Answer:
Step-by-step explanation:
A parallel line will have the same slope as the reference line. In this case, I don't see the "given line" as promised in the question. If it does appear, and it looks like y = 5x + 3, for example, the slope is 5 and the new line will have the same slope.
<h3>
<u>If this slope is correct</u>, we can start the equation for the parallel line that goes through point (-3,2) by starting with:</h3><h3 /><h3>y = 5x + b</h3><h3 /><h3>We need a value of b that forces the line to go through point (-3,2). We can do that by using the given point in the equation and solving for b:</h3><h3>y = 5x + b</h3><h3>2 = 5(-3) + b</h3><h3>b = 17</h3><h3 /><h3>The parallel line to y=5x+3 is</h3><h3>y = 5x + 17</h3><h3 /><h3>See attachment.</h3><h3 /><h3 /><h3 />
Answer:
<em>First.</em> Let us prove that the sum of three consecutive integers is divisible by 3.
Three consecutive integers can be written as k, k+1, k+2. Then, if we denote their sum as n:
n = k+(k+1)+(k+2) = 3k+3 = 3(k+1).
So, n can be written as 3 times another integer, thus n is divisible by 3.
<em>Second. </em>Let us prove that any number divisible by 3 can be written as the sum of three consecutive integers.
Assume that n is divisible by 3. The above proof suggest that we write it as
n=3(k+1)=3k+3=k + k + k +1+2 = k + (k+1) + (k+2).
As k, k+1, k+2 are three consecutive integers, we have completed our goal.
Step-by-step explanation:
Answer:
by doing it the same way you did with the first one.
Step-by-step explanation:
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})]](https://tex.z-dn.net/?f=M_%7Bn%7D%20%3D%20%5Cfrac%7Bb%20-%20a%7D%7B2%7D%20%5B%20f%28%5Cfrac%7Bx_%7B0%7D%20%2B%20x_%7B1%7D%20%7D%7B2%7D%29%20%2B%20%20f%28%5Cfrac%7Bx_%7B1%7D%20%2B%20x_%7B2%7D%20%7D%7B2%7D%29%20%2B%20...%20%2B%20%20f%28%5Cfrac%7Bx_%7Bn-1%7D%20%2B%20x_%7Bn%7D%20%7D%7B2%7D%29%5D)
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})]](https://tex.z-dn.net/?f=M_%7B4%7D%20%3D%20%5Cfrac%7B1%20-%200%7D%7B4%7D%20%5B%20f%28%5Cfrac%7B0%20%2B%20%20%5Cfrac%7B1%7D%7B4%7D%20%7D%7B2%7D%29%20%2B%20%20f%28%5Cfrac%7B%20%5Cfrac%7B1%7D%7B4%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%7D%7B2%7D%29%20%2B%20%20f%28%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%20%2B%20%5Cfrac%7B3%7D%7B4%7D%20%7D%7B2%7D%29%20%2B%20f%28%5Cfrac%7B%5Cfrac%7B3%7D%7B4%7D%20%2B%201%7D%20%7B2%7D%29%5D)
![M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]](https://tex.z-dn.net/?f=M_%7B4%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5B%20f%28%5Cfrac%7B1%7D%7B8%7D%29%20%2B%20%20f%28%5Cfrac%7B3%7D%7B8%7D%29%20%2B%20%20f%28%5Cfrac%7B5%7D%7B8%7D%29%20%2B%20f%28%5Cfrac%7B7%7D%7B8%7D%29%5D)
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})]](https://tex.z-dn.net/?f=M_%7B4%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5B%28%5Cfrac%7B1%7D%7B8%7D%20-%20%28%5Cfrac%7B1%7D%7B8%7D%29%5E%7B3%7D%29%20%2B%20%28%5Cfrac%7B3%7D%7B8%7D%20-%20%28%5Cfrac%7B3%7D%7B8%7D%29%5E%7B3%7D%29%20%2B%20%28%5Cfrac%7B5%7D%7B8%7D%20-%20%28%5Cfrac%7B5%7D%7B8%7D%29%5E%7B3%7D%29%20%2B%20%28%5Cfrac%7B7%7D%7B8%7D%20-%20%28%5Cfrac%7B7%7D%7B8%7D%29%5E%7B3%7D%29%5D)
![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})]](https://tex.z-dn.net/?f=M_%7B4%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5B%28%5Cfrac%7B1%7D%7B8%7D%20-%20%5Cfrac%7B1%7D%7B512%7D%29%20%2B%20%28%5Cfrac%7B3%7D%7B8%7D%20-%20%5Cfrac%7B27%7D%7B512%7D%29%20%2B%20%28%5Cfrac%7B5%7D%7B8%7D%20-%20%5Cfrac%7B125%7D%7B512%7D%29%20%2B%20%28%5Cfrac%7B7%7D%7B8%7D%20-%20%5Cfrac%7B343%7D%7B512%7D%29%5D)
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.
