Answer:
The area below the line is shaded.
Step-by-step explanation:
Using the zero-interval test [test point (0, 0)], we plug in 0 for both <em>y</em><em> </em>and <em>x</em><em> </em>to determine whether we shade <em>below the </em><em>line</em><em> </em>[the portion that DOES contain the origin] or <em>above</em><em> </em><em>the</em><em> </em><em>line</em><em> </em>[the portion that does not contain the origin]. This is where me must verify the inequality as false or true:
![\displaystyle 0 < 2[0] + 1 → 0 < 1](https://tex.z-dn.net/?f=%5Cdisplaystyle%200%20%3C%202%5B0%5D%20%2B%201%20%E2%86%92%200%20%3C%201)
This is a genuine statement, therefore we shade <em>below</em><em> </em><em>the</em><em> </em><em>line</em>.
I am joyous to assist you anytime.
Answer:
The correct option is;
C. (1.6, 1.3)
Step-by-step explanation:
Given that at x = 1.5 the y-values of both equations are y = 1.5 and y = 1 respectively
The x-value > The y-value
The difference in the y-values = 1.5 - 1 = 0.5
At x = 1.6 the y-values of both equations are y = 1.2 and y = 1.4 respectively
The x-value > The y-value
The difference in the y-values = 1.2 - 1.4 = -0.2
At x = 1.7 the y-values of both equations are y = 0.9 and y = 1.8 respectively
The x-value > The first y-value and the x-value < the second y-value
The difference in the y-values = 0.9 - 1.8 = 0.9
Therefore, the approximate y-value can be found by taking the average of both y-values when x = 1.6 where the difference in the y-values is least as follows;
Average y-value at x = 1.6 = (1.2 + 1.4)/2 = 1.3
Therefore, the best approximation of the exact solution is (1.6, 1.3)
By calculation, we have;
-3·x + 6 = 4·x - 5
∴ 7·x = 11
x = 11/7 ≈ 1.57
y = 4 × 11/7 - 5 ≈ 1.29
The solution is (1.57, 1.29)
I believe the answer is d y=x^3 i hope this helped
Answer:
Linear
Step-by-step explanation:
When you graph them, they do not hit each other.
Linear equation in standard form is Ax+By=C
If it is in this form it is linear.
Answer: 6 in ^2
Step-by-step explanation:
