\mathbf F(2\sin t_1,2\cos t_1,0)\cdot(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1$
${}\displaystyle\,\,\,\,\,\,\,\,+\int_0^1\mathbf F(2+t_2,3t_2,3t_2)\cdot(1,3,3)\,\mathrm dt_2$