How to calculate y=0 on to a graph

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

8 0

Https://www.themathpage.com/alg/equation-of-a-line-2.htm

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The midpoint of AB is M (2,0). If the coordinates of A are, (-3, 3), what are the coordinates of B?

gregori [183]

Given:

M is the midpoint of AB.

M(2,0) and A(-3, 3).

To find:

The coordinates of point B.

Solution:

Midpoint formula:

$Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

Let the coordinates of point B are (a,b). Then, using the midpoint formula, we get

$(2,0)=\left(\dfrac{-3+a}{2},\dfrac{3+b}{2}\right)$

On comparing both sides, we get

$\dfrac{-3+a}{2}=2$

$-3+a=2\times 2$

$a=4+3$

$a=7$

And,

$\dfrac{3+b}{2}=0$

$3+b=0$

$3+b-3=0-3$

$b=-3$

Therefore, the coordinates of point B are (7,-3).

5 0

2 years ago

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)

USPshnik [31]

See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
$M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) + f(\frac{x_{1} + x_{2} }{2}) + ... + f(\frac{x_{n-1} + x_{n} }{2})]$

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
$M_{4} = \frac{1 - 0}{4} [ f(\frac{0 + \frac{1}{4} }{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]$
$M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]$

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
$M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]$

$M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]$

M₄ = 1/4 · (2 - 478/512)
= 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.