Hi there!
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I believe your answer is:
Quadrant III
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Here’s why:
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- The plane is split into four quadrants.
- Quadrant III houses all the points with negative signs for both X and Y values.
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See the attached picture for reference.
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Hope this helps you. I apologize if it’s incorrect.
<h2>
Explanation:</h2><h3>Part A.</h3>
The boundary lines of both inequalities will be dashed, because neither includes the "or equal to" case.
The first inequality solution area is bounded by a line with slope +5 and y-intercept +5. The solution area is above the line (y is greater than ...). Since the line rises steeply, the solution area looks to be to the left of the line. (It is shaded red on the attached graph.)
The second inequality solution area is bounded by a line with slope -1/2 and y-intercept +1. The solution area for this inequality is also above the line.
The solution area is where the two solution spaces overlap, in the quadrant to the upper left of the point where the lines intersect.
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<h3>Part B.</h3>
The graph shows the point (-2, 5) to be in the solution space.
We can show this point satisfies both inequalities.
- 5 > 5(-2)+5 ⇒ 5 > -5 . . . true
- 5 > (-1/2)(-2) +1 ⇒ 5 > 2 . . . true
Answer:
9
Step-by-step explanation:
For each pizza variety, Daniel can choose 18 different toppings. Therefore, for 23 pizza varieties, Daniel can choose from 23 * 18 = 414 different combinations of varieties and toppings.
For each pizza size, Daniel can choose from 414 different combinations of varieties and toppings. Therefore, for n pizza sizes, Daniel can choose from n * 414 = 3726 combinations
414n = 3726
divide both sides by 414 to isolate the n
n = 9
Answer:
The answer is b.
5 ≤ x ≤ 50; Billy needs to make at least 5 more pizzas but no more than 50.
Step-by-step explanation:
Please mark me brainliest, if I helped!
The two consecutive odd numbers that have a product of 15 are 3 and 5
There are 100 random 2 digits numbers [00 , 99]
There are 34 divisible by 3 {00, 03, 06, 09, … 93, 96, 99}
There are 20 divisible by 5 {00, 05, 10, … 90, 95}
However we must avoid counting numbers twice so we need to subtract those divisible by 15.
There are 7 divisible by 15 {00, 15, 30, 45, 60, 75, 90}
This means there are 34 + 20 - 7 = 47 2-digit numbers that are divisible by 3 or 5.
If you pick a random 2-digit number then P(divisible by 3 or 5) = 47/100 = 0.47
