<h3>3
Answers:</h3>
- Choice A. (10, -1)
- Choice B. (-8, 9)
- Choice D. (6, -3)
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Explanation:
If we plug the coordinates of point A into the inequality, then we get
x+y > -2
10 + (-1) > -2
9 > -2
That last inequality is a true statement since 9 is to the right of -2 on the number line. That means (10,-1) is a solution. Choice A is one of the answers
Choices B and D are also answers for similar reasons.
Something like choice C is not a solution because
x+y > -2
-1+(-9) > -2
-10 > -2
which is false
You should find that choice E is false as well.
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If you graphed the inequality and all of the points mentioned (see below), then you can visually confirm the answers. Notice how points A, B and D are in the blue shaded region which is the solution set.
The point E on the boundary does not count as a solution. This is due to the lack of "or equal to" portion of the inequality sign. That visually shows point E is not a solution. Point C isn't a solution either as it's nowhere near the blue shaded region.
Answer:
6/5 or 1 1/5
Step-by-step explanation:
2/3 * 9/5
~Multiply both numerators and denominators together
18/15
~Simplify
6/5
Best of Luck!
Answer:
Steve gave his mother $75.
Step-by-step explanation:
Steve earned a total of $125. Sixty percent of that went to his mother. Convert 60% into the equivalent decimal fraction and multiply as indicated:
0.60($125) = $75
Steve gave his mother $75.
As you may already be familiar, these functions f(x) and g(x) are piecewise. They consist of multiple functions with different domains.
1. For #1, the given input is f(0). Since 0≤1, you should use the first equation to solve. f(0)=3(0)-1 ➞ f(0)=-1
2. Continue to evaluate the given input for the domains given. 1≤1, therefore f(1)=3(1)-1➞f(1)=2
3. 5>1, therefore f(5)=1-2(5)➞f(5)=-9
4. -4≤1; f(-4)=3(-4)-1➞f(-4)=-13
5. -3<0<1; g(0)=2
6. -3≤-3; g(-3)=3(-3)-1➞g(-3)=-10
7. 1≥1; g(1)=-3(1)➞g(1)=-3
8. 3≥1; g(3)=-3(3)➞g(3)=-9
9. -5≤-3; g(-5)=3(-5)-1➞g(-5)=-16
Hope this helps! Good luck!
The
first choice gives you the same result. The factor of 5 can be removed to the outside of the summation symbol.
