<span>6^3 + 5(4 − 2)
= </span><span>216 + 5(2)
</span><span>= 216 + 10
= 226
based on multiple choice options, 63 = 6^3
so answer is </span><span>c. 226</span>
The easiest way to find a parallel equation is to write your equation in slope-intercept form. The general equation for slope-intercept form is y = mx + b, where m = the slope of the equation, b = the y intercept, and x and y are your variables.
You're given 5x + 10y = -4.
1) Move 5x to the right side by subtracting 5x from both sides:
5x + 10y = -4
10y = -5x - 4
2) Divide both sides by 10 to get y by itself on the left. Simplify:

<span>Remember that for
parallel lines, the slope, m, is the same for both equations. You can make the y-intercept, b, whatever number you want.
When the equation is in slope-intercept form, </span>

, you can see that

.
A parallel equation is in the form:

Plug in anything you want for b. One example is:

-------
Answer:

(just one example)
Answer: b
Step-by-step explanation:
You can use the Pythagorean Theorem to find the length of the third side AB (Identified as "x" in the figure attached in the problem), which says that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the legs:
a² = b²+c²
As we can see the figure, the triangle does not have an angle of 90°, but it can be divided into two equal parts, leaving two triangles with a right angle. We already have the values of the hypotenuse and a leg in triangle "A" , so we can find the value of the other leg:
b = √(a²-c²) b = √(10²-4²) b = 9.16
With these values, we can find the hypotenuse in the triangle "B": x = √b²+c² x = √(9.16)²+(4)² x = 10
Start by graphing each line.
• Because the first inequality is smaller than, it will have a dotted (- - -) line.
• Because the second inequality is smaller than or equal to, it will have a solid line (---).
Then, plug in points to see where your shading will go. If the statement is true (x = x), you will shade that area along the line.



0 is less than 2.
Do the same step for the other equation. Your solution to the problem is any point that lies between the shading from both inequalities (where the blue and red meet).