The table is attached as a figure
The given equation is ⇒⇒⇒⇒ y = 2x
To solve this equation, we need to pick numbers from the table then this number will be substituted into the equation to find y
we need 3 solutions . so, we need to pick 3 numbers of x
From the table, let us choose x = 0
y = 2x
y = 2 * (0)
y = 0
From the table, let choose second value of x such as x = 1
y = 2 * (1)
y = 2
From the table, let choose third value of x such as x = -1
y = 2 * (-1)
y = -2
So, the picked three solutions are y = 0 at x = 0y = -2 at x = -1y = 2 at x = 1
<span>when flipped over a line of reflection the lengths are still the same
the point to the line of reflection is the same length as the line of reflection to the reflected position
the distance from the original point to the reflected point is twice the distance from the original point to the line of reflection. </span>
Answer:
I assume you know Arithmetic Progression .
so, we have to find the first and last 4-digit number divisible by 5
first = 1000 , last = 9990
we have a formula, 
= a + (n-1)d
here, is the last 4-digit number divisible by 5.
n is the number of 4-digit even numbers divisible by 5
d is the common difference between the numbers, which is 10 in this case
a is the first 4-digit number divisible by 5
9990 = 1000 + (n-1)*10
899 = n-1
n = 900
Hence, there are 900 4-digit even numbers divisible by 5
<h3>Answer:</h3>
(x, y) ≈ (1.49021612010, 1.22074408461)
<h3>Explanation:</h3>
This is best solved graphically or by some other machine method. The approximate solution (x=1.49, y=1.221) can be iterated by any of several approaches to refine the values to the ones given above. The values above were obtained using Newton's method iteration.
_____
Setting the y-values equal and squaring both sides of the equation gives ...
... √x = x² -1
... x = (x² -1)² = x⁴ -2x² +1 . . . . . square both sides
... x⁴ -2x² -x +1 = 0 . . . . . polynomial equation in standard form.
By Descarte's rule of signs, we know there are two positive real roots to this equation. From the graph, we know the other two roots are complex. The second positive real root is extraneous, corresponding to the negative branch of the square root function.
Answer:
Not sure if this is correct but I hope this helps :)
